Some solutions of simulation

8/7/2019 Some solutions of simulation

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Ans. 6

The numbers 00-99 are allocated in proportion to the probabilities associated with each event as given below:

Daily Demand Probability CumulativeProbability

Random NumbersAllocated

0 0.01 0.01 00-00

10 0.20 0.21 01-20

20 0.15 0.36 21-35

30 0.50 0.86 36-85

40 0.12 0.98 8697

50 0.02 1.00 98-99

Let us simulate the demand for the next 10 days using the given random numbers in order to find out the stock

position if the owner of the bakery decides to make 30 breads every day. We will also estimate the daily averagedemand for the bread on the basis of simulated data.

Day Random Number Simulated Demand Stock if 30 breadsare prepared every

day

1 48 30 0

2 78 30 0

3 19 10 20

4 51 30 20

5 56 30 20

6 77 30 20

7 15 10 40

8 14 10 60

9 68 30 60

10 9 10 80

Total 220

Daily average demand of the basis of simulated data = 22

Ans. 7:

The random numbers are established as in Table below:

Production probability cumulative Random numberPer day probability196 0.05 0.05 00-04197 0.09 0.14 05-13198 0.12 0.26 14-25199 0.14 0.40 26-39200 0.20 0.60 40-59201 0.15 0.75 60-74202 0.11 0.86 75-85203 0.08 0.94 86-93

204 0.06 1.00 94-99


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Based on the 15 random numbers given we simulate the production per day as above in table 2below.

Random No. Estimated No. of mopeds waiting No. of emptyProduction spaces in the lorryPer day

Opening current current TotalBalance excess short waiting

Production production

1 82 202 -- 2 -- 2 ---2 89 203 2 3 --- 5 ---3 78 202 5 2 --- 7 ---4 24 198 7 -- 2 5 ---5 53 200 5 --- -- 5 --6 61 201 5 1 --- 6 ---7 18 198 6 --- 2 4 ---

8 45 200 4 --- -- 4 ---9 04 196 4 --- 4 0 ---10 23 198 0 --- 2 0 211 50 200 0 -- -- -- ---12 77 202 0 2 -- 2 ---13 27 199 2 --- 1 1 ---14 54 200 1 -- -- 1 ---15 10 197 1 --- 3 _-- __2

Total 42 __4

Average number of mopeds waiting =15

42= 2.80

Average number of empty spaces in lorry =15

4= 0.266

Ans. 8:If the numbers 00-99 are allocated in proportion to the probabilities associated with eachcategory of work, then various kinds of dental work can be sampled, using random numbertable :-

Type Probability Random Numbers

Filling 0.40 00-39Crown 0.15 40-54

Cleaning 0.15 55-69Extraction 0.10 70-79Checkup 0.20 80-99

Using the given random numbers, a work sheet can now be completed as follows :-

FUTURE EVENTS

PATIENT SCHEDULED ARRIVAL RN CATEGORY SERVICE TIME

1 8.00 40 Crown 60 minutes2 8.30 82 Checkup 15 minutes

3 9.00 11 Filling 45 minutes4 9.30 34 Filling 45 minutes


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5 10.00 25 Filling 45 minutes6 10.30 66 Cleaning 15 minutes7 11.00 17 Filling 45 minutes8 11.30 79 Extraction 45 minutes

Now, let us simulate the dentists clinic for four hours starting at 8.00 A.M.

STATUS

Time Event Number of the patient Patientsbeing served (time to go) waiting

8.0 1st patient arrives 1st(60) -8.30 2nd arrives 1st(30) 2nd9.00 1st departs

3rd arrives 2nd(15) 3rd9.15 2nd departs 3rd(45) -

9.30 4

th

arrives 3

rd

(30) 4

th

10.00 3rd departs5th arrives 4th(45) 5th

10.30 6th arrives 4th(15) 5th & 6th10.45 4th departs 5th(45) 6th11.00 7th arrives 5th(30) 6th & 7th11.30 5th departs

8th arrives 6th(15) 7th & 8th11.45 6th departs 7th(45) 8th12.00 End 7th(30) 8th12.30 - 8th(45) -

The dentist was not idle during the entire simulated period :-The waiting times for the patients were as follows :-

Patient Arrival Service Starts Waiting (Minutes)

1 8.00 8.00 02 8.30 9.00 303 9.00 9.15 154 9.30 10.00 305 10.00 10.45 456 10.30 11.30 607 11.00 11.45 45

8 11.30 12.30 60Total 285

The average waiting time of a patient was285

15= 35.625 minutes.

