Chapter 2 An Introduction to Linear Programming

Instructor Dr. Mohamed Mostafa

Overview

- Linear Programming Problem
- Problem Formulation
- A Simple Maximization Problem
- Graphical Solution Procedure
- Extreme Points and the Optimal Solution
- A Simple Minimization Problem
- Special Cases

2

Linear Programming

- Linear programming has nothing to do with

computer programming. - The use of the word programming here means

choosing a course of action. - Linear programming is a problem-solving approach

developed to help managers make decisions.

Linear Programming (LP) Problem

- The maximization or minimization of some quantity

is the objective in all linear programming

problems. - All LP problems have
- Constraints that limit the objective function

value. - feasible solution satisfies all the problem's

constraints. - optimal solution is the largest possible

objective function value when maximizing (or

smallest when minimizing). - A graphical solution method can be used to solve

a linear program with two variables.

Linear Programming (LP) Problem

- If both the objective function and the

constraints are linear, the problem is referred

to as a linear programming problem. - Linear functions are functions in which each

variable appears in a separate term raised to the

first power and is multiplied by a constant

(which could be 0). - Linear constraints are linear functions that are

restricted to be "less than or equal to", "equal

to", or "greater than or equal to" a constant.

Which are NOT linear functions?

- -2x1 4x2 1x3 lt 80
- 3vx1 - 2x2 15
- 2x1x2 5x3 lt 17
- 4x1 7x3 12

6

Problem Formulation

- Problem formulation or modeling is the process of

translating a verbal statement of a problem into

a mathematical statement. - Formulating models is an art that can only be

mastered with practice and experience. - Every LP problem has some unique features, but

most problems also have common features. - General guidelines for LP model formulation are

illustrated on the slides that follow.

Guidelines for Model Formulation

- Read and Understand the problem.
- Describe the objective.
- Describe each constraint.
- Define the decision variables.
- Write the objective in terms of the decision

variables. - Write the constraints in terms of the decision

variables.

Problem Statement

- A Starbucks wants to maximize hourly profit on

sales of lattes and cappuccinos. They make 5

per latte and 7 per cappuccino. - In any given hour,
- The latte frother can blend up to 6 cups per

hour. - The maximum milk supply in each hour is 19 cups.

Lattes require 2 cups, and cappuccinos take 3. - The lid station can provide a max of 8 lids per

hour. Each latte and cappuccino must have a lid. - How can Starbucks maximize profit in each hour on

sales of lattes and cappuccinos?

9

Objective Function

Max 5x1 7x2 s.t. x1

lt 6 2x1 3x2 lt 19

x1 x2 lt 8 x1 gt 0 and x2 gt 0

Regular Constraints

Non-negativity Constraints

Graphical Solution

- Prepare a graph of the feasible solutions for

each of the constraints. - Determine the feasible region that satisfies all

the constraints simultaneously. - Type of constraint Feasible region (usually)

will be - gt above/to the right
- lt below/to the left
- the line
- Draw an objective function line.
- Move parallel objective function lines toward

larger objective function values without entirely

leaving the feasible region. - Any feasible solution on the objective function

line with the largest value is an optimal

solution.

11

Starbucks Graphical Solution

x2

8 7 6 5 4 3 2 1

x1

1 2 3 4 5 6 7

8 9 10

Starbucks Graphical Solution

- First Constraint Graphed

x2

8 7 6 5 4 3 2 1

x1 6

Shaded region contains all feasible points for

this constraint

(6, 0)

x1

1 2 3 4 5 6 7

8 9 10

- Second Constraint Graphed

x2

8 7 6 5 4 3 2 1

(0, 6 1/3)

2x1 3x2 19

Shaded region contains all feasible points for

this constraint

(9 1/2, 0)

x1

1 2 3 4 5 6 7

8 9 10

- Third Constraint Graphed

x2

(0, 8)

8 7 6 5 4 3 2 1

x1 x2 8

Shaded region contains all feasible points for

this constraint

(8, 0)

x1

1 2 3 4 5 6 7

8 9 10

- Combined-Constraint Graph
- Showing Feasible Region

x2

x1 x2 8

8 7 6 5 4 3 2 1

x1 6

2x1 3x2 19

Feasible Region

x1

1 2 3 4 5 6 7

8 9 10

- Objective Function Line

x2

8 7 6 5 4 3 2 1

(0, 5)

