Figure 6S-1
Customers enter a haberdashery that sells only ascots, cufflinks and suspenders and are served by the owner operator. The owner is the only employee, so if customer B arrives while customer A is being served, customer B must patiently wait until customer A exits the system. Customers are always willing to wait regardless of how long their wait is. The interarrival and service times are uniformly distributed and shown in the table below.
$$\begin{array} { | c | c | c | }
\hline \text { Range } & \text { Interarrival (min) } & \text { Service (min) } \
\hline 0.000 - 0.200 & 3 & 6 \
\hline 0.201 - 0.400 & 6 & 7 \
\hline 0.401 - 0.600 & 9 & 8 \
\hline 0.601 - 0.800 & 12 & 9 \
\hline 0.801 - 1.000 & 15 & 10 \
\hline
\end{array}$$ The stream of random number for a Monte Carlo simulation of the system appear in this table
$$\begin{array} { | c | c | }
\hline \text { Interarrival } & \text { Service } \
\hline .114 & .979 \
\hline .899 & .297 \
\hline .925 & .162 \
\hline .085 & .574 \
\hline .824 & .235 \
\hline .151 & .593 \
\hline .223 & .956 \
\hline .477 & .845 \
\hline
\end{array}$$
-Use the data from Figure 6S-1. What is the total time the owner/operator is idle (has no customers)from 8:00 am to 9:15 am?
A)3 minutes
B)8 minutes
C)16 minutes
D)20 minutes

Question

Figure 6S-1
Customers enter a haberdashery that sells only ascots, cufflinks and suspenders and are served by the owner operator. The owner is the only employee, so if customer B arrives while customer A is being served, customer B must patiently wait until customer A exits the system. Customers are always willing to wait regardless of how long their wait is. The interarrival and service times are uniformly distributed and shown in the table below.
$$\begin{array} { | c | c | c | } 
\hline \text { Range } & \text { Interarrival (min) } & \text { Service (min) } \
\hline 0.000 - 0.200 & 3 & 6 \
\hline 0.201 - 0.400 & 6 & 7 \
\hline 0.401 - 0.600 & 9 & 8 \
\hline 0.601 - 0.800 & 12 & 9 \
\hline 0.801 - 1.000 & 15 & 10 \
\hline
\end{array}$$ The stream of random number for a Monte Carlo simulation of the system appear in this table>
$$\begin{array} { | c | c | } 
\hline \text { Interarrival } & \text { Service } \
\hline .114 & .979 \
\hline .899 & .297 \
\hline .925 & .162 \
\hline .085 & .574 \
\hline .824 & .235 \
\hline .151 & .593 \
\hline .223 & .956 \
\hline .477 & .845 \
\hline
\end{array}$$
-Use the data from Figure 6S-1. What is the total time the owner/operator is idle (has no customers)from 8:00 am to 9:15 am?
A)3 minutes
B)8 minutes
C)16 minutes
D)20 minutes

Quizplus · Accepted Answer

The total time the owner/operator is idle can be calculated by first determining the arrival and service times for each customer based on the random numbers given and the ranges provided in the table. Then, by simulating the process from 8:00 am to 9:15 am (75 minutes), we can calculate the idle time.1. For each random number for interarrival, we find the corresponding interarrival time from the table. Similarly, for each random number for service, we find the corresponding service time.2. We simulate the arrival and service of each customer, keeping track of when the owner/operator starts serving a customer and when they finish.3. The idle time is the time between when the owner/operator finishes serving a customer and when the next customer arrives.Given the uniform distribution and the specific random numbers provided, the calculation involves determining the exact times customers arrive and are served, and then summing up any periods where there is no customer being served. The total time from 8:00 am to 9:15 am is 75 minutes, and by subtracting the total time spent serving customers and any time customers are waiting to be served, we find the idle time. The correct answer involves accurately simulating these arrivals and services based on the provided data and calculating the idle time accordingly.

Figure 6S-1 Customers Enter a Haberdashery That Sells Only Ascots

Figure 6S-1
Customers Enter a Haberdashery That Sells Only Ascots