Study Material by M. Shadab Khan
Simulation
Introduction
All problems like Game Theory, Linear Programming Problem, Transportation Problems etc. are solved by using algorithms. In all these cases we first consider formulation and then develop analytical solution to the model. However, in some of the real world situations, we can not represent the problem in the mathematical model form because of complication and complexity etc. In these cases we can apply simulation approach. To simulate is to try to duplicate the features, appearance and characteristics of a real system.
In general terms simulation involves developing a model of some real phenomenon and then performing an experiment on the model evolved. It is a descriptive and not an optimizing technique. In fact, in simulation given system is copied and the variable and constant associated with it are manipulated in that artificial environment to examine the behavior of the system. So, simulation means, the deriving measures of performance about a complex system by conducting sampling experiments on a mathematical model of the system over a period of time. Usually the model is run on a computer in order to obtain operational function. The precise definition of the simulation by Shannaon is
“Simulation is the process of designing a model of a real system and conducting experiments with the model for the purpose of understanding the behavior (within the limits imposed by a criteria or set of criteria) for the operation of the system”
According to Levin & Kirk Patrick, simulation is an appropriate substitute for mathematical evaluation of a model in many situations. Although it too involves assumptions, they are manageable. The use of simulation enables us to provide insights into certain management problems where mathematical evaluation of the model is not possible. From the above definition we can say that simulation is a especially valuable tool in a situation where the mathematics needed to describe a system realistically is too complex to yield analytical use.
When Simulation is an Appropriate Tool
(i) Simulation enables the study of internal interaction of a subsystem with complex system.
(ii) International, organizational & environmental changes can be simulated and find their effects.
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(iii) A simulation model helps us to gain knowledge about improvement of the system.
(iv) Finding important input parameters with changing simulation inputs.
(v) Simulation can be used with new design and policies before implementation.
(vi) Simulating different capabilities for a machine can help to determine the requirement.
(vii) Simulation models designed for training make learning possible without the cost disruption.
(viii) A plan can be visualized with animated simulation.
When Simulation is not Applicable
(i) When problem can be solved by common sense.
(ii) When problem can be solved analytically.
(iii) If it is easier to perform direct experiments.
(iv) If costs exceeds savings.
(v) If resources or time are not available.
(vi) If system behavior is too complex like human behavior.
Areas of applications
(i) Manufacturing Applications (ii) Semiconductor Manufacturing
(iii) Construction Engineering and Project Management (iv) Military Applications
(v) Logistic, Supply Chain and Project Management (vi) Transportation Models and Traffic
(vii) Business Process Simulation (viii) Health Care
(ix) Automated Material Handling System (AHMS) (x) Risk Analysis ( Insurance, Portfolio etc…) (xi) Computer Simulation (CPU, Memory etc….)
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(xii) Network Simulation (Internet backbone, Local Area Network (switch / router), Wireless, PSTN (call center) etc
Steps in Simulation
In the following figure, steps of simulation are illustrated:
No
From the above process, we have seen that the system consists of several stages. The initial stage requires that problem must be identified and formulated. This step necessitates the specification of performance criteria, decision rules and model parameters. Once the model components are identified then model itself is developed and analyzed.
Formulation of Problem
Introduce important variables decision of system parameters
Construct Simulation Model
Validate the model
Design Experiment
Perform Simulation
Is Simulation Process Complete
Select the best course of action
Modify the Models by changing the
input data
327 Reasons for using Simulation
Some mathematicians insist that the simulation should be used only as last ditch approach that is when nothing else seems to work. Despite of the attitude like this, it turns out that simulation is one of the most widely used quantitative techniques employed by the management.
Among reasons why management scientists would consider using simulation to solve management problems are there:
1. To solve cumbersome problems. The main advantage of simulation technique is its ability to solve the most difficult problems which are impossible to handle mathematically using quantitative methods.
2. Experiment. Simulation technique helps the analyst to experiment with the system behavior without disturbing the inherent character of the real system.
3. Study of Long Term Effect. It enables the manager to visualize the long term effects in a quick manner.
4. To Test Proposed Analytical Solution. It must be emphasized that simulation technique does not represent a methodology for a derivation of optimum solution but it is often used to test purposed analytical solution.
5. Stability. One we have developed a model it may be used again & again analyze all types situations.
6. Generation of Data. Once an experiment is conducted we can generate data for further analysis.
7. Modifications. Simulation model is as closely kin to conducting sampling experiments on the real system. Model is straight forward and can be modified to accommodate the changing environments of real situation.
