Unit ii-1-lp

Linear Programming
• Programming is another term for planning.
• In LP we study about planning and allocation
of resources.
• The allocation of resources (men, machine,
materials) is necessary because they are
limited in supply.

• In LP we are concerned with the following
definition of Economics given by Lionel
Robbins:
• Economics is the science which studies
human behaviour as a relationship
between ends and scarce means which
have alternative uses.
• ‘Ends’ are the objectives to be achieved and
resources are to be allocated so as to achieve
the objectives, that is, the resources have
alternative applications.

• LP aims to answer the question – ‘How to
allocate resources?’
• There must be a definite criterion, a measure
of effectiveness for optimum allocation of
resources.
• Resource generates a separate constraint.
• Constraints on the utilisation of resources can
also be expressed as linear equations or linear
inequalities.

• This gives us a Linear Programming Problem.
• The term ‘linear’ means a relationship of
proportionality.

LP: Origin
• The origin of LP can be traced to the early
1920’s when the celebrated economist
Leontiff discovered the Input/Output Model.
• The more recent advances can be traced to
1947 when the discovery of the Simplex
method of LP (by George Dantzig)
revolutionised the science of OR.

LPP: Essential Characteristics
• The measure of effectiveness (spells out the
objective that we are trying to achieve) which
is generally profit maximisation.
• Resources are limited in supply.
• Linearity, both of the objective function as
well as of the constraints.

• Alternative courses of action. An LPP
generates the alternatives. Only when there
are more than one solutions, will we decide
on the optimum one.
• Variables are interrelated and not
independent of each other.
• Variables are non-negative in nature, that is
they are either zero or positive.

• Divisibility or continuity of the variables: when
solving an LPP, the variables can take non-
integral values also.
• As far as uncontrollable variables are
concerned, they are supposed to be
deterministic, that is, they are known for
certain.
• In the short run, they are supposed to be
constant.

• Examples of uncontrollable variables:
• Magnitude or quantum of each resource
available.
• Technological coefficients.
• Input coefficients.
• Output coefficients.

Application of LPP
• LP is considered a very versatile technique
capable of handling a large variety of
management problems:
– Production.
– Finance.
– Marketing.
– Human Resource.

• Production.
• Optimal product mix.
• Fluid blending.
• General blending problems (blending
liquor, proteins, calories, etc., to prepare
baby food).
• Production scheduling.
• Production planning.

• Finance.
• Portfolio selection.
• Profit planning.
• Marketing.
• Media selection.
• Physical distribution of products.
• Market share.
• Deciding the channels of distribution.

• Personnel. This field has very few applications
of LP, the main of them being Staffing.
• OR also has a number of applications in the
field of agriculture – both at macro and micro
levels.

Mathematical Formulation of LP
Model for Product-Mix Problems
• Two products, namely P1 and P2 are being
manufactured. Each product has to be processed
through two machines M1 and M2. One unit of
product P1 consumes 4 hours of time on M1 and 2
hours of time on M2. Similarly, one unit of P2
consumes 2 hours of time on M1 and 4 hours of
time on M2. 60 hours of time is available on M1
and 48 hours on M2. The per unit contribution
margin of P1 is 8 and of P2 is 6. Determine the
number of units of P1 and P2 to be manufactured
so as to maximise total contribution.

A company has two grades of inspectors I and II who are
to be assigned for a quality control inspection. It is
required that at least 2000 pieces be inspected per 8
hour day. Grade I inspectors can check pieces @ 40 per
hour with an accuracy of 97%. Grade II inspectors check
@ 30 pieces per hour with an accuracy of 95%. The wage
rate of Grade I inspectors is Rs.5 per hour while that of a
Grade II inspector is Rs.4 per hour. An error made by an
inspector costs Rs.3 to the company. There are only 9
Grade I inspectors and 11 Grade II inspectors available in
the company. The company wishes to assign work to the
available inspectors so as to minimise the total cost of
inspection. Formulate the LP model for the problem and
solve it by the graphical method.

A company, makes two types of leather belts.
Belt A is a high quality belt and belt B is of lower
quality. The respective profits are Rs.40 and
Rs.30. Each belt of type A requires twice as
much time as a belt of type B and if all the belts
were of type B, the company could make 1000
belts per day. The supply of leather is sufficient
for only 800 belts per day, both A and B
combined. Belt A requires a fancy buckle of
which only 400 per day are available. There are
700 buckles per day available for belt B. Set up
the LPP and solve it by the graphical method.

