operations research 10

7/27/2019 Operations Research 10

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Operations Research Unit 10

Sikkim Manipal University 160

Unit 10 Simulation

Structure

10.1. Introduction

10.2 Basic concepts

10.3. Simulation procedure

10.3.1. Allocation of Random Numbers

10.3.2. Use of Random Number Tables

10.4. Sample Size

10.5. Application of simulation

10.5.1 Limitations:

10.6. Summary

Terminal Question

Answers to SAQs and TQs

10.1 Introduction

Generally in developing mathematical models of various systems or situations, it is assumed that

the statistical distribution of the variables conforms to a standard pattern. This, however, is not

always true. In a typical pricing problem, the management cannot risk changing the price of the

product without evaluating the various alternatives. Also, representation of the reality in terms of a

mathematical model becomes virtually impossible because of the complexity of the interaction of

several variables having a bearing on the final outcome. One approach to the problem is to

assign probabilities of achieving various sales targets under different conditions of completion

with changes in price, demand, etc. and choose the alternative which gives the maximum profit.

Where formulating a mathematical model is difficult, simulation is of great help for decision

making.

Learning Objectives

After studying this unit, you should be able to understand the following

1. What is simulation

2. How is it applied in business problems

3. Use of Monte Carlo Method


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10.2 Basic Concepts

Simulation may be called experimentation in the management laboratory. In the context of

business problems, simulation is often referred to as Monte Carlo Analysis. The expression may

be traced to two American mathematicians, Von Neumann and Ulan, who in the late 1940s founda problem in the field of nuclear physics too complex for analytical solution and too dangerous for

actual experimentation. Eventually they arrived at an approximate solution by sampling. The

method used by them somewhat resembles the manner in which gamblers develop betting

systems on the roulette table and the name Monte Carlos stuck.

Imagine a betting game in which the stakes are based on correct prediction of the number of

heads which occur when five coins are tossed. If it were only a question of one coin most people

know that there is an equal likelihood of a head or a tail occurring. i.e., the probability of a head is

. However, without the application of probability theory, it would be difficult to predict the

chances of getting various numbers of heads when five coins are tossed. We may take five coins

and toss them repeatedly. The outcomes may be noted for each toss and, say, after every ten

tosses the probabilities of various outcomes may be estimated. As we know, the values of these

probabilities will initially fluctuate but they would tend to stabilise as the number of tosses is

increased. This approach in effect is a method of sampling but is not very convenient. Instead of

actually tossing the coins, we may carry out the experiment by using random numbers. Random

numbers have the property that any number is equally likely to occur irrespective of the digit that

has already occurred.

Let us estimate the probability of tossing of different numbers of heads with five coins. We startwith set random numbers given below:

78466 71923

78722 78870

06401 61208

04754 05003

97118 95983

By following a convention that even digits signify a head (H) and the odd digits represent a tail

(T), the tossing of a coin can be simulated. The probability of occurrence of the first set of digits is and that of the other set is also - a condition corresponding to the probability of the

occurrence of a head and the probability of occurrence of a tail respectively.

It is immaterial which set of five digits should signify a head. The rule could be that the digits 0, 1,

2, 3 and 4 represent a head and the digits 5, 6, 7, 8 and 9 a tail. It is only necessary to take care


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that the set of random numbers allotted to any event matches with its probability of occurrence.

For instance, if we are interested in allotting random numbers to three events A, B and C with

respective probabilities 0.24, 0.36 and 0.40 we choose two digit random numbers 00 to 99.

The numbers 00 to 23 signify event A,

24 to 59 signify B and

60 to 99 signify C.

The first set of five random digits in the list of random numbers implies that the outcome of the

first toss of 5 coins is as follows:

Coin 1 2 3 4 5

Random number 7 8 4 6 6

Outcome T H H H H

Hence it is 4 heads and 1 tail.

Proceeding in the same way, we can tabulate the results of the first ten tosses.

