Question

The parts department of a large automobile dealership has a counter used exclusively for mechanics’ requests for parts. The time between requests can be modeled by a negative exponential distribution that has a mean of five minutes. A clerk can handle requests at a rate of 15 per hour, and this can be modeled by a Poisson distribution that has a mean of 15. Suppose there are two clerks at the counter. On average, how many mechanics would be at the counter, including those being served? What is the probability that a mechanic would have to wait for service? If a mechanic has to wait, how long would the average wait be? What percentage of time are the clerks idle? If clerks represent a cost of $20 per hour and mechanics a cost of $30 per hour, what number of clerks would be optimal in terms of minimizing total cost?

Answer 1

Avg. inter-arrival duration = 5 min. àλ= 12 / hr.

μ= 15 / hr.

M = 2 clerks

a. λ/μ = 12/ 15 = .80. , Lq= 0.152

Ls= Lq+ λ/μ= .152 + .80 = .952 mechanics

b. Wa= Lq/ λ = 0.152/12 = 0.0126 hr

c. ρ = λ/ Mμ = 12/ 2(15) = .40

% idle time = 1 - ρ= 1 - .40 = .60

d.     Clerks Costs   Lq + λ/μ = Ls Machanics costs TC

M M($20)       $Ls ($30 / hr)

1 $20 3.2+ .8 =4 $120 $120

2 $40   152 + .8 = .952   28.56 68.56

3 $60   .019 + .8 = .819 24.57 84.57

Total clerks = 2 clerks