Ans. 9: Random allocation tables are as under:

Time Arrival Arrivals Random Time Service Service Random(Mts) (Proba.) cumulative No. (Mts) (Proba.) Cumulative No.

Probability allocated Probability allocated

1 0.05 0.05 00-04 1 0.10 0.10 00-09


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2 0.20 0.25 05-24 2 0.20 0.30 10-293 0.35 0.60 25-59 3 0.40 0.70 30-694 0.25 0.85 60-84 4 0.20 0.90 70-895 0.10 0.95 85-94 5 0.10 1.00 90-996 0.05 1.00 95-99

Simulation of ten trails:

R. No. Arrival Mts. Time Start R. No. Time Mts. FinishTime

Waiting Time

Clerk Passanger

60 4 9.04 9.04 09 1 9.05 4

16 2 9.06 9.06 12 2 9.08 1

08 2 9.08 9.08 18 2 9.10

36 3 9.11 9.11 65 3 9.14 1

38 3 9.14 9.14 25 2 9.16

07 2 9.16 9.16 11 2 9.18

08 2 9.18 9.18 79 4 9.22

59 3 9.21 9.22 61 3 9.25 1

53 3 9.24 9.25 77 4 9.29 1

03 1 9.25 9.29 10 2 9.31 _ 4

Total 6 6

In half an hour trial, the clerk was idle for 6 minutes and the passengers had to wait for 6 minutes.

Ans. 10:

From the frequency distribut ion of arrivals and service times, probabilities and cumulat ive probabilities are

first worked out as shown in the following table:

Timebetweenarrivals

Frequency ProbabilityCum.Prob.

ServiceTime

Frequency

Prob.Cum.Prob.

1

2

3

4

5

6

5

20

35

25

10

5

0.05

0.20

0.35

0.25

0.10

0.05

0.05

0.25

0.60

0.85

0.95

1.00

1

2

3

4

5

6

1

2

4

2

1

0

0.10

0.20

0.40

0.20

0.10

0.00

0.10

0.30

0.70

0.90

1.00

1.00

Total 100 10

The random numbers to various intervals have been allotted in the following table:

Time betweenarrivals

Probability Randomnumbersallotted

Service Time Probability Randomnumbersallotted


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1

2

3

4

56

0.05

0.20

0.35

0.25

0.100.05

00-04

05-24

25-59

60-84

85-9495-99

1

2

3

4

56

0.10

0.20

0.40

0.20

0.100.00

00-09

10-29

30-69

70-89

90-99-

Simulation Work Sheet

RandomNumber

Time tillnext

arrival

ArrivalTimea.m.

Servicebeginsa.m.

Randomnumber

Servicetime

ServiceEndsa.m.

Clerk

Waiting

time

CustomerwaitingTime

Timespend bycustomerin system

Lengthof

waitingline

64 4 11.04 11.04 30 3 11.07 04 - 3 -

04 1 11.05 11.07 75 4 11.11 - 2 6 1

02 1 11.06 11.11 38 3 11.14 - 5 8 2

70 4 11.10 11.14 24 2 11.16 - 4 6 2

03 1 11.11 11.16 57 3 11.19 - 5 8 2

60 4 11.15 11.19 09 1 11.20 - 4 5 2

16 2 11.17 11.20 12 2 11.22 - 3 5 2

18 2 11.19 11.22 18 2 11.24 - 3 5 2

36 3 11.22 11.24 65 3 11.27 - 2 5 1

38 3 11.25 11.27 25 2 11.29 - 2 4 1

07 2 11.27 11.29 11 2 11.31 - 2 4 1

08 2 11.29 11.31 79 4 11.35 - 2 6 1

59 3 11.32 11.35 61 3 11.38 - 3 6 153 3 11.35 11.38 77 4 11.42 - 3 7 1

01 1 11.36 11.42 10 2 11.44 - 6 8 2

62 4 11.40 11.44 16 2 11.46 - 4 6 2

36 3 11.43 11.46 55 3 11.49 - 3 6 2

27 3 11.46 11.49 52 3 11.52 - 3 6 1

97 6 11.52 11.52 59 3 11.55 - - 3 -

86 5 3 2 - 3 -

20 57

11.57 11.57 63

54

12.00

6 56 26

Average queue length =Number of customers in waiting line

Number of arrivals=

261.3

20

Average waiting time per customer =56

2.820

minutes

Average service time =54

2.720

minutes

Ans. 11:

Cumulative frequency distribution for Ramu is derived below. Also fitted against it are the eight given randomnumbers. In parentheses are shown the serial numbers of random numbers.


