260423-Aula-Atividades-1

Notas de Aula - Atividades

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Exercícios Fila M/M/1/$\infty$/FIFO

Fórmulas Referência:

$\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$

Questão 1: API de autenticação do portal acadêmico

Dados: $\lambda =18$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms

a) calcule a, $\rho$, $P_0$, $E[n]$, $E[nw]$, $E[s]$, $E[w]$ e $P(N\ge 4)$; $\rho =\frac{18}{24}=\frac{3}{4}=0,75$ $P_0=(1-0,75)=0,25$ $P(N\ge 4)={\rho }^n={\left(\frac{3}{4}\right)}^4=\frac{81}{256}=0,3164$ $E[n]=\frac{\rho }{(1-\rho )}=\frac{\frac{3}{4}}{1-\frac{3}{4}}=\frac{12}{4}=3$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}=\frac{0,75^2}{0,25}=\frac{9}{16}\cdot \frac{4}{1}=\frac{36}{16}=\frac{9}{4}=2,25$ $E[s]=\frac{1}{(\mu -\lambda )}=\frac{1}{6}=0,1667$ $E[w]=\frac{1}{(\mu -\lambda )}-\frac{1}{\mu }=\frac{1}{(24-18)}-\frac{1}{24}=\frac{4-1}{24}=\frac{3}{24}=\frac{1}{8}=0,125$

b) verifique se a meta de desempenho é atendida; Sim, $E[s]<200$

c) faça análise de sensibilidade para: cenário A: aumento da taxa de chegada para 21 req/s; cenário B: otimização do serviço para 30 req/s;

Cenário A Dados: $\lambda =21$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms Cálculos: $\rho =\frac{21}{24}=\frac{7}{8}=0,875$ $P_0=(1-0,875)=0,125$ $P(N\ge 4)={\rho }^n={\left(\frac{7}{8}\right)}^4=\frac{2401}{4096}=0,586$ $E[n]=\frac{\rho }{(1-\rho )}=\frac{\frac{7}{8}}{1-\frac{7}{8}}=\frac{56}{8}=7$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}=\frac{{\left(\frac{7}{8}\right)}^2}{\frac{1}{8}}=\frac{49}{64}\cdot \frac{8}{1}=\frac{49}{8}=6,125$ $E[s]=\frac{1}{(\mu -\lambda )}=\frac{1}{3}=0,3333$ $E[w]=\frac{1}{(\mu -\lambda )}-\frac{1}{\mu }=\frac{1}{3}-\frac{1}{24}=\frac{8-1}{24}=\frac{7}{24}=0,291$ A latência excede a meta permitida.

Cenário B: Dados: $\lambda =18$ reqs/s $\mu =30$ reqs/s Meta: Tempo médio resposta inferior a 200ms Cálculos $\rho =\frac{18}{30}=\frac{9}{15}=0,6$ $P_0=(1-0,6)=0,4$ $P(N\ge 4)={\rho }^n={\left(\frac{9}{15}\right)}^4=\frac{6561}{50625}=0,1296$ $E[n]=\frac{\rho }{(1-\rho )}=\frac{\frac{6}{10}}{1-\frac{6}{10}}=\frac{6}{10}\cdot \frac{10}{4}=1,5$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}=\frac{{\left(\frac{6}{10}\right)}^2}{\frac{4}{10}}=\frac{36}{100}\cdot \frac{10}{4}=\frac{9}{10}=0,9$ $E[s]=\frac{1}{(\mu -\lambda )}=\frac{1}{12}=0,083$ $E[w]=\frac{1}{(\mu -\lambda )}-\frac{1}{\mu }=\frac{1}{12}-\frac{1}{30}=\frac{5-2}{60}=\frac{3}{60}=\frac{1}{20}=0,05\cdot 100=50ms$ A latência excede a meta permitida.

d) emita um parecer técnico e proponha uma solução. O servidor está estável mas precisa de atenção para a taxa de chegada, um pouco mais e já estoura o tempo.

