So welcome to the review session for module four. It's the same format as in the past. We're going to work through a couple of questions. This time it's going to be five questions. And we'll do this in the format that I'll present the question to you, you pause the video, and then I'll present the answer to you once you're ready to check your results. Okay, so the first question is about this firm, My-law.com, and the way My-law works is basically they have a lawyer that is answering emails. And emails come in from 8 to 6 PM, and there are 10 emails arriving per hour. There's only one lawyer on call, and it takes a lawyer five minutes to write a response email, with a standard deviation of four minutes. All right, a couple of questions are coming up. Pause me now and take a shot at these questions yourself. All right, this is obviously a waiting time problem. The magic word here, as you see, wait. How long does a customer have to wait? And we want to be careful in this question because it is asking you to consider not just the wait question, the time in the queue. But you have to also if you want to wait for the email, the customer has to wait for the agent or the lawyer to get to the email. And then the customer also has to wait for the five minutes that it takes to respond to the email. So we are also looking for the processing time. Now the hardest thing is really coming up with the wait time, and so just remember time in the queue is P times u divided by 1-u times CVa square plus CVp square divided by 2. Let me emphasize it's one lawyer, so m = 1. For bigger m's, as we'll see in a moment on the next question, for a bigger m we need a other, bigger formula. So this is a simple case for m = 1. All right, 10 emails per hour, that means that there's an email coming in every 6 minutes, and so the inter-arrival time is 6. The processing time we notice, P is 5 minutes. And so if you look then at the utilization, utilization is flow rate divided by capacity. Flow rate is emails flow into the system every 6 minutes. 1 over 5 is our capacity. And so the utilization is really 5/6. That means here I have P equals to 5 times u, 5/6, divided by 1- 5/6 times CVa square. Coefficient of variation of the inter-arrival time is 1, and so that is simply a 1. 1 squared stays a 1. Plus CVp squared, so standard deviation here is four minutes and the mean is five minutes. And so we are dealing here with 0.8, 4 divided by 5, 0.8. And we're going to square this, and then we're going to divide both of these by 2, all right? So that is basically giving you, here we have 1- 5/6, that is 1 over 6. 5 over 6 divided by 1 over 6, that is going to be 5. 5 times 5 is 25. This other ratio here, this is 0.64 plus 1 is 1.64 divided by 2 is 0.82. And that is giving you 20.5 minutes. Again, this is just the time in the queue. If you want to add up to that the actually writing time of email, then you have to add a another five minutes, and that would give you then a total of 25.5 minutes is the answer to the question. Now b and c are actually, I don't know what I did when I wrote these questions, but these are really friendly questions in many ways. Just over ten hour days. Remember 10 emails are coming in per hour, and so really we have nothing but 100 emails here that are answered. And then finally, the last question asks us the time that the lawyer can use to pursue these large settlement cases. And for that we have to look at the idle time, which is 1 minus the utilization times the 10 hours in the day. We know that the idle time is one-sixth because the utilization is five-sixths, and so that is simply 10/6 hours per day. Okay, the next question, a small computer store here with five employees that is doing repairs for people like Jim, who has his computer broken. And so this computer store, PC Fixers, or PF for short, is going to get customers walk in every 10 minutes, and they take 40 minutes to fix the computers. And there are five computer technicians to do the work. All right, same business as usual. I shut up for a while, you work on the problem, and then press on Play again so that you hear my answers. Okay, so again we're dealing with a waiting time problem. But this time because of the multiple technicians, we are dealing with a more complicated waiting formula that has TQ as a function of M. Multiple servers expressed. So let's just start writing down the formula from the lecture. So we're looking at the expected wait time P divided by m times u raised to the square root 2(m + 1). And now very careful, the square root ends after the 1, minus 1. The minus 1 is outside the square root. 1 minus u times CVa squared plus CVp squared divided by 2. All right, so now let's look for what's easy and what is hard. We clearly see that there's a customer walking in every 10 minutes, so there is this inter-arrival time a. We have the service time P of 40 minutes. So we can first compute the utilization as the flow rate, which is 1 over a divided by the capacity, which is m over P, and so that simplifies to P divided by a times m. And that is, in this case, where we have a P of 40 divided by 10 times 5, so that is 40 by 50, or 0.8 utilization. Now note that in this example here, CVa is equal to 1. Because both the mean inter-arrival time at the standard deviation is equals to ten, so that cancels out. And CVp is also equal to 1 because 40 is both the mean and the standard deviation. So this whole thing here at the end is all going to be equal to 1. That simplifies the math because now we have 40 minutes divided by 5 servers times 0.8, the utilization raised to the power of 12, which is 2m plus 1. 5 plus 1 is 6, 2 times 6 is 12, so square root of 12 minus 1 divided by 0.2 times 1. And you plug that into your calculator. And at least I get 23.08. Now the next question asks for the length of the line or how many customers will be waiting for their computer to be fixed. This is ultimately something that we can determine with Little's Law, I = R x T. Remember already we know that there's a customer coming in every 10 minutes. So 1 over 10 customers per minute is going to be our flow rate. And our flow time, well, that's the 23.08 here in the queue, plus the 40 minutes that they spend waiting for the actual service time. And so this is then going to be 63.08 minutes. And that gives you simply 6.308 customers. And that is the inventory of customers in the store. Now Real Compute is a company that offers super computers to users via the Internet, and they get jobs every 4 minutes. Standard deviation of the inter-arrival time is also 4 minutes. Takes 10 minutes to execute these jobs. And here's the trick that it's just supercomputers are busy, the job actually gets rerouted to another supercomputer of another supplier of us or partner of us, in which case, we have to pay these people actually $40 per job. So you see the question here? Test yourself, and then put me back on play. All right, the key