TABLE 1

Toss No.Heads Tails

1

2

34

5

6

7

8

9

10

4

3

43

1

1

3

4

3

1

1

2

12

4

4

2

1

2

4

Based on the ten tosses of the coins, the estimates of probabilities of occurrence of different

numbers of heads are:

0 Head 0

1 Head 10

3

2 Heads 0

Number of


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3 Heads 10

4

4 Heads 10

3

5 Heads 0

As these estimates will come closer to the theoretical value with increasing sample size, the

experiment is to be continued further.

The results for obtaining 2 heads and 3 tails for 100 throws are shown below:

In the first 10 throws 0

20 throws 6

30 throws 11

40 throws 1450 throws 18

60 throws 19

70 throws 21

80 throws 22

90 throws 24

100 throws 27

Table 2 compares the final results at the end of 100 throws with the theoretical probabilities.

TABLE 2

No. of headsEstimated

Probability

Theoretical

Probabilities

0

1

2

3

4

5

0.03

0.21

0.27

0.33

0.12

0.04

0.03

0.16

0.31

0.31

0.16

0.03

It is observed that the results obtained with the large sample of 100 compare more favourably

with the theoretical values, than with a sample of ten sets of numbers.

Self Assessment Questions 1

Fill in the blanks

1. Simulation may be called experimentation in the ________ ________.

2. Random numbers have the property that any number has ________ ________ to occur.


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3. The totality of probability assigned to the variable should always be equal to _________.

10.3 Simulation Procedure

The approach adopted for solving a problem in gambling can be extended to decision making inbusiness where risk is a common feature. The probabilities associated with the variables can be

estimated on the basis of past data if available, or by inputting subjective values.

In any simulation problems the variables to be studied will be given with associated probabilities.

The initial conditions will also be specified. We can choose random number from table. However

to get uniform results the random numbers to be used will be specified. The first step is we code

the data, i,e, we assign random numbers to the variable. We identify the relationship between the

variables and run the simulation to get the results

Let us illustrate this by a simple example of a queuing process.

Example 1:

A sample of 100 arrivals of customers at a retail sales depot is according to the following

distribution:

Time between

Arrivals (mts.)Frequency

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

2

6

10

25

20

14

10

7

4

2

A study of the time required to service customers by adding up the bills, receiving payment,

making change and placing packages in hand trucks, yields the following distribution:

Service time (mts.) Frequency.5

1.0

1.5

2.0

2.5

12

21

36

19

7


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3.0 5

Estimate the average percentage customer waiting time and average percentage idle time of the

server by simulation for the next 10 arrivals.

Solution:

Step 1 : Convert the frequency distributions of time between arrivals and service time to

cumulative probability distributions.

Step 2: Allocate random numbers 00 to 99 for each of the values of time between arrivals and

service time, the range allocated to each value corresponding to the value of cumulative

probability. (Tables 12 -3 and 12 - 4) .

Step 3 : Using random numbers from table, sample at random the tome of arrival and service

time for ten sets of random numbers.

Step 4 : Tabulate waiting time of arrivals and idle time of servers.

(Table 12-5)

Step 5: Estimate the percent waiting time of arrivals and percent idle time of servers

corresponding to the ten samples.

10.3.1. Allocation of Random Numbers

TABLE 3

Allocation of Random Number Time between arrivals

Timebetweenarrivals

(1)

Frequency

(2)

Cumulative

Frequency

(3)

Cumulative

Probability

100

(3)

(4)

Random

Number

Allocated

(5)

0.5

1.0

1.5

2.0

2.5

3.03.5

4.0

4.5

5.0

2

6

10

25

20

1410

7

4

2

2

8

18

43

63

7787

94

98

100

0.02

0.08

0.18

0.43

0.63

0.770.87

0.94

0.98

1.00

00 to 01

02 to 07

08 to 17

18 to 42

43 to 62

63 to 7677 to 86

87 to 93

94 to 97

98 and 99


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TABLE 4

Allocation of Random Numbers Service Time

Service

Time

(1)

Frequency

(2)

Cumulative

Frequency

(3)

Cumulative

Probability

(4)

RandomNos.

