Example simulation execution The Able Bakers Carhops Problem There are situation where there are more than one service channel. Consider a drive-in restaurant where carhops takes orders and bring food to the cars. Cars arrive in the manner shown in Table 3.0. There are 2 carhops- Abu and Bakar. Abu works faster and works better than Bakar. The distribution of their service is as shown in Table 3.1 and Table 3.2.

Example simulation execution The Abu Bakar Carhops Problem Because this system has two service servers, the simulation is more complex. Because Abu is a senior, in this case a simple rule is applied. When both of them (carhops) idle then Abu will get the customer. Please take note that this condition is specifically for this drive-in restaurant and can change according to situation or other applied rules

Arrival queue server1 Service node Departure Figure 3.0 Abu and Bakar Carhops node diagram server2

Arrival Event Server1 Busy ? Unit enters service Unit enters Queue for service Server2 Busy ? Figure 3.1 Arrival Event Yes No Yes

Departure Event Another Unit waiting ? Begin server idle time Remove the waiting unit from the queue Begin servicing the unit YesNo Figure 3.1 Departure Event for each service station

Service Time (minutes) ProbabilityCumulative Probability Random- Digit Assignment Table 3.1 Interarrival Distribution of cars

Service Time (minutes) ProbabilityCumulative Probability Random- Digit Assignment Table 3.2 Service Time Distribution for Abu

Service Time (minutes) ProbabilityCumulative Probability Random- Digit Assignment Table 3.3 Service Time Distribution for Bakar

1. Average waiting time ( minutes ) = total time customer wait in queue (minutes) total numbers of customers == 2. Probability (wait) = Number of customers who wait total numbers of customers ==

3. Probability of idle server = total idle time of server (minutes) total run time of simulation == 4. Average service time (minutes) for each Abu And Bakar = Total service time (minutes) total numbers of customers ==

5. Expected Service time ( minutes ) E(s) = ∞ Σ sp(s) S=0 6. Average time between arrivals (minutes) = Sum of all times between arrival (minutes) Number of arrivals - 1 = =

7. Average waiting time of those who wait ( minutes ) = total time customer wait in queue (minutes) total numbers of customers who wait == 8. Average time customer spends in the system = total time customer spend in system (minutes) total numbers of customers ==

Example simulation execution The newspaper Seller’s Problem: Concern on the purchase and sale of newspaper. The paper seller buys the newspaper for 33 cents each and sells them for 50 cents each. The newspaper not sold for the day will be sold as scrap for 5 cents each. Newspaper can be purchase in bundle of 10 and usually bought in bundle of 50,60 so on. Three type of newdays good, fair and poor with probabilities 0.35,0.45 and The distribution on demand is in Table 3.4. Try to simulate demands for 20 days and recording profits for each Day.

The newspaper Seller’s Problem Profit’s formula = (Revenue from sales) – (cost of newspapers) – (Lost of profit from excess demand) + (salvage from sale of scrap papers)

DemandGoodFairPoor Table 3.4 Distribution of newspaper demands

Type of newspaper day ProbabilityCumulative Probability Random- Digit Assignment Good Fair Poor Table 3.5 Random digits assignment for Type of Newsday

A simple model Let me go through a simple example that clarifies some concepts. Suppose we wish to procure some items for a special store promotion (call them pans). We can buy these pans for RM22 and sell them for RM35. If we have any pans left after the promotion we can sell them to a discounter for RM15. If we run out of special pans, we will sell normal pans, of which we have an unlimited supply, and which cost us RM32. We must buy the special pans in advance. How many pans should we purchase?

Clearly, the answer depends on the demand. We do not know the demand that the promotion will generate. We do have enough experience in this business to have some feel for the probabilities involved. Suppose the probabilities are:

We will attack this problem through simulation. In this model, we have captured uncertainty (demand) in a single random variable. To simulate a possible demand, we will determine a random value. This returns a value between 0 and 1 with a number of nice properties: no value is more likely to occur than another, and two successive values are not correlated.

Suppose I generate the random number.78. How can I generate a demand. Simply assign each demand to a range of values proportional to its probability and determine where the.78 occurs. One possibility is: Looking at the ranges, we see the demand is 11. The demand for.35 is 10 while that for.98 is 13.

How can we use this random demand? Suppose we have decided to procure 10 pans. We could determine the total profit for each of our random demands: the profit for the first is 133, for the second is 130, and 139 for the third. Our estimate for the profit if we order 10 is $134. We could then go on to check the profit if we order a different amount. For instance, if we order 13, our profit is estimated at $

At the end, we would have a guess at the best order quantity, and an estimate of the expected profit. We would not know for certain that we made the right move, but statistical analysis can estimate how far off we might be (by the way, to get statistical significance for this problem, you need roughly 20 runs at each demand level). Note also, for this problem there is an analytic solution.