e-Notes by D.G.Kantharaju, East Point College, Bangalore
Network Analysis • Network Network is is a graphica graphicall represent representati ation on of all the the Activi Activitie tiess and Events Events arrang arranged ed in a logical and sequential order. • Network Network analysi analysiss plays plays an impo importa rtant nt role role in proje project ct manag manageme ement. nt. • A project project is is a combin combination ation of interrel interrelated ated activi activities ties all all of which must be executed executed in a certain order for its completion. Activity: Activity is the actual performance of the job. This consumes resources (Time, human resources, money, and material) Event: An event refers to start or completion of a job. This does not consume any resources. • • • •
Analyz Analyzing ing netwo network, rk, the the plannin planning, g, schedul scheduling ing and and control control of a projec projectt becomes becomes easier. PERT PERT and CPM CPM are the the two most most popul popular ar networ networkk analysi analysiss techniq technique ue used used to assist managers in planning and controlling large scale projects. PERT PE RT-- (Progr (Program amme me Eval Evalua uati tion on Revie Review w Techni Techniqu que) e) CPM CPM - (Cri (Criti tica call Pat Pathh Met Metho hod) d)
Applications: Construction of a Residential complex, Commercial complex, Petro-chemical complex Ship building Satellite mission development Installation of a pipe line project etc... Historical Evolution. Before 1957 there was no generally accepted procedure to aid the management of a project. In 1958 PERT was developed by team of engineers working on a Polaris Missile programme of the navy. This was a large project involved 250 prime contractors and about 9000 job contractors. It had about 19 million components. In such projects it is possible that a delay in the delivery of a small component might hold the progress of entire project. PERT was used successfully and the total time of completion was reduced from 7 years to 5 years. In 1958 Du Pont Company used a technique called Critical Path Method (CPM) to schedule and control a very large project like overhauling of a chemical plant, there by reducing the shutdown period from 130hrs to 90 hrs saving the company 1 million dollar. Both of these techniques are referred to as project scheduling techniques.
1
Differences between PERT & CPM PERT
C PM
1. It is a technique for planning 1. It is a technique for planning scheduling & controlling of projects scheduling & controlling of projects whose activities are subject to whose activities not subjected to uncertainty in the performance time.any uncertainty and the Hence it is a probabilistic model performance times are fixed. Hence it is a deterministic model 2.It 2.It is is an Ev Even entt orie orient nted ed sys syste tem m
2. It is an Activ Activit ityy orie orient nted ed sys syste tem m
3.Basically does not differentiate critical and non-critical activities
Differentiates clearly the critical activities from the other activities.
4. Used in projects where resources 4. Used in projects where overall (men, materials, money) are always costs is of primarily important. available when required. Therefore better utilized resources 5. Suitable for Research and 5.Suitable for civil constructions, Development projects where times installation, ship building etc. cannot be predicted
2
Differences between PERT & CPM PERT
C PM
1. It is a technique for planning 1. It is a technique for planning scheduling & controlling of projects scheduling & controlling of projects whose activities are subject to whose activities not subjected to uncertainty in the performance time.any uncertainty and the Hence it is a probabilistic model performance times are fixed. Hence it is a deterministic model 2.It 2.It is is an Ev Even entt orie orient nted ed sys syste tem m
2. It is an Activ Activit ityy orie orient nted ed sys syste tem m
3.Basically does not differentiate critical and non-critical activities
Differentiates clearly the critical activities from the other activities.
4. Used in projects where resources 4. Used in projects where overall (men, materials, money) are always costs is of primarily important. available when required. Therefore better utilized resources 5. Suitable for Research and 5.Suitable for civil constructions, Development projects where times installation, ship building etc. cannot be predicted
2
3
4
5
Problem 2. 6
Proble Problem m 3. 7
8
9
10
11
CRITICAL PATH Meaning: The longest path in a project network which determine the duration of the project is known as critical path.
Determination 2
Step 1. List all the p –
–
–
12
Step 2.For each seq
Determination 2
13
Float (Slack) • Float (Slack ) refers to the amount of time by which a particular event or an activity can be delayed without affecting the time schedule of the network. • Float (Slack) Float (Slack) is defined as the difference between latest allowable and the earliest expected time. Event Float/Slack = LS – ES Where LS = Latest start time ES = Early start time. Earliest start : Denoted as ‘ES’ Earliest start time is the earliest possible time by which the activity can be started. Early finish time : Denoted as ‘EF’ Early finish time is the earliest possible time by which the activity can be completed. Latest start time : Denoted as ‘LS’ Latest start time is the latest possible time by which the activity can be started Late finish time : Denoted as ‘LS’ Late finish time is the latest possible time by which the activity can be completed Total float (TF) / Total slack (TS) Total float of the job is the differences between its Late start and Early start ‘or’ Late finish and Early finish i.e. TF( CA) = LS (CA) - ES (CA) Or TF( CA) = LF (CA) - EF (CA) CA = Current activity Free float (FF) Free float is the amount of time a job can be delayed without affecting the Early start time of any other job. FF(CA) = ES(SA) – EF (CA) CA = Current Activity SA = Succeeding Activity Independent Float (IF) Independent Float is the amount of time that can be delayed without affecting either predecessor or successor activities. IF = ES(SA) – LF(PA) - Duration of CA ES = Early Start LF = Late Finish SA = Succeeding Activity PA = Preceding Activity CA = Current Activity
14
Example 1 Construct the Network for the following Project and determine the following i) Critical Path ii) ES,EF,LS,LF iii) TF,FF Activity
Duration
1-2
14
1-4
3
2-3
7
2-4
0
3-5
4
4-5
3
5-6
10
Construction of t
15
Determination ) 1 1 (
Activity Duration
16
Example 2 Construct the Network for the following Project and determine the following i) Critical Path ii) ES,EF,LS,LF iii) TF,FF Activity
Duration
1-2
2
2-3
3
2-4
5
3-5
4
3-6
1
4-6
6
4-7
2
5-8
8
6-8
7
7-8
4
17
Activity
Duration
1-2
2
In PERT model, 3 time values are associated with each activity. They are i) Optimistic time = to ii) Pessimistic time = tp iii) Most likely time = tm These three times provide a measure of uncertainty associated with that activity Optimistic Time: This is the shortest possible time in which the activity can be finished. It assumes that every thing goes well. Pessimistic Time: This is the longest time that an activity could take. It assumes that every thing goes wrong. Most likely Time: It is the estimate of the normal time that an activity would take. This assumes normal delays. Expected Time ( te): ‘te’ can be calculated by the following formula te = (to + 4tm + tp) / 6 Example. If a job has to = 5 days, tp = 17 days, tm = 8 days Then Expected time for the job would be 18
te = (to + 4tm + tp) / 6 = (5 + 4 x 8 + 17) / 6 = 9 days Variability of activity times • Standard deviation and Variance are commonly used in statistics to measure the variability of number. In PERT model, to measure the variability of an activity time duration standard deviation and variance are used. A large standard deviation represents high variability and vice-versa. • Calculation of Standard Deviation and Variance Variance = (Standard deviation )2 Standard deviation =(t p – t o) / 6 • Expected length of the Critical Path = te of all the activities along the Critical Path Probability of completing the project within a given date Z = (TS – TE ) / σ Where TS = Scheduled time for project completion TE = Expected time for the project completion σ = Standard deviation for the Network σNetwork = √Sum of variances along the Critical Path = √ (σ i-j )2 Where σ i-j is the variance of a activity i-j along the critical path Example 1 Construct the Network for the following project and calculate the probability of completing the project in 25 days Activity