Objective Function 5x1 7x2 35

(7, 0)

x1

1 2 3 4 5 6 7

8 9 10

Selected Objective Function Lines

x2

8 7 6 5 4 3 2 1

5x1 7x2 35

5x1 7x2 39

5x1 7x2 42

x1

1 2 3 4 5 6 7

8 9 10

- Optimal Solution

x2

Maximum Objective Function Line 5x1 7x2 46

8 7 6 5 4 3 2 1

Optimal Solution (x1 5, x2 3)

x1

1 2 3 4 5 6 7

8 9 10

- Solve for the Extreme Point at the Intersection

of the Two Binding Constraints - 2x1 3x2 19
- x1 x2 8
- The two equations will give
- x2 3
- Substituting this into x1 x2 8 gives

x1 5

Solve for the Optimal Value of the Objective

Function 5x1 7x2 5(5) 7(3)

46

Extreme Points and the Optimal Solution

- The corners or vertices of the feasible region

are referred to as the extreme points. - An optimal solution to an LP problem can be found

at an extreme point of the feasible region. - When looking for the optimal solution, you do not

have to evaluate all feasible solution points

consider only the extreme points of the feasible

region.

Extreme Points

x2

8 7 6 5 4 3 2 1

(0, 6 1/3)

5

(5, 3)

4

Feasible Region

(6, 2)

3

(0, 0)

(6, 0)

2

1

x1

1 2 3 4 5 6 7

8 9 10

x2

Max Z 5x1 7x2

(x1 , x2 ) Z

8 7 6 5 4 3 2 1 1 2

3 4 5 6 7 8 9

10

(0, 0) 0

1

5

(6, 0) 30

2

(6, 2) 44

3

(5, 3) 46

4

(6 1/3,0) 30 5/3

4

5

Feasible Region

3

1

2

x1

23

Problem

- A woodcarving ??? ????? agency manufactures two

types of wooden toys soldiers and trains. A

soldier sells for 27 and uses 10 worth of raw

materials. Each soldier that is manufactured

increases his variable labor and overhead costs

(Examples include rent, gas, electricity, and

wages) by 14. A train sells for 21 and uses 9

worth of raw materials. Each train built

increases his variable labor and overhead costs

by 10. - The manufacture of soldier and trains requires

two types of skilled labor carpentry and

finishing. A soldier requires 2 hours of

finishing labor and 1 hour of carpentry labor. A

train requires 1 hour of finishing and 1 hour of

carpentry labor. Each week he can obtain all the

raw material he needs but only 100 hours of

finishing and 80 carpentry hours. Demand for

trains is unlimited, but at most 40 soldiers are

bought every week. - Formulate the problem that may maximize the

companys weekly profit and solve it graphically.

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Problem

- The following table summarizes the key facts

about two products, A and B, and the resources Q,

R, and S required to produce them. Formulate a

linear programming model for this problem. The

profit per unit is 3 for Product A and 2 for

Product B.

Resource Resource Usage per Unit Produced Resource Usage per Unit Produced Amount of Available Resource

Resource Product A Product B Amount of Available Resource

Q 2 1 2

R 1 2 2

S 3 3 4

Assignment

- The Primo Insurance Company is introducing two

new product lines special risk insurance and.

Mortgages ??? ????? The expected profit is 5 per

unit on special risk insurance and 2 per unit on

mortgages. Management wishes to establish sales

quotas ??? ???????? for new product lines. The

work req. are as follows. Formulate the LP model

and solve it graphically.

Department Work-hours per unit Work-hours per unit Work hours Available

Department Special Risk Mortgage Work hours Available

Underwriting ???? ????????-???????? 3 2 2,400

Admin.??????? 0 1 800

Claims ?????????-??????? 2 0 1,200

Computer Solution Windows QM

Demo version available at http//wps.prenhall.co

m/bp_weiss_software_1/0,6750,91664-,00.html

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Solving Graphically Minimization Problem

- Prepare a graph of the feasible solutions for

each of the constraints. - Determine the feasible region that satisfies all

the constraints simultaneously. - Draw an objective function line.
- Move parallel objective function lines toward

smaller objective function values without

entirely leaving the feasible region. - Any feasible solution on the objective function

line with the smallest value is an optimal

solution.