Allocated

(5)

0.5

1.0

1.5

2.0

2.5

3.0

12

21

36

19

7

5

12

33

69

88

95

100

0.12

0.33

0.69

0.88

0.95

1.00

00 to 11

12 to 32

33 to 68

69 to 87

88 to 94

95 to 99

(Note that the upper bound of random numbers allocated for each value of the parameter is one

less than the corresponding cumulative frequency since we have chosen a range of random

numbers form 00 to 99)

TABLE 5

Arrivals Service

Waiting

Timeof

Arrival

IdleTime

of

Server

123456789

10

78780604977178610595

3.53.51.01.04.53.03.52.51.04.5

3.57.08.09.0

13.516.520.022.523.528.0

54245145468458586024

1.51.01.51.51.52.01.51.51.51.0

3.57.08.09.5

13.516.520.022.524.028.0

5.08.09.5

11.015.018.521.524.025.529.0

Total

0.50.5

3.52.02.51.51.51.02.5

1.0 14.5

The service facility is made available at clock time zero and the server has to be idle for 3.5

minutes when the service for first arrival starts. The service is completed at 5.0 minutes and again

the server is idle for 2 minutes till the second arrival joins the system. The first three arrivals get

immediate service and they dont have to wait, as the server is idle when they arrive. The fourth

ArrivalNo

.

Random

No

.

Timebetween

arrivals

Timeof

arrival

Random

No

.

ServiceTime

TimeofStart

TimeofFinish


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arrival who joins at 9.0 minutes has to wait for .5 minute when the service for the third is

completed. Similarly the waiting time and idle time can be computed for further arrivals.

Total elapsed time = 29 minutes

Waiting time of arrival = 1 minute

Percentage of waiting time = 4.329

101=

Idle time for server = 14.5 minutes

Percentage of idle time = 5029

1005.14=

10.3.2. Use of Random Number Tables

The random numbers could be selected by any random process, such as drawing numbered

chips from a hat. However, it is convenient to use a table of random numbers which in fact is

prepared on the basis of some such physical phenomenon. The grouping of random numbers in

the tables has no significance and one should be concerned with individual digits only. The first

random number could be picked at random from any point in the tables and the subsequent ones

are to be selected proceeding sequentially either in a vertical or horizontal direction. Depending

upon the number of digits required, the random numbers will be chosen in sets of single digit, two

digit numbers, etc. Pseudo Random Numbers

Truly random numbers cannot be produced by an algorithm and hence random numbers

generated by using a recursive equation are referred to as pseudo Random Numbers.

There are several methods of generating pseudo random numbers but we shall briefly describe

only the Mid Square Method. Operation starts with an arbitrary four digit integer called the seed.

To obtain the first random number, the seed is squared and all digits except the middle four are

ignored. The process is then repeated each time using the previous random number as the new

seed.

Seed0U = 8695

( )220

8695U =

= 75603025

Taking the middle 4 digits,

1U = 6030

21U = 36360900

2U = 3609


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Repeating the above procedure:

3035U,3782U,0615U,0248U 6543 ====

2436U,2063U,4605U,2112U 10987 ====

One of the basic disadvantages of the mid square method is that the generated numbers may

start cycling after a short set of random numbers is obtained.

There are methods by which the seed can be chosen, so as to obtain a fairly long sequence of

numbers before cycling starts. Also statistical tests are available to check whether the generated

sequence is truly random.


Examine whether the following statements are True or False

1. In any simulation problem initial conditions are stated.

2. Assigned random numbers for cumulative probability values

3. Without identifying any relationship between variables we can solve the simulation problem.

10.4 Sample Size

As we have seen with the coin tossing experiment, the larger the number of trials, the more

confident we can be in our estimates. The question that arises is how many trials for simulation?

If the experiment is as simple as tossing a coin involving only one variable, the sample size

required for a given confidence level at a specified degree of accuracy can be worked out.

Example 2

If it is needed to be 95% certain of being correct in an experiment with marginal error of 1% of the

true value, what should be the sample size?

