Chapter 7 Linear Programming Models: Graphical and Computer Methods.

Chapter 7

Linear Programming Models: Graphical and Computer

Methods

Chapter Outline

7.17.1 Introduction7.27.2 Requirements of a Linear Programming Problem7.37.3 Formulating LP Problems7.47.4 Graphical Solution to an LP Problem7.57.5 Solving LP Problem (QM for Window and Excel)7.67.6 Solving Minimization Problems7.77.7 Four Special Cases in LP7.87.8 Sensitivity Analysis

Objectives of business decisions frequently involve maximizing profit or minimizing costs.

Linear programming is an analytical technique in which linear algebraic relationships represent a firm’s decisions, given a business objective, and resource constraints.

Steps in application: Identify problem as solvable by linear programming. Formulate a mathematical model of the unstructured

problem. Solve the model. Implementation

Linear Programming: An Overview

Requirements Decision variables - mathematical symbols representing levels of activity of a

firm. Objective function - a linear mathematical relationship describing an objective of

the firm, in terms of decision variables - this function is to be maximized or minimized.

Constraints – requirements or restrictions placed on the firm by the operating environment, stated in linear relationships of the decision variables.

Parameters - numerical coefficients and constants used in the objective function and constraints.

Assumptions Proportionality - The rate of change (slope) of the objective function and

constraint equations is constant. Additivity - Terms in the objective function and constraint equations must be

additive. Divisibility -Decision variables can take on any fractional value and are therefore

continuous as opposed to integer in nature. Certainty - Values of all the model parameters are assumed to be known with

certainty (non-probabilistic). Non-negativity - all answers or variables are greater than or equal to (≥) zero

Requirements of LP Problems and LP Basic Assumptions

Minime Restaurant

Pho

(40K VND/bowl)

Hu tieu

(50K VND/bowl)

Chef

(works 40 min./day)

Meat

(120 grs meat/day)

I can prepare 1 bowl of:- Pho in 1 min.

- Hu tieu in 2 min.1 bowl of:- Pho needs 4 gr meat- Hu tieu neads 3 gr. meat

LP Model FormulationA Maximization Example (1 of 3)

Product mix problem – Minime Noodle Restaurant

How many bowls of Pho and Hu tieu should be prepared to maximize profits given labor and materials constraints?

Product resource requirements and unit profit:

Products Resource Requirements Profits

Labor (min./bowl)

Meat (gr./bowl)

(K. VND/bowl)

Pho 1 4 40

Hu tieu 2 3 50

ResourceAvailability

40 min. 120 grs.


Products Resource Requirements Profits

Labor (min./bowl) Meat (gr./bowl) (K. VND/bowl)

Pho 1 4 40

Hu tieu 2 3 50

Avail. Resource 40 120

Trial and error

• All for Pho:

40 bowls 30 bowls 30 bowls x 40= 1200 (chef’s time surplus)

• All for Hu tieu: 20 bowls 40 bowls 20 bowls x 50= 1000 (meat surplus)

Half for Pho and Half for Hu tieu PHO 20 bowls 15 bowls 15 bowls x 40= 600 (chef’s time surplus) HU TIEU 10 bowls 20 bowls 10 bowls x 50= 500 (meat surplus)

Time Meat Profit


Decision x1 = # of bowls of Pho to prepare per dayVariables: x2 = # of bowls of Hu tieu to prepare per day

Objective Maximize Z = 40x1 + 50x2

Function: Where Z = profit per day subject to (s.t.):

Resource 1x1 + 2x2 40 minutes of laborConstraints: 4x1 + 3x2 120 grs. of meat

Non-Negativity x1 0; x2 0 Constraints:

A feasible solution does not violate any of the constraints:

Example x1 = 5 bowls of Pho

x2 = 10 bowls of Hu tieu

Z = 40x1 + 50x2 = 700 (K.VND) Labor constraint check: 1(5) + 2(10) = 25 < 40 minutes, within constraint Meat constraint check: 4(5) + 3(10) = 70 < 120 grams, within constraint

An infeasible solution violates at least one of the constraints:

Example x1 = 10 bowls of Pho

x2 = 20 bowls of Hu tieu Z = 1400 (K.VND) Labor constraint check: 1(10) + 2(20) = 50 > 40 minutes, violates constraint

Feasible and Infeasible Solutions

Graphical solution is limited to linear programming models containing only two decision variables (can be used with three variables but only with great difficulty).

