$($a$)$ Model the system using an M$/$M$/1/$b queue.

To model the system using an M$/$M$/1/$b queue, we first need to calculate the arrival rate $(\backslash$lambda $)$ and service rate $(\backslash$mu $)$ for each machine. The arrival rate for the first machine is given by the number of jobs in the buffer, which is $10$ jobs, since the buffer can hold only $10$ jobs. The arrival rate is therefore $10$ jobs$/20$ minutes $=0\mathrm{.}5$ jobs$/$minute$.$ The arrival rate for the second machine is given by the number of jobs in the buffer, which is $0$ jobs, since the buffer can hold only $10$ jobs. The arrival rate is therefore $0$ jobs$/20$ minutes $=0$ jobs$/$minute$.$ The service rate for the first machine is given by the processing time per job, which is $20$ minutes, since each job takes $20$ minutes to process. The service rate for the second machine is given by the processing time per job, which is $20$ minutes, since each job takes $20$ minutes to process.

The parameters for the M$/$M$/1/$b queue are as follows:

Arrival rate $(\backslash$lambda $)$: $0\mathrm{.}5$ jobs$/$minute for the first machine, $0$ jobs$/$minute for the second machine

Service rate $(\backslash$mu $)$: $20$ minutes$/$job for the first machine, $20$ minutes$/$job for the second machine

Number of machines $($m$)$: $2$

Number of buffers $($b$)$: $1$

To calculate the throughput, we use Little's law, which states that the average number of jobs in the system $($L$)$ is equal to the average arrival rate $(\backslash$lambda $)$ multiplied by the average time a job spends in the system $($W$)$:

L $=\backslash$lambda W

We can calculate the average time a job spends in the system as follows:

W $=20$ minutes $+20$ minutes $=40$ minutes

Substituting the values for $\backslash$lambda and W$,$ we get:

L $=\backslash$lambda W

L $=(0\mathrm{.}5$ jobs$/$minute$)*(40$ minutes$)$

L $=20$ jobs$/$hour

The throughput is therefore $20$ jobs$/$hour$.$

To calculate the partial WIP, we use Little's law again, this time with the average number of jobs in the system $($L$)$ and the average time a job spends in the system $($W$)$:

L $=\backslash$lambda W

L $=20$ jobs$/$hour $*40$ minutes

L $=800$ jobs$/$hour

The partial WIP is therefore $800$ jobs$/$hour$.$

To calculate the total cycle time, we first need to calculate the average time a job spends in the system $($W$).$ As mentioned earlier, the processing time for each job is $20$ minutes, so the average time a job spends in the system is $40$ minutes.

Next, we use Little's law to calculate the average number of jobs in the system $($L$)$:

L $=\backslash$lambda W

L $=(0\mathrm{.}5$ jobs$/$minute$)*(40$ minutes$)$

L $=20$ jobs$/$hour

Finally, we can calculate the total cycle time as the sum of the processing time for each job and the average time a job spends in the system:

Total cycle time $=$ Processing time per job $+$ Average time a job spends in the system

Total cycle time $=20$ minutes $+40$ minutes

Total cycle time $=60$ minutes

The total cycle time for the line is therefore $60$ minutes.

To calculate the total WIP, we use Little's law again, this time with the average number of jobs in the system $($L$)$ and the average time a job spends in the system $($W$)$:

L $=\backslash$lambda W

L $=20$ jobs$/$hour $*40$ minutes

L $=800$ jobs$/$hour

The total WIP is therefore $800$ jobs$/$hour$.$

$($b$)$ Reduce the buffer to one $($so that b$=3),$ and recompute the above measures. What happens to throughput, cycle time, and WIP? Comment on this as a strategy.

When we reduce the buffer from two machines to one machine, we need to adjust the arrival rate and service rate for the first machine accordingly. The arrival rate for the first machine is now given by the number of jobs in the buffer, which is $3$ jobs, since the buffer can hold only $3$ jobs. The arrival rate is therefore $3$ jobs$/20$ minutes $=0\mathrm{.}15$ jobs$/$minute$.$ The service rate for the first machine is given