1. LINEAR PROGRAMMINGLINEAR PROGRAMMING
SOLUTION TECHNIQUES:SOLUTION TECHNIQUES:
GRAPHICALAND COMPUTERGRAPHICALAND COMPUTER
METHODSMETHODS

2. LEARNING OBJECTIVESLEARNING OBJECTIVES
 Understand basic assumptions and properties of linear
programming (LP).
 Use graphical solution procedures for LP problems with
only two variables to understand how LP problems are
solved.
 Understand special situations such as redundancy,
infeasibility, unboundedness, and alternate optimal
solutions in LP problems.
 Understand how to set up LP problems on a spreadsheet
and solve them using Excel’s solver.

3. INTRODUCTIONINTRODUCTION
 Management decisions in many organizations involve trying
to make most effective use of resources (machinery, labor,
money, time, warehouse space, and raw materials) in order
to:
Produce products - such as computers, automobiles, or
clothing or
Provide services - such as package delivery, health
services, or investment decisions.
 To solve problems of resource allocation one may use
mathematical programming.

4. LINEARLINEAR PROGRAMMINGPROGRAMMING
Linear programming (LP) is the most common type of
mathematical programming.
LP seeks to maximize or minimize a linear objective
function subject to a set of linear constraints
LP assumes all relevant input data and parameters are
known with certainty (deterministic models).
Computers play an important role in the solution of LP
problems

5.  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, that is to be maximized or minimized
 Constraints - 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 constraint equations.
LP MODEL COMPONENTS ANDLP MODEL COMPONENTS AND
FORMULATIONFORMULATION

6. DDEVELOPMENT OF A LP MODELEVELOPMENT OF A LP MODEL
 LP applied extensively to problems areas -
 medical, transportation, operations,
 financial, marketing, accounting,
 human resources, and agriculture.
 Development and solution of all LP models can be
examined in a four step process:
(1) identification of the problem as solvable by LP
(2) formulation of the mathematical model.
(3) solution.
(4) interpretation.

7. BASIC STEPS OF DEVELOPING A LPBASIC STEPS OF DEVELOPING A LP
MODELMODEL
Formulation
– Process of translating problem scenario into simple LP model
framework with a set of mathematical relationships.
Solution
– Mathematical relationships resulting from formulation process are
solved to identify optimal solution.
Interpretation and What-if Analysis
– Problem solver or analyst works with the manager to
 interpret results and implications of problem solution.
 investigate changes in input parameters and model variables
and impact on problem solution results.

8. LLINEAR EQUATIONS ANDINEAR EQUATIONS AND
INEQUALITIESINEQUALITIES
 This is a linear equation:
2A + 5B = 10
 This equation is not linear:
2A2
+ 5B3
+ 3AB = 10
 LP uses, in many cases, inequalities like:
A + B ≤ C or A + B ≥ C

9. BBASIC ASSUMPTIONS OF A LP MODELASIC ASSUMPTIONS OF A LP MODEL
1. Conditions of certainty exist.
2. Proportionality in objective function and constraints (1
unit – 3 hours, 3 units- 9 hours).
3. Additivity (total of all activities equals sum of individual
activities).
4. Divisibility assumption that solutions need not necessarily
be in whole numbers (integers); ie.decision variables can
take on any fractional value.

10. FORMULATING A LP PROBLEMFORMULATING A LP PROBLEM
 A common LP application is product mix problem.
– Two or more products are usually produced using
limited resources - such as personnel, machines, raw
materials, and so on.
 Profit firm seeks to maximize is based on profit
contribution per unit of each product.
 Firm would like to determine -
– How many units of each product it should produce in
order to maximize overall profit given its limited
resources.

11. MAXIMIZATION MODEL EXAMPLES:MAXIMIZATION MODEL EXAMPLES:
 BEAVER CREEK EXAMPLEBEAVER CREEK EXAMPLE
 FLAIR FURNITURE EXAMPLEFLAIR FURNITURE EXAMPLE
 GALAXY INDUSTRIES EXAMPLEGALAXY INDUSTRIES EXAMPLE

12. Resource Requirements
Product
Labor
(hr/unit)
Clay
(lb/unit)
Profit
($/unit)
Bowl 1 4 40
Mug 2 3 50
PROBLEM DEFINITION:PROBLEM DEFINITION:
BEAVER CREEK MAXIMIZATIONBEAVER CREEK MAXIMIZATION
PROBLEM (1 of 18)PROBLEM (1 of 18)
 Product mix problem - Beaver Creek Pottery Company
 How many bowls and mugs should be produced to
maximize profits given labor and materials constraints?
 Product resource requirements and unit profit:

13. PROBLEM DEFINITION: BEAVERPROBLEM DEFINITION: BEAVER
CREEK EXAMPLE (2 of 18)CREEK EXAMPLE (2 of 18)
Resource 40 hrs of labor per day
Availability: 120 lbs of clay
Decision Variables x1 = number of bowls to produce per day
x2 = number of mugs to produce per day
Objective Maximize Z = $40x1 + $50x2
Function: Where Z = profit per day
Resource 1x1 + 2x2 ≤ 40 hours of labor
Constraints: 4x1 + 3x2 ≤ 120 pounds of clay
Non-Negativity x1 ≥ 0; x2 ≥ 0
Constraints:

14. PROBLEM DEFINITION:PROBLEM DEFINITION:
BEAVER CREEK EXAMPLE (3 of 18)BEAVER CREEK EXAMPLE (3 of 18)
Complete Linear Programming Model:
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 ≤ 40
4x1 + 3x2 ≤ 120
x1, x2 ≥ 0

15. A feasible solution does not violate any of the
constraints:
Example x1 = 5 bowls
x2 = 10 mugs
Z = $40x1 + $50x2 = $700
Labor constraint check:
1(5) + 2(10) = 25 < 40 hours, within constraint
Clay constraint check:
4(5) + 3(10) = 50 < 120 pounds, within constraint
FEASIBLE SOLUTIONS:FEASIBLE SOLUTIONS:
BEAVER CREEK EXAMPLE (4 of 18)BEAVER CREEK EXAMPLE (4 of 18)

