1. Simulation A Queuing Simulation

2. Example 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 earlierThese conditions do not meet the restrictions of queuing models developed earlier

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

4. 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

5. 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

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

7. 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 1/25-9

8. 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

9. 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 from Little’s Laws we can estimate: –L Q = W Q –L = W

10. 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.

11. 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. Will reachSteady State! Will reach Steady State!

12. 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

13. 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

14. 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

15. 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  4.9 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  9.0 min. W  90/10 = 9.0 min.

16. CALCULATIONS FOR L, L q Little’s Laws: 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 2.45L Q = W Q  (.5)(4.9) = 2.45 L4.50L = W  (.5)(9.0) = 4.50 ρ2.05ρ = est. of system utilization  4.50 -2.45 = 2.05 Est. of Average number of idle workers 0.95Est. of Average number of idle workers  3- 2.05 = 0.95

17. Review Simulation of Queuing Models to Determine System Parameters Check to See if Steady State Will Be Reached Determine random number mappings Use of pseudorandom numbers to estimate W Q and W Ignore the results from the first few arrivals Use Little’s Laws to get L, L Q Average Number of Busy Workers = L - L Q