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Problem 7

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Problem 7

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  1. Bank of San Pedro has only 1 teller. On average, 1customer comes every 6 minutes, and it takes the teller an average of 3 minutes to serve a customer. To improve customer satisfaction, the bank is going to implement a unique policy called, “We Pay While You Wait.” Once implemented, the bank will pay each customer $3 per minute while a customer waits in line. (So the clock starts when a customer joins the line, and stops when the customer begins to talk to the teller.) Bank of San Pedro hired you as a consultant and you are responsible for estimating how much the “We Pay While You Wait” program will cost. Assume linear cost. If a customer waits for ten seconds in line, Bank of San Pedro will pay $0.5. Assume that arrival follows Poisson and service time follows exponential distribution. Problem 7

  2. Problem 7 a) Compute the capacity of the teller • 10 customers/hour • 3.33 customers/hour • 20 customers/hour • 30 customers/hour • Cannot be determined It takes the teller an average of 3 minutes to serve a customer 60/3 = 20 customers per hour

  3. Problem 7 b) Calculate the proportion of the time the teller is busy. • 100% • 80% • 62% • 50% • 40% R=10, Rp= 20 R/Rp= 10/20 = 0.5 = 50%

  4. Problem 7 - Approximation c) How long, on average, does a customer wait in line? • 6 minutes • 4.8 minutes • 3 minutes • 2.6 minutes • 2 minutes Indeed Ii was even given in the problem. Ti= Ii/R Ti= 0.5/10 = .05 0.05(60) = 3 minutes

  5. Problem 7 d) Calculate the expected “hourly” cost of the “We Pay While You Wait” program. • $9 • $36 • $60 • $90 • $140 Ii =0.5. Therefore, a half of a customer is always there. For each hour one customer gets 60(3) = $180. Thus 0.5 customer gets $90. Perhaps you do not believe me. Each customer waits, on average, 3 minutes. He or she receives, on average, 3(3) =$9. There are 10 customers arriving per hour. The overall hourly cost of this program is 9*10=$90.

  6. Problem 7 e) Suppose each additional clerk costs X dollars per hour (including all other clerk related costs such as benefits, space and equipment hourly costs). Compute the maximum value of X if it is at our benefit to hire one additional clerk? If we have two clerks, Rp increases from 20 to 40, and utilization drops from 0.5 to 10/40 = 0.25 Ii reduced from 0.5 customers to 0.0447 customers = 0.4553

  7. Problem 7 The number of customers waiting in the line reduced by 0.455. It means each hour, there are 0.455 less customers waiting in line. The cost of each hour waiting per 1 customer is $180 The waiting cost of 0.455 customers is 180(0.4553) = 81.95 If the additional clerk costs less than $81.96 per hour it is at our benefit to hire her. • f) Suppose each additional clerk costs $30 per hour. How many new clerk should we hire, one or two? • Obviously, it is at our benefit to hire one clerk. • If we hire two clerks (to have 3 clerks), Rp increases to 60, and utilization drops 10/60 = 0.167

  8. Problem 7 • Ii is reduced from 0.045 customers to 0.008, a 0.037 customer reduction. • By adding the third clerk, there are 0.037 less customers waiting in line (each hour and always) • 180(0.037) = about $6-$7 • It is not at our benefit to hire the second additional clerk, pay $30 per hour capacity cost to reduce waiting cost by $6-$7 per hour. • And we will not pay more that $1-$2 to the fourth clerk.

  9. Problem 7 We did not need to do this much computations for the third clerk. With two clerks the total number of customers waiting in line was: Ii was equal to 0.045 for c =2. Even if we reduce the number of customers in the waiting line to 0,we have reduced the line by 0.045 customers. 0.045(180) = 8.1 It is not worth the cost $30 to benefit $8

  10. Problem 7 - Exact

  11. Problem 8. Effect of Pooling- M/M/2 vs. 2M/M/1 Tp R Server 1 Queue1 2M/M/1 Server 2 Queue2 R Tp Server 1 20/hr M/M/2 Queue Server 2

  12. Problem 8. Effect of Pooling- M/M/2 vs. 2M/M/1 10/hr 5 min Server 1 Queue1 2M/M/1 Server 2 Queue2 10/hr 5 min Server 1 20/hr M/M/2 Queue Server 2

  13. Problem 8. Effect of Pooling- M/M/2 vs. 2M/M/1 10/hr Tp= 5 min Tp= 5 min Server 1 Queue1 +0.833 +0.833 --------- 1.667 Ip= 0.833 Tp= 5 min Server 2 Queue2 10/hr Ip= 0.833 Design A: R = 10, Tp= 5 min., c =1. Rp = 1/5 per min. or 12 per hour U = R/Rp =10/12 = 5/6  Ip = cU = U = 0.833 or RTp =Ip  10(5/60) = 0.833

  14. Problem 8. Effect of Pooling- M/M/2 vs. 2M/M/1 Ti = 25 min. 10/hr Tp= 5 min Server 1 Queue1 Ip= 0.833 Ii = 4.167 Ti = 25 min. Tp= 5 min Server 2 Ii = 4.167 Queue2 10/hr Ip= 0.833 Ii = 4.167 RTi =Ii 10Ti=4.167  Ti = .4167 hr or 25 min.

