Notes on optimization-2

BITS PilaniK K Birla Goa Campus

aaoC zC222optimization

Dr. Anil KumarAssistant Professor, Department of Mathematics

January 11, 2013

BITS Pilani, K K Birla Goa Campus

Contents

• Formulation of Linear Programming Problems

• Solution of Linear Programming Problems

• Graphical Method

• Exceptional Cases


Three basic steps in constructing a linear programming model:

• Identify the unknown variables to be determined (decisionvariables) and represent them in terms of algebraic symbols.

• Identify all the restrictions or constraints in the problem andexpress them as linear equations or inequalities which arelinear functions of the unknown variables.

• Identify the objective or criterion and represent it as a linearfunction of the decision variables, which is to be maximized orminimized.

Formulation of LPP


A Retired person wants to invest up to anamount of Rs. 30000 in fixed incomesecurities. His broker recommends investingin two bonds: Bond A yielding 7% and Bond Byielding 10%. After some consideration, hedecides to invest at most Rs. 12000 in Bond Band at least Rs. 6000 in Bond A. He alsowants the amount invested in Bond A to be atleast equal to the amount invested in Bond B.What should the broker recommended if theinvestor wants to maximize his returns oninvestment? Formulate this problem as alinear programming problem.

Example-1


Let x1 : amount invested in Bond A and

x2 : amount invested in Bond B.

Solution


Therefore, LPP in final form

Max Z = 0.07 x1 + 0.10 x2

Subject to

x1 + x2 ≤ 30000

x1 ≥ 6000

x2 ≤ 12000

x1 –x2 ≥ 0

x1 ≥ 0, x2 ≥ 0

Example-2


Solving

Linear Programming Problems


Any values of x1, x2 that satisfy all the Constraints (mainas well as non-negativity constraints) constitute afeasible solution. Otherwise the solution is infeasible.

A feasible solution which optimizes the objective functionvalue of the given LP is called an optimum feasiblesolution.

Aim: The aim of the problem is to find the best (optimal)feasible solution. We need a systematic procedure thatwill locate the optimum solution in a finite number ofsteps.

Solution


To do that we need to know how many feasiblesolutions the problem has.

We will see that there are infinitely many solutions;which makes it impossible to solve the problem byenumeration. Instead, we need a systematic procedurethat will locate the optimum solution in a finite numberof steps.

Solution…


GRAPHICAL METHOD


The graphical procedure includes two steps:

• Determination of the feasible solution space.

– (The feasible solution space of the problemrepresents the area in the first quadrant in which allthe constraints are satisfied simultaneously.)

• Determination of the optimum solution from among allthe feasible points in the solution space.

Graphical Method…


Step 1. Determination of the feasible solution space:

Draw the variable constraints (e.g. the non negativityrestrictions on the decision variables restrict the solutionspace to the first quadrant only)

Draw the main constraints by changing the inequalitiesinto equations and graph the resulting straight lines bylocating two distinct points.

Draw an arrow in the direction of the inequality.

Solving LPP by Graphical method


Step 2: Determination of the optimum solution.

– The optimum solution lies on one of the corner pointsof the feasible solution space


Example 1

Maximize x + y

Subject to: x + 2 y 2

x 3

y 4

x 0 y 0


Solution

x30 1 2

y

0

1

2

4

3

Feasible Region

x 0 y 0

x + 2 y 2

y 4

x 3

Subject to

Maximize x + y

Optimal Solution


Example 1…

x30 1 2

y

0

1

2

4

3

Feasible Region

x 0 y 0

x + 2 y 2

y 4

x 3

Subject to:

Maximize x + y

Optimal Solution


Example 2

Minimize x + 1/3 y

Subject to: x + y 20

-2 x + 5 y 150

x 5

x 0 y 0


Solution

Feasible

Region

x 0 y 0

x + y 20

x 5

-2 x + 5 y 150

Subject to:

Minimize x + 1/3 y

Optimal Solution

x

3010 20

y

0

10

20

40

0

30

40


Determination of the optimum solution.

The optimum solution lies on one of the corner points(vertices) of the feasible region (PF).

