Operation Management Heizer Solution Module F

7/25/2019 Operation Management Heizer Solution Module F

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1 BUS P301:01

MODULE F: SIMULATIONSuggested Solutions to SelectedQuestions

Summer II, 2009

Question F.8

Time Between

Arrivals Prob. RN Interval Service Time Prob. RN Interval

1 0.2 0120 1 0.3 0130

2 0.3 2150 2 0.5 3180

3 0.3 5180 3 0.2 8100

4 0.2 8100

First arrival (RN = 14) at 11:01.

Service time = 3 (RN = 88) Leaves at 11:04.

Second arrival (RN = 74) at 11:04 (3 minutes after 1st).

Service time = 2 (RN = 32). Leaves at 11:06.

Third arrival (RN = 27) at 11:06.

Service time = 2 (RN = 36). Leaves at 11:08.

Fourth arrival (RN = 03) at 11:07. Must wait 1 minute for service to start.

Service time = 1 minute (RN = 24). Leaves at 11:09.

Question F.16

(a)

Demand for Cumulative Random NumberMercedes Freq. Probability Probability Interval*

6 3 0.083 0.083 01087 4 0.111 0.194 09198 6 0.167 0.361 20369 12 0.333 0.694 3769

10 9 0.250 0.944 709411 1 0.028 0.972 959712 1 0.028 1.000 9800

=36

=1.000

* Note that the cumulative probabilities have been rounded to two significant digits when used todevelop the random number intervals.

Lead Probability Cumulative Random No. Interval

1 0.44 0.44 01442 0.33 0.77 45773 0.16 0.93 78934 0.07 1.00 9400


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2 BUS P301:01

Let us arbitrarily choose a beginning inventory of 14 cars

Time Beg Rand Demand Sold End Lost Place Rand Lead

1 14 07 6 6 8 0 Yes 60 2

2 8 77 10 8 0 2 No

3 0 49 9 0 14 9 No

4 14 76 10 10 4 0 Yes 95 45 4 51 9 4 0 5 No

6 0 16 7 0 0 7 No

7 0 14 7 0 0 7 No

8 0 85 10 0 14 10 No

9 14 59 9 9 5 0 Yes 85 3

10 5 40 9 5 0 4 No

11 0 42 9 0 0 9 No

12 0 52 9 0 14 9 No

13 14 39 9 9 5 0 Yes 73 2

14 5 89 10 5 0 5 No

15 0 88 10 0 14 10 No

16 14 24 8 8 6 0 Yes 01 1

17 6 11 7 6 14 1 No

18 14 67 9 9 5 0 Yes 62 2

19 5 51 9 5 0 4 No

20 0 33 8 0 14 8 No

21 14 08 6 6 8 0 Yes 40 1

22 8 29 8 8 14 0 No

23 14 75 10 10 4 0 Yes 33 1

24 4 95 11 4 14 7 No

209 112 157 97 16

Some useful statistics from the simulation:

Average demand

Simulation209/24 = 8.71

Theoretical 8.75 (i.e. Average actual sales over past 36 months i.e. 315/36)

Average lead time

Simulation16/8 = 2.00

Theoretical 1.86 (i.e. Sum of delivery time x probability)

Average ending inventory = 157/24 = 6.5

Average number of lost sales = 97/24 = 4.04

(b)Total Cost, Ct = Carrying cost 24 x 600 x 6.50 = 93,600

+ Lost sale cost 4,350 x 97 = 421,950

+ Order cost 8 x 570 = 4,560520,110

=$520,110.00 or $21,671.00 per month


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3 BUS P301:01

Question F.17 We use the following random number intervals when simulating demand and leadtime. We then select column 1 of Table F.4 in the text to get the random numbersfor demand, and use column 2 of the same table to find the lead time whenever anorder is placed.

Demand Probability

Cumulative

Probability

RN

Interval0 0.20 0.20 01201 0.40 0.60 21602 0.20 0.80 61803 0.15 0.95 81954 0.05 1.00 9600

The results are:

WeekUnits

ReceivedBegin

Inventory RN DemandEnd

InventoryLostSales Order? RN

Leadtime

1 5 52 1 4 02 4 37 1 3 0

3 3 82 3 0 0 Yes 06 14 0 69 2 0 25 10 10 98 4 6 06 6 96 4 2 0 Yes 63 37 2 33 1 1 08 1 50 1 0 09 0 88 3 0 3

10 10 10 90 3 7 0Total 23 5

The total stock out cost = 5($40) = $200. The total holding cost = 23($1) = $23.

