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A computer processes jobs on a first-come, first served basis in a time-sharing environment. The jobs have Poisson arrival rates average 0.6 jobs per minute. The objective in processing these jobs is that they spend no more than 5 minutes, on average, in the system. Assuming exponential service times, how fast does the computer have to process jobs (in minutes), on average, to meet this objective

Respuesta :

ajeigbeibraheem

Answer:

0.8 minutes

Step-by-step explanation:

From the given information:

The arrival time for the jobs to the computer obeys a Poisson distribution;

Thus, the arrival rate is:

[tex]\lambda = 0.6 \ jobs \ per \ minute[/tex]

Assuming the average time spent on the jobs in the system is denoted by:

[tex]W_s= 5 \ minutes[/tex]

The average time a job process in the system can be expressed as follows:

[tex]W_s = \dfrac{1}{\mu - \lambda}[/tex]

From above formula:

[tex]\mu =[/tex] service rate

[tex]\lambda =[/tex] arrival rate

replacing the values;

[tex]5 = \dfrac{1}{\mu - 0.6}[/tex]

[tex]5(\mu - 0.6) = 1[/tex]

Open brackets

[tex]5 \mu - 3 = 1[/tex]

[tex]5 \mu = 3+ 1 \\ \\ \mu = \dfrac{4}{5}[/tex]

[tex]\mu =[/tex] 0.8 minutes