Steps for solving the LPP by Graphical

method…


A Fundamental Point

If an optimal solution exists,

there is always a corner point

optimal solution for LPP!

y

x0

1

2

3

4

0 1 2

3

x3010 20

y

0

10

20

40

0

30

40

Example 1: Optimal Solution

Example 2: Optimal Solution


Example 3

Maximize Z = x + 5y

Subject to: -x + 3y ≤ 10

x + y 6

x - y 2

x 0, y 0


Graphical Solution

x

62 4

y

0

2

4

8

0

6

8

PF

( 0, 10/3 )

( 2, 4 )

( 4, 2 )

The Vertices are :

(0, 0), (2, 0), (4, 2), (2, 4),

(0,10/3)

Optimal Solution


The value of the objective function is computed at these are:

Z = 0 at (0, 0) Z = 2 at (2, 0) Z = 14 at (4, 2)

Z = 22 at (2 ,4) Z = 50/3 at (0, 10/3)

Obviously, the maximum occurs at vertex (2, 4) with the maximum value 22. Hence,

Optimal Solution: x = 2, y = 4, z = 22.

Solution…


Example 4

Minimize Z = x - 2y

Subject to:

-x + y ≤ 1

2x + y 2

x 0 y 0


Solution

x

1 2

y

0

1

2

0

3

3

PF

( 1/3, 4/3 )

The Vertices are :

(0, 0), (1, 0), (0, 1), (1/3, 4/3)


The value of the objective function is computed at these are:

Z = 0 at (0, 0) Z = 1 at (1, 0) Z = -7/3 at (1/3, 4/3) Z = -2 at (0, 1)

Obviously, the maximum occurs at vertex (1/3, 4/3) with the maximum value -7/3. Hence,

Optimal Solution: x = 1/3, y = 4/3, z = -7/3.

Solution…


Graphical Solution to a 2-Variable LP Exceptional Cases


• Some LPPs have an infinite number of solutions(alternative or multiple optimal solutions).

• Some LPPs have no feasible solutions (infeasible LPs).

• Some LPPs are unbounded: There are points in thefeasible region with arbitrarily large (in a maximizationproblem) z-values.


Example 1: Alternate Optimal Solution

Consider the following LPP

1 2

1 2

1 2

1 2

Max 3 2

subject to

1 11

40 60

1 11

50 50

, 0

Z x x

x x

x x

x x


Solution

Any point (solution falling on the line segment AE will yield an optimal solution of Z = 120.


Infeasibility is a condition that arises when no value of thevariable satisfy all the constraints simultaneously.

i.e.

there is no unique (single) feasible region.

Remarks:

Such a problem arises due to wrong model formulation withconflicting constraints.

Infeasibility depends strictly on the constraints and has nothingto do with the objective function.

Case 2: Infeasible Solution


Example 1

Some LPP’s have no solution. For example:

1 2

1 2

1 2

1

2

1 2

Max 3 2

subject to

1 11

40 60

1 11

50 50

30

30

, 0

Z x x

x x

x x

x

x

x x


Solution


Example 2

21

1 2

1

1 2

1 2

1 2

M aximize2

s.t. 3 2 12

5 10

8

4

, 0

xZ x

x x

x

x x

x x

x x


21

1 2

1

1 2

1 2

1 2

M aximize2

s.t. 3 2 12

5 10

8

4

, 0

xZ x

x x

x

x x

x x

x x


Sometimes an LP problem will not have a finite solution i.e.

when one or more decision variable values and the value of theobjective function (max.) are permitted to increase infinitelywithout violating the feasibility condition, then the solution issaid to be unbounded.

The general cause for an unbounded LP problem is a mistake inmathematical model formulation.

Difference between a feasible region being unbounded and anLP problem being unbounded.

Case 3: Unbounded Solution


Example

1 2

1 2

1 2

1 2

Max 2

s.t. 1

2 6

, 0

Z x x

x x

x x

x x


Solution


Plotting of each of the constraints on the graph serves todetermine the feasible region of the given LPP.

If and when a constraint, when plotted, does not formpart of the boundary marking the feasible region of theproblem, it is said to be redundant.

The inclusion or exclusion of a redundant constraintdoes not affect the optimal solution to the problem.

Redundant Constraint(s)


Consider the following LPP:

Example

1 2

1 2

1 2

1 2

1 2

max 40 35

s.t. 2 3 60

4 3 96

4 3.5 105

, 0

z x x

x x

x x

x x

x x


Solution


Thanks

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Notes on optimization-2