An office employs several clerks who originate documents and one operator who enters the document information in a computer system. The group originates documents at a rate of 19 per hour. The operator can enter the information with average exponentially distributed time of 3 minutes. Assume the population is infinite, arrivals are Poisson, and queue length is infinite with FCFS discipline.

a. Calculate the percentage utilization of the operator. (Round your answer to 2 decimal places.)
Utilization of the operator in %:

b. Calculate the average number of documents in the system. (Round your answer to 1 decimal place.)
Average number of documents:

c. Calculate the average time in the system. (Round your answer to 1 decimal place.)
Average time in system in minutes:

d. Calculate the probability of four or more documents being in the system. (Round your intermediate calculations to 3 decimal places and final answer to 1 decimal place.
Probability in %:

Respuesta :

Solution and Explanation:

In this system, population is infinite, arrivals are poisson distributed, service time is exponentially distributed, queue length is infinite with FCFS discipline. Therefore M/M/1 model fits.

Arrival rate, λ = 19 customers per hour

Service rate, μ = 1 / service time =[tex]1 / 3 * 60=[/tex]20 customers per hour

a). Percentage utilization of the operator, [tex]\rho=\lambda / \mu=19 / 20[/tex] = 95.00%

b). Average number of documents in the system, [tex]\mathrm{L}=\rho /(1-\rho)=.95 /(1-.95)[/tex] = 19.00

c). Average time in system in minutes, [tex]\mathrm{W}=\mathrm{L} \lambda=19 / 19[/tex]  = 1 hour = 60 minutes

d). Probability of four or more documents being in the system = 1 - (probability of 0 document in the system + probability of 1 document in the system + probability of 2 documents in the system + probability of 3 documents in the system)

[tex]=1-\left((1-\rho)^{*} \rho^{0}+(1-\rho)^{*} \rho^{1}+(1-\rho)^{*} \rho^{2}+(1-\rho)^{*} \rho^{3}\right)[/tex]

[tex]=1-(.0500+.0475+.0451+.0429)[/tex]

[tex]=1-.1855[/tex]

= .8145 or 81.5%