to
tm
tp
1-2
2
6
10
1-3
4
8
12
2-3
2
4
6
2-4
2
3
4
3-4
0
0
0
3-5
3
6
9
4-6
6
10
14
5-6
1
3
5
19
1.Constructio
2. Calculation o
20
3. Determinati
Activity
to
tm
tp
σ2 = ((t p – t o) / 6)2
10 12
Critical activities 1-2 -
1-2 1-3
2 4
6 8
2-3 2-4
2 2
4 3
6 4
2-3 -
0.44
3-4 3-5
0 3
0 6
0 9
3-4 -
0
4-6 5-6
6 1
10 3
14 5
4-6 -
1.78
1.78
Σ σ2 = 4.00
σNetwork = √Sum of variances along the Critical Path = √ (σNetwork )2 21
=√4 =2 Probability of completing the project within a given date Z = (TS – TE ) / σ Where TS = Scheduled time for project completion TE = Expected time for the project completion σ = Standard deviation for the Network = (25 – 20) / 2 = + 2.5 From the Normal distribution Table, we get the probability of completing the project in 25 days is 99.4%
Example 2. The following table lists the jobs of a network along with their time estimates. Activity
to
tm
tp
1-4
3
9
27
1-3
3
6
15
1-2
6
12
30
4-5
1
4
07
3-5
3
9
27
3-6
2
5
08
5-6
6
12
30
2-6
4
19
28
a) Draw the project network. b) What is the probability that the job will be completed in 35 days? c) What due date has 90% chance of being met? 22
1.Construction
2. Calculation
23
3. Determinati
As there are two Critical Paths, the path which gives more variance(σ2) is taken as Critical Path Path A Activity
σ2 = ((t p – t o) / 6)2
σ2
1-2
((30 – 6)/6)2
16
2-6
((28 – 4)/6)2
16 Σ σ2 = 32.00
Path B Activity
σ2 = ((t p – t o) / 6)2
σ2
1-3
((15 – 3)/6)2
4
3-5
((27 – 3)/6)2
16
5-6
((30 – 6)/6)2
16 Σ σ2= 36.00 24
σ = √Σ σ2 = √ 36
=6
Therefore the Critical Path is 1 - 3 - 5 - 6 b) Probability of completing the project within a given date Z = (TS – TE ) / σ Where TS = Scheduled time for project completion TE = Expected time for the project completion σ = Standard deviation for the Network = (35 – 32) / 6 = + 0.5 From the Normal distribution Table, we get the probability of completing the project in 35 days is 69.15% c) The due date for 90% chance of being met. Probability of completing the project within a given date Z = (TS – TE ) / σ The value of Z from the table for a 90% probability is +1.28 TS = ? (to be calculated) ,TE = 32, σ = 6 i.e. 1.28 = (TS– 32) / 6 TS = 39.68 days CPM Model In 1958 Du Pont Company used a technique called Critical Path Method (CPM) to schedule and control a very large project like overhauling of a chemical plant, there by reducing the shutdown period from 130hrs to 90 hrs saving the company 1 million dollar. Both of these techniques are referred to as project scheduling techniques. Cost considerations in PERT / CPM The total cost of any project comprises of two costs. • Direct cost - material cost, manpower loading • Indirect cost - overheads such as managerial services, equipment rent, building rent etc.
25
t
Example 1.
26
Find the lowest cost and optimum cost schedule for the following project, given the over head expenses as Rs.45/-day. Activity
Normal duration
Crash duration
Cost of crashing per day
1-2
3
1
Rs.40
2-3
4
2
Rs.40
2-4
7
3
Rs.10
3-4
5
2
Rs.20
1.Constructio
27
1.Determinati
Step 1. 3 -1 1 28
Step 2. 3 -1 1
Step 3. 3 -1 1 29
Step 3. 3 -1 1
30
Step 4. 1 -1 1
Step 5. 3 -1 1
31
SIMULATION
Simulation is to imitate reality to represent reality. Simulation is a technique for conducting experiments Simulation is descriptive and not optimizing technique Simulation is a process often consists of repetition of an experiment in many, many times to obtain an estimate of the overall effect of certain actions. Simulation is usually called for only when the problem under investigation is too complex to be treated by analytical models or by numerical optimization techniques. In a simulation, a given system is copied and the variables and constants associated with it are manipulated in that artificial environment to examine the behaviour of the system. In general terms, Simulation involves developing a model of some real life phenomenon and then performing experiments on the model evolved. Often we do not find a mathematical technique that; a model once constructed may permit us to predict what will be the consequences of taking a certain action. In particular we could ‘experiment’ on the model by ‘trying’ alternative actions or parameters and compare their consequences. This ‘experimentation’ allow us to answer ‘what if’ questions relating the effects of your assumption on the model response. The availability of the computers makes it possible for us to deal with an extraordinary large quantity of details which can be incorporated into a model and the ability to manipulate the model over many experiments (i.e. replicating all the possibilities that may be imbedded in the external world and events would seem to recur). For example, •
Testing of an aircraft model in a wind tunnel to test the aerodynamic properties of an the model
•
A model of a traffic signal system
•
Military war games
•
Business games for training
•
Planetarium etc. 32
Simulation defined A simulation of a system or an organism is the operation of a model or simulator which is a representation of the system or organism. The model is amenable to manipulation which would be impossible, too expensive or unpractical to perform on the entity it portrays. The operation of the model can be studied and for it, properties concerning the behaviour of the actual system can be inferred. -Shubik Simulation is the process of designing a model of a real system and conducting the experiments with this model for the purpose of understanding the behaviour (within the limits imposed by a criterion or set of criterion) for the operation of the system. -
Shannon
Steps of Simulation process Step 1: Identify the problem. If an inventory system is being simulated, then the problem may concern the determination of the size of the order (number of units to be ordered) when the inventory falls up to the reorder level (point). Step 2: Identify the decision variables, performance criterion (objective) and the decision rules. In the context of the above defined inventory problem, the demand (consumption rate), lead time and safety stock are identified as the decision variable. These variables shall be responsible to measure the performance of the system in terms of total inventory cost under the decision rule- when to order. Step 3: Construct a numerical model. Numerical model is constructed to be analyzed on the computer. Some times the model is written in a particular simulation language which is suited for the problem under the analysis. Step 4: Validate the model Validation is necessary to ensure whether it is truly representing the system being analyzed and the results will be reliable. Step 5: Design the experiments
33
Conduct experiments with the simulation model by listing specific values of variables to be tested (i.e. list courses of action for testing) at each trail (run). Step 6: Run the simulation model. Run the model on the computer to get the results in form of operating characteristics. Step 7: Examine the results. Examine the results of the problem as well as their reliability and correctness. If the simulation is complete, then select the best course of action (or alternative) otherwise make the desired changes in the model decision variables, parameters or design and return to step 3. Advantages of simulation
1. Simulation is a straight forward and simple technique 2. The technique is very useful to analyze large and complex problems which are not amenable to mathematical or quantitative methods. 3. It is an interactive method, which enables the decision maker to study the changes and their effects on the performance of the system. 4. The experiments in a simulation are run on the model and not on the system itself. Limitations of simulation
1. At times simulation models can be very costly and expensive 2. It is trail and error technique to produce different solutions in repeated runs. 3. The solution obtained from the simulation may not be optimal. 4. The simulation model needs to be examined and analyzed for decision making. It only creates an alternative and not an optimal solution by itself.