Solution: Let p be the proportion as percent of success.

Standard error will be ( )n

p100p -

Where n is the sample size.

The standard normal deviate value corresponding to 95% confidence level from normal tables is

1.96.

Margin of error = 1.96 1n

)p100(p=

-


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\ ( )p100p96.1n 2 -=

p100p - is maximum for p = 50

Hence, the value of n = 9600.

Usually, a simulation model involves several variables and it may not be possible to determine the

number of trials required to obtain the desired accuracy at a specified confidence level.

One can only say that the accuracy associated with simulation improves as the square root of the

number of trials and hence there is a need for a large number of trials.

This calls for a great deal of computational effort and for most real life problems, the use of

computer becomes inevitable. In fact, special simulation languages such as GPSS and

SIMSCRIPT have been developed to save time and effort required to structure and debug

simulation models.

A practical indicator of when to stop simulation trials is given by the fact that the results which

violently fluctuate initially tend to stabilize as the simulation is continued. If the successive

cumulative results tally reasonable well, the simulation may be stopped. The degree of accuracy

required, of course, varies with the problem on hand and calls for the judgment of the analyst.


Do you agree or Not given:

1. Standard error for percentage of success = (P(1-p) / n)1/2

2. It is possible to determine number of trials.

3. The accuracy of results increases as the square of number of trials.

10.5. Application of Simulation

The range of application of simulation in business is extremely wide. Unlike the other

mathematical models, through abstract, simulation can be easily understood by the users and

thereby facilitates their active involvement. This, in turn, makes the results more reliable and also

ensures easy acceptance for implementation. The degree to which a simulation model can be

made close to reality is dependent upon the ingenuity of the O.R team who should identify the

relevant variables as well as their behaviour.

We have already seen by means of an example how simulation could be used in a queuing

system. It can also be employed for a wide variety of problems encountered in production


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systems the policy for optimal maintenance in terms of frequency of replacement of spares or

preventive maintenance, number of maintenance crews, number of equipment for handling

materials, job shop scheduling, routing problems, stock control and so forth. The other areas of

application include dock facilities, facilities at airports to minimize congestion, hospitalappointment systems and even management games.

As in the case of other O.R. models, with the help of simulation, the manager tries to strike a

balance between opposing costs of providing facilities (which usually mean long term

commitment of funds) and the opportunity and other costs of not providing them.

10.5.1 Limitations:

Simulation approach is recognized as a powerful tool for management decision making. This

does not mean that one should ignore the cost associated with a simulation study for data

collection, formation of the model and the computer time. Often this is quite significant.

A simulation application is based on the premise that the behaviour pattern of relevant variables

is known, and this very premise sometimes becomes questionable. Not always can the

probabilities be estimated with ease or desired reliability. The results of simulation should always

be compared with solutions obtained by other methods wherever possible, and tempered with

managerial judgment.

10.5.2 Some Examples

Example 3

A bread vendor buys every morning loaves of bread at 0.45 each by placing his order one day in

advance (at the time of receiving his previous order) and sells them at Rs. 0.70 each. Unsold

bread can be sold the next day at Rs. 0.20 per loaf and thereafter should be treated as of no

value. The pattern of demand for bread is given below:

Fresh Bread One day old Bread

Daily

Sales

Probability

of demand

Daily

Sales

Probability

of

demand

50

51

52

53

54

55

0.01

0.03

0.04

0.07

0.09

0.11

0

1

2

3

0.10

0.20

0.08

0.02


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56

57

58

59

60

0.15

0.21

0.18

0.09

0.02

The vendor adopts the following order rule. If there is no stock with him at the end of the previous

day, he orders 60 units. Otherwise he orders 50 or 55 whichever is nearest the actual fresh bread

sale on the previous day. Starting with zero stock and a pending order for 55 loaves, simulate for

10 days and calculate the vendors profits.

