QMB 4701 MANAGERIAL OPERATIONS

ANALYSIS

Jay Marshall TeetsDecision and Information Sciences

University of Florida

Chapter 3Linear Programming: Graphical Solution

Procedure & Spreadsheet Solution Procedure

Agenda• Graphical linear programming

– Example: Par, Inc. Problem – Graphical solution procedure– Example: Dorian problem

• Additional considerations– Degeneracy– Inconsistent problem – Multiple optimal solutions– Unbounded problem

• Spreadsheet solution basics

Example: Par, Inc. Problem

Par, Inc: Manufacturer of golf supplies

New products: Standard golf bags and Deluxe golf bags

Manufacturing operations:

Objective: Maximize total profit

Production Time (Hrs)Standard Bag Deluxe Bag

ResourceConstraints (Hrs)

Cutting &Dyeing 7/10 1 630Sewing 1/2 5/6 600

Finishing 1 2/3 708Inspection

& Packaging 1/10 1/4 135

ProfitContribution ($)

10 9

Cutting & Dyeing

Sewing

Finishing

Inspection & Packaging

Par Inc. Model Formulation

1. What is the objective in words?

2. What are the constraints in words?

Maximize the total profit

Cutting & Dyeing Cutting & Dyeing Hours Allocated Hours Available

Sewing Hours Allocated Sewing Hours Available

Finishing Hours Allocated Finishing Hours Available

Inspection & Packaging Inspection & Packaging Hours Allocated Hours Available

Par Inc. Model Formulation

3. What are the decision variables?

4. Formulate the objective function:

5. Formulate the constraints:

6. Do we need non-negative constraints?

x1 = Number of Standards Producedx2 = Number of Deluxes Produced

Profit Contribution Maximize 10 x1 + 9 x2

Subject to:Cutting & Dyeing: 7/10 x1 + x2 630Sewing 1/2 x1 + 5/6 x2 600Finishing x1 + 2/3 x2 708Inspection & Packaging 1/10 x1 + 1/4 x2 135

Logic x1 0 , x2 0

How to Solve It?Profit Contribution Maximize 10 x1 + 9 x2

Subject to:Cutting & Dyeing: 7/10 x1 + x2 630Sewing 1/2 x1 + 5/6 x2 600Finishing x1 + 2/3 x2 708Inspection & Packaging 1/10 x1 + 1/4 x2 135

Logic x1 0 , x2 0

Trial and Error?

Feasible Solution

Infeasible Solution

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x1 = 600,x2 = 100?

Profit = 6900

520383.33666.67

85

x1 = 700,x2 = 100?

Profit =7900

590433.33766.67

95

Graphical Approach

Solutions

Solution: any production plan

Feasible solution:solution that satisfies all constraintsFeasible solutions

A. Determine feasible region(draw constraint lines)

Optimal solution:feasible solution that achievesthe maximal objective value

B. Construct isoprofit lines

C. Locate the optimal solutionOptimal solutions

Constraints Determine Feasible Region

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Approach: 1. Draw lines representing the constraints solved as equalities2. Show the direction of feasibility3. The feasible region is the set of values which simultaneously satisfies all the constraints

Cutting & Dyeing 7/10 x1 + x2 630Sewing 1/2 x1 + 5/6 x2 600Finishing x1 + 2/3 x2 708Inspection & Packaging 1/10 x1 + 1/4 x2 135 x1 0, x2 0,

A. Constraints Determine Feasible Region

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A. Constraints Determine Feasible Region

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Feasible Region

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Redundantconstraint

Construct Isoprofit Lines Approach:1. Select a constant C2. Draw the “profit line” 10 x1 + 9 x2 = C3. Construct parallel profit lines

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4500

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Slope of the objective function: x2= -10/9 x1+C/9

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4500

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Locate The Optimal Solution - Observations

Optimal Solution

Observations:1. The optimal solution cannot fall in the interior of the feasible region2. The optimal solution must occur at a “corner point” of the feasible region3. In searching for the optimal solution we only have to examine the corner points

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F

C&D

I &P

Locate The Optimal Solution Find extreme points (corner points) and calculate objective valuesExtreme Points: the vertices or “corner points” of the feasible region. With two variables, extreme points are determined by the intersection of the constraint lines.

