A waiting-line problem that cannot be modelled by standard distributions has been simulated. The table below shows the result of a Monte Carlo simulation. (Assume that the simulation began at 8:00 a.m. and there is only one server.
$18:0628:0828:07108:1838:12108:2848:24118:3958:3058:44\begin{array} { | c | c | c | c | } \hline \text { Customer Number } & \text { Arrival Time } & \text { Service Time } & \text { Service Ends } \\\hline 1 & 8 : 06 & 2 & 8 : 08 \\\hline 2 & 8 : 07 & 10 & 8 : 18 \\\hline 3 & 8 : 12 & 10 & 8 : 28 \\\hline 4 & 8 : 24 & 11 & 8 : 39 \\\hline 5 & 8 : 30 & 5 & 8 : 44 \\\hline\end{array}$ a. What is the average waiting time in line?
b. What is the average time in the system?

Question

A waiting-line problem that cannot be modelled by standard distributions has been simulated. The table below shows the result of a Monte Carlo simulation. (Assume that the simulation began at 8:00 a.m. and there is only one server.

18:0628:0828:07108:1838:12108:2848:24118:3958:3058:44\begin{array} { | c | c | c | c | } \hline \text { Customer Number } & \text { Arrival Time } & \text { Service Time } & \text { Service Ends } \\\hline 1 & 8 : 06 & 2 & 8 : 08 \\\hline 2 & 8 : 07 & 10 & 8 : 18 \\\hline 3 & 8 : 12 & 10 & 8 : 28 \\\hline 4 & 8 : 24 & 11 & 8 : 39 \\\hline 5 & 8 : 30 & 5 & 8 : 44 \\\hline\end{array}

a. What is the average waiting time in line?
b. What is the average time in the system?

Alana Hoblin · Accepted Answer

Final Answer : (a) Waiting time is 0 + 1 + 6 + 4 + 9 = 20. Average waiting time is 20/5 = 4.0 min; (b) Total time in system is 2 + 11 + 16 + 15 + 14 = 58. Average time in system is 58/5 = 11.6 min.

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