quantitative tech compiled kantharaju

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e-Notes by D.G.Kantharaju, East Point College, Bangalore

Network Analysis Network is a graphical representation of all the Activities and Events arranged in

a logical and sequential order.

Network analysis plays an important role in project management. A project is a combination of interrelated activities all of which must be 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 anyresources.

Analyzing network, the planning, scheduling and control of a project becomes

easier. PERT and CPM are the two most popular network analysis technique used toassist managers in planning and controlling large scale projects.

PERT- (Programme Evaluation Review Technique) CPM - (Critical Path Method)

Applications: -Construction of a Residential complex,

Commercial complex,Petro-chemical complex

Ship buildingSatellite mission developmentInstallation of a pipe line project etc...

Historical Evolution.Before 1957 there was no generally accepted procedure to aid the management of aproject. In 1958 PERT was developed by team of engineers working on a Polaris Missileprogramme of the navy. This was a large project involved 250 prime contractors andabout 9000 job contractors. It had about 19 million components. In such projects it ispossible that a delay in the delivery of a small component might hold the progress ofentire 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) toschedule and control a very large project like overhauling of a chemical plant, there byreducing 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.

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Differences between PERT & CPM

PERT CPM

1. It is a technique for planningscheduling & controlling of projectswhose activities are subject touncertainty in the performance time.Hence it is a probabilistic model

1. It is a technique for planningscheduling & controlling of projectswhose activities not subjected toany uncertainty and theperformance times are fixed. Henceit is a deterministic model

2.It is an Event oriented system 2. It is an Activity oriented system

3.Basically does not differentiatecritical and non-critical activities

Differentiates clearly the criticalactivities from the other activities.

4. Used in projects where resources(men, materials, money) are alwaysavailable when required.

4. Used in projects where overallcosts is of primarily important.Therefore better utilized resources

5. Suitable for Research andDevelopment projects where timescannot be predicted

5.Suitable for civil constructions,installation, ship building etc.

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Problem 2.

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Problem 3.

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CRITICAL PATHMeaning: The longest path in a project network which determine the duration of theproject is known as critical path.

Determination

2

Step 1. List all the p

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Step 2.For each seq

Determination

2

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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 ESWhere LS = Latest start time

ES = Early start time.

Earliest start : Denoted as ESEarliest start time is the earliest possible time by which the activity can be started.Early finish time : Denoted as EFEarly 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 startedLate finish time : Denoted as LSLate 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 Latefinish and Early finishi.e.TF( CA) = LS (CA) - ES (CA)OrTF( 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 starttime of any other job.FF(CA) = ES(SA) EF (CA)CA = Current ActivitySA = Succeeding Activity

Independent Float (IF)Independent Float is the amount of time that can be delayed without affecting eitherpredecessor or successor activities.IF = ES(SA) LF(PA) - Duration of CAES = Early StartLF = Late FinishSA = Succeeding ActivityPA = Preceding ActivityCA = Current Activity

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Example 1Construct the Network for the followingProject and determine the followingi) Critical Pathii) 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

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(11)

Determination

DurationActivity

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Example 2Construct the Network for the followingProject and determine the followingi) Critical Path

ii) ES,EF,LS,LFiii) 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

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21-2

DurationActivity

In PERT model, 3 time values are associated

with each activity. They arei) Optimistic time = to

ii) Pessimistic time = tpiii) 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 thatevery 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 formulate = (to + 4tm + tp) / 6

Example.If a job has to = 5 days, tp = 17 days, tm = 8 daysThen Expected time for the job would be

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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 deviationand variance are used.A large standard deviation represents high variability and vice-versa.

Calculation of Standard Deviation and VarianceVariance = (Standard deviation )2Standard 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 ) / Where TS = Scheduled time for project completionTE = Expected time for the project completion = Standard deviation for the Network

Network = Sum of variances along theCritical Path

= ( i-j )2Where i-j is the variance of a activity i-j along the critical path

Example 1Construct the Network for the following project andcalculate 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

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1.Constructio

2. Calculation o

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3. Determinati

Activity to tm tp Criticalactivities

2 = ((t p t o) / 6)2

1-2 2 6 10 1-2 1.78

1-3 4 8 12 -

2-3 2 4 6 2-3 0.44

2-4 2 3 4 -

3-4 0 0 0 3-4 0

3-5 3 6 9 -

4-6 6 10 14 4-6 1.78

5-6 1 3 5 -

2= 4.00

Network = Sum of variances along theCritical Path

= (Network )2

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= 4= 2

Probability of completing the project within a given dateZ = (TS TE ) / Where TS = Scheduled time for project completion

TE = Expected time for the project completion = Standard deviation for the Network= (25 20) / 2= + 2.5

From the Normal distribution Table, we get the probability of completing the project in25 days is 99.4%

Example 2.The following table lists the jobs of a network along with theirtime 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 5 085-6 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?

