QP5013 – LINEAR PRORAMMING 1
Part I: Linear Programming Model Formulation and Graphical Solution
• Model Formulation
• A Maximization Model Example
• Graphical Solutions of Linear Programming Models
• A Minimization Model Example
• Irregular Types of Linear Programming Models
• Characteristics of Linear Programming Problems
QP5013 – LINEAR PRORAMMING 2
Linear Programming - An Overview
• Objectives of business firms frequently include maximizing profit or minimizing costs.
• Linear programming is an analysis technique in which linear algebraic relationships represent a firm’s decisions given a business objective and resource constraints.
• Steps in application:
1- Identify problem as solvable by linear programming.
2- Formulate a mathematical model of the unstructured problem.
3- Solve the model.
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Model Components and Formulation
• 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 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.
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A Maximization Model Example (1 of 2)
Problem Definition
• 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:
Resource Requirements
Product Labor (hr/unit)
Clay (lb/unit)
Profit ($/unit)
Bowl 1 4 40
Mug 2 3 50
QP5013 – LINEAR PRORAMMING 5
A Maximization Model Example (2 of 2) Resource availability: 40 hours of labor per day 120 pounds of clay Decision Variables:
x1=number of bowls to produce/day
x2= number of mugs to produce/day Objective function
maximize Z = $40x1 + 50x2
where Z= profit per day Resource Constraints:
1x1 + 2x2 40 hours of labor
4x1 + 3x2 120 pounds of clay Non-negativity Constraints:
x10; x2 0 Complete Linear Programming Model:
maximize Z=$40x1 + 50x2
subject to
1x1 + 2x2 40
4x2 + 3x2 120
x1, x2 0
QP5013 – LINEAR PRORAMMING 6
Feasible/Infeasible Solutions• A feasible solution does not violate any of the constraints:
Example x1= 5 bowls
x2= 10 mugs
Z = $40 x1 + 50x2= $700 Labor constraint check: 1(5) + 2(10) = 25 < 40 hours, within constraint Clay constraint check: 4(5) + 3(10) = 70 < 120 pounds, within constraint
• 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 constraint
QP5013 – LINEAR PRORAMMING 7
Graphical Solution of Linear Programming Models
• 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.
QP5013 – LINEAR PRORAMMING 8
Graphical Solution of a Maximization Model Coordinate Axes
maximize Z=$40x1 + 50x2
subject to 1x1 + 2x2 40 hours of labor 4x2 + 3x2 120 pounds of clay x1, x2 0
Coordinates for graphical analysis
QP5013 – LINEAR PRORAMMING 9
Graphical Solution of a Maximization Model Labor Constraint
maximize Z=$40x1 + 50x2
subject to 1x1 + 2x2 40 hours of labor 4x2 + 3x2 120 pounds of clay x1, x2 0
Graph of the labor constraint line
QP5013 – LINEAR PRORAMMING 10
Graphical Solution of a Maximization Model Labor Constraint Area
maximize Z=$40x1 + 50x2
subject to 1x1 + 2x2 40 hours of labor 4x2 + 3x2 120 pounds of clay x1, x2 0
The labor constraint area
QP5013 – LINEAR PRORAMMING 11
Graphical Solution of a Maximization Model Clay Constraint Area
maximize Z=$40x1 + 50x2
subject to 1x1 + 2x2 40 hours of labor 4x2 + 3x2 120 pounds of clay x1, x2 0
The constraint area for clay
QP5013 – LINEAR PRORAMMING 12
Graphical Solution of a Maximization Model Both Constraints
maximize Z=$40x1 + 50x2
subject to 1x1 + 2x2 40 hours of labor 4x2 + 3x2 120 pounds of clay x1, x2 0
Graph of both model Constraints
QP5013 – LINEAR PRORAMMING 13
Graphical Solution of a Maximization Model Feasible Solution Area
maximize Z=$40x1 + 50x2
subject to 1x1 + 2x2 40 hours of labor 4x2 + 3x2 120 pounds of clay x1, x2 0
The feasible solution area constraints
QP5013 – LINEAR PRORAMMING 14
Graphical Solution of a Maximization Model Objective Function = $800
