Module C10

Simulation of Inventory/Queuing Models

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

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

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 .08 0 days .05

1 .37 1 day .55

2 .33 2 days .30

3 .17 3 days .10

4 .05

RANDOM NUMBER MAPPINGSDAILY 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

SIMULATION OF Q*= 10; r* = 6

COSTSDAY BI RN DEM EI LOST ORDER RN LT ORD HOLD SHORT

1 10 33 1 9 --- --- --- --- --- 18 ---

2 9 98 4 5 --- YES 24 1 50 10 ---

3 5 26 1 4 --- --- --- 0 --- 8 ---

4 14 91 3 11 --- --- --- --- --- 22 ---

5 11 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 ---

9 11 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

SIMULATION OF Q*= 12; r* = 3 COSTS

DAY BI RN DEM EI LOST ORDER RN LT ORD HOLD SHORT

1 10 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

6 12 48 2 10 --- --- --- --- --- 20 ---

7 10 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

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

TIME BETWEEN ARRIVALS

MINUTES PROB RN

1 .40 00-39

2 .30 40-69

3 .20 70-89

4 .10 90-99

SERVICE 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 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 CARL

MINUTES PROB RN

6 .25 00-24

7 .50 25-74

8 .25 75-99

CHOICE 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 OFCOLUMNS FOR SIMULATION

EVENT COLUMN

ARRIVALS 10

CHOICE OF SERVER 15

ANN’S SERVICE 1

BOB’S SERVICE 2

CARL’S SERVICE 3

DESIRED 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 we estimate:

– Lq = Wq

– L = W

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.

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 ===> Steady State will be reached

THE 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)

# 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 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

THE CALCULATIONS FOR W, Wq

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

• Wq 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.

THE CALCULATIONS FOR L, Lq

• Lq = Wq and L = W

and W and Wq must be in the same time units

= 30/hr. = .5/min.

• Lq = Wq (.5)(4.9) = 2.45

• L = W (.5)(9.0) = 4.5

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 WQ and W

– Use Little’s Laws to get L, LQ