Graphical Method of Solving LP
Problems

Iso-Profit Line
• The line representing the objective function is
called the iso-profit line.
• We give a random value to z (objective
function) and draw an iso-profit line.
• We keep increasing this value and successive
iso-profit lines will be parallel to the first one,
and above it.

• There will ultimately be one such iso-profit
line that will just touch one of the extremities
or boundaries of the feasible polygon.
• Any other iso-profit line above this line will
not pass through or touch the feasible
polygon.
• Therefore the maximum value of z will be
reached at one of the extremity or boundary
of the feasible polygon.
• The reverse process may be adopted for a
minimisation problem.

Extreme Point Theorem
• The optimum solution to an LPP lies at one of
the extremities of the feasible polygon,
provided there exists a solution to the LPP
which is UNIQUE, FINITE and OPTIMAL.

Extreme Point Theorem:
Exceptions
• Multiple Optimal Solutions.
• Unbounded Solution.
• Infeasible Solution.

Trial and Error Method
• This and the subsequent methods are based
on another theorem – The Basis Theorem.

Basis Theorem of LP
• If in a system of equations we have ‘m’
variables and ‘n’ constraints, m being greater
than n, then the solution obtained by putting
m–n of the variables as zero results in a corner
point. The solution at the corner point is
known as a Basic Solution.

• With extreme point theorem, we examined all the
extreme points but not all the basic solutions.
• Basic solutions can be feasible or infeasible.
• Extreme point theorem gave only the feasible basic
solutions.
• x1, x2, etc., are structural variables as they constitute
the structure of the LPP.
• Any other variables used (S1, S2, etc.:
Slack/Surplus Variables; A1, A2, etc.: Artificial
Variables) are non-structural variables.
• Those variables given the value zero are known as non-
basic variables.
• The non-zero variables are known as basic variables.

Algebraic Method
• The T & E and the algebraic methods
constitute the basis of the Simplex Method.

Simplex Method of Solving LP
Problems
• Discovered by George Dantzig.
• The Initial Simplex Table serves two purposes
– gives a tabular representation of the LPP
and provides an initial basic feasible solution.
• First three columns of the simplex table are
common to all problems.
• The number of remaining columns depends
upon the number of variables involved.

• Simplex Criterion I: Used to identify the
incoming variable (Optimum Column).
• Simplex Criterion II: Used to identify the
outgoing variable (Replaced Row).
• Pivot Element: The element lying at the
intersection of the optimum column and the
replaced row.
• Criterion for optimality in a maximisation
problem: All elements in the Cj-Zj row should
be either 0 or negative.

• Augmented Matrix: The coefficient matrix is
augmented by the column vector of the RHS
constants.
• We perform ‘row operations’ in such a
manner so as to obtain an identity matrix in
place of the coefficient matrix.

The Big ‘M’ Method
• Structural and slack/surplus variables can
reenter the basis.
• An artificial variable driven out of the basis
cannot reenter it.
• Therefore, if an artificial variable has been
driven out of the basis, the terms in its column
need not be computed.

• In minimisation problem we reach the optimal
solution if all the elements of the Cj-Zj row are
either positive or zero.
• In maximisation problem, the objective
function coefficient of an artificial variable is
‘– M.’
• In minimisation problem, the objective
function coefficient of an artificial variable is
‘+M.’

The Three Exceptions
• Multiple Optimal Solutions: Cj-Zj element of a
non-basic variable is zero.
• Unbounded Solution: All elements of optimum
column are negative.
• Infeasible Solution: If the criterion for
optimality is being satisfied and yet there is an
artificial variable in the basis.

A manufacturer of 3 products tries to follow a
policy of producing those products which
contribute most to profit. However, there is also a
policy of recognising certain minimum sales
requirements. Currently these are:
Product X: 20 units/week.
Product Y: 30 units/week.
Product z: 60 units/week.
There are 3 producing departments. The
production times in hours/unit in each department
and the total time available for each week in each
department are:

Deptt. Time Reqd. (hrs.) Total Hrs.
AvailableX Y Z
1 0.25 0.20 0.15 420
2 0.30 0.40 0.50 1048
3 0.25 0.30 0.25 529

The contributions per unit of products X, Y and Z
are Rs.10, 9 and 8, respectively. The company
has scheduled 1558 units of X, 30 units of Y and
60 units of Z for production in the following
week.
You are required to state whether the present
schedule is optimum and if it is not, then what it
should be? What recommendation should be
made to the firm regarding its production
facilities?