to this question is that this is not a waiting time question but a loss question. If you see words in the question such as redirect or the job, probability that the job can or cannot be executed. We are not dealing with waiting time questions, we're dealing with loss questions. For loss question, we have the airline loss function, and so remember we start by looking at the ratio of P divided by a. So we have a processing time here that is 10 minutes, and we have an inter-arrival time that is 4 minutes, so that ratio here is 2.5. We have m=4. We then look at the table that I had mentioned in the lecture, but that is posted in the wiki for the current week. Look at the table and there you will find that the probability that we have a loss with r=2.5 and m=4 is equal to 14.99%, so 0.1499. So notice the question here is asking for what's the probability with which an incoming job can be executed? And so we have to take 1 minus that probability as the answer to the question. And so the answer to this question is that is is 85.01%. The second part of the question in some sense looks at the overflow, and it's important to keep in mind that we have 15 jobs coming in per hour. That is just 60 minutes divided by the inter-arrival time of 4 minutes here. So 15 jobs come per hour and we know from the previous question that jobs will be turned down or turned away to the partner company, OnComputer that happens with a probability of 14.99%. So in that case simply we have to turn 2.2485 expected jobs away. And if we multiply that with a penalty of $40 per job, we're going to get an expected hourly payment to our partner company of $89.94. So this is dollars per hour. Okay, next question. We have a contractor here that is doing a lots of renovation and building work. And the contractor has six projects lined up, you see their processing times listed here. And he wants to minimize the average wait time that projects are waiting for. And so that's how he sequences his work, and the question is, what will he be doing in 30 days from now? All right, ready, you go first and then I'll give you the solution. All right, the key thing here that he is not working first come first served, but he's using the shortest processing time rule. The shortest processing time rule will minimize the average wait time of these projects. Now notice that regardless in which sequence, first come first serve, shortest processing time rule or whatever else, the total amount of time it takes is simply just the sum of these various tasks. But by doing the shortest processing time rule he can reduce the average wait time. It simply makes more sense that we start with a short job so that the long job here waits for the short job as opposed to the other way around. So if he follows the shortest processing time rule, he will do this job first, this job second, third, fourth, fifth, and this one here is going to be last. And so now I know the sequence in which he's going to do the work and so the first job is going to be over after two days. The second job he's going to touch is going to be over after 2 plus 6 is equal to 8 days. The next job is going to be over after 16 days. Then he's going to take the job number four that takes 10 days, so that's going to get him to 26 days. So after 26 days, he has the first four jobs completed. And then he's going to start the fifth job in his sequence, and that is going to be the new kitchen here at Rosemont. And so with that in mind, it is the new kitchen in Rosemont that answers the question on what he will be working 30 days from now. All right, the last question is about a call center. The call center right now has a constant staff available all over the day, and there is some seasonality in the demand. In particular, there is lots of demand in the morning hours and very little demand in the afternoon hours. So utilization in the call center is high in the morning and low in the afternoon. Now which of the following action will decrease the average waiting time in the call center? Take some time for yourself, and then I share my thoughts with you. All right, so this is a qualitative question. There is really no number here that you could crunch into the waiting time formula. Nevertheless, I think it's helpful to remind ourselves of the waiting time formula. Tq is equal to P divided by m, utilization raised to the 2m plus 1 minus 1 divided by 1 minus u times CVa squared plus CVp squared divided by 2. All right, now let's look at the suggestions that I put up here. Adding more servers. Adding more servers is a good idea, right? You notice the larger m here will decrease Tq. Now I want to be careful here in reminding you that the utilization which we defined as flow rate divided by capacity and as 1 over the inter-arrival time times m over P, and hence P divided by a times m. I want to just emphasize that the utilization is also a function of m and so if you're taking the sensitivity analysis as a derivative of Tq with respect to m, you cannot just look at this term. Also u is changing as you are adding service, but either way, adding service is a good thing and will shorten the waiting time. Decrease the service coefficient of variation. That one is very straightforward. You see the marginal impact here. CVp shows up just linearly in here and everything that you do to CVp will be reducing the wait time. So this one also makes sense. Decreasing the average service time, you see this will help through P here in the formula. But, same comment as on m, P is also sitting in u and will also have an auxiliary effect by allowing you to operate at a lower level of utilization. And so that one is certainly true as well. Now the last one, option d, is a little tricky here, leveling the demand. Why would that matter and how do you think about that? For that, please remember that if you plot the time in the queue as a function of the utilization u, you have this very steep non-linear effect. And so right now in the morning hours we have a very high level of utilization. So say we are here, a high level of utilization. And in the afternoon hours we have a low level of utilization. Now conceptually you see that if we could now bring this together and level the demand, so we move some of the demand from the morning into the afternoon hours, the folks who had previously a super short wait time, they will lose a little. They wait a little longer here. You see there is a little increase in the waiting time for these guys. But the morning people will really, really benefit. You notice that on average it's nice to actually have a constant level of utilization as opposed to having half of the customers get really high level of utilization and half of the customers having a really low level of utilization. So leveling the demand, because of this convexity, the steep increase in the waiting time as a function of u, will be a good thing. And thus we can check option e, All of the above.