Monte-Carlo techniques or Monte-Carlo simulation. The Monte-Carlo method is a simulation technique in which statistical distribution functions are created by using series of random numbers Random numbers. The underlying theory in random number is that, each number has an equal opportunity of being selected.
34
There are various ways in which random numbers may be generated. These could be: result of some device like coin or die; published table of random numbers, mid-square method, or some other sophisticated method. It may be mentioned here that random numbers generated by some method may not be really random in nature. In fact such numbers are called pseudo-random-numbers. Rand corporation (of USA): A million random digits, is a random number table used in simulation situations. The numbers in these tables are in random arrangement. The Monte-Carlo simulation technique consists of the following steps. 1. Setting up a probability distribution for variables to be analyzed. 2. Building a cumulative probability distribution for each random variable. 3. Generating random numbers. Assign an appropriate set of random numbers to represent value or range (interval) of values for each random variable. 4. Conduct the simulation experiment by means of random sampling 5. Repeat Step 4 until the required number of simulation runs has been generated. 6. Design and implement a course of action and maintain control. Example 1.
A bakery keeps stock of popular brand of cake. Previous experience shows the daily demand pattern for the item with associated probabilities, as given below: Daily demand(number) Probability
0 0.01
10 0.20
20 0.15
30 0.50
40 0.12
50 0.02
Use the following sequence of random numbers to simulate the demand for next 10 days. Random Numbers
40
19
87
83
73
84
29
09
02
20
Also estimate the daily average demand for the cakes on the basis of simulated data. Solution: Using the daily demand distribution, we obtain a probability distribution as shown in the following Table.
Daily demand distribution 35
Daily Demand 0 10 20 30 40 50
Probability 0.01 0.20 0.15 0.50 0.12 0.02
Cumulative Probability 0.01 0.21 0.36 0.86 0.98 1.00
Random Number Interval 00 01-20 21-35 36-85 86-97 98-99
By conducting the simulation experiment for demand by taking a sample of 10 random numbers from the table of random numbers we get, Days 1 2 3 4 5 6 7 8 9 10
Random Number 40 19 87 83 73 84 29 09 02 20 Total
Demand 30 10 40 30 30 30 20 10 10 10 = 220
Because 0.36 < 0.40 < 0.85 Because 0.01 < 0.19 < 0.20 and so on
Expected demand = 220/10 = 22 units per day.
Example 2.
XYZ spare parts company wishes to determine the level of stock it should carry for 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
3
4
5
6
7
(Units/day) 36
Probability Carrying
0.10
0.20
0.30 Rs.2
0.30
0.10
cost(Per unit per day) Ordering
Rs.50
cost(per order) Lead time for
3 days
replenishment 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 outstanding falls below 15 units. The sequence of random numbers to be used is :0,9,1,1,5,1,8,6,3,5,7,1,1,2,9 using the first number for day one. Solution: Let us begin the simulation by assuming that i) orders are placed at the end of the day and received after 3 days at the end of the day. ii) back orders are accumulated in case of short supply and are supplied when stock is available. The cumulative probability distribution and the random number range for daily demand is shown in the table.
Daily Demand Distribution Daily Demand
Probability
Cumulative
Random Number
Probability Range 3 0.10 0.10 00 4 0.20 0.30 01-02 5 0.30 0.60 03-05 6 0.30 0.90 06-08 7 0.10 1.00 09 The results of the simulation experiment conducted are as shown below. Days Opening Random Resulting Closing
Order
Order
Average stock
37
Stock Number Demand Stock 1 20 0 3 17 2 17 9 7 10 3 10 1 4 6 4 6 1 4 2 5 2 5 5 0(-3)* 6 12 1 4 8 7 8 8 6 2 8 2 6 6 0(-4)* 9 11 3 5 6 10 6 5 5 1 *Negative figure indicates back orders.
Placed Delivered 15 15 15 15 15 -
in the evening 18.5 13.5 8 4 1 10 5 1 8.5 3.5
Average ending stock = 73/10 = 7.3 units/day Daily ordering cost = (Cost of placing one order) x (Number of orders placed per day) = 50 x 0.3 = Rs.15 Daily carrying cost = (Cost of carrying one unit per day) x (Average ending stock) = 2 x 7.3 = Rs.14.6 Total daily inventory cost = Daily ordering cost + Daily carrying cost = 15 + 14.6 = Rs.29.6 Example 3.
A small retailer deals in a perishable commodity, the daily demand and supply of which are random variables. The past 500 days data show the following: Supply Demand Available ( kg ) Number of Days Demand (kg ) Number of Days 10 40 10 50 20 50 20 110 30 190 30 200 40 150 40 100 50 70 50 40 The retailer buys the commodity at Rs.20 per kg and sells at Rs.30 per kg. If any commodity remains at the end the day it has no resale value and is a dead loss. Moreover, the loss on any unsatisfied demand is Rs.8 per kg Given the following random numbers. Simulate six day sales. 31
18
63
84
15
79
07
32
43
75
81
27
38
Using the random numbers alternatively, for example, first pair (31) to simulate supply and second pair (18) to simulate demand, etc. Solution: Probability and Random Number Interval for Daily Demand and Supply Available
Probability
(kg)
Number 40/500=0.08 00-07 50/500=0.10 08-17 190/500=0.38 18-55 150/500=0.30 56-85 70/500=0.14 85-99
10 20 30 40 50
Random
Demand
Probability
Random
50/500=0.10 110/500=0.22 200/500=0.40 100/500=0.20 40/500=0.08
Number 00-09 10-31 32-71 72-91 92-99
(kg) 10 20 30 40 50
Simulation Experimentation Sheet Day 1 2 3 4 5 6
Random Supply
Random Demand Cost
Revenue Shortage Profit
Number (kg) 31 30 63 40 15 20 07 10 43 30 81 40
Number (kg) 18 20 84 40 79 40 32 30 75 40 27 20
(Rs) 600 1,200 600 300 900 600
(Rs) 600 800 400 200 600 800
(Loss) 160 160 80 -
400 40 -60 220 -200
The above table shows that during these six days period retailer makes a net profit of Rs.400.
Example 4.
A sample of 100 arrivals of a customer at a retail sales depot is according to the following distribution. Time between arrival(Minutes) 0.5 1.0 1.5 2.0 2.5
Frequency 2 6 10 25 20 39
3.0 3.5 4.0 4.5 5.0
14 10 7 4 2
A study of the time required to service customers by adding up the bills, receiving payments and placing packages, yields the following distribution Time between service(Minutes) 0.5 1.0 1.5 2.0 2.5 3.0
Frequency 12 21 36 19 7 5
Estimate the average percentage of customer waiting time and average percentage of idle time of the server by simulation for the next 10 arrivals. Solution: Arrival Arrivals
Frequency
Probability
Cumulative
Random No.
Probability
Interval
0.5 1.0 1.5 2.0 2.5
2 6 10 25 20
0.02 0.06 0.10 0.25 0.20
0.02 0.08 0.18 0.43 0.63
00-01 02-07 08-17 18-42 43-62
3.0 3.5 4.0 4.5
14 10 7 4
0.14 0.10 0.07 0.04
0.77 0.87 0.94 0.98
63-76 77-86 87-93 94-97 40
5.0
2
0.02
1.00
98-99
Service Time between
Frequency
Probability
service(Minutes)
Probability
Interval
12 21 36
0.12 0.21 0.36
0.12 0.33 0.69
00-11 12-32 33-68
2.0
19
0.19
0.88
69-87
2.5
7
0.07
0.95
88-94
3.0
5
0.05
1.00
95-99
No.