Point Constraint Constraint x1 x2 Obj. 1 Non-negative Non-negative 0 0 0 2 Non-negative Finishing 708 0 77080 3 Cutting & Dyeing Finishing 540 252 7668 4 Cutting & Dyeing Inspection

& Packaging 300 420 6780

5 Non-negative Inspection & Packaging

0 540 4860

Locate The Optimal Solution - Find Extreme Points

Example: Find extreme point #3:

Intersection of the cutting & dyeing constraint and the finishing constraint:

Cutting & Dyeing: 7/10 x1 + x2 = 630

Finishing x1 + 2/3 x2 = 708

Solve the two equations for x1 and x2:

7/10 x1 + x2 = 630 7/10 x1 = 630 x2 x1 = 900 10/7 x2

x1 + 2/3 x2 = 708 (900 10/7 x2) +2/3 x2 = 708

900 - 30/21 x2 + 14/21 x2 = 708 16/21 x2 = 192 x2 = 21/16 192 = 252

x1 = 900 - 10/7 x2 x1 = 900 10/7 (252) x1 = 900 360 = 540

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C&D

I &P

Slack and Binding Constraints

At optimal solution (540, 252) Hrs used Hrs availableC&D: 630 = 630S: 480 < 600F: 708 = 708I&P: 117 < 135

BindingSlack = 600 480 = 120BindingSlack = 135 117 = 18

Binding constraints

Slack constraint

Summary of Graphical Approach

A. Graph Feasible Region– Draw each constraint line

• For each constraint, plot points on the x1 axis (x2=0) and x2 axis (x2=0) and connect the points with a line.

– Indicate the feasible area with arrow• Check if (0,0) is include, if so, draw arrow towards (0,0). If

not, draw arrow away from (0,0). If (0,0) is on the constraint line, you must choose another point to use as a check

– Hatch the area in common for all constraints and their respective arrows

B. Find Extreme Points– Set the intersecting constraints equal to one another - basic algebra

Summary of Graphical Approach

C. Find optimal solution

Option #1

– List all the extreme points and calculate objective function values for each extreme point.

– Choose the point with the highest (profit) or lowest (cost) value

Option #2

– Find the slope of the objective function

– Choose which extreme point is last touched by the isoprofit/isocost line as it is increased/decreased. This point is the optimal point

– Calculate the objective value for the optimal point.

Note: Simplex Method

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1. Start with a feasible corner point solution2. Check to see if a feasible neighboring corner point is better3. If not, stop; otherwise move to that better neighbor and return to step 2.

Graphing Example: Minimization Problem

MIN 50,000 x1 + 100,000 x2STHIW: 7x1 + 2 x2 28HIM: 2x1 + 12 x2 24

x1 0, x2 0

Dorian Problem

Extreme Points x1 x2 OBJ1 0 14 14002 3.6 1.4 3203 12 0 600

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HIW

HIM

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Additional Consideration: DegeneracyDegeneracy: Assume Inspection & Packaging time is 117 hours (instead of 135)

Profit Contribution Maximize 10 x1 + 9 x2

STCutting & Dyeing: 7/10 x1 + x2 630Sewing 1/2 x1 + 5/6 x2 600Finishing x1 + 2/3 x2 708Inspection & Packaging 1/10 x1 + 1/4 x2 117Logic x1 0 , x2 0

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Additional Consideration: Inconsistent Problem

Assume there is an additional constraint: need to produce at least 725 standard bags, i.e., x1 725 Profit Contribution Maximize 10 x1 + 9 x2

STCutting & Dyeing: 7/10 x1 + x2 630Sewing 1/2 x1 + 5/6 x2 600Finishing x1 + 2/3 x2 708Inspection & Packaging 1/10 x1 + 1/4 x2 135

x1 725Logic x1 0 , x2 0

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Additional Consideration: Multiple Optimal Solutions

Assume the objective function is 12 x1 + 8 x2 (instead of 10 x1 + 9 x2) Profit Contribution Maximize 12 x1 + 8 x2

STCutting & Dyeing: 7/10 x1 + x2 630Sewing 1/2 x1 + 5/6 x2 600Finishing x1 + 2/3 x2 708Inspection & Packaging 1/10 x1 + 1/4 x2 135Logic x1 0 , x2 0

Slope of the new objective function:x2 =-12/8 x1 +C/8= =-3/2 x1 +C/8

Additional Consideration: Unbounded Problem

Profit Contribution Maximize 10 x1 + 9 x2

STDemand: x1 + x2 1000

x1 725Logic x1 0 , x2 0

Assume we have the following problem:

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Excel Solver Solution Procedure

• Excel can do the following:– Solve linear programs Note: Activate the “Assume Linear Model”

option if your problem is a LINEAR program, otherwise the Solver will treat it as a NONLINEAR program.

– Solve nonlinear programs– Solve integer programs– Perform sensitivity analysis

Implementing an LP Model in a Spreadsheet

• Organize the data for the model on the spreadsheet including:

– Objective Coefficients

– Constraint Coefficients

– Right-Hand-Side (RHS) values

• clearly label data and information

• visualize a logical layout

• row and column structures of the data are a good start

Implementing an LP Model in a Spreadsheet

• Reserve separate cells to represent each decision variable – arrange them such that the structure parallels the input

data – keep them together in the same place – Right-Hand-Side (RHS) values

• Create a formula in a cell that corresponds to the Objective Function

• For each constraint, create a formula that corresponds to the left-hand-side (LHS) of the constraint

• NOTE: Once you’ve formulated a spreadsheet model for a certain type of application, it’s useful to save it as a template for future use!