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Questão 2: Servidor de arquivos do laboratório

Dados: $\lambda =18$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms

a) calcule a, $\rho$, $P_0$, $E[n]$, $E[nw]$, $E[s]$, $E[w]$ e $P(N\ge 4)$; $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$ b) verifique se a meta de desempenho é atendida;

c) faça análise de sensibilidade para:

Cenário A: aumento da taxa de chegada para 21 req/s; Dados: $\lambda =21$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$

Cenário B: otimização do serviço para 30 req/s; Dados: $\lambda =18$ reqs/s $\mu =30$ reqs/s Meta: Tempo médio resposta inferior a 200ms Cálculos $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$ d) emita um parecer técnico e proponha uma solução.

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Questão 3: Central de tickets de TI

Dados: $\lambda =18$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms

a) calcule a, $\rho$, $P_0$, $E[n]$, $E[nw]$, $E[s]$, $E[w]$ e $P(N\ge 4)$; $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$ b) verifique se a meta de desempenho é atendida;

c) faça análise de sensibilidade para:

Cenário A: aumento da taxa de chegada para 21 req/s; Dados: $\lambda =21$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$

Cenário B: otimização do serviço para 30 req/s; Dados: $\lambda =18$ reqs/s $\mu =30$ reqs/s Meta: Tempo médio resposta inferior a 200ms Cálculos $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$ d) emita um parecer técnico e proponha uma solução.

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Questão 4: Impressora compartilhada da secretaria

Dados: $\lambda =18$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms

a) calcule a, $\rho$, $P_0$, $E[n]$, $E[nw]$, $E[s]$, $E[w]$ e $P(N\ge 4)$; $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$ b) verifique se a meta de desempenho é atendida;

c) faça análise de sensibilidade para:

Cenário A: aumento da taxa de chegada para 21 req/s; Dados: $\lambda =21$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$

Cenário B: otimização do serviço para 30 req/s; Dados: $\lambda =18$ reqs/s $\mu =30$ reqs/s Meta: Tempo médio resposta inferior a 200ms Cálculos $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$ d) emita um parecer técnico e proponha uma solução.

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Questão 5: Gateway de telemetria IoT

Dados: $\lambda =18$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms

a) calcule a, $\rho$, $P_0$, $E[n]$, $E[nw]$, $E[s]$, $E[w]$ e $P(N\ge 4)$; $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$ b) verifique se a meta de desempenho é atendida;

c) faça análise de sensibilidade para:

Cenário A: aumento da taxa de chegada para 21 req/s; Dados: $\lambda =21$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$

Cenário B: otimização do serviço para 30 req/s; Dados: $\lambda =18$ reqs/s $\mu =30$ reqs/s Meta: Tempo médio resposta inferior a 200ms Cálculos $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$ d) emita um parecer técnico e proponha uma solução.

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Questão 6: Fila de renderização de mapas

Dados: $\lambda =18$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms

a) calcule a, $\rho$, $P_0$, $E[n]$, $E[nw]$, $E[s]$, $E[w]$ e $P(N\ge 4)$; $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$ b) verifique se a meta de desempenho é atendida;

c) faça análise de sensibilidade para:

Cenário A: aumento da taxa de chegada para 21 req/s; Dados: $\lambda =21$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$

Cenário B: otimização do serviço para 30 req/s; Dados: $\lambda =18$ reqs/s $\mu =30$ reqs/s Meta: Tempo médio resposta inferior a 200ms Cálculos $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$ d) emita um parecer técnico e proponha uma solução.

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Questão 7: Servidor de integração contínua

Dados: $\lambda =18$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms

a) calcule a, $\rho$, $P_0$, $E[n]$, $E[nw]$, $E[s]$, $E[w]$ e $P(N\ge 4)$; $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$ b) verifique se a meta de desempenho é atendida;

c) faça análise de sensibilidade para:

Cenário A: aumento da taxa de chegada para 21 req/s; Dados: $\lambda =21$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$

Cenário B: otimização do serviço para 30 req/s; Dados: $\lambda =18$ reqs/s $\mu =30$ reqs/s Meta: Tempo médio resposta inferior a 200ms Cálculos $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$ d) emita um parecer técnico e proponha uma solução.