93 22 53 64 39 07 10 63 76 35
Arrival Random Service Service Service Wt.time
Arrival Time Time
1 2 3 4 5 6 7 8 9 10
Random No.
0.5 1.0 1.5
Arrival Random Inter No.
Cumulative
(Mins.) 4.0 2.0 2.5 3.0 2.0 1.0 1.5 3.0 3.0 2.0
No.
(Mins) 4.0 6.0 8.5 11.5 13.5 14.5 16.0 19.0 22.0 24.0
Time
Start
End
(Mins) 78 76 58 54 74 92 38 70 96 92
2.0 2.0 1.5 1.5 2.0 2.5 1.5 2.0 3.0 2.5
4 6 8.5 11.5 13.5 15.5 18.0 19.5 22.0 25.0
Wt.time
of the
of the
end
Server
6 8 10.0 13 15.5 18 19.5 21.5 25.0 27.5
customer 1.0 2.0 0.5 1.0
4.0 0.5 1.5 0.5 0.5 -
Total
4.5
7.0
Average waiting time per customer is 4.5/10 = 0.45 minutes Average waiting time or idle time of the server = 7.0/10 = 0.7 minutes.
41
Example 5.
A tourist car operator finds that during the past few months the cars use has varied so much that the cost of maintaining the car varied considerably. During the past 200 days the demand for the car fluctuated as below. Trips per week 0 1 2 3 4 5
Frequency 16 24 30 60 40 30
Using the random numbers simulate the demand for a 10 week period. Solution: Trips per
Frequency
Probability
week/
Cumulative
Random
Probability
Number
Demand per week 0 1 2 3 4 5
Interval 16 24 30 60 40 30
0.08 0.12 0.15 0.30 0.20 0.15
0.08 0.20 0.35 0.65 0.85 1.00
00-07 08-19 20-34 35-64 65-84 85-99
The simulated demand for the cars for the next 10 weeks period is given in the table below
42
Week 1 2 3 4 5 6 7 8 9 10
Random Number 82 96 18 96 20 84 56 11 52 03 Total
Demand 4 5 1 5 2 4 3 1 3 0 28
Average Demand = 28/10 = 2.8 cars per week
Example 6.
An automobile production line turns out about 100 cars a day, but deviations occur owing to many causes. The production is more accurately described by the probability distribution given below. Production per day 95
Probability 0.03
43
96 0.05 97 0.07 98 0.10 99 0.15 100 0.20 101 0.15 102 0.10 103 0.07 104 0.05 105 0.03 Finished cars are transported across the bay at the end of each day by the ferry. If the ferry has space for only 101 cars, what will be the average numbers of waiting to be shipped and what will be the average number of empty space on the ship? Solution: Production per day
Probability
95 96 97 98 99 100 101 102 103 104 105
0.03 0.05 0.07 0.10 0.15 0.20 0.15 0.10 0.07 0.05 0.03
Cumulative
Random Number
Probability
Interval
0.03 0.08 0.15 0.25 0.40 0.60 0.75 0.85 0.92 0.97 1.00
00-02 03-07 08-14 15-24 25-39 40-59 60-74 75-84 85-91 92-96 97-99
The simulated production of cars for the next 15 days is given in the following table. Day
Random
Production per
No. of cars
No. of empty
Number
day
waiting
space in the ship
1 2 3 4 5 6 7 8 9
97 02 80 66 96 55 50 29 58
105 95 102 101 104 100 100 99 100
4 1 3 -
6 1 1 2 1 44
10 11 12 13 14 15
51 04 86 24 39 47
100 96 103 98 99 100 Total
2 10
1 5 3 2 1 23
Average number of cars waiting to be shipped = 10/15 = 0.67 per day Average number of empty space on the ship = 23/15 = 1.53 per day
Example 7.
A company manufactures 30 units per product. The sale of these items depends upon demand which has the following distribution. Sale (units) 27 28 29 30 31 32
Probability 0.10 0.15 0.20 0.35 0.15 0.05
The production cost and sale 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 met. Using the following random numbers, estimate the total profit/loss for the company for the next 10 days. Random Numbers: 10, 99, 65,99,95,01,79,11,16 and 20. If the company decides to produce 29 units per day, what is the advantage or disadvantage to the company? Sale (units) 27 28
Probability
Cumulative
Random Number
0.10 0.15
Probability 0.10 0.25
Interval 00-09 10-24 45
29 30 31 32
0.20 0.35 0.15 0.05
0.45 0.80 0.95 1.00
25-44 45-79 80-94 95-99
Set up the simulation for next 10 days Day
Random
Simulated
Number of
Number of
1 2 3 4 5 6 7 8 9 10
Number 10 99 65 99 95 01 79 11 16 20 Total
Sales 28 32 30 32 32 27 30 28 28 28
units Unsold 30-28=2 0 0 0 0 30-27=3 0 30-28=2 2 2 11
units short 0 32-30=2 0 2 2 0 0 0 0 0 6
At the production rate of 30 per day, total number of units produced in 10 days = 30 X 10 = 300 No’s, since the production cost is Rs.40 and the sales price is Rs.50, the per unit profit is Rs.10 Day
1 2 3 4 5 6 7
Demand
10 99 65 99 95 01 79
Actual Sales
28 30 30 30 30 27 30
Profit on
Loss due to
Loss due to
Actual sales
the unsold
penalty for
280 300 300 300 300 270 300
units 2 X 15 = 30 3 X 15 = 45 0
shortage 2 X 5 = 10 2 X 5 = 10 2 X 5 = 10 0 0
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8 9 10
11 16 20
28 28 28 289
280 280 280 Rs.2890
2 X 15 = 30 2 X 15 = 30 2 X 15 = 30 Rs.165
0 0 0 Rs.30
(a) Total profit for next 10 days when production is 30 units per day = (2890) – (165 +30) = 2890-195 = 2695
(b) If the company decides to produce 29 units per day. The calculation of profit and loss is as shown below. Day
1 2 3 4 5 6 7 8 9 10
Demand
28 32 30 32 32 27 30 28 28 28
Production
Actual
Profit on
Loss due to
Loss due to
units
Sales
Actual
the unsold
penalty for
28 29 29 29 29 27 29 28 28 28 284
sales 280 290 290 290 290 270 290 280 280 280 Rs.2840
units 1 X 15 = 15 0 0 0 0 1 X 15 = 15 0 1 X 15 = 15 1 X 15 = 15 1 X 15 = 15 Rs.90
shortage 0 3 X 5= 15 1X5=5 3 X 5= 15 3 X 5= 15 0 1X5=5 0 0 0 Rs.55
29 29 29 29 29 29 29 29 29 29 Total
Total profit for next 10 days when production is 29 units per day = (2840) – (90 + 55) = Rs.2695 Since the profit is same for both the cases there is no disadvantage to the company.