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1.Construction

2. Calculation

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3. Determinati

As there are two Critical Paths, the path which gives more variance(2) is taken asCritical Path

Path AActivity 2 = ((t p t o) / 6)2 2

1-2 ((30 6)/6)2 16

2-6 ((28 4)/6)2 16

2 = 32.00

Path B Activity 2 = ((t p t o) / 6)2 2

1-3 ((15 3)/6)2 4

3-5 ((27 3)/6)2 16

5-6 ((30 6)/6)2 16

2= 36.00

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= 2 = 36 = 6

Therefore the Critical Path is 1 - 3 - 5 - 6

b)Probability of completing the project within a given dateZ = (TS TE ) / Where TS = Scheduled time for project completion

TE = Expected time for the project completion = Standard deviation for the Network

= (35 32) / 6= + 0.5

From the Normal distribution Table, we get the probability of completing the project in35 days is 69.15%

c)The due date for 90% chance of being met.Probability of completing the project within a given dateZ = (TS TE ) / The value of Z from the table for a 90% probability is +1.28TS = ? (to be calculated) ,TE = 32, = 6i.e. 1.28 = (TS 32) / 6TS = 39.68 days

CPM ModelIn 1958 Du Pont Company used a technique called Critical Path Method (CPM) toschedule and control a very large project like overhauling of a chemical plant, there byreducing 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 / CPMThe 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.

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t

Example 1.

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Find the lowest cost and optimum cost schedule for the following project, given the overhead 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

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1.Determinati

13 -1

Step 1.

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1

3 -1

Step 2.

13 -1

Step 3.

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1

3 -1

Step 3.

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1

1 -1

Step 4.

13 -1

Step 5.

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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.

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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 processStep 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

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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 notamenable 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.

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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) 0 10 20 30 40 50Probability 0.01 0.20 0.15 0.50 0.12 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

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Daily Demand Probability Cumulative Probability Random Number Interval0 0.01 0.01 0010 0.20 0.21 01-2020 0.15 0.36 21-3530 0.50 0.86 36-85

40 0.12 0.98 86-9750 0.02 1.00 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 Random Number Demand1 40 30 Because 0.36 < 0.40 < 0.852 19 10 Because 0.01 < 0.19 < 0.203 87 40 and so on

4 83 305 73 306 84 307 29 208 09 109 02 1010 20 10

Total = 220

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

(Units/day)

3 4 5 6 7

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Probability 0.10 0.20 0.30 0.30 0.10Carrying

cost(Per unit

per day)

Rs.2

Orderingcost(per

order)

Rs.50

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 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

Probability

Random Number

Range

3 0.10 0.10 004 0.20 0.30 01-025 0.30 0.60 03-056 0.30 0.90 06-087 0.10 1.00 09The results of the simulation experiment conducted are as shown below.

Days Opening Random Resulting Closing Order Order Average stock

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Stock Number Demand Stock Placed Delivered in the evening1 20 0 3 17 - - 18.52 17 9 7 10 15 - 13.53 10 1 4 6 - - 84 6 1 4 2 - - 4

5 2 5 5 0(-3)* 15 15 16 12 1 4 8 - - 107 8 8 6 2 - - 58 2 6 6 0(-4)* 15 15 19 11 3 5 6 - - 8.5

10 6 5 5 1 - - 3.5*Negative figure indicates back orders.

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 DemandAvailable ( kg ) Number of Days Demand (kg ) Number of Days

10 40 10 5020 50 20 11030 190 30 20040 150 40 10050 70 50 40

The retailer buys the commodity at Rs.20 per kg and sells at Rs.30 per kg. If anycommodity 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

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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

(kg)

Probability Random

Number

Demand

(kg)

Probability Random

Number10 40/500=0.08 00-07 10 50/500=0.10 00-0920 50/500=0.10 08-17 20 110/500=0.22 10-3130 190/500=0.38 18-55 30 200/500=0.40 32-7140 150/500=0.30 56-85 40 100/500=0.20 72-9150 70/500=0.14 85-99 50 40/500=0.08 92-99

Simulation Experimentation Sheet

Day Random

Number

Supply

(kg)

Random

Number

Demand

(kg)

Cost

(Rs)

Revenue

(Rs)

Shortage

(Loss)

Profit

1 31 30 18 20 600 600 - -2 63 40 84 40 800 1,200 - 4003 15 20 79 40 400 600 160 404 07 10 32 30 200 300 160 -605 43 30 75 40 600 900 80 2206 81 40 27 20 800 600 - -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) Frequency0.5 21.0 61.5 102.0 252.5 20

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3.0 143.5 104.0 74.5 45.0 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) Frequency0.5 121.0 211.5 362.0 192.5 73.0 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

Probability

Random No.