A Simple Minimization Problem

- LP Formulation

Min 5x1 2x2 s.t. 2x1 5x2 gt

10 4x1 - x2 gt 12

x1 x2 gt 4 x1, x2 gt 0

Graphical Solution

- Constraints Graphed

x2

6 5 4 3 2 1

Feasible Region

4x1 - x2 gt 12

x1 x2 gt 4

2x1 5x2 gt 10

x1

1 2 3 4 5 6

Objective Function Graphed

x2

Min 5x1 2x2

6 5 4 3 2 1

4x1 - x2 gt 12

x1 x2 gt 4

2x1 5x2 gt 10

x1

1 2 3 4 5 6

- Solve for the Extreme Point at the Intersection

of the Two Binding Constraints - 4x1 - x2 12
- x1 x2 4
- Adding these two equations gives
- 5x1 16 or x1 16/5
- Substituting this into x1 x2 4 gives

x2 4/5

Solve for the Optimal Value of the Objective

Function 5x1 2x2 5(16/5)

2(4/5) 88/5

LP in Standard Form

- A linear program in which all the variables are

non-negative and all the constraints are

equalities is said to be in standard form - To attain ????standard form you must
- lt constraints Add slack variable ????? ?????? to

constraint (coefficient of 0 in obj function) - gt constraints Subtract surplus ???? from

constraint (coefficient of 0 in obj function) - constraints Add artificial variable ?????

????????? to constraint (coefficient of M in obj

function)

- Slack and surplus variables represent the

difference between the left and right sides of

the constraints. Slack is any unused resource,

while surplus is the amount over some required

minimum level. - The objective function coefficient for slack and

surplus variables is equal to 0. - If slack/surplus variables are equal to 0 for a

constraint, the constraint is said to be binding

??????.

Slack Variables (for lt constraints)

- Max 5x1 7x2 0s1 0s2 0s3
- s.t. x1 s1 6
- 2x1 3x2 s2 19
- x1 x2 s3 8
- x1, x2 , s1 , s2 ,

s3 gt 0

s1 , s2 , and s3 are slack variables

Example in Standard Form

Optimal Solution

x2

Third Constraint x1 x2 8

First Constraint x1 6

8 7 6 5 4 3 2 1

s3 0

s1 1

Second Constraint 2x1 3x2 19

Optimal Solution (x1 5, x2 3)

s2 0

x1

1 2 3 4 5 6 7

8 9 10

Surplus Variables

Minimization Example in Standard Form

Min 5x1 2x2 0s1 0s2 0s3 s.t.

2x1 5x2 - s1 10

4x1 - x2 - s2 12

x1 x2 - s3

4 x1, x2, s1,

s2, s3 gt 0

s1 , s2 , and s3 are surplus variables

LPs Special Case Alternative Optimal Solutions

Max 4x1 6x2 s.t. x1

lt 6 2x1 3x2 lt 18

x1 x2 lt 7 x1 gt 0 and x2

gt 0

Boundary constraint 2x1 3x2 lt 18 and objective

function Max 4x1 6x2 are parallel. All points

on line segment A B are optimal solutions.

x2

x1 x2 lt 7

7 6 5 4 3 2 1

Max 4x1 6x2

A

B

x1 lt 6

2x1 3x2 lt 18

x1

1 2 3 4 5 6 7

8 9 10

Infeasibility

Max 2x1 6x2 s.t. 4x1 3x2 lt 12

2x1 x2 gt 8

x1, x2 gt 0

- There are no points that satisfy both

constraints, so there is no feasible region (and

no feasible solution).

x2

10

2x1 x2 gt 8

8

6

4x1 3x2 lt 12

4

2

x1

2 4 6 8 10

Unbounded Problem

Max 4x1 5x2 s.t. x1 x2 gt 5

3x1 x2 gt 8

x1, x2 gt 0

- The feasible region is unbounded and the

objective function line can be moved outward from

the origin without bound, infinitely increasing

the objective function.

x2

10

3x1 x2 gt 8

8

6

Max 4x1 5x2

4

x1 x2 gt 5

2

x1

2 4 6 8 10

Assignment

- The Sanders Garden Shop mixes two types of grass

seed into a blend. Each type of grass has been

rated (per pound) according to its shade

tolerance, ability to stand up to traffic, and

drought resistance, as shown in the table. Type A

seed costs 1 and Type B seed costs 2. If the

blend needs to score at least 300 points for

shade tolerance, 400 points for traffic

resistance, and 750 points for drought

resistance, how many pounds of each seed should

be in the blend? How much will the blend cost?