SimulationSimulation

A Queuing Simulation

ExampleExample

• 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 These conditions do not meet the restrictions of queuing models developed earlierof queuing models developed earlier

TIME BETWEEN ARRIVALSTIME BETWEEN ARRIVALS

MINUTES PROB RN

1 .40 00-39

2 .30 40-69

3 .20 70-89

4 .10 90-99

SERVICE TIME FOR ANNSERVICE TIME FOR ANN

MINUTES PROB RN

3 .10 00-09

4 .20 10-29

5 .35 30-64

6 .15 65-79

7 .10 80-89

8 .05 90-94

9 .05 95-99

SERVICE TIME FOR BOBSERVICE TIME FOR BOB

MINUTES PROB RN

2 .05 00-04

3 .10 05-14

4 .15 15-29

5 .20 30-49

6 .20 50-69

7 .15 70-84

8 .10 85-94

9 .05 95-99

SERVICE TIME FOR CARLSERVICE TIME FOR CARL

MINUTES PROB RN

6 .25 00-24

7 .50 25-74

8 .25 75-99

CHOICE OF SERVERCHOICE OF SERVERALL THREE SERVERS IDLE

CHOICE PROB RN

ANN 1/3 0000-3332

BOB 1/3 3333-6665

CARL 1/3 6666-9999*

(* 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

ARBITRARY CHOICE OFARBITRARY CHOICE OFCOLUMNS FOR SIMULATIONCOLUMNS FOR SIMULATION

EVENT COLUMN

ARRIVALS 10

CHOICE OF SERVER 15

ANN’S SERVICE 1

BOB’S SERVICE 2

CARL’S SERVICE 3

DESIRED QUANTITIESDESIRED QUANTITIES

• WQ -- the average waiting time in queue

• W -- the average waiting time in system

• LQ -- the average # customers in the queue

• L -- the average # customers in the system

• If we get estimates for Wq and W, then from Little’s Laws we can estimate:– LQ = WQ

– L = W

WILL WE REACH STEADY WILL WE REACH STEADY STATE?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.

WILL WE REACH STEADY STATE?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.

< A + B + C ===> Will reachWill reach Steady State!Steady State!

THE SIMULATIONTHE SIMULATION# RN IAT AT WQ 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 8:03 98 8 8:11 8

3 99 4 8:07 0 9 B 8:07 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 8:13 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

SIMULATION (CONT’D)SIMULATION (CONT’D)

# RN IAT AT WQ 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

CALCULATING THE STEADY CALCULATING THE STEADY STATE QUANTITIESSTATE 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

CALCULATIONS FOR W, WCALCULATIONS FOR W, Wqq

• Total Wait in the queue of the last 10 customers = (1+1+5+5+6+7+6+7+4+7) = 49 min.

WWQQ 49/10 = 4.9 min. 4.9 min.

• Total Wait in the system of the last 10 customers = (10+6+10+12+13+13+14+13+9+10) = 90 min.

WW 90/10 = 9.0 min. 9.0 min.

CALCULATIONS FOR L, LCALCULATIONS FOR L, Lqq

• Little’s Laws: LQ = WQ and L = W and W and Wq must be in the same time units

= 30/hr. = .5/min.

• LLQQ= WQ (.5)(4.9) = 2.452.45

• LL = W (.5)(9.0) = 4.504.50• Est. of Average number of busy workersEst. of Average number of busy workers 4.50-

2.45 = 2.05• ρρ = est. of system utilization 2.05/3 = 0.68330.6833• Est. of Average number of idle workersEst. of Average number of idle workers 3- 2.05

= 0.950.95

Mapping for Continuous Random Mapping for Continuous Random VariablesVariables

• The Explicit inverse distribution method can be used to map a random number to exponential distribution variables.

P(x<t) = 1-e-µt

e-µt = 1-P(x<t)LN(e-µt) = LN(1-P(x<t))

-µt = LN(1-P(x<t))t = -LN(1-P(x<t))(1/µ)

Mapping for Continuous Random Mapping for Continuous Random Variables (continued)Variables (continued)

• Assume that service time is exponential distribution with = 2 (i.e. average of 2 customers per minute).• Random number generated =

0.333801• This RN is mapped to a service time

of 0.203 minutes

= - LN(1-0.333801)*(1/2)

Mapping for Continuous Random Mapping for Continuous Random Variables – Using ExcelVariables – Using Excel

=RAND()Drag to cell B13

=-LN(1-B4)/$B$1Drag to cell C13

• To map random number to normal distribution variables, use NORMINV function in Excel

• Assume that service time is normal distribution with µ = 5 minutes and = 1 minute

• Random number generated = 0.368734

• This RN is mapped to a service time of 4.66 minutes

=NORMINV(0.368734,5,1)

Random numbers and ExcelRandom numbers and Excel

ReviewReview• 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

WQ and W

• Ignore the results from the first few arrivals

• Use Little’s Laws to get L, LQ

• Average Number of Busy Workers = L - LQ