  15. Problem 8. Effect of Pooling- M/M/2 vs. 2M/M/1 Ti = 25 min. Tp=5 10 Server 1 Queue1 2M/M/1 Server 2 Queue2 10 Ii = 2*4.167 = 8. 334 Tp Server 1 20/hr M/M/2 Queue Server 2

  16. Problem 8. Effect of Pooling- M/M/2 vs. 2M/M/1 Server 1 20/hr M/M/2 Queue Server 2 Ii = 3.839 RTi =Ii 20Ti=3.839  Ti = 0.1919 hr or 11.52.

  17. Problem 8. Effect of Pooling- M/M/2 vs. 2M/M/1 Server 1 20/hr M/M/2 Queue Server 2 Ii = 3.839 RTi =Ii 20Ti=3.839  Ti = 0.1919 hr or 11.52.

  18. Effect of Pooling • Why is Design B better than A? • Design A the waiting time of customer is dependent on the processing time of those ahead in the queue • Design B, the waiting time of customer is only partially dependent on each preceding customer’s processing time • Combining queues reduces variability and leads to reduce waiting times

  19. Problem 9. Exponential Probability Distribution In a single phase single server service process and exponentially distributed interarrival time and service times, the actual total time that a customer spends in the process is also exponentially. Suppose total time the customers spend in a pharmacy is exponentially distributed with mean of 15 minutes. The pharmacy has promised to fill all prescriptions in 30 minutes. What percentage of the customers cannot be served within this time limit? P(x≥30) = EXP(-30/15) = 0.1353 13.53% of customers will wait more than 30 minutes. =P(x≤30) = EXPON.DIST(30,1/15,1) = P(x≤30) = 0.864665 P(x≥ 30) = 1- P(x≤30) = 1- 0.864665 = 0.1353

  20. Problem 9. Exponential Probability Distribution 90% of customers are served in less than what time interval? 1-e-X0/= 0.9 Find X0 SOLVER

  21. Problem 9. Exponential & Poisson The Poisson distribution provides an appropriate Description of the number of occurrences per interval The exponential distribution provides an appropriate description of the length of the interval between occurrences Compute the probability of arrival of 3 customers in 30 minutes. One customer arrives per 15 minutes. The average number of customers arriving in 30 mins is 2. This is Poisson distribution. =POISSON.DIST(3,2,1) =0.857123

  22. Problem 10.a. Exponential Probability Distribution Average trade time in Ameritrade is one second. Ameritrade has promised its customers if trade time exceeds 5 second it is free (a $10.99 cost saving. The same promises have been practiced by Damion Pizza (A free regular pizza) and Wells Fargo ($5 if waiting time exceeds 5 minutes). There are 150,000 average daily trade. What is the cost to Ameritrade” P(x≥ X0) = e-X0/= e-5/1 = EXP(-5) = 0.006738 Probability of not meeting the promise is 0.6738% 0.006738*150,000* = 1011 orders @10.99 per order = 10.99*1011 = $11111 per day

  23. Problem 10.b. Exponential Probability Distribution What was the cost if they had improved their service level by 50%? that is to make it free for transactions exceeding 2.5 secs. P(x≥ X0) = e-X0/= e-2.5/1 = EXP(-2.5) = 0.082085 8.2085%*150,000*10.99 = $135317 per day We cut the promised time by half, our cost increased 12 times.

  24. Bank of San Pedro has only 1 teller. On average, 1customer comes every 6 minutes, and it takes the teller an average of 3 minutes to serve a customer. To improve customer satisfaction, the bank is going to implement a unique policy called, “We Pay If You Wait Long.” Once implemented, the bank will pay $10 to any customer being in the bank for more than 10 minutes. Ta and Tp both have exponential distribution. Compute Ti. Problem 10.c Ti= Ii/R Ti= 0.5/10 = .05 Ti = 0.05(60) = 3 minutes Tp = 3 minutes T = Ti +Tp = 6 min

  25. Problem 10.c. Exponential Probability Distribution P(x≥ X0) = e-X0/= e-10/6 = EXP(-10/6) = 0.188876 Probability of not meeting the promise is = 18.89% 0.180*10 = $1.89 Each hour 10 customers arrive Total cost = 1.89*10 = $19.9 per hour.

  26. Problem 11 American Vending Inc. (AVI) supplies vending food to a large university. Because students often kick the machines out of anger and frustration, management has a constant repair problem. The machines break down on an average of 3/hr, and the breakdowns are distributed in a Poisson manner. Downtime costs the company $25/hr/machine, and each maintenance worker gets $4 per hr. One worker can service machines at an average rate of 5/hr, distributed exponentially; 2 workers working together can service 7/hr, distributed exponentially; and a team of 3 workers can do 8/hr, distributed exponentially. What is the optimal maintenance crew size for servicing the machines?

  27. Problem 11 • c=???? • c = 1 • U=?? Down time cost = 25Ii Capacity Cost= 4(# of team members) Total Cost

  28. Problem 11 Have I made any mistakes? Downtime costs the company $25 /hr/machine. When the machine is down? Until it is up. In the waiting line it is down. In the processor until the end of the process it is down. There fore, besides Ii, I also need Ip

  29. Problem 11 Lets check by using Ti and Tp instead of Ii and Ip

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