Z= $800 = $40x1 + 50x2
subject to 1x1 + 2x2 40 hours of labor 4x2 + 3x2 120 pounds of clay x1, x2 0
Objective function line for Z 5 $800
QP5013 – LINEAR PRORAMMING 15
Graphical Solution of a Maximization Model Alternative Objective Functions
Z=$800, $1200, $1600 = $40x1 + 50x2
subject to 1x1 + 2x2 40 hours of labor 4x2 + 3x2 120 pounds of clay x1, x2 0
Alternative objective function lines for profits, Z, of $800, $1,200, and $1,600
QP5013 – LINEAR PRORAMMING 16
Graphical Solution of a Maximization Model Optimal Solution
Z= $800 =$40x1 + 50x2
subject to 1x1 + 2x2 40 hours of labor 4x2 + 3x2 120 pounds of clay x1, x2 0
Identification of optimal solution point
QP5013 – LINEAR PRORAMMING 17
Graphical Solution of a Maximization Model Optimal Solution Coordinates
maximize Z=$40x1 + 50x2
subject to 1x1 + 2x2 40 hours of labor 4x2 + 3x2 120 pounds of clay x1, x2 0
Optimal solution coordinates
QP5013 – LINEAR PRORAMMING 18
Graphical Solution of a Maximization Model Corner Point Solutions
maximize Z=$40x1 + 50x2
subject to 1x1 + 2x2 40 hours of labor 4x2 + 3x2 120 pounds of clay x1, x2 0
Solutions at all corner points
QP5013 – LINEAR PRORAMMING 19
Graphical Solution of a Maximization Model Optimal Solution for New Objective Function
maximize Z=$40x1 + 50x2
subject to 1x1 + 2x2 40 hours of labor 4x2 + 3x2 120 pounds of clay x1, x2 0
The optimal solution with Z 5 70x1 1 20x2
QP5013 – LINEAR PRORAMMING 20
Slack Variables
• 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.
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Complete Linear Programming Model in Standard Form
maximize Z=$40x1 + 50x2 + 0s1 + 0s2
subject to 1x1 + 2x2 + s1 = 40 4x2 + 3x2 + s2 = 120 x1,x2,s1,s2 = 0
where x1 = number of bowls x2 = number of mugs s1, s2 are slack variables
Solutions at points A, B, and C with slack
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A Minimization Model Example Problem Definition
• 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 ?
Chemical Contribution
Brand
Nitrogen (lb/bag)
Phosphate (lb/bag)
Super-gro
2
4
Crop-quick
4
3
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A Minimization Model Example Model Construction
Decision variables x1 = bags of Super-gro
x2 = bags of Crop-quick
The objective function:
minimize Z = $6x1 + 3x2
where $6x1 = cost of bags of Super-gro
3x2 = cost of bags of Crop-quick
Model constraints:
2x1 + 4x2 16 lb (nitrogen constraint)
4x1 + 3x2 24 lb (phosphate constraint)
x1, x2 0 (nonnegativity constraint)
QP5013 – LINEAR PRORAMMING 24
A Minimization Model ExampleComplete Model Formulation and Constraint Graph
Complete model formulation:
minimize Z = $6x1 + 3x2
subject to
2x1 + 4x2 16 lb of nitrogen
4x1 + 3x2 24 lb of phosphate
x1, x2 0
QP5013 – LINEAR PRORAMMING 25
A Minimization Model ExampleFeasible Solution Area
minimize Z = $6x1 + 3x2
subject to
2x1 + 4x2 16 lb of nitrogen
4x1 + 3x2 24 lb of phosphate
x1, x2 0
Feasible solution area
QP5013 – LINEAR PRORAMMING 26
A Minimization Model Example Optimal Solution Point
minimize Z = $6x1 + 3x2
subject to
2x1 + 4x2 16 lb of nitrogen
4x1 + 3x2 24 lb of phosphate
x1, x2 0
The optimal solution point
QP5013 – LINEAR PRORAMMING 27
A Minimization Model Example Surplus Variables
• 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 slack variables in the farmer problem constraints:
2x1 + 4x2 - s1 = 16 (nitrogen)
4x1 + 3x2 - s2 = 24 (phosphate)
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A Minimization Model Example Graphical Solutions
minimize Z = $6x1 + 3x2 + 0s1 + 0s2
subject to
2x1 + 4x2 - s1 = 16
4x1 + 3x2 - s2 = 24
x1, x2, s1, s2 = 0
Graph of the fertilizer example
QP5013 – LINEAR PRORAMMING 29
Irregular Types of Linear Programming Problems
• For some linear programming models, the general rules do not apply.