Graphical methods provide visualization of how a solution for a linear programming problem is obtained.

Graphical Solution of LP Models

Coordinate AxesGraphical Solution of Maximization Model (1 of 11)

Maximize Z = 40x1 + 50x2

subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.2 Coordinates for Graphical Analysis

Labor ConstraintGraphical Solution of Maximization Model (2 of 11)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.3 Graph of Labor Constraint

Labor Constraint AreaGraphical Solution of Maximization Model (3 of 11)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.4 Labor Constraint Area

Meat Constraint AreaGraphical Solution of Maximization Model (4 of 11)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.5 Clay Constraint Area

Both ConstraintsGraphical Solution of Maximization Model (5 of 11)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.6 Graph of Both Model Constraints

Feasible Solution AreaGraphical Solution of Maximization Model (6 of 11)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.7 Feasible Solution Area

Graphical Solution Methods

ISOPROFIT METHOD1. Graph all constraints and find the feasible region.

2. Select a specific profit (or cost) line and graph it to find the slope.

3. Move the objective function line in the direction of increasing profit (or decreasing cost) while maintaining the slope. The last point it touches in the feasible region is the optimal solution.

4. Find the values of the decision variables at this last point and compute the profit (or cost).

CORNER POINT METHOD1. Graph all constraints and find the feasible region.

2. Find the corner points of the feasible reason.

3. Compute the profit (or cost) at each of the feasible corner points.

4. select the corner point with the best value of the objective function found in Step 3. This is the optimal solution.

Isoprofit Line MethodGraphical Solution of Maximization Model (7 of 11)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.8 Objection Function Line for Z = $800

Construct the isoprofit line:

Z = 40x1 + 50x2

x2 = Z/50 - 40/50x1

= Z/50 -4/5x1



subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.9 Alternative Objective Function Lines



subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.10 Identification of Optimal Solution

Optimal Solution CoordinatesGraphical Solution of Maximization Model (10 of 11)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.11 Optimal Solution Coordinates

Extreme (Corner) Point MethodGraphical Solution of Maximization Model (11 of 11)


subject to: 1x1 + 2x2 40 4x1 + 3x2 120

x1, x2 0

Figure 2.12 Solutions at All Corner Points

Using Excel’s Solver to Solve LP Problems

The Solver tool in Excel can be used to find solutions to LP problems Integer programming problems Noninteger programming problems

Solver may be sensitive to the initial values it uses Solver is limited to 200 variables and 100

constraints It can be used for small real world problems Add-ins like What’s Best! Can be used to expand

Solver’s capabilities

Click on “Tools” to invoke “Solver.”

Objective function

Decision variables – Pho(x1)=B10; Hu tieu (x2)=B11

=C6*B10+D6*B11

=C7*B10+D7*B11

=E6-F6

=E7-F7

Maximize Z = 40 x1 + 50 x2

Subject to

x1 + 2x240 mins. (labor constraint)

4x1 + 3x2120 grs. (meat constraint)

x1 , x2 0


After all parameters and constraints have been input, click on “Solve.”