Interval

0.5 2 0.02 0.02 00-01

1.0 6 0.06 0.08 02-07

1.5 10 0.10 0.18 08-17

2.0 25 0.25 0.43 18-42

2.5 20 0.20 0.63 43-62

3.0 14 0.14 0.77 63-76

3.5 10 0.10 0.87 77-86

4.0 7 0.07 0.94 87-93

4.5 4 0.04 0.98 94-97

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5.0 2 0.02 1.00 98-99

Service

Time between

service(Minutes)

Frequency Probability Cumulative

Probability

Random No.

Interval

0.5 12 0.12 0.12 00-11

1.0 21 0.21 0.33 12-32

1.5 36 0.36 0.69 33-68

2.0 19 0.19 0.88 69-872.5 7 0.07 0.95 88-94

3.0 5 0.05 1.00 95-99

Arrival

No.

Random

No.

Inter

Arrival

Time

(Mins.)

Arrival

Time

(Mins)

Random

No.

Service

Time

(Mins)

Service

Start

Service

End

Wt.time

of the

end

customer

Wt.time

of the

Server

1 93 4.0 4.0 78 2.0 4 6 - 4.02 22 2.0 6.0 76 2.0 6 8 - -3 53 2.5 8.5 58 1.5 8.5 10.0 - 0.54 64 3.0 11.5 54 1.5 11.5 13 - 1.55 39 2.0 13.5 74 2.0 13.5 15.5 - 0.56 07 1.0 14.5 92 2.5 15.5 18 1.0 -7 10 1.5 16.0 38 1.5 18.0 19.5 2.0 -8 63 3.0 19.0 70 2.0 19.5 21.5 0.5 -9 76 3.0 22.0 96 3.0 22.0 25.0 - 0.510 35 2.0 24.0 92 2.5 25.0 27.5 1.0 -

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.

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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 Frequency0 161 242 303 604 405 30

Using the random numbers simulate the demand for a 10 week period.

Solution:

Trips per

week/

Demand per

week

Frequency Probability Cumulative

Probability

Random

Number

Interval

0 16 0.08 0.08 00-071 24 0.12 0.20 08-192 30 0.15 0.35 20-343 60 0.30 0.65 35-644 40 0.20 0.85 65-845 30 0.15 1.00 85-99

The simulated demand for the cars for the next 10 weeks period is given in the table

below

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Week Random Number Demand1 82 42 96 53 18 14 96 55 20 26 84 47 56 38 11 19 52 310 03 0

Total 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 Probability95 0.03

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96 0.0597 0.0798 0.1099 0.15100 0.20

101 0.15102 0.10103 0.07104 0.05105 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 CumulativeProbability

Random NumberInterval

95 0.03 0.03 00-0296 0.05 0.08 03-0797 0.07 0.15 08-1498 0.10 0.25 15-2499 0.15 0.40 25-39100 0.20 0.60 40-59101 0.15 0.75 60-74102 0.10 0.85 75-84

103 0.07 0.92 85-91104 0.05 0.97 92-96105 0.03 1.00 97-99

The simulated production of cars for the next 15 days is given in the following table.

Day Random

Number

Production per

day

No. of cars

waiting

No. of empty

space in the

ship1 97 105 4 -

2 02 95 - 63 80 102 1 -4 66 101 - -5 96 104 3 -6 55 100 - 17 50 100 - 18 29 99 - 29 58 100 - 1

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10 51 100 - 111 04 96 - 512 86 103 2 -13 24 98 - 314 39 99 - 2

15 47 100 - 1Total 10 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) Probability27 0.1028 0.1529 0.2030 0.35

31 0.1532 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 ordisadvantage to the company?