Solution

TABLE 6

Allocation of random numbers

Fresh Bread One day old bread

50

51

52

53

54

55

56

57

58

59

60

0.01

0.03

0.04

0.07

0.09

0.11

0.15

0.21

0.18

0.09

0.02

0.01

0.04

0.08

0.15

0.24

0.35

0.50

0.71

0.89

0.98

1.00

00

01 03

04 07

08 14

15 23

24 34

35 49

50 70

71 88

89 97

98 99

0

1

2

3

0.70

0.20

0.08

0.02

0.70

0.90

0.98

1.00

00 to

69

70 to

89

90 to

97

98 and

99

We can now construct a table to see, through simulation how the stocks and sales fluctuate.

Dailysale

Probability

ofdemand

Cumulative

Probability

ofdemand

Random

numbers

allocated

Cumulative

Probability

ofdemand

Random

numbers

allocated

Probability

ofdemand

Dailysale


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TABLE 7

Results of simulation

FRESH BREAD ONE DAY OLD BREAD

Day

Receiptof the

Start ofday

Random No

SaleClosin

gstock

Orderfor

nextday

Opening

Stock

Random

NoSale

1

2

3

4

5

6

7

8

9

10

Total

55

60

60

50

55

60

60

55

55

60

570

72

06

12

14

79

70

85

71

21

98

15

58

52

53

50 5

8

55

58

57

58

55 5

8

54

60

549

0

8

7

0

0

3

2

0

1

0

60

60

50

55

60

60

55

55

60

55

0

0

8

7

0

0

3

2

0

1

21

86

54

88

58

48

0

0

1

0

0

0

1

0

0

0

2

* Represents lost sales as stock is limited

@ previous days closing stock is zero

Estimated profit = (549 x 0.70 + 2 x 0.20) 570 x 0.45 = Rs. 128.20

Example 4

The maintenance manager of a chemical company is interested in determining a rational policy for

maintenance of a pneumatic conveying equipment. The equipment is a part of the process line and

hence production holding. It has one bearing each on the inlet side (A) and the outlet side (B).

Whenever there is a failure of any bearing, it has to be replaced immediately. The company has a

good system of maintaining records on performance of the equipment and the following data is

available:


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No. of failures

Age at failure

(hours)

Inlet sidebearing (A)

Outlet sidebearing (B)

150

300

450

600

750

900

1,050

1,200

1,350

7

16

18

23

14

10

7

5

0

0

3

9

12

16

18

33

7

2

The cost of bearing is Rs. 300 each for A and Rs. 500 each for B. The cost of downtime of

equipment is Rs. 700 per hour, and it takes 2 hours to replace one bearing either at inlet or outlet

side and 3 hours to replace both the bearings.

The three maintenance policies to be evaluated are

i) replace a bearing only when it fails,

ii) replace both the bearings if one fails,

iii) replace the bearing which fails plus the other one if it has been in use for more than its

iv) estimated average service life i.e. 600 hours for bearing A and 860 hours for bearing B.

Find the best alternative through simulation.

Solution

It is assumed that the failure of a bearing is independent of the maintenance policy followed.

Random numbers are allocated for different failure times:


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TABLE 8

Bearing A Bearing B

Life (hrs)Cumulative

probability

Random

Nos.

Cumulative

probability

Random

Nos.150

300

450

600

750

900

1050

1200

1350

0.07

0.23

0.41

0.64

0.78

0.88

0.95

1.00

00 to 96

07 to 22

23 to 40

41 to 63

64 to 77

78 to 87

88 to 94

95 to 99

0.00

0.03

0.12

0.24

0.40

0.58

0.91

0.98

1.00

00 to 02

03 to 11

12 to 23

24 to 39

40 to 57

58 to 90

91 to 97

98 and 99

We can now select random numbers form the tables and generate a set of 12 bearing lives, for

each of the bearings.TABLE 9

1 10 300 300 99 1350 1350

2 22 300 600 96 1200 2550

3 24 450 1050 18 600 3150

4 42 600 1650 36 750 3900

5 37 450 2100 50 900 4800

6 77 750 2850 79 1050 5850

7 99 1200 4050 80 1050 6900

8 96 1200 5250 96 1200 81009 89 1050 6300 34 750 8850

10 85 900 7200 07 450 9300

11 28 450 7650 62 1050 10,350

12 63 600 8250 77 1050 11,400

Let us compare the costs of the three policies for the first 7200 hours.