Objective function

Decision variables

C6*B10+D6*B11≤40

C7*B10+D7*B11≤120

Click on “Add” to insert constraints


Using QM for Windows and Excel Solving Flair Furniture’s LP Problem

Most organizations have access to software to solve big LP problems

While there are differences between software implementations, the approach each takes towards handling LP is basically the same

Once you are experienced in dealing with computerized LP algorithms, you can easily adjust to minor changes

Maximize profit = $70T + $50C

subject to

4T + 3C ≤240 (carpentry constraint)2T + 1C ≤100 (painting and varnishing constraint)T, C ≥ 0 (nonnegativity constraint)

Using QM for Windows

First select the Linear Programming module Specify the number of constraints (non-negativity is assumed) Specify the number of decision variables Specify whether the objective is to be maximized or minimized For the Flair Furniture problem there are two constraints, two

decision variables, and the objective is to maximize profit

Computer screen for input of data


Computer screen for input of data

Computer screen for output of solution


Graphical output of solution

LP Model FormulationA Minimization Example (1 of 6)

Chemical Contribution

Brand Nitrogen (lb/bag)

Phosphate (lb/bag)

Super-gro 2 4

Crop-quick 4 3

Two brands of fertilizer available - Super-Gro, Crop-Quick.

Field requires at least 16 pounds of nitrogen and 24 pounds of phosphate.

Super-Gro costs $6 per bag, Crop-Quick $3 per bag. Problem: How much of each brand to purchase to

minimize total cost of fertilizer given following data ?

Fertilizing Farmer’s Field

LP Model FormulationA Minimization Example (2 of 6)

Decision Variables: x1 = bags of Super-Grox2 = bags of Crop-Quick

The Objective Function:Minimize Z = 6x1 + 3x2

Model Constraints:2x1 + 4x2 16 lb (nitrogen constraint)4x1 + 3x2 24 lb (phosphate constraint)x1, x2 0 (non-negativity constraint)

LP Model Formulation and Constraint GraphA Minimization Example (3 of 6)

Minimize Z = $6x1 + $3x2

subject to: 2x1 + 4x2 16 4x1 + 3x2 24

x1, x2 0

Figure 2.16 Graph of Both Model Constraints

Feasible Solution AreaA Minimization Example (4 of 6)


subject to: 2x1 + 4x2 16 4x1 + 3x2 24

x1, x2 0

Figure 2.17 Feasible Solution Area

Optimal Solution PointA Minimization Example (5 of 6)


subject to: 2x1 + 4x2 16 4x1 + 3x2 24

x1, x2 0

Figure 2.18 Optimum Solution Point

Construct the objective line:

Z = 6x1 + 3x2

x2 = Z/3 – 6/3x1

x2 = Z/3 -2x1

Figure 2.19 Graph of Fertilizer Example

Graphical SolutionsA Minimization Example (6 of 6)


subject to: 2x1 + 4x2 16 4x1 + 3x2 24

x1, x2 0

Four Special Cases in LP

Four special cases and difficulties arise at times when using the graphical approach to solving LP problems

Infeasibility Unboundedness Redundancy Alternate Optimal Solutions

Four Special Cases in LP Infeasibility-A problem with no feasible solution

Exists when there is no solution to the problem that satisfies all the constraint equations

No feasible solution region exists

8 –

–

6 –

–

4 –

–

2 –

–

0 –

X2

| | | | | | | | | |

2 4 6 8 X1

Region Satisfying First Two ConstraintsRegion Satisfying First Two Constraints

Region Satisfying Third Constraint

Four Special Cases in LP Unboundedness- A solution region unbounded to the right

In a graphical solution, the feasible region will be open ended

This usually means the problem has been formulated improperly

15 –

10 –

5 –

0 –

X2

| | | | |

5 10 15 X1

Feasible Region

X1 ≥ 5

X2 ≤ 10

X1 + 2X2 ≥ 15

Four Special Cases in LP Redundancy- A problem with a redundant constraint

A redundant constraint is one that does not affect the feasible solution region

30 –

25 –

20 –

15 –

10 –

5 –

0 –

X2

| | | | | |

5 10 15 20 25 30 X1

Redundant Constraint

Feasible Region

X1 ≤ 25

2X1 + X2 ≤ 30

X1 + X2 ≤ 20

Four Special Cases in LP Alternate optimal solutions

Occasionally two or more optimal solutions may exist Graphically this occurs when the objective function’s

isoprofit or isocost line runs perfectly parallel to one of the constraints8 –

7 –

6 –

5 –

4 –

3 –

2 –

1 –

0 –

X2

| | | | | | | |

1 2 3 4 5 6 7 8 X1

Feasible Region

Isoprofit Line for $8

Optimal Solution Consists of All Combinations of X1 and X2 Along the AB Segment

Isoprofit Line for $12 Overlays Line Segment AB

B

A

Sensitivity Analysis

Sensitivity analysis is used to determine effects on the optimal solution within specified ranges for the objective function coefficients, constraint coefficients, and right hand side (RHS) values.