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Questão 8: Validador de notas fiscais eletrônicas

Dados: $\lambda =18$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms

a) calcule a, $\rho$, $P_0$, $E[n]$, $E[nw]$, $E[s]$, $E[w]$ e $P(N\ge 4)$; $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$ b) verifique se a meta de desempenho é atendida;

c) faça análise de sensibilidade para:

Cenário A: aumento da taxa de chegada para 21 req/s; Dados: $\lambda =21$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$

Cenário B: otimização do serviço para 30 req/s; Dados: $\lambda =18$ reqs/s $\mu =30$ reqs/s Meta: Tempo médio resposta inferior a 200ms Cálculos $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$ d) emita um parecer técnico e proponha uma solução.

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Questão 9: Transcodificador de videoaulas

Dados: $\lambda =18$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms

a) calcule a, $\rho$, $P_0$, $E[n]$, $E[nw]$, $E[s]$, $E[w]$ e $P(N\ge 4)$; $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$ b) verifique se a meta de desempenho é atendida;

c) faça análise de sensibilidade para:

Cenário A: aumento da taxa de chegada para 21 req/s; Dados: $\lambda =21$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$

Cenário B: otimização do serviço para 30 req/s; Dados: $\lambda =18$ reqs/s $\mu =30$ reqs/s Meta: Tempo médio resposta inferior a 200ms Cálculos $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$ d) emita um parecer técnico e proponha uma solução.

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Questão 10: Servidor de consultas ao acervo acadêmico

Dados: $\lambda =18$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms

a) calcule a, $\rho$, $P_0$, $E[n]$, $E[nw]$, $E[s]$, $E[w]$ e $P(N\ge 4)$; $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$ b) verifique se a meta de desempenho é atendida;

c) faça análise de sensibilidade para:

Cenário A: aumento da taxa de chegada para 21 req/s; Dados: $\lambda =21$ reqs/s $\mu =24$ reqs/s Meta: Tempo médio resposta inferior a 200ms $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$

Cenário B: otimização do serviço para 30 req/s; Dados: $\lambda =18$ reqs/s $\mu =30$ reqs/s Meta: Tempo médio resposta inferior a 200ms Cálculos $\rho =\frac{\lambda }{\mu }$ $P_0=1-\rho$ $P_n=(1-\rho ){\rho }^n$ $P(N\ge n)={\rho }^n$ $E[n]=\frac{\rho }{(1-\rho )}$ $E[n_w]=\frac{{\rho }^2}{(1-\rho )}$ $E[s]=\frac{1}{(\mu -\lambda )}$ $E[w]=E[s]-\frac{1}{\mu }$ d) emita um parecer técnico e proponha uma solução.

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Exercícios Fila M/M/m/$\infty$/FIFO

Fórmulas de referência

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

Questão 1: Cluster de autenticação do portal

Dados: $m=4$ $\lambda =18$ $\mu =6$ Meta: Tempo médio na fila inferior a 100ms

a) calcule a, ρ, P0, C(m,ρ), E[nw], E[ns], E[n], E[s] e E[w]

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

b) verifique se a meta é atendida

c) faça análise de sensibilidade para:

Cenário A: aumento da carga para 22 reqs/s Dados: $m=4$ servidores $\lambda =22$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

Cenário B: adição de mais um servidor ($m=5$) Dados: $m=5$ servidores $\lambda =18$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

d) Emita parecer técnico.

