Question

A television repairman finds that the time spent on TV sets has an exponential distribution with mean of 20 minutes. If he repairs sets in the order in which they came in, and if the arrival of sets follows a Poisson distribution approximately with an average rate of 15 per 8-hour day, what is the repairman’s expected idle time each day in hours?
1. 3
2. 2
3. 5
4. 4

Answer 1

Correct Answer - Option 1 : 3

Concept:

Utilization factor, \({\rm{ρ }} = \frac{{\rm{λ }}}{{\rm{μ }}}\)

Where, λ = Arrival rate, μ = Service rate

Idle time = 1 – ρ   

Calculation: 

Given:

Arrival of sets follows a Poisson distribution

Arrival rate (λ) = \(\frac{15}{8}\) sets per hour

Service time for one TV set is 20 min

Therefore, Service rate (μ) \( = \frac{1}{20}\;\times\;60 = 3\) sets per hour

Utilization factor, \({\rm{ρ }} = \frac{{\rm{λ }}}{{\rm{μ }}} = \frac{5}{8}\)

Idle time = 1 – ρ = 1 - \(\frac{5}{{8}} = \;\frac{3}{{8}}{\rm{\;hour}}\) 

Therefore, Idle time for an 8-hour shift

= \(\frac{3}{{8}} \times 8 \) = 3 hours

The repairman’s expected idle time is 3 hours