• Special types of problems include those with:
1. Multiple optimal solutions
2. Infeasible solutions
3. Unbounded solutions
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Multiple Optimal Solutions
Objective function is parallel to a
constraint line:
maximize Z=$40x1 + 30x2
subject to 1x1 + 2x2 40 hours of labor 4x2 + 3x2 120 pounds of clay x1, x2 0 where x1 = number of bowls x2 = number of mugs
Graph of the Beaver Creek Pottery Company example with multiple optimal solutions
QP5013 – LINEAR PRORAMMING 31
An Infeasible Problem
Every possible solution violates at least one constraint:
maximize Z = 5x1 + 3x2
subject to
4x1 + 2x2 8
x1 4
x2 6
x1, x2 0
Graph of an infeasible problem
QP5013 – LINEAR PRORAMMING 32
An Unbounded Problem
Value of objective function increases indefinitely:
maximize Z = 4x1 + 2x2
subject to
x1 4
x2 2
x1, x2 0
An unbounded problem
QP5013 – LINEAR PRORAMMING 33
Characteristics of Linear Programming Problems
• A linear programming problem requires a decision - a choice amongst alternative courses of action.
• The decision is represented in the model by decision variables.
• The problem encompasses a goal, expressed as an objective function, that the decision maker wants to achieve.
• Constraints exist that limit the extent of achievement of the objective.
• The objective and constraints must be definable by linear mathematical functional relationships.
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Properties of Linear Programming Models
• 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.
• Divisability -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).
QP5013 – LINEAR PRORAMMING 35
Example Problem No. 1 Problem Statement
- Hot dog mixture in 1000-pound batches.
- Two ingredients, chicken ($3/lb) and beef ($5/lb),
- Recipe requirements:
at least 500 pounds of chicken
at least 200 pounds of beef.
- Ratio of chicken to beef must be at least 2 to 1.
- Determine optimal mixture of ingredients that will minimize costs.
QP5013 – LINEAR PRORAMMING 36
Example Problem No. 1 Solution
Step 1: Identify decision variables.
x1 = lb of chicken
x2 = lb of beef
Step 2: Formulate the objective function.
minimize Z = $3x1 + 5x2
where Z = cost per 1,000-lb batch
$3x1 = cost of chicken
5x2 = cost of beef
QP5013 – LINEAR PRORAMMING 37
Example Problem No.1 Solution (continued)
Step 3: Establish Model Constraints
x1 + x2 = 1,000 lb
x1 500 lb of chicken
x2 200 lb of beef
x1/x2 2/1 or x1 - 2x2 0
x1,x2 0
The model: minimize Z = $3x1 + 5x2
subject to
x1 + x2 = 1,000 lb
x1 50
x2 200
x1 - 2x2 0
x1,x2 0
QP5013 – LINEAR PRORAMMING 38
Example Problem No.2
Solve the following model graphically:
maximize Z = 4x1 + 5x2
subject to
x1 + 2x2 10
6x1 + 6x2 36
x1 4
x1,x2 0
Step 1: Plot the constraint s as equations:
The constraint equations
QP5013 – LINEAR PRORAMMING 39
Example Problem No.2
maximize Z = 4x1 + 5x2
subject to
x1 + 2x2 1
6x1 + 6x2 36
x1 4
x1,x2 0
Step 2: Determine the feasible solution area:
The feasible solution space and extreme points
QP5013 – LINEAR PRORAMMING 40
Example Problem No.2
maximize Z = 4x1 + 5x2
subject to
x1 + 2x2 10
6x1 + 6x2 36
x1 4
x1,x2 0
Steps 3 and 4:
Determine the solution points and optimal solution.
Optimal solution point
QP5013 – LINEAR PRORAMMING 41
Part II: Linear Programming Modeling Examples
• A Product Mix Example• A Diet Example• An Investment Example• A Marketing Example• A Transportation Example• A Blend Example • A Multiperiod Scheduling Example• A Data Envelopment Analysis Example
QP5013 – LINEAR PRORAMMING 42
Product Mix Example Problem Definition
- Four-product T-shirt/sweatshirt manufacturing company.
- Must complete production within 72 hours
- Truck capacity = 1,200 standard sized boxes.
- Standard size box holds12 T-shirts.
- One-dozen sweatshirts box is three times size of standard box.
- $25,000 available for a production run.
