In this case, the throughput \(\mu\) of the network is \(1/T_0\): only one server at a time is serving a job, the rest of the servers are forced to be idle as we only allow one job at a time in the network.

If there is one job in the system, \(\mu(1) = 1/T_0\). If there are two, both jobs can be simultaneously worked on, each at a different station, hence \(\mu(2)=2/T_0\). And so on, until all stations are filled, at the same time, by a job, in which case \(\mu(n) = n/T_0\). When there are yet more jobs in the network, only \(n\) can be simultaneously in service, the rest must wait at some station.

\begin{equation*} \mu(w) = \frac{n}{T_0} = \frac{n}{n S} = \frac 1S. \end{equation*}

\begin{equation*} \begin{split} \E W = \frac{\E L}{\lambda} = \frac{w}{\mu(w)} = \frac{w}{\min\{w, n\}} T_0 = \begin{cases} T_0=n S, & \text{ if } w\leq n, \\ \frac{w}{n} T_0 = w S, & \text{ if } w\geq n. \end{cases} \end{split} \end{equation*}

When \(w>n\), there must be \(w-n\) jobs in queue and the other \(n\) jobs are in service. So, in conclusion, in deterministic tandem networks adding more than \(n\) jobs does not increase the network's throughput, but only increases the waiting times. On the other hand, setting \(w<n\) minimizes the waiting times, but also negatively affects the throughput of the network.

The processing rates are \(\mu_1=1/1\) and \(\mu_2 = 1/2\) per hour, respectively. The utilizations \(\rho_i=\lambda_i/\mu_i\) are \(1/3\) and \(2/3\), respectively. Thus, machine 2 is the bottleneck: it has the highest utilization.

Since these machines are not related by a routing (jobs going from one machine to another), computing \(T_0\) and so on does not make much sense for the network. We can just consider each machine on its own. As we need \(2\) jobs to keep each machine busy, it follows that \(W_0=2\).

We can use Little's law to find \(T_0\). The rate at which jobs can be served is \(\mu = 1+1/2 = 3/2\). Thus,

\begin{equation*} T_0 = W_0/\mu = 2/(3/2)=4/3. \end{equation*}

Clearly, the average time a job spends in this system is \(4/3\) time units if jobs arrive at the rate they are served. (Recall, everything is deterministic.)

Can we see this in another way? Yes, because \(2\) out of \(3\) jobs have service time of \(1\), and \(1\) out of \(3\) has a service time of \(2\),

\begin{equation*} \frac23 1 + \frac 1 3 2 = \frac 4 3. \end{equation*}

The raw processing time is just the sum of the processing times at each station. All stations have the same arrival rate, hence the second machine, being the slowest, is the bottleneck. Hence,

\begin{align*} T_0 &= 2 + 3 + 2 = 7\\ r_b &= \mu_2 = \frac13 \\ W_0 &= r_b T_0 = \frac73. \end{align*}

Realize that machines at a station are in parallel, not in series. Thus, when we add capacity to station 2, we don't add an extra processing step to station 2, we increase capacity.

Now,

\begin{equation*} r_2 = \frac13 + \frac13 = \frac 23, \end{equation*}

so that station 2 is no longer the bottleneck. Station 1 and station 3 are the bottlenecks.

\begin{align*} r_b &= \frac12 \\ T_0 &= 2 + 3 + 2 = 7\\ W_0 &= r_b T_0 = \frac72. \end{align*}

Often, students think that the processing rate is the inverse of the processing time, i.e., \(t=r^{-1}\). This is not the case for multi-server stations. Thus, realize that the raw processing time at station 2 is still \(3\) hours, not \(3/2\).

Now,

\begin{equation*} r_2 = \frac13 + \frac 14 = \frac 7{12} > \frac12, \end{equation*}

hence, stations 1 and 3 are bottlenecks.

However, \(t_2\) cannot be \(3\) hours anymore: some jobs will spend 4 hours at station 2 rather than 3 hours. In fact, the computation of the raw processing time at station 2, \(t_2\) say, is a bit tricky. Some students reason that both machines serve half of the jobs, so that \(t_2=(3+4)/2\) hours. This, however, is not quite realistic. This way, you would load the slow machine all of the time, and the faster machine would be idle quite a bit of the time.

Rather than spreading the jobs evenly over the two machines, we can also assume that each machine takes in work in proportion to its processing rate. Then we reason as follows. Consider 12 hours. In those 12 hours, the slow machine processes 3 jobs with duration 4 and the fast machine processes 4 jobs with duration 3. Therefore:

\begin{equation*} t_2 = \frac 3 7 4 + \frac 4 7 3 = \frac{24}7. \end{equation*}

Another way to get this is like this. We can `stuff' station 2 with jobs. The most jobs that fit in simultaneously is 2, one job per machine. Then, with Little's law,

\begin{equation*} t_2 = \frac{W}{r_2} = \frac{2}{7/12} = \frac{24}7, \end{equation*}

i.e., the same as the other answer.