Sale (units) Probability Cumulative

Probability

Random Number

Interval27 0.10 0.10 00-0928 0.15 0.25 10-24

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29 0.20 0.45 25-4430 0.35 0.80 45-7931 0.15 0.95 80-9432 0.05 1.00 95-99

Set up the simulation for next 10 days

Day Random

Number

Simulated

Sales

Number of

units Unsold

Number of

units short1 10 28 30-28=2 02 99 32 0 32-30=23 65 30 0 04 99 32 0 25 95 32 0 26 01 27 30-27=3 07 79 30 0 08 11 28 30-28=2 09 16 28 2 0

10 20 28 2 0Total 11 6

At the production rate of 30 per day, total number of units produced in 10 days = 30 X 10

= 300 Nos, since the production cost is Rs.40 and the sales price is Rs.50, the per unit

profit is Rs.10

Day Demand Actual Sales Profit on

Actual sales

Loss due to

the unsold

units

Loss due to

penalty for

shortage

1 10 28 280 2 X 15 = 30 -2 99 30 300 - 2 X 5 = 103 65 30 300 - -4 99 30 300 - 2 X 5 = 105 95 30 300 - 2 X 5 = 106 01 27 270 3 X 15 = 45 07 79 30 300 0 0

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8 11 28 280 2 X 15 = 30 09 16 28 280 2 X 15 = 30 010 20 28 280 2 X 15 = 30 0

289 Rs.2890 Rs.165 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 Demand Production

units

Actual

Sales

Profit on

Actual

sales

Loss due to

the unsold

units

Loss due to

penalty for

shortage1 28 29 28 280 1 X 15 = 15 02 32 29 29 290 0 3 X 5= 153 30 29 29 290 0 1 X 5 = 54 32 29 29 290 0 3 X 5= 155 32 29 29 290 0 3 X 5= 156 27 29 27 270 1 X 15 = 15 07 30 29 29 290 0 1 X 5 = 58 28 29 28 280 1 X 15 = 15 09 28 29 28 280 1 X 15 = 15 010 28 29 28 280 1 X 15 = 15 0

Total 284 Rs.2840 Rs.90 Rs.55

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 orscarce 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 givencriteria of optimality.

The term Linear is used to describe the proportionate relationship of two or morevariables in a model. The given change in one variable will always cause a resultingproportional change in another variable.

The word , programming is used to specify a sort of planning that involves theeconomic allocation of limited resources by adopting a particular course of action orstrategy among various alternatives strategies to achieve the desired objective.

Hence, Linear Programming is a mathematical technique for optimum allocation oflimited 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 consistsof three components.

i) The activities (variables) and their relationshipsii) The objective function andiii) 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 mathematicalrepresentation 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 imposerestriction of the limited resources

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Assumptions of Linear Programming

Certainty.

In all LP models it is assumed that, all the model parameters such as availability ofresources, profit (or cost) contribution of a unit of decision variable and consumption ofresources 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 wholenumbers (integers) or mixed numbers (integer or fractional). However, if only integervariables 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 totalsum 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 decisionvariable respectively. /The objective function is the direct sum of the individualcontributions of the different variables

LinearityAll relationships in the LP model (i.e. in both objective function and constraints) must belinear.

General Mathematical Model of an LPP

Optimize (Maximize or Minimize) Z=C1 X1 + C2 X2 ++CnXnSubject 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 () bmand X1, X2 .Xn >

Guidelines for formulating Linear Programming model

i) Identify and define the decision variable of the problemii) Define the objective functioniii) 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

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Example 1.A manufacturer produces two types of models M1 and M2.Each model of the type M1requires 4 hours of grinding and 2 hours of polishing; where as each model of M2requires 2 hours of grinding and 5 hours of polishing. The manufacturer has 2 grindersand 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.Whateverproduced in a week is sold in the market. How should the manufacturer allocate hisproduction capacity to the two types of models, so that he makes maximum profit in aweek?

i) Identify and define the decision variable of the problemLet X1 and X2 be the number of units of M1 and M2 model.

ii) Define the objective functionSince the profits on both the models are given, the objective functionis 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 by4X1 + 2X2 < 80No 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 by2X1 + 5X2 < 180No 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 2kg of raw material and 4 labour hours for processing, where as each unit of B requires 3kg of raw materials and 3 labour hours for the same type. Every week, the firm has anavailability of 60 kg of raw material and 96 labour hours. One unit of product A soldyields 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 ofeach of the products should be produced per week so that the firm can earn maximumprofit.