16. An infeasible solution violates at least one of the
constraints:
Example x1 = 10 bowls
x2 = 20 mugs
Z = $1400
Labor constraint check:
1(10) + 2(20) = 50 > 40 hours, violates the constraint
INFEASIBLE SOLUTIONS:INFEASIBLE SOLUTIONS:
BEAVER CREEK EXAMPLE (5 of 18)BEAVER CREEK EXAMPLE (5 of 18)

17. The set of all points that satisfy all the constraints of
the model is called
a
FEASIBLE REGIONFEASIBLE REGION

18.  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 LINEARGRAPHICAL SOLUTION OF LINEAR
PROGRAMMING MODELSPROGRAMMING MODELS

19.  Primary advantage of two-variable LP models (such as
Beaver Creek problem) is their solution can be graphically
illustrated using two-dimensional graph.
 Allows one to provide an intuitive explanation of how
more complex solution procedures work for larger LP
models.

20. GRAPHICAL REPRESENTATION OF LPGRAPHICAL REPRESENTATION OF LP
MODELSMODELS
Coordinates
for graphical
analysis
10
20
30
40
50
60
10 20 30 40 50 600

21. GRAPHICAL REPRESENTATION OFGRAPHICAL REPRESENTATION OF
CONSTRAINTS:CONSTRAINTS:
COORDINATE AXES-BEAVER CREEKCOORDINATE AXES-BEAVER CREEK
EXAMPLE (6 of 18)EXAMPLE (6 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 ≤ 40
4x1 + 3x2 ≤ 120
x1, x2 ≥ 0
Coordinates for Graphical Analysis

22. GRAPHICAL REPRESENTATION OFGRAPHICAL REPRESENTATION OF
CONSTRAINTS:CONSTRAINTS:
-BEAVER CREEK EXAMPLE-LABOR-BEAVER CREEK EXAMPLE-LABOR
CONSTRAINT (7 of 18)CONSTRAINT (7 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 ≤ 40
4x1 + 3x2 ≤ 120
x1, x2 ≥ 0
Graph of Labor Constraint

23. GRAPHICAL REPRESENTATION OFGRAPHICAL REPRESENTATION OF
CONSTRAINTS:CONSTRAINTS:
-BEAVER CREEK EXAMPLE-LABOR-BEAVER CREEK EXAMPLE-LABOR
CONSTRAINT AREA(8 of 18)CONSTRAINT AREA(8 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 ≤ 40
4x1 + 3x2 ≤ 120
x1, x2 ≥ 0
Labor Constraint Area

24. GRAPHICAL REPRESENTATION OFGRAPHICAL REPRESENTATION OF
CONSTRAINTS:CONSTRAINTS:
BEAVER CREEK EXAMPLE-CLAYBEAVER CREEK EXAMPLE-CLAY
CONSTRAINT AREA(9 of 18)CONSTRAINT AREA(9 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 ≤ 40
4x1 + 3x2 ≤ 120
x1, x2 ≥ 0
Clay Constraint Area

25. Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 ≤ 40
4x1 + 3x2 ≤ 120
x1, x2 ≥ 0
Graph of Both Model Constraints
GRAPHICAL REPRESENTATION OFGRAPHICAL REPRESENTATION OF
CONSTRAINTS:CONSTRAINTS:
BEAVER CREEK EXAMPLE- BOTHBEAVER CREEK EXAMPLE- BOTH
CONSTRAINTS (10 of 18)CONSTRAINTS (10 of 18)

26. FEASIBLE SOLUTION AREA:FEASIBLE SOLUTION AREA:
BEAVER CREEK EXAMPLE (11 of 18)BEAVER CREEK EXAMPLE (11 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 ≤ 40
4x1 + 3x2 ≤ 120
x1, x2 ≥ 0
Feasible Solution Area

27. GGRAPHICAL SOLUTION:RAPHICAL SOLUTION:
ISOPROFIT LINE SOLUTION METHODISOPROFIT LINE SOLUTION METHOD
 Optimal solution is the point in feasible region that
produces highest profit
 There are many possible solution points in region.
 How do we go about selecting the best one, one
yielding highest profit?
 Let objective function (that is, $$40x1 + $50x2) guide
one towards optimal point in feasible region.
 Plot line representing objective function on graph as a
straight line.

28. ISOPROFIT LINE METHOD:ISOPROFIT LINE METHOD:
BEAVER CREEK EXAMPLE (12 of 18)BEAVER CREEK EXAMPLE (12 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 ≤ 40
4x1 + 3x2 ≤ 120
x1, x2 ≥ 0
Objective Function Line for Z = $800
Set Objective Function = 800

29. ISOPROFIT LINE METHOD -ISOPROFIT LINE METHOD -
ALTERNATIVE OBJECTIVE FUNCTIONALTERNATIVE OBJECTIVE FUNCTION
SOLUTION LINES:SOLUTION LINES:
BEAVER CREEK EXAMPLE (13 of 18)BEAVER CREEK EXAMPLE (13 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 ≤ 40
4x1 + 3x2 ≤ 120
x1, x2 ≥ 0
Alternative Objective Function Lines

30. ISOPROFIT LINE METHOD-OPTIMALISOPROFIT LINE METHOD-OPTIMAL
SOLUTION):SOLUTION):
BEAVER CREEK EXAMPLE (14 of 18)BEAVER CREEK EXAMPLE (14 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 ≤ 40
4x1 + 3x2 ≤ 120
x1, x2 ≥ 0
Identification of the Optimal Solution

31. ISOPROFIT LINE METHOD - OPTIMALISOPROFIT LINE METHOD - OPTIMAL
SOLUTION COORDINATES:SOLUTION COORDINATES:
BEAVER CREEK EXAMPLE (15 of 18)BEAVER CREEK EXAMPLE (15 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 ≤ 40
4x1 + 3x2 ≤ 120
x1, x2 ≥ 0
Optimal Solution Coordinates

32. CCORNER POINT PROPERTYORNER POINT PROPERTY
It is a very important property of Linear
Programming problems:
This property states optimal solution to LP problem
will always occur at a corner point.