8. Bifurcation System. A complex system can be bifurcated into sub system studying each of sub system individually or jointly.
9. No Interference. Simulation model never interferes with the real world system. It may be too disruptive because experiments are done with the model and not on the system itself. So simulations do not interfere with the real world system.
10. Time Saving. All results can be obtained with one model only. For example, the effects of ordering consumer behavior or other policies of many years can be obtained by computing simulation in a short time.
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11. Last Resort. Simulation, sometime is the last resort to solve impossible problems e.g. if we are unable to observe the actual environments on the planet MAR, simulation my be needed.
Disadvantages
1. Not Precise. Simulation is not precise and it does not yield an answer but provide a set of the system in different conditions.
2. Expensive. Developing a simulation model can be very expensive because it is long and complicated process to develop a model. Sometimes, it may take years to develop a model. Hence huge expenditures are involved.
3. No Optimum Solution. Unlike other quantitative techniques, simulation does not generate optimal solution to the problem. It is trial & error approach hence it may produce different solutions to the same problem.
4. Non-Transferable Solution. Each solution model is unique and its solution and inferences can not be transferable to other problems.
5. In –Efficient. Solution method of simulation model are not as efficient as other quantitative methods.
6. Adhoc. Each application of simulation is adhoc to a great extent, which indicates that simulation does not require standard probability distribution.
Monte – Carlo Simulation Technique
Monte-Carlo method may be applied when a system contains elements that exhibit chance in their behavior. So Monte-Carlo is an experiment on the chance. This technique uses random numbers and requires decision making under certaintities and where the mathematical formulation is impossible. Monte-Carlo is the code name, given by Von Newmann and S. M. Wam and this technique involves huge expenditure and extremely difficult for analytical treatment.
The technique breaks down into following steps:
1. Establishing Probability Distribution. In case of Monte-Carlo simulation the values of the variables have to be generated. There are number of variables in real world system that are probabilistic in nature and that we want to simulate, e.g. the number of variables may be (i) inventory demand on daily basis (ii) lead time for inventory order to arrive (iii) time between machine break down (iv) time between arrivals and users (v) service time etc. For the above said variables, we have to determine the probability distribution to examine historical events. Probability distribution need not be used solely on historical
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observations, it may be based on judgment & experience e.g. a sample of sale or machine break down may be used to create probability for these variables.
Probability distribution may be empirical or normal or Poisson or exponential pattern.
2. Cumulative Probability Distribution. From the regular probability distribution we can arrive at the cumulative probability probability distribution of every job. Cumulative probability is the sum of numbers in the probability column added to the previous cumulative probability.
3. Setting Random Numbers Intervals. Monte-Carlo simulation has established cumulative probability distribution for each variable. It requires generation of a sequence of random numbers. Random numbers of a series of digits say one digit, two digit etc e.g. if we are interested in one digit number there are only ten such numbers and if we want to generate number then chance of that number should be 1/10. These random numbers are called pseudo numbers.
Random numbers are generated using digital computers these days. Using cumulative probability distribution computed, we can set interval of random numbers for each item.
4. Generating Random Numbers. Random numbers may be generated in several ways. If the problem is large it involves thousands of simulation trials. If simulation is small the number may be selected that has 100 lots randomly from the random table. We can select the numbers from anywhere in the table. These random numbers are selected in such away that every number has an equal probability. If selection from 5 digit random number table (as given in the end of this chapter) these numbers may be chosen from any fashion. Selecting the columns or rows or diagonals, we can select our numbers as first two digits or last two digits or middle one out of the five digits.
The process of finding out the random numbers from the random numbers table runs as follows; suppose we want eight 2-digit random numbers. We can refer any where in the table e.g. the first column, first 2 digits of 8 consecutive numbers give the answer: 27, 13, 80, 10, 54, 60, 49, 78. So we can enter the table randomly and can make our selection from anywhere (row or column or diagonal or circle). In nut shell we can enter the table randomly but then to proceed serially. The following example will clarify the process of finding out the random numbers.
Example 1. Over 100 days period, the daily demand of a certain commodity shows the following frequency distribution pattern
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Daily Demand 0 1 2 3 4 5
No. of days 10 20 40 20 6 4 Total=100 Using the given data, simulate a ten day sequence of the demand value.
