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Sagot :
Answer:
Following are the solution to the given points:
Explanation:
When the arrival rate matches the distribution of Poisson, the rate of delivery varies exponentially as well as the server exists. The paradigm, therefore, is W/M/I, an exponential time interval concept, and one server.
Arrival rate [tex](\lambda) = 14\ per\ hour[/tex]
Service rate[tex]( \mu) =\frac{3\ minutes}{customer} = \frac{60 minutes (1 hour)}{3 minutes}= 20 \ \frac{customers}{hour}[/tex]
For point A:
System utilization[tex]=\frac{\lambda}{\mu}=\frac{14}{20}=0.7[/tex]
For point B:
Average number in line:
[tex](L_q )=\frac{\lambda^2}{\mu \times (\mu-\lambda)}\\\\=\frac{14^2}{20\times (20-14)}\\\\= 1.633[/tex]
For point C:
Average time in line:
[tex](W_q)=\frac{L_q}{\lambda}\\\\=\frac{1.633}{14} \ hours\\\\=\frac{1.633}{14} \times 60 \ minutes\\\\= 7\ minutes[/tex]
For point D:
Average time in the system:
[tex](W_s)=w_q+\frac{1}{\mu}\\\\=7\ minutes + \frac{1}{20} \ hours\\\\=7 \ minutes + \frac{60}{20} \ hours\\\\= 7 \ minutes + 3\ minutes\\\\= 10 \ minutes\\[/tex]
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