Sensitivity analysis provides answers to certain what-if questions.

Range of Optimality A range of optimality of an objective function coefficient is

found by determining an interval for the objective function coefficient in which the original optimal solution remains optimal while keeping all other data of the problem constant. The value of the objective function may change in this range.

Graphically, the limits of a range of optimality are found by changing the slope of the objective function line within the limits of the slopes of the binding constraint lines. (This would also apply to simultaneous changes in the objective coefficients.)

The slope of an objective function line, Max c1x1 + c2x2, is

-c1/c2

The slope of a constraint, a1x1 + a2x2 = b, is -a1/a2

Shadow Price

A shadow price for a RHS value (or resource limit) is the amount the objective function will change per unit increase in the right hand side value of a constraint.

Graphically, a shadow price is determined by adding +1 to the right hand side value in question and then resolving for the optimal solution in terms of the same two binding constraints.

The shadow price is equal to the difference in the values of the objective functions between the new and original problems.

The shadow price for a nonbinding constraint is 0.

Solve graphically for the optimal solution:

Max z = 5x1 + 7x2

s.t. x1 < 6

2x1 + 3x2 < 19

x1 + x2 < 8

x1, x2 > 0

Example: Sensitivity Analysis

Graphical Solution

8

7

6

5

4

3

2

1

1 2 3 4 5 6 7 8 9 10

22xx11 + 3 + 3xx22 << 19 19

xx22

xx11

xx11 + + xx22 << 8 8Max 5x1 + 7x2Max 5x1 + 7x2

xx11 << 6 6

Optimal Optimal xx11 = 5, = 5, xx22 = 3 = 3 zz = 46 = 46


Example: Sensitivity Analysis Range of Optimality for c1

The slope of the objective function line is -c1/c2. The slope of the 1st binding constraint, x1 + x2 = 8, is -1, the slope of the 2nd binding constraint, 2x1 + 3x2 = 19, is -2/3.

Find the range of values for c1 (with c2 staying 7) such that

the objective function line slope lies between that of the two binding constraints: -1 < -c1/7 < -2/3

Multiplying through by -7 (and reversing the inequalities):

14/3 < c1 < 7


Range of Optimality for c2

Find the range of values for c2 ( with c1 staying 5) such that the objective function line slope lies between that of the two binding constraints:

-1 < -5/c2 < -2/3

Multiplying by -1: 1 > 5/c2 > 2/3

Inverting, 1 < c2/5 < 3/2

Multiplying by 5: 5 < c2 < 15/2

Example: Sensitivity Analysis Shadow Prices

Constraint 1: Since x1 < 6 is not a binding constraint, its shadow price is 0.

Constraint 2: Change the RHS value of the 2nd constraint to 20 and resolve for the optimal point determined by the last two constraints: 2x1 + 3x2 = 20 and x1 + x2 = 8.

The solution is x1 = 4, x2 = 4, z = 48.

Hence, the shadow price = znew - zold = 48 - 46 = 2.

Constraint 3: Change the RHS value of the 3rd constraint to 9 and resolve for the optimal point determined by the last two constraints: 2x1 + 3x2 = 19 and x1 + x2 = 9.

The solution is: x1 = 8, x2 = 1, z = 47.

Hence, the shadow price = znew - zold = 47 - 46 = 1.

Homework 06

7.15, 7.16, 7.20, 7.22, 7.24, 7.26, 7.28

Chapter 7 Linear Programming Models: Graphical and Computer Methods.

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Transcript of Chapter 7 Linear Programming Models: Graphical and Computer Methods.