Question

Jobs arrive at a facility at an average rate of 5 in an 8 hour shift. The arrival of the jobs follows Poisson distribution. The average service time of a job on the facility is 40 minutes. The service time follows exponential distribution. Idle time (in hours) at the facility per shift will be
1. \(\frac{5}{7}\)
2. \(\frac{{14}}{3}\)
3. \(\frac{7}{5}\)
4. \(\frac{{10}}{3}\)

Answer 1

Correct Answer - Option 2 : \(\frac{{14}}{3}\)

Concept:

Job arrival rate follows Poisson distribution

Arrival rate (λ) = \(\frac{5}{8}\frac{{{\rm{Jobs}}}}{{{\rm{Hour}}}}\)

Service time for one job is 40 min

Therefore, Service rate (μ) \( = \frac{3}{2}\frac{{{\rm{Jobs}}}}{{{\rm{Hour}}}}\)

 Utilization factor, \({\rm{\rho }} = \frac{{\rm{\lambda }}}{{\rm{\mu }}}\)

Idle time = 1 – ρ

Calculation: \({\rm{\lambda }} = \frac{5}{8}\frac{{{\rm{Jobs}}}}{{{\rm{Hour}}}},{\rm{\;\mu }} = \frac{3}{2}\frac{{{\rm{Jobs\;}}}}{{{\rm{Hour}}}}{\rm{\;\rho }} = \frac{{\rm{\lambda }}}{{\rm{\mu }}} = \frac{{\frac{5}{8}}}{{\frac{3}{2}}} = \frac{5}{{12}}\)

Idle time = 1 – ρ = 1 - \(\frac{5}{{12}} = \;\frac{7}{{12}}{\rm{\;hour}}\) 

Therefore, Idle time for 8-hour shift:

\(\frac{7}{{12}} \times 8 = \frac{{14}}{3}{\rm{\;hours}}\)