Linear programming

Linear Programming
Dr. Varun Kumar
Dr. Varun Kumar ( IIIT Surat) Linear Programming 1 / 15

Outlines
1 Fundamental of Operation Research
2 Introduction to Linear Programming
Problem Formulation
Generalized form of Linear Programming
Solution by Graphical Method

Introduction to operation research
Operation research
⇒ It is scientific approach for decision making.
⇒ It operates on a system under the condition requiring the allocation of
scarce resource.
⇒ Origin during world war II, when British military asked to scientists to
analyze the military problems.
⇒ Application of mathematics + Scientific approach → Operation
research

Fundamental of operation research
1 Linear Programming- Formulation
2 Linear Programming- Solutions
3 Duality and Sensitivity Analysis
4 Transportation Problem
5 Assignment Problem
6 Dynamic Programming
7 Deterministic Inventory Models

Introduction to Linear Programming
Programming problem flow chart
Linear Programming
⇒ Linear programming is a method to achieve the best outcome in a
mathematical model whose requirements are represented by linear
relationships.
⇒ Linear programming is a technique for the optimization of a linear
objective function, subject to linear equality and linear inequality
constraints.

Example 1: Maximization of the profit
Example 1
⇒ Consider a small manufacturer making two products A and B.
⇒ Two resources R1 and R2 are required for making these products.
⇒ Each unit of A required 1 unit of R1 and 3 unit of R2.
⇒ Each unit of B required 1 unit of R1 and 4 unit of R2.
⇒ Manufacturer has 20 units of R1 and 72 unit R2.
⇒ The manufacturer also makes a profit of
Rs 4 per unit of product A sold
Rs 5 per unit of product B sold

Problem formulation
⇒ Let manufacturer sold x1 unit of product A and x2 unit of product B.
⇒ Here objective function f (x1, x2) = Z is to maximize the profit.
⇒ As per the question
Rs 4 per unit of product A sold
Rs 5 per unit of product B sold
Hence, the aim of manufacturer is to maximize the profit, i.e
f (x1, x2) = Z = max{4x1 + 5x2}
Condition
x1 + x2 ≤ 20, 3x1 + 4x2 ≤ 72, x1 ≥ 0, x2 ≥ 0
Final formulated problem (maximization)
f (x1, x2) = Z = 4x1 + 5x2
s.t x1 + x2 ≤ 20, 3x1 + 4x2 ≤ 72, x1 ≥ 0, x2 ≥ 0

Terminology and essential condition
Terminology
⇒ Decision variable → x1 and x2 (As per above example)
⇒ Objective function → f (x1, x2) = Z (As per above example)
⇒ Constraints
Inequality → x1 + x2 ≤ 20, 3x1 + 4x2 ≤ 72 (As per above example)
Equality
Essential condition
⇒ Non-negative restriction
x1 ≥ 0, x2 ≥ 0
⇒ Objective function → f (x1, x2) and constraints both should be linear.

Example 2
Problem related to product manufacturing
⇒ A company makes a single product.
⇒ The estimated demand for product for next four months are 10000,
8800, 12000, 9000.
⇒ The company has regular time (RT) capacity of 8000 per month and
over-time (OT) capacity of 2000 per month.
⇒ The cost of RT is Rs 200 per unit and OT is Rs 250.
⇒ The company can carry inventory to the next month is Rs 30/unit.
⇒ The demand has to meet every month.

Problem formulation
⇒ Expected demand from 1st to 4th month → 10000, 8800, 12000,
9000.
⇒ Production cost/capacity:
Rs 200 → RT→ 8000/month
Rs 250 → OT→ 2000/month.
Inventory cost: Rs 30/unit/month
⇒ Decision variable:
Let Xj be the quantity produced in RT at jth
month.
Let Yj be the quantity produced in OT at jth
month.
Let Ij be the quantity carried at the end of month jth
.
⇒ Objective function
minimize
n
Z = 200
4
X
j=1
Xj + 250
4
X
j=1
Yj + 30
3
X
j=1
Ij
o

Continued–
⇒ Constraints
X1 + X2 = 10000 + I1 (For 1st
month)
I1 + X3 + X4 = 8800 + I2 (For 2nd
month)
I2 + X5 + X6 = 12000 + I3 (For 3rd
month)
I3 + X7 + X8 = 9000 (For 4th
month)
Xj ≤ 8000 ∀ j = {1, 2, 3, 4}
Yj ≤ 2000 ∀ j = {1, 2, 3, 4}
⇒ Non-negative restriction
Xj , Yj , Ij ≥ 0 ∀ j

Linear programming problem generalization
In general linear programming problem having n decision variable
x1, x2, ....., xn optimize the objective function. Ex- Let F is a linear
objective function. Mathematically,
F = W1x1 + W2x2 + ...... + Wnxn → Linear objective function
Linear equality/inequality constraints
C11x1 + C12x2 + ...... + C1nxn (≤ = ≥) b1
C21x1 + C22x2 + ...... + C2nxn (≤ = ≥) b2
.
.
.
Cm1x1 + Cm2x2 + ...... + Cmnxn (≤ = ≥) bm
and
x1, x2, ....., xn ≥ 0
In matrix form, above linear programming problem can be expressed
as

Continued–
Objective function
F = WT
x
Linearity equality/inequality constraints
Cx(≤ = ≥)b
⇒ W is weight vector of size n × 1.
⇒ C is matrix of size m × n
⇒ x is vector of size n × 1
⇒ b is a vector of size m × 1
Constraints on decision variable:
x 4 or < 0
⇒ 4 or < notation is used for the element-wise inequality.
⇒

Linear programming solution-
Graphical method
Q Formulated problem is as follow
f (x1, x2) = Z = 4x1 + 5x2
s.t x1 + x2 ≤ 20, 3x1 + 4x2 ≤ 72, x1 ≥ 0, x2 ≥ 0
Graphical method:

Linear programming

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