Now, finally,

\begin{align*} r_b &= \frac12 \\ T_0 &= 2 + \frac{24}7 + 2 = 4 + 3 + \frac37 = 7 \frac37\\ W_0 &= r_b T_0. \end{align*}

Finally, the most realistic assumption is that the fast machine is always working, and the slow machine covers the rest. In that case, suppose that the fast machine processes jobs at rate \(r_1\) with processing time \(s_1=3\) and the slow at rate \(r_2\) with processing time \(s_2=4\). Then, since jobs arrive with rate \(r_a\), the fast machine processes a fraction of \(r_1/r_a\) of the jobs, and the slow machine processes the leftovers, i.e., a fraction of \((r_a-r_1)/r_a\) of the jobs. The raw processing time then becomes

\begin{equation*} \begin{split} t_2 &= \frac{r_1}{r_a} s_1 + \frac{r_a-r_1}{r_a} s_2 \\ &= \frac{1/3}{1/2} 3 + \frac{1/2-1/3}{1/2} 4 \\ &= 2 + \frac{1/6}{1/2} 4 \\ &= 2 + \frac{4}{3} = 3\frac13, \end{split} \end{equation*}

and then,

\begin{equation*} T_0 = 2 + t_2 + 2 = 7\frac13, \end{equation*}

and this is a tiny bit less than the previous way of organizing things.

So, why are the above answers different? One reason is that the objective is not explicitly stated. It might be that, to minimize time in a multi-server station, it is optimal to let a fast machine idle and wait until the next job comes in. If we were to assign this next job to a slow machine, the time in the system may be longer. In general, it might be that schedules that minimize \(T_0\) (on average) are not work-conserving.

Because \(T_0\) increases and/or the bottleneck rate increases. In both cases, we need more \(W_0\) to fill the network and achieve that the throughput can be equal to \(r_b\) in the best case.

Now the slow machine at station 2 becomes superfluous. The fast machine at station 2 can cope with all jobs that arrive from station 1.

\begin{align*} r_b &= \frac12 \\ T_0 &= 2 + 1 + 2 = 5\\ W_0 &= r_b T_0. \end{align*}

Thus, why would you assign part of the jobs to the slow machine at station 2? There is no queue, the fast machine can cope with all the jobs. Moreover, assigning any job to the slow machine just adds to the cycle time.

Taking \(t_2=(1+3)/2=2\) is plainly silly.

See above. I would not remove the slow machine. In real life there is variability: the fast machine might fail for instance. Spare capacity is useful. However, if it is very costly to keep the slow machine operational, I would scrap it.

\begin{align*} r_2 &= \frac13 + \frac17 = \frac{10}{21} < \frac12,\\ \intertext{hence station 2 is still the bottleneck} t_2 &= \frac{3}{10}7 + \frac{7}{10}3 = \frac{42}{10}=\frac{21}{5}\text{hour}\\ T_0&=2+\frac{21}{5} +2\\ r_b &= \frac{10}{21}\\ W_0 &= r_b T_0. \end{align*}

Note that we add capacity, so that we can process more jobs per unit time, but the raw processing time increases!

Now there are two loops. One-third of the jobs pass all three stations. Station 1 can process these jobs at rate

\begin{equation*} \frac13 r_1 = \frac 13 \frac 12=\frac 16 \end{equation*}

per hour. Since this is smaller than \(r_2\) and \(r_3\), station 1 is the bottleneck rate for this loop. Thus, \(W_0\) in this loop is \(r_b T_0 = 1/6\cdot 7 = 7/6\).

The other loop only contains station 1. The bottleneck rate in this loop is

\begin{equation*} r_1\frac23 = \frac12\frac23 = \frac13. \end{equation*}

Thus, \(W_0 = r_b T_0 = 1/3\cdot 2 = 2/3\) for this loop.

Hence, the total WIP is \(7/6 + 2/3 = 11/6\) and

\begin{equation*} T_0 = \frac13 7 + \frac 23 2=\frac{11}3, \end{equation*}

since \(1/3\) take the long loop with \(T=7\) and \(2/3\) the short loop with \(T=2\). Thus, \(T_0\) is the weighted average raw processing time.

Another way to analyze this situation is like this. Station 1 is the bottleneck, because its utilization is 1, while

\begin{align*} \rho_2 &= \lambda_2 t_2 = \frac12\frac13 3 = \frac 12\\ \rho_3 &= \lambda_3 t_3 = \frac12\frac13 2 = \frac 13. \end{align*}

Since station 1 is the bottleneck in this network, \(r_b=1/2\). Since \(T_0 = 11/3\), which we obtain as the weighted average computation above, we must have that

\begin{equation*} W_0 = r_b T_0 = \frac12\frac{11}3 = \frac{11}6, \end{equation*}

which agrees with our earlier result.

Now stations 1 and 3 are the bottlenecks with \(r_b = 1/2\). For \(T_0\) we have

\begin{equation*} T_0 = \frac15 7 + \frac45 4= \frac{23}5. \end{equation*}

Thus,

\begin{equation*} W_0 = r_b T_0 = \frac 12 \frac{23}5 = \frac{23}{10}. \end{equation*}

Now stations 1 and 3 are the bottlenecks. Thus, both stations need to be part of the conwip loop. For that matter, it is just as easy to also include station 2, and just include all work on the floor in the WIP measurements.