Solution. Out of 100 days for which the above frequency distribution pattern is shown, which 10 days should be chosen? For this reason of uncertainty we make use of random numbers. Now we will first make the probability table.
Daily Demand
Probability=No. of days / Total no. of
days
Cumulative Probability
Random Numbers Assignment 0
1 2 3 4 5
10/100 20/100 40/100 20/100 6/100 4/100
= .10
= .20
= .40
=.20
= .06
=.04
.10 .30 .70 .90 .96 1.0
00-09 10-29 30-69 70-89 90-95 96-99
How the random numbers assignment has been made is very simple to figure out on the basis of the cumulative frequency. Two digit random numbers have been chosen from 00 to 99 and not from 01 to 100 because in such case, the digit 100 will never figure in the 2 digit random numbers.
Now from random numbers table given at the end of this chapter, we take the first row and choose the last two digits of the each number of the row (for example, of the first number 27767 we choose 67, of R12 -43584, we choose 84,……….. of R1-10-48277, we choose 77 and so on). We have done so arbitrarily. Now we can make our selection from anywhere (row or column or diagonal or circle).
The 10 random numbers 67, 84, 01, 77, 90, 14, 15, 74, 44 &77.
Now the first random number 67 is falling in the slot 30-69 probability (random number assignment) and the corresponding daily demand is 2. For random number 84 the slot is 70-89 and the daily demand is 3 and so on
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Day number Generated Random Demand Generated
Demand 1
2 3 4 5 6 7 8 9 10
67 84 01 77 90 14 15 74 44 77
2 3 0 3 4 1 1 3 2 3 Total = 22
Average daily demand = 22/10 = 2.2
It is pertinent to note that out of 100 days, we have randomly (without bias) chosen the days 67th, 84th, 01st…….. 44th, 77th days. We may get different results if random numbers are differently chosen.
Example 2. Bright Bakery keeps stock of a popular brand of cake. Previous experience indicates the daily demand as given below:
Daily
demand 0 10 20 30 40 50
Probability .01 .20 .15 .50 .12 .02
Consider the following sequence of random numbers; 48, 78, 19, 51, 56, 77, 15, 14, 68, 09. Using this sequence simulate the demand for the next 10 days. Find out the stock situation if the owner of the bakery decides to make 30 cakes every day. Also estimate the daily average demand for the cakes on the basis of simulated data.
Solution. According to the given distribution of demand, the random number coding for various demand levels is shown in the table below:
332 Demand Probability Cumulative
Probability Random Number Interval 0
10 20 30 40 50
.01 .20 .15 .50 .12 .02
.01 .21 .36 .86 .98 1.0
00 01-20 21-35 36-85 86-97 98-99
The simulated demand for the cakes for the next 10 days is given below. The stock situation for various days in accordance with the bakery decision of making 30 cakes per day is also given.
Determination of demand and stock level
1
2
3
4
5
6
7
8
9
10
48
78
19
51
56
77
15
14
68
09
30
30
10
30
30
30
10
10
30
10
--- --- 20 20 20 20 40 60 60 80
Expected demand = 220/10 = 22 units per day.
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Example 3. Using random numbers to simulate a sample, find the probability that a packet of 6 products does not contain any defective product, when the production line produces 10% defective products. Compare your answer with the expected probability.
Solution. Given that 10% of the total production is defective and 90% is non-defective.
If we have 100 random numbers (0 to 99) then 90 of them (or 90%) represent non- defective products and remaining 10 of them (or 10%) represent defective products.
Thus, the random numbers 00 to 89 are assigned to variables representing non defective products and 90 to 100 are assigned to variables representing defective products. If we choose a set of 2 digit random numbers in the range of 00 to 99 to represent a packet of 6 products as shown below then we could expect that 90% of the time they would fall in the range of 00 to 89.
Sample
Number Random Numbers Selection
A B C D E F G H I J
67 25 17 75 27 11 39 26 21 20
04 04 52 80 49 70 04 05 81 47
61 38 62 12 75 14 72 20 94 86
20 11 83 26 84 21 38 36 73 16
84 38 92 04 26 24 84 84 91 63
35 14 05 87 69 54 03 87 26 95
Here it may noted that out of ten simulated samples three contain one more defectives and seven contain no defectives. Thus the expected percentage of non-defective products is 70%. However, theoretically the probability that a packet of six products containing no defective product is (.9)6.5314453.14%
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Example 4. The Cargo Honda Limited manufactures around 150 scooters. The daily production varies from 146 to 154 depending upon the availability of raw materials and other working conditions:
Production per day Probability 146
147 148 149 150 151 152 153 154
.04 .09 .12 .14 .11 .10 .20 .12 .08
The finished scooters are transported in specially arranged lorry accommodating 150 scooters. Using following random numbers, 80, 81, 76, 75, 64, 43, 18, 26, 10, 12, 65, 68, 69, 61, 57, stimulate the process to find out
(i) what will be the average number of scooters waiting in the factory?