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 functionSince 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 labourconstraint..The raw material constraint is given by

2X1 + 3X2 < 60The labour hours constraint is given by4X1 + 3X2 < 96

Finally we have,MaxZ = 40X1 + 35X2Subject to constraints,2X1 + 3X2 < 604X1 + 3X2 < 96X1,X2 > 0

Example 3.The agricultural research institute suggested the farmer to spread out atleast 4800 kg ofspecial phosphate fertilizer and not less than 7200 kg of a special nitrogen fertilizer toraise 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 andthey cost Rs.40 and Rs.24 respectively. Mixture A contains phosphate and nitrogenequivalent of 20kg and 80 kg respectively, while mixture B contains these ingredientsequivalent of 50 kg each. Write this as an LPP and determine how many bags of eachtype the farmer should buy in order to obtain the required fertilizer at minimum cost.

i) Identify and define the decision variable of the problemLet X1 and X2 be the number of bags of mixture A and mixture B.ii) Define the objective functionThe cost of mixture A and mixture B are given ;the objective function is to minimize the costMin.Z = 40X1 + 24X2

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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20X1 + 50X2 >4800 Phosphate requirement80X1 + 50X2 >7200 Nitrogen requirementX1, X2 >0

Finally we have,

Min.Z = 40X1 + 24X2is subjected to three constraints20X1 + 50X2 >480080X1 + 50X2 >7200X1, X2 >0

Example 4.A firm can produce 3 types of cloth, A , B and C.3 kinds of wool are required Red, Greenand 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 wooland 2 meters of blue wool.1 unit type of C cloth needs 5 meters of green wool and 4meters of blue wool. The firm has a stock of 8 meters of red, 10 meters of green and 15meters of blue. It is assumed that the income obtained from 1 unit of type A is Rs.3, fromB 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 problemLet X1, X2 and X3 are the quantity produced of cloth type A,B and C respectively.ii) Define the objective functionThe incomes obtained for all the three types of cloths are given; the objective function isto maximize the income.Max Z = 3X1 + 5X2 + 4X3iii) 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 < 103X1 + 2X2 + 4X3 < 15X1, X2 X3 >0

Finally we have,Max Z = 3X1 + 5X2 + 4X3is subjected to three constraints2X1 + 3X2 < 82X2 + 5X3 < 103X1 + 2X2 + 4X3 < 15X1, 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 Byielding 10%. After some consideration, he decides to invest at most of Rs.12,000 inbond 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 ifthe investor wants to maximize his return on investment? Solve graphically.(January/February 2004)

i) Identify and define the decision variable of the problemLet X1 and X2 be the amount invested in Bonds A and B.ii) Define the objective functionYielding for investment from two Bonds are given; the objective function is to maximizethe 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 < 30,000X1 > 6,000X2 < 12,000X1 -- X2 >0X1, X2 >0

Finally we have,MaxZ = 0.07X1 + 0.1X2is subjected to three constraintsX1 + X2 < 30,000X1 > 6,000X2 < 12,000X1 -- X2 >0X1, 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 dryproduct 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 therequirements?

i) Identify and define the decision variable of the problemLet X1 and X2 be the number of units of liquid and dry products.ii) Define the objective functionThe cost of Liquid and Dry products are given ; the objective function is to minimize thecost

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Min. Z = 3X1 + 2X2iii) State the constraints to which the objective function should be optimized.The above objective function is subjected to following three constraints.5X1 + X2 >102X1 + 2X2 >12

X1 + 4X2 >12X1, X2 >0

Finally we have,Min. Z = 3X1 + 2X2is subjected to three constraints5X1 + X2 >102X1 + 2X2 >12X1 + 4X2 >12X1, X2 >0

Example 6.A Scrap metal dealer has received a bulk order from a customer for a supply of atleast2000 kg of scrap metal. The consumer has specified that atleast 1000 kgs of the ordermust 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 kgsof scrap which are unfit for commercial purposes. The scrap metal dealer purchases thescrap from two different sources in an unlimited quantity with the following percentages(by weight) of high quality of copper and unfit scrap

Source A Source B

Copper 40% 75%

Unfit Scrap 7.5% 10%

The cost of metal purchased from source A and source B are Rs.12.50 and Rs.14.50 perkg respectively. Determine the optimum quantities of metal to be purchased from the twosources by the metal scrap dealer so as to minimize the total cost (February 2002)

i) Identify and define the decision variable of the problemLet X1 and X2 be the quantities of metal to be purchased from the two sources A and B.ii) Define the objective functionThe cost of metal to be purchased by the metal scrap dealer are given;

the objective function is to minimize the costMin. Z = 12.5X1 + 14.5X2iii) 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 >1,0000.075X1 + 0.1X2 + 4X3 < 200X1, X2 >0

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Finally we have,Min. Z = 12.5X1 + 14.5X2is subjected to three constraintsX1 + X2 >2,0000.4X1 + 0.75X2 >1,000