Policy I. Replace a bearing only when it fails.

A requires replacement 10 times, and B 7 times during this period as seen

from the lives of successive bearings.Total cost = Cost of bearing + cost of downtime

= (300 x 10 + 500 x 7) + (17 x 2 x 700)

= Rs. 30, 300

Policy II: Replace both the bearing if one fails.

Serial Random No. Bearing A Cum. Life Random No. Bearing B Cum. Life


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TABLE 10

300 A 300 1200

600 A 450 600

1050 A 600 750

1650 A 450 900

2100 A 750 1050

2850 A 1200 1050

3900 B 1200 1200

5100 A & B 1050 750

5850 B 900 450

6300 B 450 1050

6750 A 600 1050

7350 A -

There are 11 replacements during the period.

Total cost = 11 x (300 + 500) + 11 x 3 x 700

= Rs. 31, 900

Policy III: Replace the bearing which fails plus the other one in use for 600 or more hours for A

and 860 or more hours in case of B.

Table 12 11 gives the analysis.

There are 4 replacements of A only on failure and 7 replacements both bearings.

Total cost = 4(300 + 2 x 700) + 7(800 + 3 x 700)

= Rs. 27,100

Policy III is the cheapest.

The simulation is limited to 7200 hours of operation since the purpose is only to illustrate the

method. With such small number of trials the results may turn out to be erratic. Consider Policy I

again. With a different set of random numbers and extended simulation there may be occasions

of both bearing failing at the same time, thus affecting the cost of downtime for replacement.

Simulation with sufficiently large number of trials only can lead to dependable decisions.

Elapsed time(hrs)

Bearing whichfails first

Life (hrs) ofnewly fitted bearing A

from Table 12 9

Life (hrs) ofnewly fitted bearing B

from Table 12 9


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TABLE 11

(1) (2) (3) (4) (5) (6) (7) (8) (9)

0 300 1350 A 300 300 B No 1050 Replace

A

300 300 1050 A 600 600 B No 750 Replace

A

600 450 750 A 1050 1050 B Yes Replace

A& B

1050 600 1200 A 1650 600 B No 600 Replace

A

1650 450 600 A 2100 1050 B Yes

Replace A& B

2100 750 600 B 2700 600 A Yes Replace

A& B

27001200 750 B 3450 750 A Yes Replace

A& B

34501200 900 B 4350 900 A Yes Replace

A& B43501050 1050 A&B 5400 Replace

A& B

5400 900 1050 A 6300 900 B Yes Replace

A& B

6300 450 1200 A 6750 450 B No 750 Replace

A

6750 600 750 A 7350 1050 B Yes

ReplaceA& B

This will be the total life(from Table 12 9) if newly fitted. For survivors from previous

replacement, this will be the balance life (col.8)

Start of theperiod

Life of bearing

in service *

(hrs)

Bearingwhich

fails first

Timeelapsed

(hrs)

Age ofsurvivingbearing

Did the survivingbearing (A|B)complete its

estimated

average life?

If No incol 7 balance

lifeof surviving

Replacementpolicy

A B


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Example 5:

A factory produces 150 scooters. But the production rate varies with the following distribution.

Production

Rate

147 148 149 150 151 152 153

Probability 0.05 0.10 0.15 0.20 0.30 0.15 0.05

At present it calls for a track which will hold 150 scooters. Using the following random numbers

determine the average number of scooters waiting for shipment in the factory and average

number of empty space in the truck.

Random Numbers 82, 54, 50, 96, 85, 34, 30, 02, 64, 47

Solution:

Production Rate Probability Cumulative Probability Random Number assigned147 0.05 0.05 00 04

148 0.10 0.15 05 14

149 0.15 0.30 15 29

150 0.20 0.50 30 49

151 0.30 0.80 50 79

152 0.15 0.95 80 94

153 0.05 1.00 95 99

Simulation work sheet

Trial No

Random

No.