- 500 dozen blank T-shirts and sweatshirts in stock.
- How many dozens (boxes) of each type of shirt to produce?
QP5013 – LINEAR PRORAMMING 43
Product Mix Example Data
Processing Time (hr) Per dozen
Cost ($)
per dozen
Profit ($)
per dozen Sweatshirt - F
0.10
36
90
Sweatshirt – B/F
0.25
48
125
T-shirt - F
0.08
25
45
T-shirt - B/F
0.21
35
65
QP5013 – LINEAR PRORAMMING 44
Product Mix Example Model Construction
Decision variables:
x1 = sweatshirts, front printing
x2 = sweatshirts, back and front printing
x3 = T-shirts, front printing
x4 = T-shirts, back and front printing
Objective function:
maximize Z = $90x1 + 125x2 + 45x3 + 65x4
Model constraints:
0.10x1 + 0.25x2+ 0.08x3 + 0.21x4 72 hr
3x1 + 3x2 + x3 + x4 1,200 boxes
$36x1 + 48x2 + 25x3 + 35x4 $25,000
x1 + x2 500 dozen sweatshirts
x3 + x4 500 dozen T-shirts
QP5013 – LINEAR PRORAMMING 45
Product Mix Example Computer Solution with QM for Windows
maximize Z = $90x1 + 125x2 + 45x3 + 65x4
subject to:
0.10x1 + 0.25x2+ 0.08x3 + 0.21x4 72
3x1 + 3x2 + x3 + x4 1,200 boxes
$36x1 + 48x2 + 25x3 + 35x4 $25,000
x1 + x2 500 dozed sweatshirts
x3 + x4 500 dozen T-shirts
x1, x2, x3, x4 0
QP5013 – LINEAR PRORAMMING 46
Product Mix Example Computer Solution with QM for Windows (continued)
maximize Z = $90x1 + 125x2 + 45x3 + 65x4
subject to:
0.10x1 + 0.25x2+ 0.08x3 + 0.21x4 72
3x1 + 3x2 + x3 + x4 1,200 boxes
$36x1 + 48x2 + 25x3 + 35x4 $25,000
x1 + x2 500 dozed sweatshirts
x3 + x4 500 dozen T-shirts
x1, x2, x3, x4 0
QP5013 – LINEAR PRORAMMING 47
Diet ExampleData and Problem Definition
Breakfast to include at least 420 calaries, 5 milligrams of iron, 400 milligrams of calcium, 20 grams of protein, 12 grams of fiber, and must have no more than 20 grams of fat and 30 milligrams of cholesterol.
Breakfast Food
Calories
Fat (g)
Cholesterol (mg)
Iron (mg)
Calcium (mg)
Protein (g)
Fiber (g)
Cost ($)
1. Bran cereal (cup) 2. Dry cereal (cup) 3. Oatmeal (cup) 4. Oat bran (cup) 5. Egg 6. Bacon (slice) 7. Orange 8. Milk-2% (cup) 9. Orange juice (cup) 10. Wheat toast (slice)
90 110 100
90 75 35 65
100 120
65
0 2 2 2 5 3 0 4 0 1
0 0 0 0
270 8 0
12 0 0
6 4 2 3 1 0 1 0 0 1
20 48 12
8 30
0 52
250 3
26
3 4 5 6 7 2 1 9 1 3
5 2 3 4 0 0 1 0 0 3
0.18 0.22 0.10 0.12 0.10 0.09 0.40 0.16 0.50 0.07
QP5013 – LINEAR PRORAMMING 48
Diet ExampleModel Construction: Decision Variables
x1 = cups of bran cereal
x2 = cups of dry cereal
x3 = cups of oatmeal
x4 = cups of oat bran
x5 = eggs
x6 = slices of bacon
x7 = oranges
x8 = cups of milk
x9 = cups of orange juice
x10 = slices of wheat toast
QP5013 – LINEAR PRORAMMING 49
Diet ExampleModel Summary
minimize Z =0.18x1 + 0.22x2 + 0.10x3 + 0.12x4 + 0.10x5 + 0.09x6+ 0.40x7 + 0.16x8 + 0.50x9 0.07x10
subject to
90x1 + 110x2 + 100x3 + 90x4 + 75x5 + 35x6 + 65x7 + 100x8 + 120x9 + 65x10 420
2x2 + 2x3 + 2x4 + 5x5 + 3x6 + 4x8 + x10 20
270x5 + 8x6 + 12x8 30
6x1 + 4x2 + 2x3 + 3x4+ x5 + x7 + x10 5
20x1 + 48x2 + 12x3 + 8x4+ 30x5 + 52x7 + 250x8 + 3x9 + 26x10 400
3x1 + 4x2 + 5x3 + 6x4 + 7x5 + 2x6 + x7+ 9x8+ x9 + 3x10 20