Average Weekly stock out cost = $200/10 = $20; Weekly holding cost = $23/10 =$2.30

Question F.21 Use the following random number intervals to simulate arrival and service time.

Time Between Random NumberArrivals Probability Interval

1 0.20 01202 0.25 21453 0.30 46754 0.15 76905 0.10 9100

Random NumberService Time Probability Interval

1 0.10 01102 0.15 11253 0.35 26604 0.15 61755 0.15 76906 0.10 9100


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4 BUS P301:01

(a) Simulation of a one-teller drive-through:

Arrivals Service

Rand Time Till Arrive at Rand Time Wait Time

Num Next Time Num Needed Begin End (Mins)

52 3 1:03 60 3 1:03 1:06 0

37 2 1:05 60 3 1:06 1:09 1

82 4 1:09 80 5 1:09 1:14 069 3 1:12 53 3 1:14 1:17 2

98 5 1:17 69 4 1:17 1:21 0

96 5 1:22 37 3 1:22 1:25 0

33 2 1:24 06 1 1:25 1:26 1

50 3 1:27 63 4 1:27 1:31 0

88 4 1:31 57 3 1:31 1:34 0

90 4 1:35 02 1 1:35 1:36 0

50 3 1:38 94 6 1:38 1:44 0

27 2 1:40 52 3 1:44 1:47 4

45 2 1:42 69 4 1:47 1:51 5

81 4 1:46 33 3 1:51 1:54 5

66 3 1:49 32 3 1:54 1:57 5

74 3 1:52 30 3 1:57 2:00 5

30 2 1:54 48 3 2:00 2:03 6

59 3 1:57 88 5 2:03 2:08 667 3 2:00 40

Yearly waiting costs are given by:

Waiting cost = (40 min/hr)(7 hr/day)(200 days)($1/min) = $56,000.00

(b) Simulation of two one-teller drive-throughs:

Arrivals Service

Time Teller 1 Teller 2

Wait Time

(Mins)

Rand Till In At Rand Time

Num Next Time Num Needed Beg End Beg End52 3 1:03 60 3 1:03 1:06 0

37 2 1:05 60 3 1:05 1:08 0

82 4 1:09 80 5 1:09 1:14 0

69 3 1:12 53 3 1:12 1:15 0

98 5 1:17 69 4 1:17 1:21 0

96 5 1:22 37 3 1:22 1:25 0

33 2 1:24 06 1 1:24 1:25 0

50 3 1:27 63 4 1:27 1:31 0

88 4 1:31 57 3 1:31 1:34 0

90 4 1:35 02 1 1:35 1:36 0

50 3 1:38 94 6 1:38 1:44 0

27 2 1:40 52 3 1:40 1:43 0

45 2 1:42 69 4 1:43 1:47 1

81 4 1:46 33 3 1:46 1:49 0

66 3 1:49 32 3 1:49 1:52 074 3 1:52 30 3 1:52 1:55 0

30 2 1:54 48 3 1:54 1:57 0

59 3 1:57 88 5 1:57 2:02 0

67 3 2:00 1

Waiting cost = (1 min/hr)(7 hrs/day)(200 days)($1/min) = $1,400.00


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5 BUS P301:01

(c)Cost alternatives:

Wait Drive-through Labor (Teller)Cost/year

Cost/year Amortization/year Cost/year

Cost for single one-teller drive-through:$56,000.00 + $12,000.00 + $16,000.00 = $84,000.00

Cost for two one-teller drive-throughs:

$1,400.00 + $20,000.00 +$32,000.00 = $53,400.00

Cost savings achieved by using two booths:

$84,000.00 - $53,400.00 = $30,600.00

The conclusion is to place two teller booths in use. It should be noted, however, it iscritical to replicate the above simulation for a much longer time period before onedraws any firm conclusions.

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Operation Management Heizer Solution Module F