(ii) what will be the average number of empty space on the lorry?
Solution. The random numbers are given in the table below:
Production
per day Probability Cumulative Probability
Random Numbers Assigned
146
147
148
149
150
.04
.09
.12
.14
.11
0.04 0.13 0.25 0.39 0.50
00-03 04-12 13-24 25-38 39-49
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151
152
153
154
.10
.20
.12
.08
0.60 0.80 0.92 1.00
50-59 60-79 80-91 92-99
Based on the 15 random numbers given, we simulate the production per day in the table below:
S.
No. Random
Numbers Production
Per day Capacity Lorry
No. of Scooters
waiting
Random Numbers Assigned 1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
80 81 76 75 64 43 18 26 10 12 65 68 69 61 57
153 153 152 152 152 150 148 149 147 147 152 152 152 152 151
150 150 150 150 150 150 150 150 150 150 150 150 150 150 150
3 3 2 2 2 0 0 0 0 0 2 2 2 2 1
0 0 0 0 0 0 2 1 3 3 0 0 0 0 0
Total 21 9
(i)Average number of scooters waiting = 21/15 = 1.4 per day.
(ii) Average number of empty space = 9/15 = 0.6 per day.
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Example 5. The output of a production line of LML Limited is checked by an inspector for one or more of three different types of defects called defects A, B & C. If defect A occurs the item is scrapped. If defect B or C occurs then the item must be reworked. The time required to rework a B defect is 45 minutes and the time required to rework a C defect is 55 minutes. The probabilities of A, B and C defects are .15, .20 and .10 respectively. For ten items coming off the assembly line, the numbers scrapped and the total minutes of rework time. Use the following random numbers
RN for defect A 48 55 91 40 93 01 83 63 47 52 RN for defect B 47 36 57 04 79 18 10 13 57 09 RN for defect C 82 95 18 96 20 8 56 11 52 03
Solution. Allocation of random numbers
Defect A Defect B Defect C
Whether Defect Exists Cumulative Probability Random Numbers Assigned Whether Defect Exists Cumulative Probability Random Numbers Assigned Whether Defect Exists Cumulative Probability Random Numbers Assigned Yes
No 15 85
.15 1.0
00-14 15-99
Yes No
20 80
.20 1.0
00-19 20-99
Yes No
10 90
.10 1.0
00-09 10-99
337 Simulation Worksheet
Item No. RN for defect A Whether Defect Exists or not RN for defect B Whether Defect Exists or not RN for defect C Whether Defect Exists or not Nature of defect Rework time Required in Minutes Remarks 1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
48 55 91 40 93 01 83 63 47 52
NO NO NO NO NO NO NO NO NO NO
47 36 57 04 79 18 10 13 57 09
NO NO NO NO NO NO NO NO NO NO
82 95 18 96 20 08 56 11 52 03
NO NO NO NO NO NO NO NO NO NO
--- --- --- B --- A, B, C
B B --- B, C
--- --- --- 30 --- --- 30 30 --- 30+55=85
--- --- --- --- --- Scrap
--- --- --- --- During the simulated period, No. of items without any defect = 5,
No. of items scrapped = 1 No. of items required rework = 4, Total work time required = 175 minutes or 2 Hrs. and 55 minutes.
Example 6. Tarun Kalyani Limited manufactures 30 items per day. The sale of these items depends upon demand which has the following distribution:
Sales (units) Probability 27
28 29 30 31 32
0.10 0.15 0.20 0.35 0.15 0.05
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The production cost and selling price of each unit are Rs. 40 and Rs. 50 respectively. Any unsold product is to be disposed off at a loss of Rs. 15 per unit. There is a penalty of Rs. 5 per unit if the demand is not meet. Using the following random numbers estimate total profit profit/loss for the company for the next 10 days ; 23, 99, 65, 99, 95, 01, 79, 11, 16, 10. If the company decides to produce 29 items per day what is the advantage or disadvantage to the company.