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10 4 01 (2) 00 (7) 03 (8)

20 10

30 20 14 (1)

40 40

50 80 44 (4) 61 (5)

60 91 82 (6)

70 96 95 (3)

80 100

Thus the eight times are: 30, 10, 70, 50, 60, 10 and 10 respectively.

Like wise we can derive eight times for Raju also.

Col-1 Col-2 Col-3 (2 Col-2)

10 4 8

20 9 1830 15 30 25 (4)

40 22 44 36 (1) 34 (8) 41 (6)

50 32 64 55 (3) 56 (7)

60 40 80 76 (2)

70 46 92

80 50 100 97 (5)

(Note that cumulative frequency has been multiplied by 2 in column 3 so that all the given random numbers areutilized).

Thus, Rajus times are: 40, 60, 50, 30, 80 40, 50 and 40 seconds respectively.

Ramus and Rajus times are shown below to observe for waiting time, if any.

1 2 3 4

Ramu Cum. Times Raju Initial Rajus cumulative time with 30 secondsincluded

30 30 40 70

10 40 60 130

70 110 50 180

50 160 30 210

50 210 80 290

60 270 40 330

10 280 70 400

10 290 40 440

Since col. 4 is consistently greater than Co.2, no subsequent waiting is involved.

Ans. 12: The numbers 00-99 are allocated in proportion to the probabilities associated with each

event. If it rained on the previous day, the rain distribution & the random no allocation are given

below:


7/13

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Event Probability Cumulative RandomProbability numbers

Assigned

No rain 0.50 0.50 00-491 cm rain 0.25 0.75 50-74

2 cm rain 0.15 0.90 75-893 cm rain 0.05 0.95 90-944 cm rain 0.03 0.98 95-975 cm rain 0.02 1.00 98-99

Table 1 Rain on previous daySimilarly, if it did not rain the previous day, the necessary distribution and the random numberallocation is given below:

Event Probability Cumulative RandomProbability numbers

Assigned

No rain 0.75 0.75 00-741 cm rain 0.15 0.90 75-892 cm rain 0.06 0.96 90-953 0.04 1.00 96-99

Table 2- No rain on previous dayLet us now simulate the rain fall for 10 days using the given random numbers. For the first day it isassumed that it had not rained the day before:

Day Random Numbers Event1 67 No rain (from table 2)2 63 No rain (from table 2)

3 39 No rain (from table 2)4 55 No rain (from table 2)5 29 No rain (from table 2)6 78 1 cm rain (from table 2)7 70 1 cm rain (from table 1)8 06 No rain (from table 1)9 78 1 cm rain (from table 2)10 76 2 cm rain (from table 1)

Hence, during the simulated period, it did not rain on 6 days out of 10 days. The total rain fall duringthe period was 5 cm.

Ans.13:

The probabilities of occurrence of A, B and C defects are 0.15, 0.20 and 0.10 respectively. So, tilenumbers 00-99 are allocated in proportion to the probabilities associated with each of the threedefects

Defect-A Defect-B Defect-C

Exists Random Exists? Random Exists? Random

Numbers numbers numbers

Assigned assignedassigned

Yes 00-14 yes 00-19 yes 00-09

No 15-99 No 20-99 no 10-99


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Let us now simulate the output of the assembly line for 10 items using the given random numbers inorder to determine the number of items without any defect, the number of items scrapped and thetotal minutes of rework time required:

Item RN for RN for RN for whether ReworkRemarks

No. defect A defect B defect C any defect time (inExists minutes)

1 48 47 82 none -- --

2 555 36 95 none -- --

3 91 57 18 none -- --

4 40 04 96 B 15 --

5 93 79 20 None -- --

6 01 55 84 A -- Scrap

7 83 10 56 B 15 ---

8 63 13 11 B 15 ---

9 47 57 52 None -- --10 52 09 03 B,C 15+30 =45 --

During the simulated period, 5 out of the ten items had no defects, one item was scrapped and 90minutes of total rework time was required by 3 items.

Answer 14:

The question is not happily worded, if we go by the language of the question, the following solution can be worked

out:

First of all, random numbers 00-99 are allocated in proportion to the probabilities associated with demand as givenbelow:

Demand Probability Cum. Probability Random Nos.

0 0.05 0.05 00-04

1 0.10 0.15 05-14

2 0.30 0.45 15-44

3 0.45 0.90 45-89

4 0.10 1.00 90-99

Based on the ten random numbers given, we simulate the demand per day in the table given below.