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Questão 2: Servidor de imagens de máquinas virtuais

Dados: $m=4$ $\lambda =18$ $\mu =6$ Meta: Tempo médio na fila inferior a 100ms

a) calcule a, ρ, P0, C(m,ρ), E[nw], E[ns], E[n], E[s] e E[w]

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

b) verifique se a meta é atendida

c) faça análise de sensibilidade para:

Cenário A: aumento da carga para 22 reqs/s Dados: $m=4$ servidores $\lambda =22$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

Cenário B: adição de mais um servidor ($m=5$) Dados: $m=5$ servidores $\lambda =18$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

d) Emita parecer técnico.

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Questão 3: Triagem automática de chamados

Dados: $m=4$ $\lambda =18$ $\mu =6$ Meta: Tempo médio na fila inferior a 100ms

a) calcule a, ρ, P0, C(m,ρ), E[nw], E[ns], E[n], E[s] e E[w]

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

b) verifique se a meta é atendida

c) faça análise de sensibilidade para:

Cenário A: aumento da carga para 22 reqs/s Dados: $m=4$ servidores $\lambda =22$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

Cenário B: adição de mais um servidor ($m=5$) Dados: $m=5$ servidores $\lambda =18$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

d) Emita parecer técnico.

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Questão 4: Serviço de renderização de mapas

Dados: $m=4$ $\lambda =18$ $\mu =6$ Meta: Tempo médio na fila inferior a 100ms

a) calcule a, ρ, P0, C(m,ρ), E[nw], E[ns], E[n], E[s] e E[w]

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

b) verifique se a meta é atendida

c) faça análise de sensibilidade para:

Cenário A: aumento da carga para 22 reqs/s Dados: $m=4$ servidores $\lambda =22$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

Cenário B: adição de mais um servidor ($m=5$) Dados: $m=5$ servidores $\lambda =18$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

d) Emita parecer técnico.

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Questão 5: Cluster de consultas SQL

Dados: $m=4$ $\lambda =18$ $\mu =6$ Meta: Tempo médio na fila inferior a 100ms

a) calcule a, ρ, P0, C(m,ρ), E[nw], E[ns], E[n], E[s] e E[w]

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

b) verifique se a meta é atendida

c) faça análise de sensibilidade para:

Cenário A: aumento da carga para 22 reqs/s Dados: $m=4$ servidores $\lambda =22$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

Cenário B: adição de mais um servidor ($m=5$) Dados: $m=5$ servidores $\lambda =18$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

d) Emita parecer técnico.

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Questão 6: Sistema de moderação de mensagens

Dados: $m=4$ $\lambda =18$ $\mu =6$ Meta: Tempo médio na fila inferior a 100ms

a) calcule a, ρ, P0, C(m,ρ), E[nw], E[ns], E[n], E[s] e E[w]

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

b) verifique se a meta é atendida

c) faça análise de sensibilidade para:

Cenário A: aumento da carga para 22 reqs/s Dados: $m=4$ servidores $\lambda =22$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

Cenário B: adição de mais um servidor ($m=5$) Dados: $m=5$ servidores $\lambda =18$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

d) Emita parecer técnico.

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Questão 7: Runners de integração contínua

Dados: $m=4$ $\lambda =18$ $\mu =6$ Meta: Tempo médio na fila inferior a 100ms

a) calcule a, ρ, P0, C(m,ρ), E[nw], E[ns], E[n], E[s] e E[w]

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

b) verifique se a meta é atendida

c) faça análise de sensibilidade para:

Cenário A: aumento da carga para 22 reqs/s Dados: $m=4$ servidores $\lambda =22$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

Cenário B: adição de mais um servidor ($m=5$) Dados: $m=5$ servidores $\lambda =18$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

d) Emita parecer técnico.

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Questão 8: Transcodificação de videoaulas

Dados: $m=4$ $\lambda =18$ $\mu =6$ Meta: Tempo médio na fila inferior a 100ms

a) calcule a, ρ, P0, C(m,ρ), E[nw], E[ns], E[n], E[s] e E[w]

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

b) verifique se a meta é atendida

c) faça análise de sensibilidade para:

Cenário A: aumento da carga para 22 reqs/s Dados: $m=4$ servidores $\lambda =22$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

Cenário B: adição de mais um servidor ($m=5$) Dados: $m=5$ servidores $\lambda =18$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

d) Emita parecer técnico.