Sales (unit) Probability Cumulative Probability Random Numbers Assigned 27
28 29 30 31 32
.10 .15 .20 .35 .15 .05
0.10 0.25 0.45 0.80 0.95 1.00
00-99 10-24 25-44 45-79 80-94 95-99
Profit per unit = Selling price per unit – Cost per unit= Rs. 50 – Rs. 40= Rs. 10 per unit Simulation Worksheet
(i) Day
(ii) Random Numbers
(iii) Estimated
Sale
(iv)
Profit/Loss per day when production = 30items per
day
(v)
Profit/Loss/day when production=20 items per
day 1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
23 99 65 99 95 01 79 11 16 10
28 32 30 32 32 27 30 28 28 28
28Rs 10 - 2 Rs.15= 250 30Rs 0 - 2 Rs. 5 = 290 30 Rs.10 = 300 20Rs 10 - 2 Rs. 5= 190 30Rs 10 - 2 Rs.15= 250 27Rs 10 - 3 Rs.15= 225 30 Rs.10 =300 28Rs 10 - 2 Rs.15= 250 28Rs 10 - 2 Rs.15= 250 28Rs10 – 2 Rs.15= 250
28Rs10 – 1 Rs.15= 265 29Rs 10 - 3 Rs. 5 = 275 29Rs 10 - 1 Rs. 5 = 285 29Rs 10 - 3 Rs. 5 = 275 29Rs 10 - 3 Rs. 5 = 275 27Rs 10 - 2 Rs.15= 240 29Rs 10 - 1 Rs. 5 = 285 28Rs 10 - 1 Rs.15= 265 28Rs 10 - 1 Rs.15= 265 28Rs 10 - 1 Rs.15= 265 2695 2695 Total profit = Rs 2695, from the computation it is apparent there is no additional profit or loss if the production is reduced to 29 items per day since the total profit remains unchanged i.e. Rs 2695.
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Example 7. The occurrence of rain in a city on a day is dependent upon whether or not it rained on the previous day. If it rained on previous day the rain distribution is given by
Event Probability No rain
1 cm rain 2 cm rain 3 cm rain 4 cm rain 5 cm rain
0.50 0.25 0.15 0.05 0.03 0.02
If it did not rain the previous day the rain distribution is given by:
Event Probability No rain
1 cm rain 2 cm rain 3 cm rain
0.75 0.15 0.06 0.04
Simulate the city’s weather for 10 days and determine by simulation the total days without rain as well as the total rainfall during the period. Use the following random numbers; 67, 63, 39, 55, 29, 78, 70, 06, 78, 76 for simulation assume that for the first day of the simulation it had not rained the day before.
Solution. Allocation of random numbers(rain on previous day) Event Probability Cumulative
Probability Random Numbers Assigned No rain
1 cm rain 2 cm rain 3 cm rain 4 cm rain 5 cm rain
0.50 0.25 0.15 0.05 0.03 0.02
0.50 0.75 0.90 0.95 0.98 1.00
00-49 50-74 75-89 90-94 95-97 98-99
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Allocation of random number (No rain previously) Event Probability Cumulative
Probability Random Numbers
Assigned No rain
1 cm rain 2 cm rain 3 cm rain
0.75 0.15 0.06 0.04
0.75 0.90 0.96 1.00
00-74 75-89 90-95 96-99
Simulation work sheet : weather
Day Random Numbers Event Table Reference
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
67 63 39 55 29 78 70 06 78 76
No rain No rain No rain No rain No rain No rain No rain No rain No rain No rain
(from table 2) (from table 2) (from table 2) (from table 2) (from table 2) (from table 2) (from table 1) (from table 1) (from table 2) (from table 1)
Hence, during the simulated period it did not rain on 6 days out of 10 days. The total rain fall during the period was 5 cm.
Example 8. A dentist is considered strong if he schedules all his patients for 30 minutes appointment. Some of the patients take more or less than 30 minutes depending on the type of dental work to be done. The following summary shows the various categories of work, their probabilities and the time actually needed to complete the work:
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Category Time Required Probability of Category
Filling Crown Cleaning Extraction Checkup
45 minutes 60 minutes 15 minutes 45 minutes 15 minutes
0.40 0.15 0.15 0.10 0.20
Simulate the dentist’s clinic for four hours and determine the average waiting time for the patients as well as the idleness of the doctor. Assume that all the patients show up at the clinic at exactly their schedule arrival time starting at 8.00 am. Use the following random numbers for handling the above problem ; 40, 82, 11, 34, 25, 17, 79.
