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Sharp Discounts Wholesale Club has two service desks, one at each entrance of the store. Customers arrive at each service desk at an average of one every six minutes. The service rate at each service desk is four minutes per customer.

Required:
a. How often (what percentage of time) is each service desk idle? (Round your answer to 2 decimal places.)
b. What is the probability that both service clerks are busy? (Round your answer to 4 decimal places.)
c. What is the probability that both service clerks are idle? (Do not round intermediate calculations. Round your answer to 4 decimal places.)
d. How many customers, on average, are waiting in line in front of each service desk? (Round your answer to 2 decimal places.)
e. How much time does a customer spend at the service desk (waiting plus service time)? (Do not round intermediate calculations. Round your answer to 2 decimal places.)

2 Answers

6 votes

Answer:

  • Service desk idle= 60%
  • Probability that both service clerks are busy= 0.16
  • Probability that both service clerks are idle= 0.36
  • Customers waiting in line in front of each service desk= 0.267
  • Time does a customer spend at the service desk= 3.34 minutes

Step-by-step explanation:

a) service desk idle

Lambda=1/5 =0.2

Mu= 1/2 = 0.5

Utilization factor= lambda/mu

= 0.2/0.5 = 0.4

% idle = 1-0.4 = 0.6*100

= 60%

b) Probability that both are busy = 0.4*0.4=0.16

c) Probability that both are idle = 0.6*0.6 = 0.36

d) Lq= lambda^2/mu*(mu-lambda)

= 0.2^2/0.5*(0.5-0.2)

= 0.267

e) Ls= lambda/(mu-lambda)

= 0.667

Ws= Ls/lambda= 0.667/0.2 = 3.34 minutes

User Mike Miller
by
3.7k points
4 votes

Answer:

Step-by-step explanation:

a)

Lambda=1/5 =0.2

Mu= 1/2 = 0.5

Utilization factor= lambda/mu

= 0.2/0.5 = 0.4

The percengtage of service desk idleness = 1-0.4 = 0.6*100

= 60%

b)

The probability that both service clerks are busy = 0.4*0.4=0.16

c)

The probability that both service clerks are idle = 0.6*0.6 = 0.36

d)

Lq= lambda^2/mu*(mu-lambda)

= 0.2^2/0.5*(0.5-0.2)

= 0.267

e)

Ls= lambda/(mu-lambda)

= 0.667

Ws= Ls/lambda= 0.667/0.2 = 3.34 minutes

User Charles Johnson
by
3.9k points