LINEAR PROGRAMMING

Presented By –

Meenakshi Tripathi

Linear Programming• Linear programming (LP, or linear optimization) is a problem of

maximizing or minimizing a linear function (objective function) in presence of linear

inequality and/or equality constraints.• Standard & Canonical Form

• Objective function & Constraints • Corner Point : x=(x1, x2,……,xn) is a vertex iff columns of Ai are Linearly

Independent rank(A)=n and xi≠0. Also called “ Basic feasible solution”

• Surplus/Slack Variable: Used to transform and inequality into equality eg.

1) as

2) as

BASIC and BASIC FEASIBLE SOLUTION

• System , . If rank(A)=rank(A,b)=m then

.

Then

• If xB≥0 then x is “Basic feasible solution”, xB are basic variables and xN nonbasic matrix.

• Extreme Points: A point in a convex set is called an extreme point of X, if x cannot be represented as a strict convex combination of two distinct points in X. i.e. If

• Bounded Set: A set is bounded if

• Half-spaces:

• Polyhedral Set/Polyhedron: Intersection of finite-number

of half-spaces. A bounded polyhedral set is called “Polytope”

• Polytope:

x1

x2

x3

Geometric Solution• Interchanging Maximization & Minimization Problems:

• Types of Geometric Solution:

1) Unique Optimal Solution- unique optimal solution

2) Alternative Optimal Solutions- optimal solution set is unbounded

3) Unbounded Optimal Objective Value – both feasible region & optimal value unbounded, hence no optimal solution exists

4) Empty Feasible Region – inconsistent system of equations

Definitions• For positive variables since they represent physical quantities, if unrestricted sign

variable j, replace it with x’j – x”j, where x’j ≥ 0 and x”j ≥ 0.

• Work from vertex to vertex of the polyhedron and in each step improve the objective function value.

• Degeneracy: A LP is degenerate if s.t. more than ‘n’ constraints of that are active at x*

• Optimal Bases: Basis B is optimal if it is feasible & unique with

=0, i B . Then x*=A-1BbB is optimal solution of LP.

• Adjacent Vertices: Vertices x1 & x2 of } are adjacent if (n-1) Linearly Independent inequalities active at both x1 & x2

x* - 3 constraints active

SIMPLEX ALGORITHM

Basic Idea:

Start with vertex x*

While x* is not optimal

Find vertex x’ adjacent to x* with *

update x*:=x’

or assert that LP is unbounded

SIMPLEX ALGORITHM: Basis notation

Basic Idea:

Start with feasible basis B

While B is not optimal

Let iB be index with i<0

Compute with aTj=0, j B\{i} & aT

j d=-1

Determine

if K= assert LP is unbounded

else

Let kK index where is attained

Update B:=B\{i}{k}

SIMPLEX • Standard Maximization form of Objective function, constraints in form of less than or

equal to equations, RHS values always positive

• Table constructed with basic variables, coefficients of variables, RHS constants & ratio columns

•

• New Pivot =(old/leaving new element)/key element

• Ratio= constant/entry column coefficient

• Old element (OE)=OE-(starting column coefficient corresponding new pivot element)

Example• Z=30x1+20x2

s.t. 2x1+x2+s1=100

• X1+x2+s2=80

• X1+s3=40

• X1,x2,s1,s2,s3≥0

Iteration 1: column = minimum negative entry = -30; Row : min{ 100/2, 80/1,40/1}=40 => s3 leaving, x1 entering

BV z x1 x2 s1 s2 s3 B

Z 1 -30 -20 0 0 0 0

S1 0 2 1 1 0 0 100

S2 0 1 1 0 1 0 80

S3 0 1 0 0 0 1 40

BV z x1 x2 s1 s2 s3 B

Z 1 0 -20 0 0 30 1200

S1 0 0 1 1 0 -2 20

S2 0 0 1 0 1 -1 40

x1 0 1 0 0 0 1 40

Example

BV z x1 x2 s1 s2 s3 B

Z 1 0 0 20 0 -10 1600

x2 0 0 1 1 0 -2 20

S2 0 0 0 -1 1 1 20

x1 0 1 0 0 0 1 40

Iteration 2: column = minimum negative entry = -20; Row : min{ 20/1, 40/1,-}=20 => s1 leaving, x2 entering

Iteration 3: column = minimum negative entry = -10; Row : min{-, 20/1, 40/1,-}=20 => s2 leaving, s3 entering

BV z x1 x2 s1 s2 s3 B

Z 1 0 0 10 10 0 1800

x2 0 0 1 -1 2 0 60

S3 0 0 0 -1 1 1 20

x1 0 1 0 0 -1 1 20

All nonbasic variables with non-negative coefficients in row 0, Optimal solution : x1=20, x2=60 & Z=1800

PRIMAL-DUAL• With every LP associated LP called dual

Dual : or

Primal : Dual:

Theorems:• Dual of dual is primal

• Weak Duality: LP & its dual . If x* & y* are primal & dual feasible respectively, then cTx* bTy*

• Strong Duality: LP & its dual . If primal is feasible and bounded then primal feasible x* & dual feasible y* respectively, then cTx* =bTy*

Primal Dual • Problem 1: Maximize Z

Z=3x1+4x2

4x1+2x280

3x1+5x2 180

• Dual of Problem1: Minimize Z

Z=80y1+180y2

4y1+3y2≥3

2y1+5y2 ≥4

Reference• Linear Programming and network flows, M.S. Bazaraa,

J.J.Jarvis & H.D.Sherali• Coursera lectures,”Linear Optimization”• Wikipedia