It is given that stock n hand = 8 and stock on order = 6 (expected next day). Let us now consider both the options

stated in the question.

Option A: Order 5 Books, when the inventory at the beginning of the day plus orders outstanding is less than 8

books:

Day RandomNo.

SalesDemand

Op.Stock in

hand

Qty.Order

Qty.Recd. Atend ofthe day

TotalQty. onorder

ClosingStock

1 89 3 8 - - 6 5

2 34 2 5 - 6 - 9

3 78 3 9 - - - 6

4 63 3 6 5 - 5 3


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- 9 -

5 61 3 3 - - 5 0

6 81 3 0 0

7 39 2

8 16 2

9 13 110 73 3

Now on day 6, there is stock out position since 5 units will be received at the end of the day and demand occurringduring the day can not be met. Hence, it will into be possible to proceed further and we will have to leave the answer

at this stage.

RandomNo.

SalesDemand

OpeningStock in

hand

Qty.Order


TotalQty. onorder

ClosingStock

1 89 3 8 -- -- 6 5

2 34 2 5 -- 6 -- 9

3 78 3 9 -- -- -- 6

4 63 3 6 8 -- 8 3

5 61 3 3 -- -- 8 0

6 81 3 0 -- 8 --

7 39 2

8 16 2

9 13 1

10 73 3

Now on day 6, there is stock out position since 8 units will be received at the end of the day and demand occurring

during the day can not be met. Hence, it is not possible to proceed further and we may leave the answer at thisstage.

Alternatively, if we assume that the demand occurring during the day can be met out of stock received at the end of

the day, the solution will be as follows:

Stock in hand = 8 and stock on order = 6 (expected next day)

RandomNo.

SalesDemand

OpeningStock in

hand

Qty.Order

Qty.Recd. Atend of

the day

TotalQty. onorder

ClosingStock

1 89 3 8 -- -- 6 5

2 34 2 5 -- 6 -- 9

3 78 3 9 -- -- -- 6

4 63 3 6 5 -- 5 3

5 61 3 3 -- -- 5 0

6 81 3 0 5 5 5 2

7 39 2 2 5 -- 10 0

8 16 2 0 -- 5 5 3

9 13 1 3 -- 5 -- 7

10 73 3 7 5 -- 5 4


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- 10 -

Carrying Cost = 39 0.50 = Rs.19.50

Ordering Cost = 4 10 = Rs.40.00

Total Cost = Rs.59.50

Option B: Order 8 Books, when the inventory at the beginning of the day plus orders outstanding is less than 8

books:

RandomNo.

SalesDemand

OpeningStock in

hand

Qty.Order


TotalQty. onorder

ClosingStock

1 89 3 8 -- -- 6 5

2 34 2 5 -- 6 -- 9

3 78 3 9 -- -- -- 6

4 63 3 6 8 -- 8 3

5 61 3 3 -- -- 8 06 81 3 0 -- 8 -- 5

7 39 2 5 8 -- 8 3

8 16 2 3 -- -- 8 1

9 13 1 1 -- 8 -- 8

10 73 3 8 -- -- -- 5

Carrying Cost = 45 0.50 = Rs.22.50

Ordering Cost = 2 10 = Rs.20.00

Total Cost = Rs.42.50

Since Option B has lower cost, Manager should order 8 books.

Ans.15

Demand (Tons) Probability Cumulative Probability Random Nos. Allocated

1 0.15 0.15 00-14

2 0.30 0.45 15-44

3 0.45 0.90 45-89

4 0.10 1.00 90-99

Option-I

RN Demand Opening

Stock

Receipts Closing

Stock

Op.Stock

on Order

Order Cl.Stock on

Order

88 3 8 - 5 - - 6

41 2 5 6 9 - - -

67 3 9 - 6 - 5 5

63 3 6 - 3 5 - 5

48 3 3 - 0 5 5 10

74 3 0 5 2 5 5 10

27 2 2 - 0 10 - 10

16 2 0 5 3 5 - 5

11 1 3 5 7 - 5 5

64 3 7 - 4 5 - 5

49 3 4 - 1 5 5 1021 2 1 5 4 5 - 5


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- 11 -

44

(Rs.)