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Questão 9: Validador de documentos fiscais:

Dados: $m=4$ $\lambda =18$ $\mu =6$ Meta: Tempo médio na fila inferior a 100ms

a) calcule a, ρ, P0, C(m,ρ), E[nw], E[ns], E[n], E[s] e E[w]

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

b) verifique se a meta é atendida

c) faça análise de sensibilidade para:

Cenário A: aumento da carga para 22 reqs/s Dados: $m=4$ servidores $\lambda =22$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

Cenário B: adição de mais um servidor ($m=5$) Dados: $m=5$ servidores $\lambda =18$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

d) Emita parecer técnico.

---

Questão 10: Serviço de busca do acervo

Dados: $m=4$ $\lambda =18$ $\mu =6$ Meta: Tempo médio na fila inferior a 100ms

a) calcule a, ρ, P0, C(m,ρ), E[nw], E[ns], E[n], E[s] e E[w]

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

b) verifique se a meta é atendida

c) faça análise de sensibilidade para:

Cenário A: aumento da carga para 22 reqs/s Dados: $m=4$ servidores $\lambda =22$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

Cenário B: adição de mais um servidor ($m=5$) Dados: $m=5$ servidores $\lambda =18$ reqs/s $\mu =6$ reqs/s Meta: Tempo médio na fila inferior a 100ms

$\rho =\frac{\lambda }{m\mu }$ $U=\frac{\lambda }{\mu }$ $P_n=\{\begin{array}{l}\frac{1}{n!}{U}^n\cdot P_0\text{se}1\le n<m\\ \frac{1}{m!m^{n-m}}{U}^n\cdot P_0\text{se}n\ge m\end{array}$ $P_0={\left(\frac{U^m}{m!(1-\rho )}+{\sum }_{n=0}^{m-1}\frac{U^n}{n!}\right)}^{-1}$ $C(m,\rho )={\text{P}}_0\cdot \frac{U^m}{m!(1-\rho )}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }\cdot C(m,\rho )$ $\text{E}[n_s]=m\rho$ $\text{E}[n]=\text{E}[n_w]+\text{E}[n_s]$ $\text{E}[s]=\frac{\text{E}[n]}{\lambda }$ $\text{E}[w]=\frac{\text{E}[n_w]}{\lambda }$

d) Emita parecer técnico.

---

Exercícios Fila M/M/1/B/FIFO

Fórmulas de referência

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

Questão 1: Gateway de pagamentos instantâneos

Dados: $\lambda =15$ $\mu =20$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

a) calcule $\rho$, $P_0$, $P_B$, ${\rho }^{\prime }$, ${\lambda }^{\prime }$, ${\lambda }^{\prime }$, $E[n]$, $E[n_w]$, $E[s]$ e $E[w]$;

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=6$; Dados: $\lambda =15$ $\mu =20$ $B=6$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

cenário B: otimização do serviço para $\mu =24$ req/s; Dados: $\lambda =15$ $\mu =24$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

d) emita parecer técnico.

---

Questão 2: Spooler de impressão do laboratório

Dados: $\lambda =12$/min $\mu =16$/min $B=5$

Meta: A probabilidade de bloqueio deve ser inferior a $6$s

a) calcule $\rho$, $P_0$, $P_B$, ${\rho }^{\prime }$, ${\lambda }^{\prime }$, ${\lambda }^{\prime }$, $E[n]$, $E[n_w]$, $E[s]$ e $E[w]$;

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=6$; Dados: $\lambda =15$ $\mu =20$ $B=6$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

cenário B: otimização do serviço para $\mu =24$ req/s; Dados: $\lambda =15$ $\mu =24$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

d) emita parecer técnico.