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Linear Programming Problems – Formulation
Linear Programming is a mathematical technique for optimum allocation of limited or scarce resources, such as labour, material, machine, money, energy and so on , to several competing activities such as products, services, jobs and so on, on the basis of a given criteria of optimality. The term ‘Linear’ is used to describe the proportionate relationship of two or more variables in a model. The given change in one variable will always cause a resulting proportional change in another variable. The word , ‘programming’ is used to specify a sort of planning that involves the economic allocation of limited resources by adopting a particular course of action or strategy among various alternatives strategies to achieve the desired objective. Hence, Linear Programming is a mathematical technique for optimum allocation of limited or scarce resources, such as labour, material, machine, money energy etc. Structure of Linear Programming model.
The general structure of the Linear Programming model essentially consists of three components. i) The activities (variables) and their relationships ii) The objective function and iii) The constraints The activities are represented by X1, X2, X3 ……..Xn. These are known as Decision variables. The objective function of an LPP (Linear Programming Problem) is a mathematical representation of the objective in terms a measurable quantity such as profit, cost, revenue, etc. Optimize (Maximize or Minimize) Z=C1X1 +C2X2+ ………..Cn Xn Where Z is the measure of performance variable X1, X2, X3, X4…..Xn are the decision variables And C1, C2, …Cn are the parameters that give contribution to decision variables. The constraints These are the set of linear inequalities and/or equalities which impose restriction of the limited resources
48
Assumptions of Linear Programming Certainty.
In all LP models it is assumed that, all the model parameters such as availability of resources, profit (or cost) contribution of a unit of decision variable and consumption of resources by a unit of decision variable must be known and constant. Divisibility (Continuity)
The solution values of decision variables and resources are assumed to have either whole numbers (integers) or mixed numbers (integer or fractional). However, if only integer variables are desired, then Integer programming method may be employed. Additivity
The value of the objective function for the given value of decision variables and the total sum of resources used, must be equal to the sum of the contributions (Profit or Cost) earned from each decision variable and sum of the resources used by each decision variable respectively. /The objective function is the direct sum of the individual contributions of the different variables Linearity
All relationships in the LP model (i.e. in both objective function and constraints) must be linear. General Mathematical Model of an LPP
Optimize (Maximize or Minimize) Z=C1 X1 + C2 X2 +……+CnXn Subject to constraints, a11X1+ a 12X2+………………+ a 1nXn (<,=,>) b1 a21X1+ a 22X2+………………+ a 2nXn (<,=,>) b2 a31X1+ a 32X2+………………+ a 3nXn (<,=,>) b3 am1X1+ a m2X2+………………+ a mnXn (<,=,>) bm and X1, X2 ….Xn > Guidelines for formulating Linear Programming model
i) Identify and define the decision variable of the problem ii) Define the objective function iii) State the constraints to which the objective function should be optimized (i.e. Maximization or Minimization) iv) Add the non-negative constraints from the consideration that the negative values of the decision variables do not have any valid physical interpretation
49
Example 1. A manufacturer produces two types of models M1 and M2.Each model of the type M1 requires 4 hours of grinding and 2 hours of polishing; where as each model of M2 requires 2 hours of grinding and 5 hours of polishing. The manufacturer has 2 grinders and 3 polishers. Each grinder works for 40 hours a week and each polisher works 60 hours a week. Profit on M1 model is Rs.3.00 and on model M2 is Rs.4.00.Whatever produced in a week is sold in the market. How should the manufacturer allocate his production capacity to the two types of models, so that he makes maximum profit in a week? i) Identify and define the decision variable of the problem Let X1 and X2 be the number of units of M1 and M2 model. ii) Define the objective function Since the profits on both the models are given, the objective function is to maximize the profit. Max Z = 3X1 + 4X2 iii) State the constraints to which the objective function should be optimized (i.e. Maximization or Minimization) There are two constraints one for grinding and the other for polishing. The grinding constraint is given by 4X1 + 2X2 < 80 No of hours available on grinding machine per week is 40 hrs. There are two grinders. Hence the total grinding hour available is 40 X 2 = 80 hours. The polishing constraint is given by 2X1 + 5X2 < 180 No of hours available on polishing machine per week is 60 hrs. There are three grinders. Hence the total grinding hour available is 60 X 3 = 180 hours. Finally we have, Max Z = 3X1 + 4X2 Subject to constraints, 4X1 + 2X2 < 80 2X1 + 5X2 < 180 X1, X2 > 0 Example 2. A firm is engaged in producing two products. A and B. Each unit of product A requires 2 kg of raw material and 4 labour hours for processing, where as each unit of B requires 3 kg of raw materials and 3 labour hours for the same type. Every week, the firm has an availability of 60 kg of raw material and 96 labour hours. One unit of product A sold yields Rs.40 and one unit of product B sold gives Rs.35 as profit.
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Formulate this as an Linear Programming Problem to determine as to how many units of each of the products should be produced per week so that the firm can earn maximum profit. i) Identify and define the decision variable of the problem Let X1 and X2 be the number of units of product A and product B produced per week. ii) Define the objective function Since the profits of both the products are given, the objective function is to maximize the profit. MaxZ = 40X1 + 35X2 iii) State the constraints to which the objective function should be optimized (i.e. Maximization or Minimization) There are two constraints one is raw material constraint and the other one is labour constraint.. The raw material constraint is given by 2X1 + 3X2 < 60 The labour hours constraint is given by 4X1 + 3X2 < 96 Finally we have, MaxZ = 40X1 + 35X2 Subject to constraints, 2X1 + 3X2 < 60 4X1 + 3X2 < 96 X1,X2 > 0 Example 3. The agricultural research institute suggested the farmer to spread out atleast 4800 kg of special phosphate fertilizer and not less than 7200 kg of a special nitrogen fertilizer to raise the productivity of crops in his fields. There are two sources for obtaining these – mixtures A and mixtures B. Both of these are available in bags weighing 100kg each and they cost Rs.40 and Rs.24 respectively. Mixture A contains phosphate and nitrogen equivalent of 20kg and 80 kg respectively, while mixture B contains these ingredients equivalent of 50 kg each. Write this as an LPP and determine how many bags of each type the farmer should buy in order to obtain the required fertilizer at minimum cost. i) Identify and define the decision variable of the problem Let X1 and X2 be the number of bags of mixture A and mixture B. ii) Define the objective function The cost of mixture A and mixture B are given ; the objective function is to minimize the cost Min.Z = 40X1 + 24X2 iii) State the constraints to which the objective function should be optimized. The above objective function is subjected to following constraints.