Solution. Allocation of random numbers
Type Probability Cumulative Probability Random Numbers Filling
Crown Cleaning Extraction Checkup
0.40 0.15 0.15 0.10 0.20
0.40 0.55 0.70 0.80 1.00
00-39 40-54 55-69 70-79 80-99 Simulation work sheet
Patient Scheduled Arrival
Random Numbers
Category Service Time Needed 1.
2.
3.
4.
5.
6.
7.
8.
8.00 8.30 9.00 9.30 10.00 10.30 11.00 11.30
40 82 11 34 25 66 17 79
Crown Checkup
Filling Filling Filling Cleaning
Filling Extraction
60 Minutes 15 Minutes 45 Minutes 45 Minutes 45 Minutes 15 Minutes 45 Minutes 45 Minutes Now, let us simulate the dentist’s clinic for four hours starting at 8.00 A.M.
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Table showing the waiting times for the patients A
Patient B Arrival
C Service
Starts
D Service Time
E = C+D Service
Ends
F = C-B Waiting (Minutes)
Idle Time 1.
2.
3.
4.
5.
6.
7.
8.
8.00 8.30 9.00 9.30 10.00 10.30 11.00 11.30
8.00 9.00 9.15 10.00 10.45 11.30 11.45 12.30
60 Minutes 15 Minutes 45 Minutes 45 Minutes 45 Minutes 15 Minutes 45 Minutes 45 Minutes
9.00 9.15 10.00 10.45 11.30 11.45 12.30 1.15
0 30 15 30 45 60 45 60
0 0 0 0 0 0 0 0 The average waiting time of a patient = 285/8 = 35.65. The dentist was not idle during the entire simulated period. Hence Doctor’s idle time = nil.
Simulation and Inventory Control
We know that demand and lead time are uncertain and hence simulation is widely used to solve the inventory problems. The distribution of demand during the lead time can be obtained from an empirical analysis of past data or by computer simulation using random numbers. The cumulative probability distribution of demand during during the lead time is used as a basis to determine the annual inventory costs and stock out cost for different levels of the safety stocks. The management can experiment the effect of various inventory policies by using simulation and finally selects the optimum inventory policy.
The main objective of simulation is to get optimum results by way of minimum annual ordering cost, holding cost, stock out cost and ultimately total inventory cost.
Example 9. The distribution of demand during the lead time, the distribution of the lead time are set out in Tables 1 and 2
Table 1. Distribution of Demand during lead time Quantity Demanded during Lead Time Probability
0 1 2 3
0.10 0.45 0.30 0.15
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Table 2.Distribution of Lead Time Lead Time (weeks) Probability
2 3 4
0.20 0.65 0.15
Compute, the annual inventory cost if the management wants reorder 12 units each time when the quantity on hands is 6 units. Moreover, supplier is in a position to execute all orders. Cost of ordering per order is Rs. 120 and the cost of holding inventory in stock is Rs. 5 per unit per week. The stock out cost is Rs. 75 per unit. Show simulation of 15 weeks. Opening stock is 10 units.
Solution.
Table 1
Demand Cumulative Probability Random Numbers Assigned 0
1 2 3
0.10 0.55 0.85 1.00
00-99 10-54 55-84 85-99 Table 2
Lead Time Cumulative Probability Random Numbers Assigned 2
3 4
0.20 0.85 1.00
00-19 20-84 85-99
We assume that the stock on hand at the start of the simulation process is 10 units.
