1. Module C10 Simulation of Inventory/Queuing Models

2. INVENTORY SIMULATIONS Daily demand for refrigerators at Hotpoint City has a probability distribution Lead time is not fixed but has a probability distribution Customers who arrive and find Hotpoint out of stock will shop elsewhere and Hotpoint will lose the sale These conditions do not meet the restrictions of inventory models developed earlier

3. Simulation Approach Simulation cannot determine the best inventory policy But it can compare policies Compare the following: –Reordering 10 when supply reaches 6 or less –Reordering 12 when supply reaches 3 or less

4. Hotpoint Input Data Current inventory = 10 Holding costs: $2/refrigerator/day Order costs: $50 per order Shortage costs: $30 per occurrence (sale is lost) Demand/day Prob Lead Time Prob 0.080 days.05 1.371 day.55 2.332 days.30 3.173 days.10 4.05

5. RANDOM NUMBER MAPPINGS DAILY DEMAND -- Use column 2 0 1 2 3 4 PROB.08.37.33.17.05 RN 00-07 08-44 45-77 78-94 95-99 LEAD TIME (DAYS) -- Use column 14 0 1 2 3 PROB.05.55.30.10 RN 00-04 05-59 60-89 90-99

6. SIMULATION OF Q*= 10; r* = 6 COSTS DAY BI RN DEM EI LOST ORDER RN LT ORD HOLD SHORT 110 33 1 9 --- 18 --- 2 9 98 4 5 --- YES 24 1 50 10 --- 3 5 26 1 4 --- 0 8 414 91 3 11 --- 22 --- 511 96 4 7 --- 14 --- 6 7 48 2 5 --- YES 63 2 50 10 --- 7 5 82 3 2 --- 1 4 8 2 27 1 1 --- 0 2 911 96 4 7 --- 14 --- 10 7 46 2 5 --- YES 99 3 50 10 --- 150 112 0 Based on this one 10-day simulation average daily cost = $26.20

7. SIMULATION OF Q*= 12; r* = 3 COSTS DAY BI RN DEM EI LOST ORDER RN LT ORD HOLD SHORT 110 33 1 9 --- --- --- --- --- 18 --- 2 9 98 4 5 --- --- --- --- --- 10 --- 3 5 26 1 4 --- --- --- --- --- 8 --- 4 4 91 3 1 --- YES 37 1 50 2 --- 5 1 96 4 0 3 --- --- 0 ---- 0 90 612 48 2 10 --- --- --- --- --- 20 --- 710 82 3 7 --- --- --- --- --- 14 --- 8 7 27 1 6 --- --- --- --- --- 12 --- 9 6 96 4 2 --- YES 84 2 50 4 --- 10 2 46 2 0 --- --- --- --- --- 0 --- 100 88 90 Based on this one 10-day simulation average daily cost = $27.80 THE OTHER POLICY APPEARS BETTER

8. QUEUING SIMULATIONS The arrival pattern to a bank is not Poisson There are three clerks with different service rates A customer must choose which idle server to go to These conditions do not meet the restrictions of queuing models developed earlier

9. TIME BETWEEN ARRIVALS MINUTES PROB RN 1.40 00-39 2.30 40-69 3.20 70-89 4.10 90-99

10. SERVICE TIME FOR ANN MINUTES PROB RN 3.1000-09 4.2010-29 5.3530-64 6.1565-79 7.1080-89 8.0590-94 9.0595-99

11. SERVICE TIME FOR BOB MINUTES PROB RN 2.0500-04 3.1005-14 4.1515-29 5.2030-49 6.2050-69 7.1570-84 8.1085-94 9.0595-99

12. SERVICE TIME FOR CARL MINUTES PROB RN 6.2500-24 7.5025-74 8.2575-99

13. CHOICE OF SERVER ALL THREE SERVERS IDLE CHOICE PROB RN ANN1/30000-3332 BOB1/33333-6665 CARL1/36666-9999* (* Carl’s prob. is.0001 more than 1/3) TWO SERVERS IDLE (A/B), (A/C), (B,C) CHOICE: A/B A/C B/C PROB RN Ann Ann Bob 1/2 0-4 Bob Carl Carl 1/2 5-9