---

Questão 3: Coletor de mensagens de sensores

Dados: $\lambda =15$ $\mu =20$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

a) calcule $\rho$, $P_0$, $P_B$, ${\rho }^{\prime }$, ${\lambda }^{\prime }$, ${\lambda }^{\prime }$, $E[n]$, $E[n_w]$, $E[s]$ e $E[w]$;

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=6$; Dados: $\lambda =15$ $\mu =20$ $B=6$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

cenário B: otimização do serviço para $\mu =24$ req/s; Dados: $\lambda =15$ $\mu =24$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

d) emita parecer técnico.

---

Questão 4: Proxy reverso do portal institucional

Dados: $\lambda =15$ $\mu =20$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

a) calcule $\rho$, $P_0$, $P_B$, ${\rho }^{\prime }$, ${\lambda }^{\prime }$, ${\lambda }^{\prime }$, $E[n]$, $E[n_w]$, $E[s]$ e $E[w]$;

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=6$; Dados: $\lambda =15$ $\mu =20$ $B=6$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

cenário B: otimização do serviço para $\mu =24$ req/s; Dados: $\lambda =15$ $\mu =24$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

d) emita parecer técnico.

---

Questão 5: Triagem automática de chamados

Dados: $\lambda =15$ $\mu =20$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

a) calcule $\rho$, $P_0$, $P_B$, ${\rho }^{\prime }$, ${\lambda }^{\prime }$, ${\lambda }^{\prime }$, $E[n]$, $E[n_w]$, $E[s]$ e $E[w]$;

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=6$; Dados: $\lambda =15$ $\mu =20$ $B=6$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

cenário B: otimização do serviço para $\mu =24$ req/s; Dados: $\lambda =15$ $\mu =24$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

d) emita parecer técnico.

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Questão 6: Exercício 6 — Controlador de consultas SQL

Dados: $\lambda =15$ $\mu =20$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

a) calcule $\rho$, $P_0$, $P_B$, ${\rho }^{\prime }$, ${\lambda }^{\prime }$, ${\lambda }^{\prime }$, $E[n]$, $E[n_w]$, $E[s]$ e $E[w]$;

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=6$; Dados: $\lambda =15$ $\mu =20$ $B=6$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

cenário B: otimização do serviço para $\mu =24$ req/s; Dados: $\lambda =15$ $\mu =24$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

d) emita parecer técnico.

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Questão 7: Servidor de integração contínua com buffer finito

Dados: $\lambda =15$ $\mu =20$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

a) calcule $\rho$, $P_0$, $P_B$, ${\rho }^{\prime }$, ${\lambda }^{\prime }$, ${\lambda }^{\prime }$, $E[n]$, $E[n_w]$, $E[s]$ e $E[w]$;

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=6$; Dados: $\lambda =15$ $\mu =20$ $B=6$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

cenário B: otimização do serviço para $\mu =24$ req/s; Dados: $\lambda =15$ $\mu =24$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

d) emita parecer técnico.

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Questão 8: Validador de notas fiscais eletrônicas

Dados: $\lambda =15$ $\mu =20$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

a) calcule $\rho$, $P_0$, $P_B$, ${\rho }^{\prime }$, ${\lambda }^{\prime }$, ${\lambda }^{\prime }$, $E[n]$, $E[n_w]$, $E[s]$ e $E[w]$;

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=6$; Dados: $\lambda =15$ $\mu =20$ $B=6$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

cenário B: otimização do serviço para $\mu =24$ req/s; Dados: $\lambda =15$ $\mu =24$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

d) emita parecer técnico.

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Questão 9: Transcodificador de videoaulas com fila limitada

Dados: $\lambda =15$ $\mu =20$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

a) calcule $\rho$, $P_0$, $P_B$, ${\rho }^{\prime }$, ${\lambda }^{\prime }$, ${\lambda }^{\prime }$, $E[n]$, $E[n_w]$, $E[s]$ e $E[w]$;

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=6$; Dados: $\lambda =15$ $\mu =20$ $B=6$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

cenário B: otimização do serviço para $\mu =24$ req/s; Dados: $\lambda =15$ $\mu =24$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

d) emita parecer técnico.