33. GRAPHICAL SOLUTION - CORNERGRAPHICAL SOLUTION - CORNER
POINT SOLUTION METHOD :POINT SOLUTION METHOD :
BEAVER CREEK EXAMPLE(16 of 18)BEAVER CREEK EXAMPLE(16 of 18)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 ≤ 40
4x1 + 3x2 ≤ 120
x1, x2 ≥ 0
Solution at All Corner Points

34. OPTIMAL SOLUTION FOR A NEWOPTIMAL SOLUTION FOR A NEW
OBJECTIVE FUNCTION:OBJECTIVE FUNCTION:
BEAVER CREEK EXAMPLE (17 of 18)BEAVER CREEK EXAMPLE (17 of 18)
Maximize Z = $70x1 + $20x2
subject to: 1x1 + 2x2 ≤ 40
4x1 + 3x2 ≤ 120
x1, x2 ≥ 0
Optimal Solution with Z = 70x1 + 20x2

35.  Standard form requires that all constraints be in the form
of equations.
 A slack variable is added to a ≤ constraint to convert it
to an equation (=).
 A slack variable represents unused resources.
 A slack variable contributes nothing to the objective
function value.
SLACK VARIABLESSLACK VARIABLES

36. STANDARD FORM OF LINEARSTANDARD FORM OF LINEAR
PROGRAMMING MODEL:PROGRAMMING MODEL:
BEAVER CREEK EXAMPLE (18 of 18)BEAVER CREEK EXAMPLE (18 of 18)
Max Z = 40x1 + 50x2 + s1 +s2
subject to:1x1 + 2x2 + s1 = 40
4x1 + 3x2 + s2 = 120
x1, x2, s1, s2 ≥ 0
Where:
x1 = number of bowls
x2 = number of mugs
s1, s2 are slack variables
Solution Points A, B, and C with Slack

37. PROBLEM DEFINITION:PROBLEM DEFINITION:
FLAIR FURNITURE MAXIMIZATIONFLAIR FURNITURE MAXIMIZATION
EXAMPLEEXAMPLE (1 of 19)(1 of 19)
Company Data and Constraints -
 Flair Furniture Company produces tables and chairs.
 Each table requires: 4 hours of carpentry and 2 hours of painting.
 Each chair requires: 3 hours of carpentry and 1 hour of painting.
 Available production capacity: 240 hours of carpentry time and 100 hours of
painting time.
 Due to existing inventory of chairs, Flair is to make no more than 60 new
chairs.
 Each table sold results in $7 profit, while each chair produced yields $5
profit.
Flair Furniture’s problem:
 Determine the best possible combination of tables and chairs to manufacture

38. DECISION VARIABLES:DECISION VARIABLES:
FLAIR FURNITURE EXAMPLEFLAIR FURNITURE EXAMPLE (2 of 19)(2 of 19)
 Problem facing Flair is to determine how many chairs
and tables to produce to yield maximum profit?
 In Flair Furniture problem, there are two unknown
entities:
 T- number of tables to be produced.
 C- number of chairs to be produced.

39. OBJECTIVE FUNCTION:OBJECTIVE FUNCTION:
FLAIR FURNITURE EXAMPLEFLAIR FURNITURE EXAMPLE (3 of 19)(3 of 19)
 Objective function states the goal of problem.
What major objective is to be solved?
Maximize profit!
 An LP model must have a single objective function.
In Flair’s problem, total profit may be expressed as:
Using decision variables T and C -
Maximize $7 T + $5 C
($7 profit per table) x (number of tables produced) +
($5 profit per chair) x (number of chairs produced)

40. CONSTRAINTS:CONSTRAINTS:
FLAIR FURNITURE EXAMPLEFLAIR FURNITURE EXAMPLE (4 of 19)(4 of 19)
 Denote conditions that prevent one from selecting any
specific subjective value for decision variables.
 In Flair Furniture’s problem, there are three
restrictions on solution.
Restrictions 1 and 2 have to do with available
carpentry and painting times, respectively.
Restriction 3 is concerned with upper limit on the
number of chairs.

41. CONSTRAINTS:CONSTRAINTS:
FLAIR FURNITURE EXAMPLEFLAIR FURNITURE EXAMPLE (5 of 19)(5 of 19)
 There are 240 carpentry hours available.
 4T + 3C < 240
 There are 100 painting hours available.
 2T + 1C ≤ 100
 The marketing specified chairs limit constraint.
 C ≤ 60
 The non-negativity constraints.
 T ≥ 0 (number of tables produced is ≥ 0)
 C ≥ 0 (number of chairs produced is ≥ 0)

42. BUILDING THE COMPLETEBUILDING THE COMPLETE
MATHEMATICAL MODEL:MATHEMATICAL MODEL:
FLAIR FURNITURE EXAMPLE (6 oFLAIR FURNITURE EXAMPLE (6 of 19)f 19)
Maximize profit = $7T + $5C (objective function)
Subject to constraints -
4T + 3C ≤ 240 (carpentry constraint)
2T + 1C ≤ 100 (painting constraint)
C ≤ 60 (chairs limit constraint)
T ≥ 0 (non-negativity constraint on tables)
C ≥ 0 (non-negativity constraint on chairs)

43. CONVERTING INEQUALITIES INTOCONVERTING INEQUALITIES INTO
EQUALITIES BY USING SLACK:EQUALITIES BY USING SLACK:
FLAIR FURNITURE EXAMPLE (7 of 19)FLAIR FURNITURE EXAMPLE (7 of 19)
Maximize profit = $7T + $5C + 0s1+ 0s2 + 0s3
Subject to constraints -
4T + 3C + s1= 240 (carpentry constraint)
2T + 1C + s2 = 100 (painting constraint)
C + s3 = 60 (chairs limit constraint)
T ≥ 0 (non-negativity constraint on tables)
C ≥ 0 (non-negativity constraint on chairs)
s1 s2 s3≥ 0 (non-negativity constraints on slacks)