No of order placed 5

Ordering cost (5x1000)

Closing Stock 44

Carrying cost (44x50)Total

5,000

2,2007,200

Option-II

RN Demand Opening

Stock

Receipts Closing

Stock

Op.Stock

on Order

Order Cl.Stock on

Order

88 3 8 - 5 - - 6

41 2 5 6 9 - - -

67 3 9 - 6 - 8 8

63 3 6 - 3 8 - 8

48 3 3 - 0 8 - 874 3 0 8 5 - 8 8

27 2 5 - 3 8 - 8

16 2 3 - 1 8 - 8

11 1 1 8 8 - - -

64 3 8 - 5 - 8 8

49 3 5 - 2 8 - 8

21 2 2 - 0

47

8 - 8

(Rs.)

No of orders 3 Ordering cost 3 x 1000Closing stock 47 Carrying cost 47x50

Total

3,0002,350

5,350

Analysis: Since the cost of inventory is less in Option II, it is suggested to implement.

Ans. 16

(i) Allocation of random numbers

Demand Probability Cumulative probability Allocated RN

0


12/13

- 12 -

1 27 450 450 300 3375 12000 2175

2 15 150 150 600 1125 2400 -1275

3 56 750 750 -- 5625 -- 5625

4 17 150 150 600 1125 2400 -1275

5 98 1650 750 -- 5625 --- 5625 9006 71 750 750 -- 5625 -- 5625

7 51 750 750 -- 5625 -- 2175

8 32 450 450 300 3375 1200 5625

9 62 750 750 -- 5625 -- 5625 300

10 83 1050 750 -- 5625 -- 5625 900

11 96 1650 750 -- 5625 -- 5625

12 69 750 750 -- 5625 5625

54000 7200 46800 2100

(iii) Loss on lost sale 21007.5 = Rs15750.

Ans. 17

The demand and supply patterns yield the following probability distribution. The numbers 00-99 are

allocated in proportion to the probabilities associated with each event.

Availability(Kg.)

Prob. Cum.Prob.

RandomNumbersallocated

Demand(Kg)

Prob. Cum.Prob.

Randomnumber

allocated

10 0.08 0.08 00-07 10 0.10 0.10 00-0920 0.10 0.18 08-17 20 0.22 0.32 10-31

30 0.38 0.56 18-55 30 0.40 0.72 32-71

40 0.30 0.86 56-85 40 0.20 0.92 72-91

50 0.14 1.00 86-99 50 0.08 1.00 92-99

Let us simulate the supply and demand for the next six days using the given random numbers in order to find the

profit if the cost of the commodity is Rs.20 per kg, the selling price is Rs.30 per kg, loss on any unsatisfied demandis Rs.8 per kg and unsold commodities at the end of the day have no saleable value.

Day Random

no.

Supply

availability

Random

no.

Demand Buying

costRs.

Selling

costRs.

Loss for

unsatisfieddemand

Profit

1 31 30 18 20 600 600 -- --

2 63 40 84 40 800 1200 -- 400

3 15 20 32 40 400 600 160 40

4 07 10 32 30 200 300 160 -60

5 43 30 75 40 600 900 80 220

6 81 40 27 20 800 600 -- -200

During the simulated period of six days, the net profit of the retailer is

= (400 + 40 + 220) (60 + 200)


13/13

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= 660 260

= Rs.400

Ans. 19:

Random No. Coding Table - ReceiptsAmount (Rs. In crores) Probability Cum. Probability Random No. Interval

30 0.20 0.20 00-1942 0.40 0.60 20-5936 0.25 0.85 60-8499 0.15 1.00 85-99

Random No. Coding Table - PaymentsAmount (Rs. In crores) Probability Cum. Probability Random No. Interval

33 0.15 0.15 00-1460 0.20 0.35 15-3439 0.40 0.75 35-74

57 0.25 1.00 75-99

Simulation TableWeek Op.Balance Receipts Payments Cl.Balance

RandomNo.

Amount(in crores)

RandomNo.

Amount(in crores)

1 15 17 30 78 57 -122 -12 43 42 16 60 -303 -30 74 36 35 39 -334 -33 31 42 23 60 -515 -51 72 36 44 39 -546 -54 46 42 92 57 -697 -69 51 42 58 39 -66

8 -66 68 36 8 33 -639 -63 93 99 58 39 -310 -3 54 42 78 57 -1811 -18 96 99 54 39 4212 42 9 30 77 57 15

(i) Probability is = 10 12 = 0.83

(ii) Total Shortfall is Rs. 399 crores. Therefore average shortfall is 399 12 = Rs. 33.25 croresAlternatively, average shortfall is 399 10 = Rs. 39.90 crores

(iii) There will be a shortfall in 5 months i.e. 4,5,6,7,8. therefore the probability is 5 12 = 0.42

Documents

Some solutions of simulation