Degeneracy
• Degeneracy is revealed when a basic variable
acquires a zero value (rather than a negative
or a positive value).

Sensitivity Analysis
• Also known as Post-Optimality Analysis.
• Undertaken to explore the effect of changes in
the LP parameters on the optimal solution.
• For example, to study the effect of costs and
price change, diminishing resources becoming
readily available or technological advances, on
the optimal LP solution, sensitivity analysis
suggests the changes in the data used to build
the model...

• Concerned with the extent of sensitivity of the
optimum solution to an LP for change in one
or more of:
– The profit or cost coefficients of the objective
function,
– The LHS coefficients of the variables in the
constraints, and
– The RHS quantities of the constraints.

A firm produces three products A, B, C. Unit contributions
of the products are Rs.5, 10, 8 respectively. Each unit of
product A requires 3kg of material, 5 machine hours and 2
labour hours; each unit of product B requires 4kg of
material, 4 machine hours and 4 labour hours and each unit
of product C requires 2kg of material, 4 machine hours and
5 labour hours.
Everyday 60kg of material, 72 machine hours and 100
labour hours are available. Find out the best production
strategy. Investigate the effect on the solution of the
following:
i.An increase of 12 machine hours.
ii.A decrease of 6kg of material.
iii.3 units of product A are to be produced.

Cj 5 10 8 0 0 0
Cb BV SV x1 x2 x3 S1 S2 S3
0S1 60 3 5 2 1 0 0 12 ←
0S2 72 4 4 4 0 1 0 18
0S3 100 2 4 5 0 0 1 25
Zj 0 0 0 0 0 0 0
Cj-Zj 5 10 8 0 0 0
↑
10x2 12 3/5 1 2/5 1/5 0 0 30 R1=R1/5
0S2 24 8/5 0 12/5 - 4/5 1 0 10 ← R2=R2-4R1
0S3 52 - 2/5 0 17/5 - 4/5 0 1 260/17 R3=R3-4R1
Zj 120 6 10 4 2 0 0
Cj-Zj -1 0 4 -2 0 0
↑
10x2 8 1/3 1 0 1/3 - 1/6 0 R1=R1-2/5R2
8x3 10 2/3 0 1 - 1/3 5/12 0 R2=5/12R2
0S3 18 -8/3 0 0 1/3 -17/12 1 R3=R3-17/5R2
Zj 160 26/3 10 8 2/3 5/3 0
Cj-Zj -11/3 0 0 - 2/3 -5/3 0

Cj 5 10 8 0 0 0
10x2 8 1/3 1 0 1/3 - 1/6 0
8x3 10 2/3 0 1 - 1/3 5/12 0
0S3 18 -8/3 0 0 1/3 -17/12 1
Zj 160 26/3 10 8 2/3 5/3 0
Cj-Zj -11/3 0 0 - 2/3 -5/3 0
An increase of 12
machine hours
(slack variable S2)

Cj 5 10 8 0 0 0
10x2 8 1/3 1 0 1/3 - 1/6 0
8x3 10 2/3 0 1 - 1/3 5/12 0
0S3 18 -8/3 0 0 1/3 -17/12 1
Zj 160 26/3 10 8 2/3 5/3 0
Cj-Zj -11/3 0 0 - 2/3 -5/3 0
A decrease of 6kg of
material
(slack variable S1)

Cj 5 10 8 0 0 0
10x2 8 1/3 1 0 1/3 - 1/6 0
8x3 10 2/3 0 1 - 1/3 5/12 0
0S3 18 -8/3 0 0 1/3 -17/12 1
Zj 160 26/3 10 8 2/3 5/3 0
Cj-Zj -11/3 0 0 - 2/3 -5/3 0
3 units of product A
are to be produced
(variable x1)

Duality
• LPPs always exist in pairs.
• Associated to every maximisation problem
there is a complementary minimisation
problem and vice versa.
• The original problem is known as primal.
• Its complementary problem is known as dual.
• The names primal and dual are
interchangeable.

Reasons for Studying Duality
• Dual may provide a computational shortcut to
the primal.
• Solution to primal also gives the solution to its
dual.
• Economic interpretations can be drawn from
dual’s solution.

Rules for Formulating Dual
1. If primal is a maximisation problem, its dual
will be a minimisation problem, and vice
versa.
2. No. of dual variables = no. of primal
constraints.
3. No. of dual constraints = no. of primal
variables.