51
20X1 + 50X2 >4800 80X1 + 50X2 >7200 X1, X2 >0
Phosphate requirement Nitrogen requirement
Finally we have, Min.Z = 40X1 + 24X2 is subjected to three constraints 20X1 + 50X2 >4800 80X1 + 50X2 >7200 X1, X2 >0
Example 4. A firm can produce 3 types of cloth, A , B and C.3 kinds of wool are required Red, Green and Blue.1 unit of length of type A cloth needs 2 meters of red wool and 3 meters of blue wool.1 unit of length of type B cloth needs 3 meters of red wool, 2 meters of green wool and 2 meters of blue wool.1 unit type of C cloth needs 5 meters of green wool and 4 meters of blue wool. The firm has a stock of 8 meters of red, 10 meters of green and 15 meters of blue. It is assumed that the income obtained from 1 unit of type A is Rs.3, from B is Rs.5 and from C is Rs.4.Formulate this as an LPP.( December2005/January 2006) i) Identify and define the decision variable of the problem Let X1, X2 and X3 are the quantity produced of cloth type A,B and C respectively. ii) Define the objective function The incomes obtained for all the three types of cloths are given; the objective function is to maximize the income. Max Z = 3X1 + 5X2 + 4X3 iii) State the constraints to which the objective function should be optimized. The above objective function is subjected to following three constraints. 2X1 + 3X2 < 8 2X2 + 5X3 < 10 3X1 + 2X2 + 4X3 < 15 X1, X2 X3 >0 Finally we have, Max Z = 3X1 + 5X2 + 4X3 is subjected to three constraints 2X1 + 3X2 < 8 2X2 + 5X3 < 10 3X1 + 2X2 + 4X3 < 15 X1, X2 X3 >0
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Example 5. A Retired person wants to invest upto an amount of Rs.30,000 in fixed income securities. His broker recommends investing in two Bonds: Bond A yielding 7% and Bond B yielding 10%. After some consideration, he decides to invest at most of Rs.12,000 in bond B and atleast Rs.6,000 in Bond A. He also wants the amount invested in Bond A to be atleast equal to the amount invested in Bond B. What should the broker recommend if the investor wants to maximize his return on investment? Solve graphically. (January/February 2004)
i) Identify and define the decision variable of the problem Let X1 and X2 be the amount invested in Bonds A and B. ii) Define the objective function Yielding for investment from two Bonds are given; the objective function is to maximize the yielding. Max Z = 0.07X1 + 0.1X2 iii) State the constraints to which the objective function should be optimized. The above objective function is subjected to following three constraints. X1 + X2 < 30,000 X1 > 6,000 X2 < 12,000 X1 -- X2 >0 X1, X2 >0 Finally we have, MaxZ = 0.07X1 + 0.1X2 is subjected to three constraints X1 + X2 < 30,000 X1 > 6,000 X2 < 12,000 X1 -- X2 >0 X1, X2 >0 Minimization problems
Example 5. A person requires 10, 12, and 12 units chemicals A, B and C respectively for his garden. A liquid product contains 5, 2 and 1 units of A,B and C respectively per jar. A dry product contains 1,2 and 4 units of A,B and C per carton. If the liquid product sells for Rs.3 per jar and the dry product sells for Rs.2 per carton, how many of each should be purchased, in order to minimize the cost and meet the requirements? i) Identify and define the decision variable of the problem Let X1 and X2 be the number of units of liquid and dry products. ii) Define the objective function The cost of Liquid and Dry products are given ; the objective function is to minimize the cost
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Min. Z = 3X1 + 2X2 iii) State the constraints to which the objective function should be optimized. The above objective function is subjected to following three constraints. 5X1 + X2 >10 2X1 + 2X2 >12 X1 + 4X2 >12 X1, X2 >0 Finally we have, Min. Z = 3X1 + 2X2 is subjected to three constraints 5X1 + X2 >10 2X1 + 2X2 >12 X1 + 4X2 >12 X1, X2 >0 Example 6. A Scrap metal dealer has received a bulk order from a customer for a supply of atleast 2000 kg of scrap metal. The consumer has specified that atleast 1000 kgs of the order must be high quality copper that can be melted easily and can be used to produce tubes. Further, the customer has specified that the order should not contain more than 200 kgs of scrap which are unfit for commercial purposes. The scrap metal dealer purchases the scrap from two different sources in an unlimited quantity with the following percentages (by weight) of high quality of copper and unfit scrap
Copper Unfit Scrap
Source A
Source B
40% 7.5%
75% 10%
The cost of metal purchased from source A and source B are Rs.12.50 and Rs.14.50 per kg respectively. Determine the optimum quantities of metal to be purchased from the two sources by the metal scrap dealer so as to minimize the total cost (February 2002) i) Identify and define the decision variable of the problem Let X1 and X2 be the quantities of metal to be purchased from the two sources A and B. ii) Define the objective function The cost of metal to be purchased by the metal scrap dealer are given; the objective function is to minimize the cost Min. Z = 12.5X1 + 14.5X2 iii) State the constraints to which the objective function should be optimized. The above objective function is subjected to following three constraints. X1 + X2 >2,000 0.4X1 + 0.75X2 >1,000 0.075X1 + 0.1X2 + 4X3 < 200 X1, X2 >0 54
Finally we have, Min. Z = 12.5X1 + 14.5X2 is subjected to three constraints X1 + X2 >2,000 0.4X1 + 0.75X2 >1,000 0.075X1 + 0.1X2 + 4X3 < 200 X1, X2 >0 Example 7. A farmer has a 100 acre farm. He can sell all tomatoes, lettuce or radishes and can raise the price to obtain Rs.1.00 per kg. for tomatoes , Rs.0.75 a head for lettuce and Rs.2.00 per kg for radishes. The average yield per acre is 2000kg.of tomatoes, 3000 heads of lettuce and 1000 kgs of radishes. Fertilizers are available at Rs.0.50 per kg and the amount required per acre is 100 kgs for each tomatoes and lettuce and 50kgs for radishes. Labour required for sowing, cultivating and harvesting per acre is 5 man-days for tomatoes and radishes and 6 man-days for lettuce. A total of 400 man-days of labour are available at Rs.20.00 per man-day. Formulate this problem as LP model to maximize the farmers profit. i) Identify and define the decision variable of the problem Let X1 and X2 and X3 be number acres the farmer grows tomatoes, lettuce and radishes respectively. ii) Define the objective function The objective of the given problem is to maximize the profit. The profit can be calculated by subtracting total expenditure from the total sales Profit = Total sales – Total expenditure The farmer produces 2000X1 kgs of tomatoes, 3000X2 heads of lettuce, 1000X3 kgs of radishes. Therefore the total sales of the farmer will be = Rs. (1 x 2000X1 + 0.75 x 3000X2 + 2 x 100X3) Total expenditure (fertilizer expenditure) will be = Rs.20 ( 5X1 + 6X2 + 5X3 ) Farmer’s profit will be Z = (1 x 2000X1 + 0.75 x 3000X2 + 2 x 100X3) – { [0.5 x 100 x X1+0.5 x 100 x X2 + 50X3]+ [20 x 5 x X1+20 x 6 x X2 + 20 x 5 x X3]} =1850X1 + 2080X2 + 1875X3 Therefore the objective function is Maximise Z = 1850X1 + 2080X2 + 1875X3 iii) State the constraints to which the objective function should be optimized. The above objective function is subjected to following constraints. Since the total area of the firm is 100 acres X1 + X2 + X3 < 100 The total man-days labour is 400 man-days
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5X1 + 6X2 + 5X3 < 400 Finally we have, Maximise Z = 1850X1 + 2080X2 + 1875X3 is subjected to three constraints X1 + X2 + X3 < 100 5X1 + 6X2 + 5X3 < 400 X1, X2 X3 >0 Example 8. An electronics company produces three types of parts for automatic washing machines .It purchases castings of the parts from a local foundry and then finishes the part on drilling, shaping and polishing machines. The selling prices of parts A, B, and C respectively are Rs 8, Rs.10 and Rs.14.All parts made can be sold. Castings for parts A, B and C respectively cost Rs.5, Rs.6 and Rs.10. The shop possesses only one of each type of machine. Cost per hour to run each of the three machines are Rs.20 for drilling, Rs.30 for shaping and Rs.30 for polishing. The capacities (parts per hour) for each part on each machine are shown in the following table. Machine
Capacities Per Hour Part A
Part B
Part C
Drilling
25
40
25
Shaping
25
20
20
Polishing
40
30
40
The management of the shop wants to know how many parts of each type it should produce per hour in order to maximize profit for an hour’s run. Formulate this problem as an LP model so as to maximize total profit to the company. i) Identify and define the decision variable of the problem Let X1 and X2 and X3 be the number of types A, B and C parts produced per hour respectively . ii) Define the objective function With the information given, the hourly profit for part A, B, and C would be as follows Profit per type A part = (8 – 5) – (20/25 +30/25 + 30/40) = 0.25 Profit per type B part = (10 – 6) – (20/40 + 30/20 + 30/30) = 1 Profit per type C part = (14 – 10) – (20/25 + 30/20 + 30/40) = 0.95 Then, Maximize Z = 0.25 X1 + 1X2 + 0.95X3 iii) State the constraints to which the objective function should be optimized. The above objective function is subjected to following constraints.