44. GRAPHICAL REPRESENTATION OFGRAPHICAL REPRESENTATION OF
CONSTRAINTS:CONSTRAINTS:
FLAIR FURNITURE EXAMPLE (8 of 19)FLAIR FURNITURE EXAMPLE (8 of 19)
4T + 3C ≤ 240
Carpentry time
constaint

45. GRAPHICAL REPRESENTATION OFGRAPHICAL REPRESENTATION OF
CONSTRAINTS:CONSTRAINTS:
FLAIR FURNITURE EXAMPLE (9 of 19)FLAIR FURNITURE EXAMPLE (9 of 19)
Carpentry TimeCarpentry Time
ConstraintConstraint
(feasible area)(feasible area)
Carpentry time and
the feasible region
Any point below
line satisfies
constraint.

46. GRAPHICAL REPRESENTATION OFGRAPHICAL REPRESENTATION OF
CONSTRAINTS:CONSTRAINTS:
FLAIR FURNITURE EXAMPLE (10 of 19)FLAIR FURNITURE EXAMPLE (10 of 19)
Painting Time
Constraint and the
Feasible Area
2T + 1C ≤ 100
Any point on line
satisfies equation:
2T + 1C = 100
(30,40) yields 100.

47. GRAPHICAL REPRESENTATION OFGRAPHICAL REPRESENTATION OF
CONSTRAINTS:CONSTRAINTS:
FLAIR FURNITURE EXAMPLE (11 of 19)FLAIR FURNITURE EXAMPLE (11 of 19)
Chair Limit
Constraint and the
Feasible Solution
Area
Feasible solution
area is constrained
by three limiting
lines

48. IISOPROFIT LINE SOLUTION METHOD:SOPROFIT LINE SOLUTION METHOD:
FLAIR FURNITURE EXAMPLE (12 of 19)FLAIR FURNITURE EXAMPLE (12 of 19)
 Let objective function (that is, $7T + $5C) guide one
towards an optimal point in the feasible region.
 Plot line representing objective function on graph.
 One does not know what $7T + $5C equals at an optimal
solution.
 Without knowing this value, how does one plot
relationship?

49. IISOPROFIT LINE SOLUTION METHOD:SOPROFIT LINE SOLUTION METHOD:
FLAIR FURNITURE EXAMPLE (13 of 19)FLAIR FURNITURE EXAMPLE (13 of 19)
 Write objective function: $7 T + $5 C = Z
 Select any arbitrary value for Z.
For example, one may choose a profit ( Z ) of $210.
 Z is written as: $7 T + $5 C = $210.
 To plot this profit line:
Set T = 0 and solve objective function for C.
Let T = 0, then $7(0) + $5C = $210, or C = 42.
 Set C = 0 and solve objective function for T.
Let C = 0, then $7T + $5(0) = $210, or T = 30.

50. IISOPROFIT LINE SOLUTION METHOD:SOPROFIT LINE SOLUTION METHOD:
FLAIR FURNITURE EXAMPLE (14 of 19)FLAIR FURNITURE EXAMPLE (14 of 19)
One can check for
higher values of Z
to find an optimal
solution.
210 is not the
highest possible
value.

51. IISOPROFIT LINE SOLUTION METHOD:SOPROFIT LINE SOLUTION METHOD:
FLAIR FURNITURE EXAMPLE (15 of 19)FLAIR FURNITURE EXAMPLE (15 of 19)
Isoprofit lines
($210, $280,
$350) are all
parallel.

52. IISOPROFIT LINE SOLUTION METHOD:SOPROFIT LINE SOLUTION METHOD:
FLAIR FURNITURE EXAMPLE (16 of 19)FLAIR FURNITURE EXAMPLE (16 of 19)
Optimal Solution:
Corner Point 4: T=30 (tables) and C=40 (chairs) with $410 profit

53. IISOPROFIT LINE SOLUTION METHOD:SOPROFIT LINE SOLUTION METHOD:
FLAIR FURNITURE EXAMPLE (17 of 19)FLAIR FURNITURE EXAMPLE (17 of 19)
 Optimal solution occurs at the maximum point in the
feasible region.
 It occurs at the intersection of carpentry and painting
constraints:
- Carpentry constraint equation: 4T + 3C = 240
- Painting constraint equation : 2T + 1C = 100
If one solves these two equations with two unknowns for T
and C (for Corner Point 4), Optimal Solution is found:
T=30 (tables) and C=40 (chairs) with $410 profit.

54. CORNER POINT SOLUTION METHOD:CORNER POINT SOLUTION METHOD:
FLAIR FURNITURE EXAMPLEFLAIR FURNITURE EXAMPLE (18 of 19)(18 of 19)
From the figure one knows
that the feasible region
for Flair’s problem has
five corner points,
namely, 1, 2, 3, 4, and 5,
respectively.
To find the point yielding
the maximum profit,
one finds coordinates
of each corner point and
computes profit level at
each point.

55. CORNER POINT SOLUTION METHOD:CORNER POINT SOLUTION METHOD:
FLAIR FURNITURE EXAMPLEFLAIR FURNITURE EXAMPLE (19 of 19)(19 of 19)
 Point 1 (T = 0, C = 0)
profit = $7(0) + $5(0) = $0
 Point 2 (T = 0, C = 60)
profit = $7(0) + $5(60) = $300
 Point 3 (T = 15, C = 60)
profit = $7(15) + $5(60) = $405
 Point 4 (T = 30, C = 40)
profit = $7(30) + $5(40) = $410
 Point 5 (T = 50, C = 0)
profit = $7(50) + $5(0) = $350 .