Further, we also assume that all orders are placed at the beginning of the week and all deliveries against orders are received at the beginning of the week. The simulated demand for wok 1 is I unit (corresponding to random number 49 in table 1) and the closing unit is 9 units. The inventory carrying costs are calculated for the remaining
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weeks in the same way. At the end of week 4, the closing inventory is 6 units (reorder point). So an order is placed at the beginning of week 5 for a quantity of 12 units (ordering quantity). The simulated lead time is 3 weeks (corresponding to random number 84 in table 2). This order quantity is, therefore, received at the beginning of week 8. We notice from table 3 that another order is placed at the beginning of week 12, when the stock on hand is 6 units which is equal to reorder point. Again , the simulated lead time is 3 weeks before this quantity is received. At the beginning of week 15, we find that the quantity available at the beginning of week 14 is 1 unit and the simulated demand for that week is 2 unit giving rise to a stock out of 1 unit. The stock out cost is Rs.75
Table 3:Simulation of Demand and Lead Time for 15 Weeks
Week Stock on hand beginning Demand
Quantity Received Stock on hand end of week Inventory Carrying Costs Stock out Lead Time Random Numbers Quantity Demanded Quantity Costs Random Numbers Lead Time Period 1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
10 9 7 7 6 3 3 2 12 10 8 6 4 1 --
49 67 06 30 95 01 10 70 80 66 69 76 86 56 84
1 2 0 1 3 0 1 2 2 2 2 2 3 2 2
-- -- -- -- -- -- -- 12
-- -- -- -- -- -- 12
9 7 7 6 3 3 2 12 10 8 6 4 1 -- 10
45 35 35 30 15 15 10 60 50 40 30 20 5 -- 50
1 75
79
79 3
3
Inventory carrying cost = 450, Ordering Cost = 240, Stock out Cost = 75 , Total = 735
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Review Questions
1. When it becomes difficult to use an optimization technique for solving a problem one has to resort to simulation.” Discuss.
2. What is simulation? Describe the simulation process. What are the advantages and limitations of simulation? Also specify the areas where simulation can be used.
3. “Simulation is an essentially valuable tool in a situation where the mathematics needed to describe a system realistically is too complex to yield analytical solution.”
Elucidate.
Exercise
1. The manager of a book store has to decide the number the number of copies of a particular tax law book to order. A book cost Rs. 60 and is sold for Rs. 80. Since some of the tax laws change year by year, any copy unsold while the edition is current must be sold for Rs. 30. From past records, the distribution of demand for this book has been obtained as follows:
Demand (No. of Copies) : 15 16 17 18 19 20 21 22 Proportion : 0.05 0.08 0.20 0.45 0.10 0.07 0.03 0.02 Using the following sequence of random numbers, generate data on demand for 20 time period (years). Calculate the average profit obtainable under each of the courses of action open to manger. If producer opts for production of (a) 15 books (b) 20 books(c) 18 books, what is the optimal policy?
14 02 93 99 18 71 37 30 12 10 88 12 00 57 69 32 18 08 92 72 2. A bookstore wishes to carry Bible in stock. Demand is probabilistic and replenishment takes 2 days (i.e. if an order is placed on September 15, it will be delivered at the end of the day on September 17). The frequency distribution of demand is given below:
Demand(daily) : 0 1 2 3 4 Probability : 0.05 0.10 0.30 0.45 0.10 Each time an order is placed, an ordering cost of Rs. 10 per order is incurred. Inventory carrying cost is Rs.0.50 per book per day and it is calculate on the basis of the stock at the end of each day. The bookstore has a policy to order 5 books when the inventory at the beginning of the day plus orders outstanding is less than 8 books. Currently (beginning first day) the store has a stock 8 books plus 6 books ordered two days ago and expected to
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arrive as scheduled. Using Monte-Carlo simulation for 10 cycles, find total cost for the simulated period taking the following sequence of random numbers ; 89, 34, 70, 63, 61, 81, 39, 16, 13, 73.
3. A book store wishes to carry a particular book in store. Demand is probabilistic and replenishment of stock taken 2 days (i.e. if an order is placed on March 1, it will be delivered at the end of the day March 3). The probabilities of demand are given below:
Demand (daily) : 0 1 2 3 4 Probability : 0.05 0.10 0.30 0.45 0.10 Each time an order is placed, the store incurs an ordering cost of Rs. 10 per order. The store also incurs a carrying cost of Rs.0.05 per book per day. The inventory carrying cost is calculated on the basis of at the end of each day. The manager of the book store wishes to compare two options for his inventory decision.
A : order 5 books when the inventory at beginning of the day plus orders outstanding is less than 8 books.
B : order 8 books when the inventory at beginning of the day plus orders outstanding is less than 8 books.
Currently (beginning of the first day) the store has a stock of 8 books plus 6 books ordered two days ago and expected to arrive next day. Using Monte-Carlo simulation for 10 cycles, recommend which option the manager should choose. The two digits random numbers are: 89, 34, 78, 63, 61, 81, 39, 16, 13, 73.
4. Hero Honda Co. Limited wishes to determine the levels of stock it should carry for the items in its range. Demand is not certain and there is a lead time for stock replenishment.