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i) The drilling machine constraint X1/25 + X2/40 + X3/24 < 1 ii) The shaping machine constraint X1/25 + X2/20 + X3/20 ≤ 1 iii) The polishing machine constraint X1/40 + X2/30 + X3/40 ≤ 1 X1, X2, X3 ≥ 0 Finally we have, Maximize Z = 0.25 X1 + 1X2 + 0.95X3 Subject to constraints X1/25 + X2/40 + X3/24 < 1 ii) The shaping machine constraint X1/25 + X2/20 + X3/20 ≤ 1 iii) The polishing machine constraint X1/40 + X2/30 + X3/40 ≤ 1 X1, X2, X3 ≥ 0 Example 9. A city hospital has the following minimal daily requirements for nurses. Period 1 2 3 4 5 6
Clock time (24 Minimum hours day) number of nurses required 6 a.m. – 10 a.m. 2 10 a.m. – 2 p.m. 7 2 p.m. – 6 p.m. 15 6 p.m. – 10 p.m. 8 10 p.m. – 2 a.m. 20 2 a.m. – 6 a.m. 6
Nurses report at the hospital at the beginning of each period and work for 8 consecutive hours. The hospital wants to determine the minimal number of nurses to be employed so that there will be a sufficient number of nurses available for each period. Formulate this as a linear programming problem by setting up appropriate constraints and objective function. i) Identify and define the decision variable of the problem Let X1, X2, X3, X4, X5 and X6 be the number of nurses joining duty at the beginning of periods 1, 2, 3, 4, 5 and 6 respectively. ii) Define the objective function Minimize Z = X1 + X2 + X3 + X4 + X5 + X6 iii) State the constraints to which the objective function should be optimized. The above objective function is subjected to following constraints. X1 + X2 ≥ 7 57
X2 + X3 ≥ 15 X3 + X4 ≥ 8 X4 + X5 ≥ 20 X5 + X6 ≥ 6 X6 + X1 ≥ 2 X1, X2, X3, X4, X5, X6 ≥ 0 Linear Programming: Graphical Solution
Example 1. Solve the following LPP by graphical method Maximize Z = 5X1 + 3X2 Subject to constraints 2X1 + X2 ≤ 1000 X1 ≤ 400 X1 ≤ 700 X1, X2 ≥ 0 Solution:
The first constraint 2X1 + X2 ≤ 1000 can be represented as follows. We set 2X1 + X2 = 1000 When X1 = 0 in the above constraint, we get, 2 x 0 + X2 = 1000 X2 = 1000 Similarly when X2 = 0 in the above constraint, we get, 2X1 + 0 = 1000 X1 = 1000/2 = 500 The second constraint X1 ≤ 400 can be represented as follows, We set X1 = 400 The third constraint X2 ≤ 700 can be represented as follows, We set X2 = 700
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The constraints are shown plotted in the above figure Point 0 A
X1 0 0
X2 0 700
B
150
700
C D
400 400
200 0
Z = 5X1 +3X2 0 Z = 5 x 0 + 3 x 700 = 2,100 Z = 5 x 150 + 3 x 700 = 2,850* Maximum Z = 5 x 400 + 3 x 200 = 2,600 Z = 5 x 400 + 3 x 0 = 2,000
The Maximum profit is at point B When X1 = 150 and X2 = 700 Z = 2850 Example 2. Solve the following LPP by graphical method Maximize Z = 400X1 + 200X2 Subject to constraints 18X1 + 3X2 ≤ 800 9X1 + 4X2 ≤ 600 X2 ≤ 150 X1, X2 ≥ 0 Solution:
The first constraint 18X1 + 3X2 ≤ 800 can be represented as follows. We set 18X1 + 3X2 = 800 When X1 = 0 in the above constraint, we get, 18 x 0 + 3X2 = 800 X2 = 800/3 = 266.67 Similarly when X2 = 0 in the above constraint, we get, 18X1 + 3 x 0 = 800 X1 = 800/18 = 44.44 The second constraint 9X1 + 4X2 ≤ 600 can be represented as follows, We set 9X1 + 4X2 = 600 When X1 = 0 in the above constraint, we get, 9 x 0 + 4X2 = 600 X2 = 600/4 = 150 Similarly when X2 = 0 in the above constraint, we get, 9X1 + 4 x 0 = 600 X1 = 600/9 = 66.67 The third constraint X2 ≤ 150 can be represented as follows, We set X2 = 150
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Point 0
X1 0
X2 0
A
0
150
B C
31.11 44.44
80 0
Z = 400X1 + 200X2 0 Z = 400 x 0+ 200 x 150 = 30,000* Maximum Z = 400 x 31.1 + 200 x 80 = 28,444.4 Z = 400 x 44.44 + 200 x 0 = 17,777.8
The Maximum profit is at point A When X1 = 150 and X2 = 0 Z = 30,000
Example 3. Solve the following LPP by graphical method Minimize Z = 20X1 + 40X2 Subject to constraints 36X1 + 6X2 ≥ 108 3X1 + 12X2 ≥ 36 20X1 + 10X2 ≥ 100 X1 X2 ≥ 0 Solution:
The first constraint 36X1 + 6X2 ≥ 108 can be represented as follows. We set 36X1 + 6X2 = 108 When X1 = 0 in the above constraint, we get, 36 x 0 + 6X2 = 108 60
X2 = 108/6 = 18 Similarly when X2 = 0 in the above constraint, we get, 36X1 + 6 x 0 = 108 X1 = 108/36 = 3 The second constraint3X1 + 12X2 ≥ 36 can be represented as follows, We set 3X1 + 12X2 = 36 When X1 = 0 in the above constraint, we get, 3 x 0 + 12X2 = 36 X2 = 36/12 = 3 Similarly when X2 = 0 in the above constraint, we get, 3X1 + 12 x 0 = 36 X1 = 36/3 = 12 The third constraint20X1 + 10X2 ≥ 100 can be represented as follows, We set 20X1 + 10X2 = 100 When X1 = 0 in the above constraint, we get, 20 x 0 + 10X2 = 100 X2 = 100/10 = 10 Similarly when X2 = 0 in the above constraint, we get, 20X1 + 10 x 0 = 100 X1 = 100/20 = 5
Point 0 A B C D
X1 0 0 2 4 12
X2 0 18 6 2 0
Z = 20X1 + 40X2 0 Z = 20 x 0 + 40 x 18 = 720 Z = 20 x2 + 40 x 6 = 280 Z = 20 x 4 + 40 x 2 = 160* Minimum Z = 20 x 12 + 40 x 0 = 240
The Minimum cost is at point C 61