Simulated Production

Rate

Scooter Waiting in

the factory

Number of

example spacesin the truck

1 82 152 2 -

2 54 150 - -

3 50 150 - -

4 96 153 3 -

5 85 152 2 -

6 34 150 - -

7 30 150 - -

8 02 147 - 3

9 64 151 1 -

10 47 150 2 -

Total 8 3

\Average number of scooters waiting = 8 / 10 = 0.8 / day

Average number of empty space = 3 / 10 = 0.3 / day

Example 6:

Dr.Strung is a dentist. He gives appointment to patients every half-an hour. However he does


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not know the nature if illness of patients arriving at this clinic. From past record he has the

following probability distribution and also know the exact treatment timings. He starts his clinic at

8.00 am using the following information determine the average waiting time of the customers and

idle time of the doctor.

Nature of illness Probability Time taken fortreatment (mts)

Filling 0.10 50

Check up 0.30 15

Crowning 0.15 40Cleaning 0.30 15

Extraction 0.15 30

Random Numbers 56, 40, 26, 66, 87, 48, 17, 22, 04, 15

Solution:

Illness Probability Cum.Prob Random No.assigned

Filling 0.10 0.10 00 09

Check up 0.30 0.40 10 39

Crowning 0.15 0.55 40 54Cleaning 0.30 0.85 55 84

Extraction 0.15 1.00 85 99

Simulation work sheet

TrialNo

RandomNo.

Nature ofillness

TimeTaken

PatientsArrivalTime

Treatment Doctoridentification Patientswaitingtime

Starts Finisher1 56 Cleaning 15 8.00 8.00 8.15 - -

2 40 Crowning 40 8.30 8.30 9.10 15 -

3 26 Check up 15 9.00 9.10 9.25 - 10

4 66 Cleaning 15 9.30 9.30 9.45 5 -

5 87 Extraction 30 10.00 10.00 10.30 15 -

6 48 Crowing 40 10.30 10.30 11.10 - -

7 17 Check up 15 11.00 11.10 11.25 - 10

8 22 Check up 15 11.30 11.30 11.45 5 -

9 04 Filling 50 12.00 12.00 12.50 15 -

10 15 Check up 15 12.30 12.50 13.05 - 20Total 45 40

Doctors idle time = 45 mts

Patients Average waiting time = 40 / 10 = 4 mts


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Examine whether following statements are True or False

1. Simulation gives optimum solution.

2. Simulation interrupts real system activities.

3. This technique can be easily understood by non technical mangers.

10.6 Summary

In this unit we studied the basics concepts concerned with simulations and then we continued

further to find simulations procedure. The allocation of random numbers was discussed next with

the use of random number tables. We studied the sample size and the application of simulation at

the end.

Terminal Questions

1) Briefly write down the basic concepts concerned with simulation.

2) What do you mean by simulation procedure?

3) Briefly explain the use of random number tables.

4) Discuss the different applications of simulation.

5) Two components have to be produced in M/C A and M/C B and then finally assembled. The

time taken to assemble on two machines varies with the following probability distribution.

Using simulation technique and the ordered pair of random numbers, first for M/C A and

second for M/C B. Find the average time taken to produced.

M/C - A M/C - B

Production

time in mts

Probability Production

time in mts

Probability

22 0.15 30 0.05

23 0.20 31 0.15

24 0.30 32 0.25

25 0.20 34 0.25

26 0.15 35 0.20

36 0.10

Random Numbers (10, 92) (25, 83) (36, 76) (44, 15) (57, 25) (62, 67) (04, 99) (72, 53) (81,

35) (94, 07)


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Answers To Self Assessment Questions


1. Management laboratory

2. Equally likely3. 1


1. True 2. True 3. False


1. Agree 2. Agree 3. Disagree


1. False 2. False 3. True

Answer For Terminal Questions

1) Refer Section 10.2


3) Refer Section 10.3.3


5) 57.7 minutes.

operations research 10

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