0.075X1 + 0.1X2 + 4X3 < 200X1, X2 >0

Example 7.A farmer has a 100 acre farm. He can sell all tomatoes, lettuce or radishes and can raisethe price to obtain Rs.1.00 per kg. for tomatoes , Rs.0.75 a head for lettuce and Rs.2.00per kg for radishes. The average yield per acre is 2000kg.of tomatoes, 3000 heads oflettuce and 1000 kgs of radishes. Fertilizers are available at Rs.0.50 per kg and theamount required per acre is 100 kgs for each tomatoes and lettuce and 50kgs forradishes. Labour required for sowing, cultivating and harvesting per acre is 5 man-daysfor 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 maximizethe farmers profit.

i) Identify and define the decision variable of the problemLet X1 and X2 and X3 be number acres the farmer grows tomatoes, lettuce and radishesrespectively.ii) Define the objective functionThe objective of the given problem is to maximize the profit.

The profit can be calculated by subtracting total expenditure from the total salesProfit = Total sales Total expenditure

The farmer produces 2000X1 kgs of tomatoes, 3000X2 heads of lettuce, 1000X3 kgs ofradishes.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 )Farmers profit will beZ = (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 isMaximise 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 acresX1 + X2 + X3 < 100The total man-days labour is 400 man-days

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5X1 + 6X2 + 5X3 < 400Finally we have,Maximise Z = 1850X1 + 2080X2 + 1875X3is subjected to three constraintsX1 + X2 + X3 < 100

5X1 + 6X2 + 5X3 < 400X1, X2 X3 >0

Example 8.An electronics company produces three types of parts for automatic washing machines .Itpurchases 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 areRs 8, Rs.10 and Rs.14.All parts made can be sold. Castings for parts A, B and Crespectively 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 thethree 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 followingtable.

MachineCapacities 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 itshould produce per hour in order to maximize profit for an hours run. Formulatethis problem as an LP model so as to maximize total profit to the company.

i) Identify and define the decision variable of the problemLet X1 and X2 and X3 be the number of types A, B and C parts produced per hourrespectively .ii) Define the objective functionWith the information given, the hourly profit for part A, B, and C would be as followsProfit per type A part = (8 5) (20/25 +30/25 + 30/40) = 0.25Profit per type B part = (10 6) (20/40 + 30/20 + 30/30) = 1Profit per type C part = (14 10) (20/25 + 30/20 + 30/40) = 0.95Then,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 constraintX1/25 + X2/40 + X3/24 < 1ii) The shaping machine constraintX1/25 + X2/20 + X3/20 1iii) The polishing machine constraint

X1/40 + X2/30 + X3/40 1X1, X2, X3 0

Finally we have,Maximize Z = 0.25 X1 + 1X2 + 0.95X3Subject to constraintsX1/25 + X2/40 + X3/24 < 1ii) The shaping machine constraintX1/25 + X2/20 + X3/20 1iii) The polishing machine constraintX1/40 + X2/30 + X3/40 1

X1, X2, X3 0Example 9.A city hospital has the following minimal daily requirements for nurses.

PeriodClock time (24

hours day)Minimum

number of nursesrequired

1 6 a.m. 10 a.m. 2

2 10 a.m. 2 p.m. 7

3 2 p.m. 6 p.m. 15

4 6 p.m. 10 p.m. 85 10 p.m. 2 a.m. 20

6 2 a.m. 6 a.m. 6

Nurses report at the hospital at the beginning of each period and work for 8 consecutivehours. The hospital wants to determine the minimal number of nurses to be employed sothat there will be a sufficient number of nurses available for each period.Formulate this as a linear programming problem by setting up appropriate constraintsand objective function.

i) Identify and define the decision variable of the problemLet X1, X2, X3, X4, X5 and X6 be the number of nurses joining duty at thebeginning of periods 1, 2, 3, 4, 5 and 6 respectively.

ii) Define the objective functionMinimize Z = X1 + X2 + X3 + X4 + X5 + X6iii) State the constraints to which the objective function should be optimized.The above objective function is subjected to following constraints.X1 + X2 7

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X2 + X3 15X3 + X4 8X4 + X5 20X5 + X6 6X6 + X1 2

X1, X2, X3, X4, X5, X6 0

Linear Programming: Graphical Solution

Example 1.Solve the following LPP by graphical methodMaximize Z = 5X1 + 3X2Subject to constraints2X1 + X2 1000X1 400X1 700

X1, X2 0Solution:

The first constraint 2X1 + X2 1000 can be represented as follows.We set 2X1 + X2 = 1000When X1 = 0 in the above constraint, we get,2 x 0 + X2 = 1000X2 = 1000

Similarly when X2 = 0 in the above constraint, we get,2X1 + 0 = 1000X1 = 1000/2 = 500The second constraint X1 400 can be represented as follows,We set X1 = 400The 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 X1 X2 Z = 5X1 +3X2

0 0 0 0

A 0 700 Z = 5 x 0 + 3 x 700 = 2,100

B 150 700 Z = 5 x 150 + 3 x 700 = 2,850*Maximum

C 400 200 Z = 5 x 400 + 3 x 200 = 2,600

D 400 0 Z = 5 x 400 + 3 x 0 = 2,000

The Maximum profit is at point BWhen X1 = 150 and X2 = 700Z = 2850

Example 2.

Solve the following LPP by graphical methodMaximize Z = 400X1 + 200X2Subject to constraints18X1 + 3X2 8009X1 + 4X2 600X2 150X1, 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 = 800X2 = 800/3 = 266.67Similarly when X2 = 0 in the above constraint, we get,18X1 + 3 x 0 = 800X1 = 800/18 = 44.44

The second constraint 9X1 + 4X2 600 can be represented as follows,We set 9X1 + 4X2 = 600When X1 = 0 in the above constraint, we get,9 x 0 + 4X2 = 600

X2 = 600/4 = 150Similarly when X2 = 0 in the above constraint, we get,9X1 + 4 x 0 = 600X1 = 600/9 = 66.67The third constraint X2 150 can be represented as follows,We set X2 = 150

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Point X1 X2 Z = 400X1 + 200X2

0 0 0 0

A 0 150Z = 400 x 0+ 200 x 150 = 30,000*

MaximumB 31.11 80 Z = 400 x 31.1 + 200 x 80 = 28,444.4

C 44.44 0 Z = 400 x 44.44 + 200 x 0 = 17,777.8

The Maximum profit is at point AWhen X1 = 150 and X2 = 0

Z = 30,000

Example 3.Solve the following LPP by graphical methodMinimize Z = 20X1 + 40X2Subject to constraints36X1 + 6X2 1083X1 + 12X2 36

20X1 + 10X2 100X1 X2 0

Solution:

The first constraint 36X1 + 6X2 108 can be represented as follows.We set 36X1 + 6X2 = 108When X1 = 0 in the above constraint, we get,36 x 0 + 6X2 = 108

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X2 = 108/6 = 18Similarly when X2 = 0 in the above constraint, we get,36X1 + 6 x 0 = 108X1 = 108/36 = 3

The second constraint3X1 + 12X2 36 can be represented as follows,We set 3X1 + 12X2 = 36When X1 = 0 in the above constraint, we get,3 x 0 + 12X2 = 36X2 = 36/12 = 3Similarly when X2 = 0 in the above constraint, we get,3X1 + 12 x 0 = 36X1 = 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 = 100X2 = 100/10 = 10Similarly when X2 = 0 in the above constraint, we get,20X1 + 10 x 0 = 100X1 = 100/20 = 5

Point X1 X2 Z = 20X1 + 40X2

0 0 0 0A 0 18 Z = 20 x 0 + 40 x 18 = 720

B 2 6 Z = 20 x2 + 40 x 6 = 280

C 4 2 Z = 20 x 4 + 40 x 2 = 160* Minimum

D 12 0 Z = 20 x 12 + 40 x 0 = 240

The Minimum cost is at point C

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When X1 = 4 and X2 = 2Z = 160

Example 4.Solve the following LPP by graphical method

Maximize Z = 2.80X1 + 2.20X2Subject to constraintsX1 20,000X2 40,0000.003X1 + 0.001X2 66X1 + X2 45,000X1 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 = 66X2 = 66/0.001 = 66,000Similarly when X2 = 0 in the above constraint, we get,0.003X1 + 0.001 x 0 = 66X1 = 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 = 45,000X2 = 45,000Similarly when X2 = 0 in the above constraint, we get,X1 + 0 = 45,000X1 =45,000

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Point X1 X2 Z = 2.80X1 + 2.20X2