When X1 = 4 and X2 = 2 Z = 160 Example 4. Solve the following LPP by graphical method Maximize Z = 2.80X1 + 2.20X2 Subject to constraints X1 ≤ 20,000 X2 ≤ 40,000 0.003X1 + 0.001X2 ≤ 66 X1 + X2 ≤ 45,000 X1 X2 ≥ 0 Solution:
The first constraint X1 ≤ 20,000 can be represented as follows. We set X1 = 20,000 The second constraint X2 ≤ 40,000 can be represented as follows, We set X2 = 40,000 The third constraint 0.003X1 + 0.001X2 ≤ 66 can be represented as follows, We set 0.003X1 + 0.001X2 = 66 When X1 = 0 in the above constraint, we get, 0.003 x 0 + 0.001X2 = 66 X2 = 66/0.001 = 66,000 Similarly when X2 = 0 in the above constraint, we get, 0.003X1 + 0.001 x 0 = 66 X1 = 66/0.003 = 22,000 The fourth constraint X1 + X2 ≤ 45,000 can be represented as follows, We set X1 + X2 = 45,000 When X1 = 0 in the above constraint, we get, 0 + X2 = 45,000 X2 = 45,000 Similarly when X2 = 0 in the above constraint, we get, X1 + 0 = 45,000 X1 =45,000
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Point 0 A
X1 0 0
X2 0 40,000
B
5,000
40,000
C
10,500
34,500
D
20,000
6,000
E
20,000
0
Z = 2.80X1 + 2.20X2 0 Z = 2.80 x 0 + 2.20 x 40,000 = 88,000 Z = 2.80 x 5,000 + 2.20 x 40,000 = 1,02,000 Z = 2.80 x 10,500 + 2.20 x 34,500 = 1,05,300* Maximum Z = 2.80 x 20,000 + 2.20 x 6,000 = 69,200 Z = 2.80 x 20,000 + 2.20 x 0 = 56,000
The Maximum profit is at point C When X1 = 10,500 and X2 = 34,500 Z = 1,05,300 Example 5. Solve the following LPP by graphical method Maximize Z = 10X1 + 8X2 Subject to constraints 2X1 + X2 ≤ 20 X1 + 3X2 ≤ 30 X1 - 2X2 ≥ -15 X1 X2 ≥ 0 Solution:
The first constraint 2X1 + X2 ≤ 20 can be represented as follows. We set 2X1 + X2 = 20 When X1 = 0 in the above constraint, we get, 2 x 0 + X2 = 20 X2 = 20 Similarly when X2 = 0 in the above constraint, we get, 2X1 + 0 = 20 63
X1 = 20/2 = 10 The second constraint X1 + 3X2 ≤ 30 can be represented as follows, We set X1 + 3X2 = 30 When X1 = 0 in the above constraint, we get, 0 + 3X2 = 30 X2 = 30/3 = 10 Similarly when X2 = 0 in the above constraint, we get, X1 + 3 x 0 = 30 X1 = 30 The third constraint X1 - 2X2 ≥ -15 can be represented as follows, We set X1 - 2X2 = -15 When X1 = 0 in the above constraint, we get, 0 - 2X2 = -15 X2 = -15/2 = 7.5 Similarly when X2 = 0 in the above constraint, we get, X1 – 2 x 0 = -15 X1 = -15
Point 0 A B C D
X1 0 0 3 6 10
X2 0 7.5 9 8 0
Z = 10X1 + 8X2 0 Z = 10 x 0 + 8 x 7.5 = 60 Z = 10 x 3 + 8 x 9 = 102 Z = 10 x 6 + 8 x 8 = 124* Minimum Z = 10 x 10 + 8 x 0 = 100
The Maximum profit is at point C When X1 = 6 and X2 = 8 Z = 124 Duality in Linear Programming
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Duality in Linear Programming For every LPP there is a unique LPP associated with it involving the same data and closely related optimal solution. The original problem is then called primal problem while the other is called its Dual problem Let the primal problem be Maximize Z = C1 X1 + C2 X2 + …… + CnXn Subject to constraints, a11X1+ a 12X2+………………+ a 1nXn <, b1 a21X1+ a 22X2+………………+ a 2nXn <,b2 a31X1+ a 32X2+………………+ a 3nXn <,b3 am1 X1+ a m2X2+………………+ a mnXn <,bm and X1, X2 ….Xn > 0 Let the primal problem be Maximize Z = C1 X1 + C2 X2 + …… + CnXn Subject to constraints, a11X1+ a 12X2+………………+ a 1nXn <, b1 a21X1+ a 22X2+………………+ a 2nXn <,b2 a31X1+ a 32X2+………………+ a 3nXn <,b3 am1 X1+ a m2X2+………………+ a mnXn <,bm and X1, X2 ….Xn > 0 Then its Dual is Minimize G = b1W1 + b2W2+ b3W3 + .......+ bmWm Subject to constraints, a11W 1 + a21W2 + a31W3 +................+ am1Wm > C1 a12W 1 + a22W2 + a32W3 +................+ am2Wm > C2 a13W 1 + a23W2 + a33W3 +................+ am3Wm > C3 a1nW 1 + a2nW2 + a3nW3 +................+ amnWm > Cn W 1, W2, W3 ......Wm > 0 Example.1 Write the Dual of the following LPP Min Z = 2X2+ 5X3 X1+ X2 > 2 2X1+ X2 + 6X3 < 6 X1- X2 +3X3 = 4 and X1, X2,X3 > 0 Rearrange the constraints into a standard form, we get Min Z = 0X1 + 2X2+ 5X3 Subject to constraints, X1+ X2 + 0X3> 2 -2X1- X2 - 6X3 > -6 X1- X2 +3X3 > 4
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-X1 + X2 -3X3 > -4 and X1, X2,X3 > 0 The Dual of the above primal is as follows Max.G = 2W1 -6W2+ 4W3 – 4W4 Subject to constraints, W 1 -2W2 + W3 –W4< 0 W 1 - W2 - W3 + W4 < 2 0W 1 - 6W2 + 3W3- 3W4 < 5 W 1, W2, W3,W4 > 0 Max.G = 2W1 -6W2+ 4(W3 – W4) Subject to constraints, W 1 -2W2 + (W3 –W4 ) < 0 W 1 - W2 - W3 + W4 < 2 0W 1 - 6W2 – 3(W3- W4 ) < 5 W 1, W2, W3,W4 > 0 Max.G = 2W1 -6W2+ 4W5 Subject to constraints, W 1 -2W2 + W5 < 0 W 1 - W2 – W5 < 2 0W 1 - 6W2 – 3W5 < 5 W 1, W2, > 0 , W5 is unrestricted in sign Example.2 Write the Dual of the following LPP Min Z = 4X1 + 5X2- 3X3 Subject to constraints, X1+ X2 + X3 = 22 3X1+ 5X2 - 2X3 < 65 X1+ 7X2 +4X3 > 120 X1 , X2 > 0 and X3 is unrestricted Since X3 is Unrestricted, replace X3 with (X4 - X5 ) and bring the problem into standard form Min Z = 4X1 + 5X2- 3(X4 - X5) Subject to constraints, X1+ X2 + (X4 - X5) >22 -X1- X2 - (X4 - X5) >- 22 -3X1- 5X2 + 2(X4 - X5) > -65 X1+ 7X2 +4(X4 - X5) > 120 X1 , X2 , X4 ,X5 > 0 The Dual of the above primal is as follows Max.G = 22(W1 -W2)- 65W3 + 120W4
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