56. PROBLEM DEFINITION:PROBLEM DEFINITION:
THE GALAXY INDUSTRIES EXAMPLETHE GALAXY INDUSTRIES EXAMPLE
(1 of 9)(1 of 9)
 Galaxy manufactures two toy models:
– Space Ray.
– Zapper.
 Resources are limited to
– 1200 pounds of special plastic.
– 40 hours of production time per week.

57. Marketing requirement
– Total production cannot exceed 800 dozens.
– Number of dozens of Space Rays cannot exceed
number of dozens of Zappers by more than 450.
Technological input
– Space Rays requires 2 pounds of plastic and
3 minutes of labor per dozen.
– Zappers requires 1 pound of plastic and
4 minutes of labor per dozen.
PROBLEM DEFINITION:PROBLEM DEFINITION:
THE GALAXY INDUSTRIES EXAMPLETHE GALAXY INDUSTRIES EXAMPLE
(2 of 9)(2 of 9)

58. Current production plan calls for:
– Producing as much as possible of the more profitable
product, Space Ray ($8 profit per dozen).
– Use resources left over to produce Zappers ($5 profit
per dozen).
The current production plan consists of:
Space Rays = 550 dozens
Zapper = 100 dozens
Profit = 4900 dollars per week
PROBLEM DEFINITION:PROBLEM DEFINITION:
THE GALAXY INDUSTRIES EXAMPLETHE GALAXY INDUSTRIES EXAMPLE
(3 of 9)(3 of 9)

59. Management is seeking a production
schedule that will increase the company’s
profit.

60.  Decision variables:
– X1 = Production level of Space Rays (in dozens per
week).
– X2 = Production level of Zappers (in dozens per week).
 Objective Function:
- Weekly profit, to be maximized
DECISION VARIABLES:DECISION VARIABLES:
GALAXY INDUSTRIES EXAMPLEGALAXY INDUSTRIES EXAMPLE
(4 of 9)(4 of 9)

61. Max. 8X1 + 5X2 (Weekly profit)
subject to
2X1 + 1X2 < = 1200 (Plastic)
3X1 + 4X2 < = 2400 (Production Time)
X1 + X2 < = 800 (Total production)
X1 - X2 < = 450 (Mix)
Xj> = 0, j = 1,2 (Nonnegativity)
BUILDING THE COMPLETEBUILDING THE COMPLETE
MATHEMATICAL MODEL:MATHEMATICAL MODEL:
GALAXY INDUSTRIES EXAMPLEGALAXY INDUSTRIES EXAMPLE
(5 of 9)(5 of 9)

62. 1200
600
The Plastic constraint
Feasible
The plastic constraint:
2X1+X2<=1200
X2
Infeasible
Production
Time
3X1+4X2<=2400
Total production constraint:
X1+X2<=800
600
800
Production mix
constraint:
X1-X2<=450
• There are three types of feasible points
Interior points.
Boundary points.
Extreme points.
X1
GRAPHICAL REPRESENTATION OFGRAPHICAL REPRESENTATION OF
CONSTRAINTS:CONSTRAINTS:
GALAXY INDISTRIES EXAMPLE (6 of 9)GALAXY INDISTRIES EXAMPLE (6 of 9)

63. SOLVING GRAPHICALLY FOR ANSOLVING GRAPHICALLY FOR AN
OPTIMAL SOLUTIONOPTIMAL SOLUTION

64. Recall the feasible Region
600
800
1200
400 600 800
X2
X1
We now demonstrate the search for an optimal solution
Start at some arbitrary profit, say profit = $2,000...
Profit = $
000
2,
Then increase the profit, if possible...
3,4,
...and continue until it becomes infeasible
Profit =$5040
ISOPROFIT LINE SOLUTION METHOD:ISOPROFIT LINE SOLUTION METHOD:
GALAXY INDUSTRIES EXAMPLE (7 of 9)GALAXY INDUSTRIES EXAMPLE (7 of 9)

65. 600
800
1200
400 600 800
X2
X1
Let’s take a closer look at
the optimal point
Feasible
region
Feasible
region
Infeasible
ISOPROFIT LINE SOLUTION METHOD:ISOPROFIT LINE SOLUTION METHOD:
GALAXY INDUSTRIES EXAMPLE (8 of 9)GALAXY INDUSTRIES EXAMPLE (8 of 9)

66. OPTIMAL SOLUTION:OPTIMAL SOLUTION:
GALAXY INDUSTRIES EXAMPLE (9 of 9)GALAXY INDUSTRIES EXAMPLE (9 of 9)
Space Rays = 480 dozens
Zappers = 240 dozens
Profit = $5040
– This solution utilizes all the plastic and all the production
hours.
– Total production is only 720 (not 800).
– Space Rays production exceeds Zapper by only 240
dozens (not 450).

67. MINIMIZATION MODEL EXAMPLES:MINIMIZATION MODEL EXAMPLES:
 FERTILIZER MIX PROBLEMFERTILIZER MIX PROBLEM
 HOLIDAY MEAL CHICKEN RANCHHOLIDAY MEAL CHICKEN RANCH
EXAMPLEEXAMPLE
 NAVY SEA RATIONS EXAMPLENAVY SEA RATIONS EXAMPLE

68. AA MINIMIZATION LP PROBLEMMINIMIZATION LP PROBLEM
Many LP problems involve minimizing objective such as cost
instead of maximizing profit function.
Examples:
– Restaurant may wish to develop work schedule to meet staffing
needs while minimizing total number of employees.
– Manufacturer may seek to distribute its products from several
factories to its many regional warehouses in such a way as to
minimize total shipping costs.
– Hospital may want to provide its patients with a daily meal plan
that meets certain nutritional standards while minimizing food
purchase costs.

69. PROBLEM DEFINITION:PROBLEM DEFINITION:
FERTILIZER MIX EXAMPLE (1 of 7)FERTILIZER MIX EXAMPLE (1 of 7)
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 the following data ?