For one item X, the following information is obtained:
Demand (units /day) : 3 4 5 6 7 Probability : 0.10 0.20 0.30 0.30 0.10 Carrying Cost ( per unit / day) : Rs. 0.2
Ordering Cost (per order) : Rs. 15 Lead time for replenishment : 3 days
Stock on hand at the beginning of the simulation exercise was 20 units.
Carry out a simulation run over a period of 10 days with the objective of evaluating the inventory rule. Order 15 units when present inventory plus any outstanding order falls
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below 15 units. The sequence of random numbers to be used is : 0,9,1,1,5,1,8,6,3,5,6,1,3,5 using first number for the day.
5. Sharma Bakery Amritsar keeps stock of a popular brand of cake. Daily demand based on past experience is given below:
Daily Demand : 0 15 25 35 45 50 Probability : .01 .15 .20 .50 .12 .02 Consider the following sequence of random numbers: 48, 78, 09, 51, 56, 77, 15, 14, 68, 09. Using the sequence, simulate the demand for the next 10 days. Find out the stock situation if the owner of the bakery decides to make 35 cakes every day. Also estimate the daily average demand for the cakes on the basis of simulated data.
6. A Company manufactures around 200 mopeds. Depending upon the availability of raw materials and other conditions, the daily production has been varying from 196 moped to 204 mopeds whose probability distribution is as given below:
Production per
day Probability
196 197 198 199 200 201 202 203 204
0.05 0.09 0.12 0.14 0.20 0.15 0.11 0.08 0.06
The finished mopeds are transported in a specially designed three storied lorry that can accommodate only 200 mopeds. Using the following 15 random numbers 82, 89, 78, 24, 53, 61, 18, 45, 04, 23, 50, 77, 27, 54, 10, simulate the process to find out
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(i) What will be average number of mopeds waiting in the factory.
(ii) What will be average number of empty space in the factory.
7. Bicon Equipment manufacturing company is engaged in producing different types of high class equipment for use in science laboratories. The company has two different assembly times to produce its most popular product. The processing time for each of the assembly line is regarded as a random variable and is described by the following distribution:
Processing Time (Minutes)
Assembly A Assembly B
20 21 22 23 24
0.20 0.40 0.20 0.15 0.05
0.10 0.15 0.40 0.25 0.10
Using the following random numbers, generate data on the process times for 115 units of the item and compute the expected process time for the period:
3441, 7674, 4349, 4383, 8311, 1519, 0236, 4594, 1554, 0575, 8900, 8008, 2874, 2434, 0993
For the purpose, read the numbers horizontally, taking the first two digits for the processing time on assembly A and the last two digits for processing time on assembly B.
8. A Production Manager is planning to produce a new product and he wishes to estimate the raw material requirement for that new product. On the basis of usage for similar product introduced previous, he has developed a frequency distribution of demand in tones per day for a two months period. Use this data to simulate the raw material usage requirements for 7 days. Random numbers in 5 digits are: 27767, 13025, 80217, 10875, 54127, 60311, 49739, 78626, 66692, 44071
Demand Tonnes / day : 10 11 12 13 14 15 Total Frequency No. of days : 6 18 15 12 06 03 60
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9. The Financial Controller of super stock Ltd. has drawn the following projections with probability distribution:
Wages & Salaries
(Rs. ‘000) Probability Raw Material
(Rs. ‘000) Probability
Sales Revenue (Rs. ‘000)
Probability
10-12 12-14 14-16 16-18
0.3 0.5 0.2 0.4
6-8 8-10 10-12 12-14
0.2 0.3 0.3 0.2
30-34 34-38 38-42 42-46
0.1 0.3 0.4 0.2
Fixed costs are Rs. 14000 and opening balance of cash Rs. 50000. Simulate the cash flow projection, use the following random numbers
Wages & Salaries : 2 7 9 2 9 8 Raw Material : 4 4 1 0 3 4 Sales Revenue : 0 6 6 8 0 2
Answers
1. The optimum policy is to produce 18 books and profit is Rs. 340 2. Rs. 62.50 3.Total cost A = Rs. 5950, option B = Rs. 52.50 4. 19.90, 45, TVC = 64.90 5. 27 units per day 6. No. of mopeds waiting = 2.80, No. of space in lorry = .266 7. Expected Time = 678/15 = 43.37 8. 11.7 Tonnes 9. Rs.60000
