The Robotics Manufacturing Company operates an equipment repair business where emergency jobs arrive randomly at the rate of three jobs per 8-hour day. The company’s repair facility is a single-channel system operated by a repair technician. The service time varies, with a mean repair time of 2 hours and a standard deviation of 1.5 hours. The company’s cost of the repair operation is $28 per hour. In the economic analysis of the waiting line system, Robotics uses $35 per hour cost for customers waiting during the repair process.

a. What are the arrival rate and service rate in jobs per hour?

b. Show the operating characteristics including the total cost per hour.

c. The company is considering purchasing a computer-based equipment repair system
that would enable a constant repair time of 2 hours. For practical purposes, the standard
deviation is 0. Because of the computer-based system, the company’s cost of the
new operation would be $32 per hour. The firm’s director of operations said no to the
request for the new system because the hourly cost is $4 higher and the mean repair
time is the same. Do you agree? What effect will the new system have on the waiting
line characteristics of the repair service?

d. Does paying for the computer-based system to reduce the variation in service time
make economic sense? How much will the new system save the company during a
40-hour work week?

See the explanation for the answers

Explanation:

a.

Arrival rate =  jobs per hour

But Jobs per hour  =8hours/3 jobs

                               = 2.666 hours per job

                                =1/2.666

                               =0.375

Therefore, arrival rate (λ) = 0.375

Calculating the service rate as follows:

Service rate(μ) =  jobs per hour

                            = 1 repair job /2 hours

                             = 0.5

b.

Calculating the operating characteristics as follows:

Lq =   λ 2 62 + (λ / μ)2 / (2(1- λ / μ)

        =(.375x.375)(1.5x1.5)+(.375/.5)2/(2(1-.375/.5)

        =0.1406 x 2.25 + 0.5625 / 0.5

       =0.8789/0.5

       =  1.7578/hour

L     = Lq + λ / μ

      =1.7578+0.375/0.5

      = 2.5078/ hour

Wq=Lq/ λ

     = 1.7578/0.375

    =4.688 hour or 281.28 minutes

W=L/ λ

  = 2.5078/0.375

  = 6.688 hour or 401.28 minutes

Pw= λ /μ

    = 0.375/0.5

    = 0.75

TC=CwL+Csk

    =35x2.508+28x1

    =$ 115.78

TC = $ 115.78

(c)

Company is considering purchasing a computer-based equipment repair system  that would enable a constant repair time of 2 hours.

            Current system                                  new system

        standard deviation  σ = 1.5                standard deviation σ = 0

Lq                1.7578                                                    1.125

L                   2.5078                                                 1.875

Wq                4.6875                                                3.00

W                  6.6875                                                 5.00

Tc                 $115.77

TC   =cw L+csk

      = 35 (1.875) + 32 (1)

       = $97.63

d. Yes; Savings = 40 ($115.77 - $97.63)

                           = $725.60

Even with the advantages of the new system, W q= 3 shows an average waiting time of 3 hours. The company should consider a second channel or other ways of improving the emergency repair service

15 1/2 years old in california