70. PROBLEM DEFINITION:PROBLEM DEFINITION:
FERTILIZER MIX EXAMPLE (2 of 7)FERTILIZER MIX EXAMPLE (2 of 7)
Decision Variables:
x1 = bags of Super-Gro
x2 = 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)
x , x ≥ 0 (non-negativity constraint)

71. GRAPHICAL REPRESENTATION OFGRAPHICAL REPRESENTATION OF
CONSTRAINTS:CONSTRAINTS:
FERTILIZER MIX EXAMPLE (3 of 7)FERTILIZER MIX EXAMPLE (3 of 7)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2 ≥ 16
4x2 + 3x2 ≥ 24
x1, x2 ≥ 0
Graph of Both Model Constraints

72. FEASIBLE SOLUTION AREA:FEASIBLE SOLUTION AREA:
FERTILIZER MIX EXAMPLE (4 of 7)FERTILIZER MIX EXAMPLE (4 of 7)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2 ≥ 16
4x2 + 3x2 ≥ 24
x1, x2 ≥ 0
Feasible Solution Area

73. OPTIMAL SOLUTION POINT:OPTIMAL SOLUTION POINT:
FERTILIZER MIX EXAMPLE (5 of 7)FERTILIZER MIX EXAMPLE (5 of 7)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2 ≥ 16
4x2 + 3x2 ≥ 24
x1, x2 ≥ 0
Optimum Solution Point

74. SURPLUS VARIABLES:SURPLUS VARIABLES:
FERTILIZER MIX EXAMPLE (6 of 7)FERTILIZER MIX EXAMPLE (6 of 7)
 A surplus variable is subtracted from a ≥ constraint to
convert it to an equation (=).
 A surplus variable represents an excess above a
constraint requirement level.
 Surplus variables contribute nothing to the calculated
value of the objective function.
 Subtracting surplus variables in the farmer problem
constraints:
2x1 + 4x2 - s1 = 16 (nitrogen)
4x1 + 3x2 - s2 = 24 (phosphate)

75. Minimize Z = $6x1 + $3x2 + 0s1
+ 0s2
subject to: 2x1 + 4x2 – s1 = 16
4x2 + 3x2 – s2 = 24
x1, x2, s1, s2 ≥ 0
Graph of Fertilizer Example
GRAPHICAL SOLUTION:GRAPHICAL SOLUTION:
FERTILIZER MIX EXAMPLE (7 of 7)FERTILIZER MIX EXAMPLE (7 of 7)

76. PROBLEM DEFINITION:PROBLEM DEFINITION:
HOLIDAY MEAL CHICKEN RANCHHOLIDAY MEAL CHICKEN RANCH
(HMCR) EXAMPLE (1 of 10)(HMCR) EXAMPLE (1 of 10)
 Buy two brands of feed for good, low-cost diet for chickens.
 Each feed may contain three nutritional ingredients (protein, vitamin,
and iron).
 One pound of Brand A contains:
 5 units of protein,
4 units of vitamin, and
0.5 units of iron.
 One pound of Brand B contains:
10 units of protein,
3 units of vitamin, and
0 units of iron.

77. PROBLEM DEFINITION:PROBLEM DEFINITION:
HMCR EXAMPLE (2 of 10)HMCR EXAMPLE (2 of 10)
Brand A feed costs ranch $0.02 per pound, while Brand B
feed costs $0.03 per pound.
Ranch owner would like lowest-cost diet that meets
minimum monthly intake requirements for each
nutritional ingredient.

78. PROBLEM DEFINITION:PROBLEM DEFINITION:
HMCR EXAMPLE (3 of 10)HMCR EXAMPLE (3 of 10)

79. BUILDING THE COMPLETEBUILDING THE COMPLETE
MATHEMATICAL MODEL:MATHEMATICAL MODEL:
HMCR EXAMPLE (4 of 10)HMCR EXAMPLE (4 of 10)
Minimize cost (in cents) = 2A + 3B
Subject to:
5A + 10B ≥ 90 (protein constraint)
4A + 3B ≥ 48 (vitamin constraint)
½A ≥ 1½ (iron constraint)
A ≥ 0, B ≥ 0 (nonnegativity constraint)
Where:
A denotes number of pounds of Brand A feed,
and B denote number of pounds of Brand B feed.

80. BUILDING THE STANDARD LP MODEL:BUILDING THE STANDARD LP MODEL:
HMCR EXAMPLE (5 of 10)HMCR EXAMPLE (5 of 10)
Minimize cost (in cents)=2A+3B+0s1+0s2+0s3
subject to constraints:
5A + 10B - s1 = 90 (protein constraint)
4A + 3B - s2 = 48 (vitamin constraint)
½A - s3 = 1½ (iron constraint)
A, B, s1,s2 s3 ≥ 0 (nonnegativity)

81. GRAPHICAL REPRESENTATION OFGRAPHICAL REPRESENTATION OF
CONSTRAINTS:CONSTRAINTS:
HMCR EXAMPLE (6 of 10)HMCR EXAMPLE (6 of 10)
Drawing
Constraints:
5A + 10B ≥ 90
4A + 3B ≥ 48
½A ≥ 1½
Nonnegativity Constraint
A ≥ 0, B ≥ 0

82. GRAPHICAL SOLUTION METHOD-ISOCOSTGRAPHICAL SOLUTION METHOD-ISOCOST
LINE METHOD: HMCR EXAMPLE (7 of 10)LINE METHOD: HMCR EXAMPLE (7 of 10)
One can start by drawing a 54-cent cost
line : 2A + 3B. = 54

83. ISOCOST LINE METHOD:ISOCOST LINE METHOD:
HMCR EXAMPLE (8 of 10)HMCR EXAMPLE (8 of 10)
 Isocost line is moved parallel to 54-cent solution line toward lower
left origin.
 Last point to touch the isocost line while still in contact with the
feasible region, is corner point 2.