0 0 0 0

A 0 40,000Z = 2.80 x 0 + 2.20 x 40,000 = 88,000

B 5,000 40,000Z = 2.80 x 5,000 + 2.20 x 40,000 =

1,02,000

C 10,500 34,500Z = 2.80 x 10,500 + 2.20 x 34,500 =

1,05,300* Maximum

D 20,000 6,000Z = 2.80 x 20,000 + 2.20 x 6,000 =

69,200

E 20,000 0 Z = 2.80 x 20,000 + 2.20 x 0 = 56,000

The Maximum profit is at point CWhen X1 = 10,500 and X2 = 34,500Z = 1,05,300

Example 5.Solve the following LPP by graphical methodMaximize Z = 10X1 + 8X2Subject to constraints2X1 + X2 20X1 + 3X2 30X1 - 2X2 -15X1 X2 0

Solution:

The first constraint 2X1 + X2 20 can be represented as follows.We set 2X1 + X2 = 20When X1 = 0 in the above constraint, we get,2 x 0 + X2 = 20X2 = 20Similarly when X2 = 0 in the above constraint, we get,2X1 + 0 = 20

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X1 = 20/2 = 10

The second constraint X1 + 3X2 30 can be represented as follows,We set X1 + 3X2 = 30When X1 = 0 in the above constraint, we get,

0 + 3X2 = 30X2 = 30/3 = 10Similarly when X2 = 0 in the above constraint, we get,X1 + 3 x 0 = 30X1 = 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 = -15X2 = -15/2 = 7.5

Similarly when X2 = 0 in the above constraint, we get,X1 2 x 0 = -15X1 = -15

Point X1 X2 Z = 10X1 + 8X2

0 0 0 0

A 0 7.5 Z = 10 x 0 + 8 x 7.5 = 60

B 3 9 Z = 10 x 3 + 8 x 9 = 102

C 6 8 Z = 10 x 6 + 8 x 8 = 124* Minimum

D 10 0 Z = 10 x 10 + 8 x 0 = 100

The Maximum profit is at point CWhen X1 = 6 and X2 = 8Z = 124

Duality in Linear Programming

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Duality in Linear Programming For every LPP there is a unique LPP associated with itinvolving the same data and closely related optimal solution. The original problem is thencalled primal problem while the other is called its Dual problem

Let the primal problem beMaximize Z = C1 X1 + C2 X2 + + CnXnSubject to constraints,a11X1+ a 12X2++ a 1nXn 22X1+ X2 + 6X3 < 6X1- X2 +3X3 = 4

and X1, X2,X3 > 0

Rearrange the constraints into a standard form, we getMin Z = 0X1 + 2X2+ 5X3

Subject to constraints,X1+ X2 + 0X3> 2-2X1- X2 - 6X3 > -6X1- X2 +3X3 > 4

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-X1 + X2 -3X3 > -4and X1, X2,X3 > 0

The Dual of the above primal is as followsMax.G = 2W1 -6W2+ 4W3 4W4

Subject to constraints,W 1 -2W2 + W3 W4< 0W 1 - W2 - W3 + W4 < 20W 1 - 6W2 + 3W3- 3W4 < 5W 1, W2, W3,W4 > 0

Max.G = 2W1 -6W2+ 4(W3 W4)Subject to constraints,W 1 -2W2 + (W3 W4 ) < 0W 1 - W2 - W3 + W4 < 20W 1 - 6W2 3(W3- W4 ) < 5

W 1, W2, W3,W4 > 0Max.G = 2W1 -6W2+ 4W5Subject to constraints,W 1 -2W2 + W5 < 0W 1 - W2 W5 < 20W 1 - 6W2 3W5 < 5W 1, W2, > 0 , W5 is unrestricted in sign

Example.2Write the Dual of the following LPPMin Z = 4X1 + 5X2- 3X3Subject to constraints,X1+ X2 + X3 = 223X1+ 5X2 - 2X3 < 65X1+ 7X2 +4X3 > 120X1 , X2 > 0 and X3 is unrestricted

Since X3 is Unrestricted, replace X3 with (X4 - X5 ) andbring 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) > -65X1+ 7X2 +4(X4 - X5) > 120X1 , X2 , X4 ,X5 > 0

The Dual of the above primal is as followsMax.G = 22(W1 -W2)- 65W3 + 120W4

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Subject to constraints,W 1 -W2 - 3W3 +W4< 4W 1 - W2 - 5W3 + 7W4 < 5W 1 - W2 + 2W3+ 4W4 < -3-W 1 + W2 - 2W3- 4W4 < 3

W 1, W2, W3,W4 > 0Max.G = 22W5 - 65W3 + 120W4Subject to constraints,W 5- 3W3 +W4< 4W5 -5W3 + 7W4 < 5W 1 - W2 + 2W3+ 4W4 < -3-W 1 + W2 - 2W3- 4W4 < 3W 1, W2, W3,W4 > 0

quantitative tech compiled kantharaju

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