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Questão 10: Terminal de autoatendimento da biblioteca

Dados: $\lambda =15$ $\mu =20$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

a) calcule $\rho$, $P_0$, $P_B$, ${\rho }^{\prime }$, ${\lambda }^{\prime }$, ${\lambda }^{\prime }$, $E[n]$, $E[n_w]$, $E[s]$ e $E[w]$;

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=6$; Dados: $\lambda =15$ $\mu =20$ $B=6$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

cenário B: otimização do serviço para $\mu =24$ req/s; Dados: $\lambda =15$ $\mu =24$ $B=4$

Meta: A probabilidade de bloqueio deve ser inferior a $8\%$

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }n<B\\ 0& \text{se }n\ge B\end{array}$$

$\rho =\frac{\lambda }{\mu }$

${\rho }^{\prime }=\rho (1-P_B)$

$$P_n=\{\begin{array}{llc}& \frac{1-\rho }{1-{\rho }^{B+1}}{\rho }^n,& \text{se }0\le n<B\\ & 0& \text{se }n>B\end{array}$$

$P_B={\left(\frac{(1-\rho )}{(1-{\rho }^{(B+1)})}\right)}^{{\rho }^B}$ ${\lambda }^{\prime }=\lambda (1-P_B)$

${\lambda }_{\text{perda}}=\lambda {\text{P}}_B$ $\text{E}[n]=\frac{\rho }{1-\rho }-\frac{(B+1){\rho }^{B+1}}{1-{\rho }^{B+1}}$ $\text{E}[n_w]=\frac{\rho }{1-\rho }-\rho \frac{1+B{\rho }^B}{1-{\rho }^{B+1}}$

$\text{E}[s]=\frac{E[n]}{\lambda (1-P_B)}$ $\text{E}[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

d) emita parecer técnico.

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Exercícios Fila M/M/m/B/FIFO

Fórmulas Referência:

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

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Questão 1: Cluster de autenticação com capacidade finita

Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

a) calcule a, ρ, P_0, P_B, ρ’, λ’, λ_perda, E[n], E[n_w], E[s] e E[w];

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

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b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=8$; Dados: $m=3$ servidores $B=8$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

---

cenário B: aumento da taxa de serviço para $\mu =7$ req/s por servidor; Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =7$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

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d) emita parecer técnico.

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Questão 2: Gateway SQL com fila limitada

Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

a) calcule a, ρ, P_0, P_B, ρ’, λ’, λ_perda, E[n], E[n_w], E[s] e E[w];

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

---

b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=8$; Dados: $m=3$ servidores $B=8$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

---

cenário B: aumento da taxa de serviço para $\mu =7$ req/s por servidor; Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =7$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

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d) emita parecer técnico.

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Questão 3: Sistema de triagem de chamados

Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

a) calcule a, ρ, P_0, P_B, ρ’, λ’, λ_perda, E[n], E[n_w], E[s] e E[w];

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

---

b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=8$; Dados: $m=3$ servidores $B=8$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

---

cenário B: aumento da taxa de serviço para $\mu =7$ req/s por servidor; Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =7$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

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d) emita parecer técnico.

Questão 4: Renderização de mapas com buffer limitado

Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

a) calcule a, ρ, P_0, P_B, ρ’, λ’, λ_perda, E[n], E[n_w], E[s] e E[w];

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

---

b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=8$; Dados: $m=3$ servidores $B=8$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

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cenário B: aumento da taxa de serviço para $\mu =7$ req/s por servidor; Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =7$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

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d) emita parecer técnico.

Questão 5: Moderação automática de mensagens

Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

a) calcule a, ρ, P_0, P_B, ρ’, λ’, λ_perda, E[n], E[n_w], E[s] e E[w];

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

---

b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=8$; Dados: $m=3$ servidores $B=8$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

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cenário B: aumento da taxa de serviço para $\mu =7$ req/s por servidor; Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =7$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

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d) emita parecer técnico.