84. • Solving for corner point 2 with two equations produces values 8.4 for
A and 4.8 for B, minimum optimal cost solution is:
2A + 3B = (2)(8.4) + (3)(4.8) = 31.2
IISOCOST LINE METHOD:SOCOST LINE METHOD:
HMCR EXAMPLE (9 of 10)HMCR EXAMPLE (9 of 10)

85. CORNER POINT SOLUTIONCORNER POINT SOLUTION
METHOD:HMCR EXAMPLE (10 of 10)METHOD:HMCR EXAMPLE (10 of 10)
 Point 1 - coordinates (A = 3, B =
12)
– cost of 2(3) + 3(12) = 42 cents.
 Point 2 - coordinates (A = 8.4, b =
4.8)
– cost of 2(8.4) + 3(4.8) = 31.2
cents
 Point 3 - coordinates (A = 18, B =
0)
– cost of (2)(18) + (3)(0) = 36
cents.
 Optimal minimal cost solution:
Corner Point 2, cost = 31.2 cents

86. PROBLEM DEFINITION:PROBLEM DEFINITION:
NAVY SEA RATIONS EXAMPLE (1 of 4)NAVY SEA RATIONS EXAMPLE (1 of 4)
• A cost minimization diet problem
– Mix two sea ration products: Texfoods, Calration.
– Minimize the total cost of the mix.
– Meet the minimum requirements of
Vitamin A, Vitamin D, and Iron.

87.  Decision variables:
– X1 (X2) -- The number of portions of Texfoods
(Calration) product used in a
serving.
 The Model:
Minimize 0.60X1 + 0.50X2
Subject to
20X1 + 50X2 ≥ 100 Vitamin A
25X1 + 25X2 ≥ 100 Vitamin D
50X1 + 10X2 ≥ 100 Iron
X1, X2 ≥ 0
COMPLETE MODEL:COMPLETE MODEL:
NAVY SEA RATIONS EXAMPLE (2 of 4)NAVY SEA RATIONS EXAMPLE (2 of 4)

88. GRAPHICAL SOLUTION:GRAPHICAL SOLUTION:
NAVY SEA RATIONS EXAMPLE (3 of 4)NAVY SEA RATIONS EXAMPLE (3 of 4)
5
4
2
2 44 5
Feasible Region
Vitamin “D” constraint
Vitamin “A” constraint
The Iron constraint

89. SUMMARY OF THE OPTIMALSUMMARY OF THE OPTIMAL
SOLUTION: NAVY SEA RATIONSSOLUTION: NAVY SEA RATIONS
EXAMPLE (4 of 4)EXAMPLE (4 of 4)
– Texfood product = 1.5 portions
– Calration product = 2.5 portions
– Cost =$ 2.15 per serving.
– The minimum requirement for Vitamin D and iron
are met with no surplus.
– The mixture provides 155% of the requirement for
Vitamine A.

90. SSUMMARY OF THE GRAPHICALUMMARY OF THE GRAPHICAL
SOLUTION METHODS (1 of 3)SOLUTION METHODS (1 of 3)
1. Plot the model constraints accepting them as equalities,
2. Considering the inequalities of the constraints identify the
feasible solution region, that is, the area that satisfies all
constraints simultaneously.
3. Select one of two following graphical solution techniques
and proceed to solve problem.
- Isoprofit or Isocost Method.
- Corner Point Method

91. SSUMMARY OF THE GRAPHICALUMMARY OF THE GRAPHICAL
SOLUTION METHODSSOLUTION METHODS ((2 of 32 of 3))
Corner Point Method
 Determine the coordinates of each of the corner points of
the feasible region by solving simultaneous equations at
each point.
 Compute the profit or cost at each point by substituting the
values of coordinates into the objective function and
solving for results.
 Identify the optimal solution as a corner point with highest
profit (maximization), or lowest cost (minimization).

92. SSUMMARY OF THE GRAPHICALUMMARY OF THE GRAPHICAL
SOLUTION METHODS (3 of 3SOLUTION METHODS (3 of 3))
Isoprofit or Isocost Method
 Select an arbitrary value for profit or cost, and plot an
isoprofit / isocost line to reveal its slope.
 Maintain the same slope and move the line up or down
until it touches the feasible region at one point. While
moving the line up or down consider whether the problem
is a maximization or a minimization problem
 Identify the optimal solution as coordinates of the point
that is touched by the highest possible isoprofit line or
lowest possible isocost line (by solving the simultaneous
equations)
 Read optimal coordinates and compute the optimal profit

93. SSPECIAL SITUATIONS IN SOLVING LPPECIAL SITUATIONS IN SOLVING LP
PROBLEMSPROBLEMS
(IRREGULAR TYPES OF LP PROBLEMS)(IRREGULAR TYPES OF LP PROBLEMS)

94.  For some linear programming models, the
general rules do not apply.
 Special types of problems include those with:
Redundancy
Infeasible solutions
Unbounded solutions
Multiple optimal solutions
IRREGULAR TYPES OF LINEARIRREGULAR TYPES OF LINEAR
PROGRAMMING PROBLEMSPROGRAMMING PROBLEMS

95. Redundancy: A redundant constraint is a constraint that
does not affect the feasible region in any way.
Maximize Profit
= 2X + 3Y
subject to:
X + Y ≤ 20
2X + Y ≤ 30
X ≤ 25
X, Y ≥ 0

96. Infeasibility: A condition that arises when an LP
problem has no solution that satisfies all of its
constraints.
X + 2Y ≤ 6
2X + Y ≤ 8
X ≥ 7

97. Unboundedness: Sometimes an LP model will not
have a finite solution
Maximize profit
= $3X + $5Y
subject to:
X ≥ 5
Y ≤ 10
X + 2Y ≥ 10
X, Y ≥ 0

98. MULTIPLE OPTIMAL SOLUTIONSMULTIPLE OPTIMAL SOLUTIONS
 An LP problem may have more than one optimal
solution.
– Graphically, when the isoprofit (or isocost) line
runs parallel to a constraint in the problem
which lies in the direction in which isoprofit (or
isocost) line is located.
– In other words, when they have the same slope.