Rules of Formulating Dual
4. The transpose of the coefficient matrix of the
primal is the coefficient matrix of the dual.
5. The direction of constraints of the dual is the
reverse of the direction of constraints in the
primal.
6. If the kth
primal constraint is a strict equality,
then the kth
dual variable will be unrestricted
in sign.

Rules of Formulating Dual
7. If the ith
primal variable is unrestricted in sign,
then the ith
dual constraint will be a strict
equality.
8. If the primal is a maximisation problem, then
before formulating the dual all the
constraints should be converted into ‘≤ type’.
If the primal is a minimisation problem, then
before formulating the dual all the
constraints should be converted into ‘≥ type’.

Rules for Formulating Dual
9. The objective function coefficients of the
primal become the RHS constants of the
constraints of the dual and the RHS constants
of the primal constraints become the
objective function coefficients of the
variables in the dual.
10.If a variable is unrestricted in sign, it can be
written as the difference between two non-
negative variables.

Cj 8 6 0 0
Cb BV SV x1 x2 S1 S2
8x1 12 1 0 1/3 -
6x2 6 0 1 - 1/3
Zj 13 8 6 5/3 2/3

Final Simplex Table
Cj 6048 0 0 M M
Cb BV SV y1 y2 S3 S4 A1 A2
60 y1 5/ 1 0 - 1/6 - -
48 y2 2/ 0 1 1/6 - - -

Cj 8 6 0 0
Cb BV SV x1 x2 S1 S2
8x1 12 1 0 1/3 -
1/6
6x2 6 0 1 -
1/6
1/3
Zj 13 8 6 5/3 2/3
Cj-Zj 0 0 -5/3 -
Cj 60 48 0 0 M M
Cb BV SV y1 y2 S3 S4 A1 A2
60 y1 5/ 1 0 - 1/6 - -
48 y 2/ 0 1 1/6 - - -

• If primal is from sellers point of view
(maximise profits), dual is from the buyers
point of view (minimise expenses).
• Slack variables in the primal (or the structural
variables of the dual) would give the imputed
values of the resources or shadow
prices/artificial accounting prices/opportunity
cost.
• Slack variables of the primal correspond to the
structural variables of the dual and vice versa.

• If the primal has m structural and n slack
variables, then in the dual there will be n
structural and m surplus variables.
• Cj-Zj row of the primal gives the solution
values of the dual variables.

Economic Interpretation of Dual

A housewife asks a butcher to grind up several
cuts of beef to form a blend of not less than
17.6% protein and 14.8% fat. The butcher may
use several available cuts. What cuts and in
what quantities should he use if he wants to
minimise the cost of the cuts he buys and satisfy
the housewife’s requirements.
The protein and fat contents in percentage in
each pound of meat are given below:

A B C
Protein (%) 20 12 14
Fat (%) 10 18 22
Cost/lb (cents) 110 150 122

A foundry is faced with the problem of scheduling production and subcontracting of
three products. Each of the products required casting, machining and assembly
operations. Casting operations for products 1 and 2 could be subcontracted but the
casting for product 3 required special equipment which precludes the use of
subcontractors. If cast in the foundry, each of product 1 requires 6 minutes of casting
time and product 3 requires 8 minutes. Machining times required per unit of products
1, 2 and 3 are 6, 3 and 8 minutes respectively and the assembly times are 3, 2 and 2
minutes respectively.
The foundry has available 8000 minutes of casting time; 12000 minutes of machining
time and 10000 minutes of assembly time per week. The overall profit obtained from
unit of products 1, 2 and 3 sold are Rs.7, Rs.10 and Rs.11 respectively with castings
produced in the foundry. With castings obtained from subcontractors, the profits per
unit of products 1 and 2 are Rs.5 and Rs.9 respectively. There is a restriction that the
foundry must supply at least 3000 units of products 1, 2 and 3 taken together, per
week. How should the foundry schedule its production and subcontracting so as to
maximise its profits?

XYZ Chemicals Ltd. must produce 10000 kg of a
special mixture for a customer. The mix consists
of ingredients X1, X2 and X3. X1 costs Rs.8 per kg, X2
costs Rs.10 per kg and X3 costs Rs.11 per kg. No
more than 3500 kg of X1 can be used and at least
1500 kg of X2 must be used. Also, at least 2000 kg
of X3 is required.
1.Calculate the number of kg for each ingredient
to use in order to minimise total cost for 10000
kg.
2.Calculate the lowest possible total cost.

Unit ii-1-lp

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Unit ii-1-lp