Questão 6: Runners de integração contínua com capacidade finita

Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

a) calcule a, ρ, P_0, P_B, ρ’, λ’, λ_perda, E[n], E[n_w], E[s] e E[w];

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

---

b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=8$; Dados: $m=3$ servidores $B=8$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

---

cenário B: aumento da taxa de serviço para $\mu =7$ req/s por servidor; Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =7$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

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d) emita parecer técnico.

Questão 7: Transcodificação de videoaulas

Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

a) calcule a, ρ, P_0, P_B, ρ’, λ’, λ_perda, E[n], E[n_w], E[s] e E[w];

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

---

b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=8$; Dados: $m=3$ servidores $B=8$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

---

cenário B: aumento da taxa de serviço para $\mu =7$ req/s por servidor; Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =7$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

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d) emita parecer técnico.

Questão 8: Validação fiscal com cluster finito

Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

a) calcule a, ρ, P_0, P_B, ρ’, λ’, λ_perda, E[n], E[n_w], E[s] e E[w];

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

---

b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=8$; Dados: $m=3$ servidores $B=8$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

---

cenário B: aumento da taxa de serviço para $\mu =7$ req/s por servidor; Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =7$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

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d) emita parecer técnico.

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Questão 9: Busca do acervo com fila limitada

Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

a) calcule a, ρ, P_0, P_B, ρ’, λ’, λ_perda, E[n], E[n_w], E[s] e E[w];

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

---

b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=8$; Dados: $m=3$ servidores $B=8$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

---

cenário B: aumento da taxa de serviço para $\mu =7$ req/s por servidor; Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =7$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

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d) emita parecer técnico.

Questão 10: Processamento de eventos IoT

Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

a) calcule a, ρ, P_0, P_B, ρ’, λ’, λ_perda, E[n], E[n_w], E[s] e E[w];

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

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b) verifique se a meta é atendida;

c) faça análise de sensibilidade para:

cenário A: aumento da capacidade para $B=8$; Dados: $m=3$ servidores $B=8$ reqs $\lambda =14$ reqs/s $\mu =6$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

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cenário B: aumento da taxa de serviço para $\mu =7$ req/s por servidor; Dados: $m=3$ servidores $B=6$ reqs $\lambda =14$ reqs/s $\mu =7$ reqs/s Meta: Probabilidade de bloqueio inferior a 5%

$${\lambda }_n=\{\begin{array}{ll}\lambda & \text{se }0\le n<B\\ 0& \text{se }n\ge B\end{array}$$ $${\mu }_n=\{\begin{array}{ll}n\mu & \text{se }1\le n<B\\ m\mu & \text{se }m\le n\le B\end{array}$$

$\rho =\frac{\lambda }{m\mu }$

$$\{\begin{array}{ll}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }1\le n<B-1\\ \frac{1}{m^{n-m}m!}{\left(\frac{\lambda }{\mu }\right)}^n{P}_0& \text{se }m\le n\le B\end{array}$$ $$P_n=\{\begin{array}{ll}{\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{B-m+1}{m!}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }=1\\ {\left[{\sum }_{n=0}^{m-1}\frac{1}{n!}{\left(\frac{\lambda }{\mu }\right)}^n+\left(\frac{\lambda }{\mu }\right)\frac{(1-\rho {)}^{B-m+1}}{m!(1-\rho )}\right]}^{-1}& \text{se }\rho =\frac{\lambda }{m\mu }\ne 1\end{array}$$

$U=\rho (1-P_0)$ ${\lambda }^{\prime }=\lambda (1-P_B)$ ${\lambda }_{\text{perda}}=\lambda P_B$ $E[n]={\sum }_{n=1}^{B}n{P}_n$ $E[n_w]={\sum }_{n=m+1}^B(n-m)P_n$ $E[s]=\frac{E[n]}{\lambda (1-P_B)}$ $E[w]=\frac{E[n_w]}{\lambda (1-P_B)}$

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d) emita parecer técnico.