99. EXAMPLE: MULTIPLE OPTIMALEXAMPLE: MULTIPLE OPTIMAL
SOLUTIONSSOLUTIONS
Maximize profit =
$3x + $2y
Subject to:
6X + 4Y ≤ 24
X ≤ 3
X, Y ≥ 0

100. EXAMPLE: MULTIPLE OPTIMALEXAMPLE: MULTIPLE OPTIMAL
SOLUTIONSSOLUTIONS
 At profit level of $12, isoprofit line will rest directly on top
of first constraint line.
 This means that any point along the line between corner
points 1 and 2 provides an optimal X and Y combination.

101. SSETTING UP AND SOLVING LPETTING UP AND SOLVING LP
PROBLEMS USING EXCEL’SPROBLEMS USING EXCEL’S
SOLVERSOLVER
Using solver to solve Flair Furniture problem:
Recall decision variables T ( Tables ) and
C ( Chairs ) in Flair Furniture problem:
Maximize profit = $7T + $5C
Subject to constraints
4T + 3C ≤ 240 (carpentry constraint)
2T + 1C ≤ 100 (painting constraint)
C ≤ 60 (chairs limit constraint)
T, C ≥ 0 (non-negativity)

102. SSOLVER SPREADSHEET SETUPOLVER SPREADSHEET SETUP

103. LP EXCEL AND SOLVER PARTSLP EXCEL AND SOLVER PARTS
Changing cells
 Solver refers to decision variables as changing cells.
 In Flair Furniture example, there are two decision
variables cells B5 and C5 to represent number of tables
to make (T) and number of chairs to make (C),
respectively.

104. LP EXCEL AND SOLVER PARTSLP EXCEL AND SOLVER PARTS
Changing Cells
 In each of excel layouts, for clarity, changing cells
(decision variables) have been shaded yellow.
Changing Cells ( B5 and C5 )

105. LP EXCEL AND SOLVER PARTSLP EXCEL AND SOLVER PARTS
Target Cell
Objective function, referred to as target cell by solver,
= SUMPRODUCT(B6:C6,$B$5:$C$5)
This is equivalent to =B6*B5+C6*C5
Target Cell

106. LP EXCEL AND SOLVER PARTSLP EXCEL AND SOLVER PARTS
Constraints
Each constraint has three parts -
(1) A left hand side (LHS) part consisting of every term
to the left of the equality or inequality sign.
(2) A right hand side (RHS) part consisting of all terms to
the right of the equality or inequality sign.
(3) An equality or inequality sign.

107. LP EXCEL AND SOLVER PARTSLP EXCEL AND SOLVER PARTS
Constraints
Each constraint has three parts
1. A left hand side (LHS) part.
2. A right hand side (RHS) part.
3. Equality or inequality sign.
1 23

108. ENTERING INFORMATION IN SOLVERENTERING INFORMATION IN SOLVER
Invoke Solver by clicking Tools|Solver
 Specify Target Cell (D6)
 Specify Changing Cells (highlight B5, C5)
Flair Furniture
T C
Tables Chairs
Number Of Units
Profit 7 5 0 <-Objective
Constraints:
Carpentry Hours 4 3 0 <= 240
Painting Hours 2 1 0 <= 100
Chairs Limit 1 0 <= 60
LHS Sign RHS

109. CONSTRAINTSCONSTRAINTS
Specifying Constraints
 Use "Add" constraints to enter relevant cell references for
LHS and RHS.
 Either add constraints one at a time or add blocks of
constraints having same sign (<=, >=, or =) at the same
time.
 Since all constraints have same <= sign one chose to
highlight all LHS D8:D10 on left and F8:F10 on right
with <= sign.

110. CONSTRAINTSCONSTRAINTS
Specifying Constraints

111. SOLVER OPTIONSSOLVER OPTIONS
 Click on Options
button to get Solver
Options window
 One must check
boxes titled
– Assume Linear
Model
– Assume Non-
Negative

112. SOLVING MODELSOLVING MODEL
When Solve button is clicked, Solver executes
model and results appear as shown.
Solver Results
window also
indicates the
availability of
three reports –
- Answer.
- Sensitivity.
- Limits.

113. SSOLUTIONOLUTION
Optimal solution indicated that one should make 30 Tables
and 40 chairs with an optimal profit of $ 410.
Flair Furniture
T C
Tables Chairs
Number Of Units 30 40
Profit 7 5 410 <-Objective
Constraints:
Carpentry Hours 4 3 240 <= 240
Painting Hours 2 1 100 <= 100
Chairs Limit 1 40 <= 60
LHS Sign RHS

114. PPOSSIBLE MESSAGES INOSSIBLE MESSAGES IN
RESULTS WINDOWRESULTS WINDOW

115. FFLAIR FURNITURE SOLVER ANSWERLAIR FURNITURE SOLVER ANSWER
REPORTREPORT

116. UUSING SOLVER TO SOLVESING SOLVER TO SOLVE
HOLIDAY MEAL CHICKEN RANCHHOLIDAY MEAL CHICKEN RANCH
PROBLEMPROBLEM
LP formulation for this problem is as follows:
Minimize cost (in cents) = 2A + 3B
subject to constraints
5A + 10B ≥ 90 (protein constraint)
4A + 3B ≥ 48 (vitamin constraint)
½A ≥ 1½ (iron constraint)
A, B ≥ 0 (nonnegativity)

117. HOLIDAY MEAL CHICKEN RANCHHOLIDAY MEAL CHICKEN RANCH
PROBLEM SPREADSHEETPROBLEM SPREADSHEET

118. EEXCEL LAYOUT AND SOLVERXCEL LAYOUT AND SOLVER
ENTRIESENTRIES

119. SSOLVER ANSWER REPORTOLVER ANSWER REPORT

120. SUMMARYSUMMARY
 A mathematical modeling technique called linear
programming (LP) is introduced
 LP models are used to find an optimal solution to
problems that have a series of constraints binding the
objective value.
 How models with only two decision variables can be
solved graphically is shown
 To solve LP models with numerous decision variables
and constraints, one need a solution procedure such as
simplex algorithm.
 How LP models can be set up on Excel and solved using
Solver is demonstrated