14. ARBITRARY CHOICE OF COLUMNS FOR SIMULATION EVENT COLUMN ARRIVALS 10 CHOICE OF SERVER 15 ANN’S SERVICE 1 BOB’S SERVICE 2 CARL’S SERVICE 3

15. DESIRED QUANTITIES W q -- the average waiting time in queue W -- the average waiting time in system L q -- the average # customers in the queue L -- the average # customers in the system If we get estimates for W q and W, then we estimate: –L q = W q –L = W

16. WILL WE REACH STEADY STATE? Average time between arrivals = 1/ =.4(1) +.3(2) +.2(3) +.1(4) = 2.0 minutes = 60/2 = 30/hr. Ann’s average service time = 1/  A =.1(3) +.2(4) + …+.05(9) = 5.3 minutes  A = 60/5.3 = 11.32/hr.

17. WILL WE REACH STEADY STATE? Bob’s average service time = 1/  B =.05(2) +.1(3) + …+.05(9) = 5.5 minutes  B = 60/5.5 = 10.91/hr. Carl’s average service time = 1/  C =.25(6) +.50(7) +.25(8) = 7 minutes  C = 60/7 = 8.57/hr. = 30/hr.  A +  B +  C = 11.32 + 10.91 + 8.57 = 30.8/hr. Steady State will be reached

18. THE SIMULATION # RN IAT AT W q RN SERV SB RN ST SE W 1 36 1 8:01 0 4231 B 8:01 33 5 8:06 5 2 52 2 8:03 0 7 C 98 8 8:11 8 3 99 4 8:07 0 9 B 26 4 8:11 4 4 54 2 8:09 0 ------ A 8:09 88 7 8:16 7 5 96 4 8:13 0 8 C 00 6 8:19 6 6 20 1 8:14 0 ------ B 8:14 48 5 8:19 5 7 41 2 8:16 0 ------ A 8:16 11 4 8:20 4 8 31 1 8:17 2 6 C 8:19 61 7 8:26 9 9 33 1 8:18 1 ------ B 8:19 96 9 8:28 10

19. SIMULATION (CONT’D) # RN IAT AT W q RN SERV SB RN ST SE W 10 07 1 8:19 1 ------ A 8:20 62 5 8:25 6 11 21 1 8:20 5 ------ A 8:25 54 5 8:30 10 12 01 1 8:21 5 ------ C 8:26 49 7 8:33 12 13 20 1 8:22 6 ------ B 8:28 84 7 8:35 13 14 18 1 8:23 7 ------ A 8:30 69 6 8:36 13 15 92 4 8:27 6 ------ C 8:33 95 8 8:41 14 16 10 1 8:28 7 ------ B 8:35 63 6 8:41 13 17 90 4 8:32 4 ------ A 8:36 31 5 8:41 9 18 66 2 8:34 7 3711 B 8:41 05 3 8:44 10

20. CALCULATING THE STEADY STATE QUANTITIES The quantities we want are steady state quantities -- –The system must be allowed to settle down to steady state –Throw out the results from the first n customers Here we use n = 8 –Average the results of the rest Here we average the results of customers 9 -18

21. THE CALCULATIONS FOR W, W q Total Wait in the queue of the last 10 customers = (1+1+5+5+6+7+6+7+4+7) = 49 min. W q  49/10 = 4.9 min. Total Wait in the queue of the last 10 customers = (10+6+10+12+13+13+14+13+9+10) = 90 min. W  90/10 = 9.0 min.

22. THE CALCULATIONS FOR L, L q L q = W q and L = W and W and W q must be in the same time units – = 30/hr. =.5/min. L q = W q  (.5)(4.9) = 2.45 L = W  (.5)(9.0) = 4.5

23. Module C10 Review Simulation of Inventory Models –Determine System Parameters –Simulate Cost –Replicate Experiment or Longer Simulation for better results Simulation of Queuing Models –Determine System Parameters –Check to See if Steady State Will Be Reached –Simulate to get W Q and W –Use Little’s Laws to get L, L Q