Solver Terminology• Target cell

– Represents the objective function– Must indicate max or min

• Changing cells– Represents the decision variables

• Constraint cells– the cells in the spreadsheet that represent the LHS formulas of the

constraints in the model – the cells in the spreadsheet that represent the RHS values of the

constraints in the model – TIP: If you scale the constraints such that they are all "" or all

"", then it's easier to input the constraints as a block. Then, you avoid having to enter each constraint as a separate line.

Solver Example: Par, Inc. ModelA B C D E F

12 Operation Standard Bag Deluxe Bag Hrs Used Hrs Available3 Cutting & Dyeing 0.7 1 0 <= 6304 Sewing 0.5 0.83333333 0 <= 6005 Finishing 1 0.66666667 0 <= 7086 Inspection & Packaging 0.1 0.25 0 <= 13578 Total Profit9 Profit per Unit 10 9 0

1011 No. of Units Produced 0 0

Production Time (hrs/unit)

A B C D E F12 Operation Standard Bag Deluxe Bag Hrs Used Hrs Available3 Cutting & Dyeing 0.7 1 =B3*$B$11+C3*$C$11 <= 6304 Sewing 0.5 =5/6 =B4*$B$11+C4*$C$11 <= 6005 Finishing 1 =2/3 =B5*$B$11+C5*$C$11 <= 7086 Inspection & Packaging 0.1 0.25 =B6*$B$11+C6*$C$11 <= 13578 Total Profit9 Profit per Unit 10 9 =B9*B11+C9*C111011 No. of Units Produced 0 0

Production Time (hrs/unit)

Par, Inc. Model - Answer Report

Solver Parameters: Objective: MAX F9 Variables: B11:C11 Constraints: D3:D6 <= F3:F6 Options: Assume Linear Model Assume Non-Negative

Worksheet: [LP.xls]ParReport Created: 1/27/00 11:09:15 PM

Target Cell (Max)Cell Name Original Value Final Value

$F$9 Profit per Unit Total Profit 0 7668

Adjustable CellsCell Name Original Value Final Value

$B$11 No. of Untis Produced Standard Bag 0 540$C$11 No. of Untis Produced Deluxe Bag 0 252

ConstraintsCell Name Cell Value Formula Status Slack

$D$3 Cutting & Dyeing Hrs Used 630 $D$3<=$F$3 Binding 0$D$4 Sewing Hrs Used 480 $D$4<=$F$4 Not Binding 120$D$5 Finishing Hrs Used 708 $D$5<=$F$5 Binding 0$D$6 Inspection & Packaging Hrs Used 117 $D$6<=$F$6 Not Binding 18

Optimal objective value

Optimal solution

Constraints Binding & Slack

Information

Summary of Excel Solver Procedure

LHS Inequality RHS=>=<=

Objective Function

Right Hand Side Coefficients

Formulas

Decision Variables

Coefficients of Decision Variables

Constraints Left Hand Side Coefficients

Solver Example: Minimization ExampleMIN 50,000 x1 + 100,000 x2

ST7x1 + 2 x2 282x1 + 12 x2 24x1 0, x2 0

Dorian Problem

A B C D E F12 Comedy Ads Football Ads LHS Requirement3 High-income women 7 2 0 >= 284 High-income men 2 12 0 >= 2456 Total Cost7 Cost per minute($M) 50 100 089 # of Ads paid 0 0

Population (M) reached by Ads

Solver Example: Minimization ExampleA B C D E F

12 Commedy Ads Football Ads LHS Requirement3 High-income women 7 2 =SUMPRODUCT(B3:C3,$B$9:$C$9) >= 284 High-income men 2 12 =SUMPRODUCT(B4:C4,$B$9:$C$9) >= 2456 Total Cost7 Cost per minute($M) 50 100 =SUMPRODUCT(B7:C7,B9:C9)89 # of Ads paid 0 0

Population (M) reached by Ads

Summary of Additional Considerations

ADDITIONAL CONSIDERATIONS

GRAPHICAL EXCEL SOLVER

MANAGERIAL IMPLICATION

Degeneracy More than two constraint lines intersect at one

point

Inconsistent Problem Feasible region does not exist

Could not find a feasible solution

Too many restrictions

Multiple Optimal Solutions

Slope of objective

function is same as that of one

constraint

Output shows only a single

optimal solution

Unbounded Problem Optimal value is infinite

The set of cell values do not

converge

Problem is improperly formulated

Summary

• Graphical solution procedure– Feasible region– Extreme points– Optimal solution– Binding and Slack Constraints

• Additional considerations– Degeneracy– Inconsistent problem– Multiple optimal solutions– Unbounded problem

• Spreadsheet solution basics