5x1 + 2x2 + 3x3 + 4x4+ x7 + 3x10 12
xi 0
QP5013 – LINEAR PRORAMMING 50
An Investment Example Model Summary
maximize Z = $0.085x1 + 0.05x2 + 0.065 x3+ 0.130x4
subject to
x1 14,000
x2 - x1 - x3- x4 0
x2 + x3 21,000
-1.2x1 + x2 + x3 - 1.2 x4 0
x1 + x2 + x3 + x4 = 70,000
x1, x2, x3, x4 0
where
x1 = amount invested in municipal bonds ($)
x2 = amount invested in certificates of deposit ($)
x3 = amount invested in treasury bills ($)
x4 = amount invested in growth stock fund($)
QP5013 – LINEAR PRORAMMING 51
A Marketing Example Data and Problem Definition
- Budget limit $100,000
- Television time for four commercials
- Radio time for 10 commercials
- Newspaper space for 7 ads
- Resources for no more than 15 commercials and/or ads.
Exposure (people/ad or commercial)
Cost
Television commercial
20,000 $15,000
Radio commercial 12,000 6,000
Newspaper ad 9,000 4,000
QP5013 – LINEAR PRORAMMING 52
Transportation ExampleProblem Definition and Data
Warehouse supply of televisions sets: Retail store demand for television sets:
1- Cinncinnati 300 A. - New York 150
2- Atlanta 200 B. - Dallas 250
3- Pittsburgh 200 C. - Detroit 200
total 700 total 600
To Store From Warehouse
A B C
1 $16 $18 $11
2 14 12 13
3 13 15 17
QP5013 – LINEAR PRORAMMING 53
A Blend Example Problem Definition and Data
Determine the optimal mix of the three components in each grade of motor oil that will maximize profit. Company wants to produce at least 3,000 barrels of each grade of motor oil.
Component
Maximum Barrels
Available/day
Cost/barrel
1 4,500 $12
2 2,700 10
3 3,500 14
Grade
Component Specifications
Selling Price ($/bbl)
Super
At least 50% of 1
Not more than 30% of 2
$23
Premium
At least 40% of 1
Not more than 25% of 3
20
Extra
At least 60% of 1 At least 10% of 2
18
QP5013 – LINEAR PRORAMMING 54
A Blend Example Decision Variables and Model Summary
Decision variables: The quantity of each of the three components used in each grade of gasoline (9 decision variables); xij = barrels of component i used in motor oil grade j per day, where i = 1, 2, 3 and j = s (super), p(premium), and e(extra).
Model Summary: maximize Z = 11x1s + 13x2s + 9x3s + 8x1p + 10x2p + 6x3p + 6x1e + 8x2e + 4x3e
subject to
x1s + x1p + x1e 4,500
x2s + x2p + x2e 2,700
x3s + x3p + x3e 3,500
0.50x1s - 0.50x2s - 0.50x3s 0
0.70x2s - 0.30x1s - 0.30x3s 0
0.60x1p - 0.40x2p - 0.40x3p 0
0.75x3p - 0.25x1p - 0.25x2p 0
0.40x1e- 0.60x2e- - 0.60x3e 0
0.90x2e - 0.10x1e - 0.10x3e 0
x1s + x2s + x3s 3,000
x1p+ x2p + x3p 3,000
x1e+ x2e + x3e 3,000
xij 0
QP5013 – LINEAR PRORAMMING 55
A Multiperiod Scheduling Example Problem Definition and Data
Production capacity : 160 computers per week
Additional 50 computers with overtime
Assembly costs: $190/comp. regular time; $260/comp. overtime
Inventory cost: $10/comp. per week
Order schedule: Week Computer Orders
1 105
2 170
3 230
4 180
5 150
6 250
QP5013 – LINEAR PRORAMMING 56
A Multiperiod Scheduling Example Decision Variables and Model Summary
Decision variables:
rj = regular production of computers per week j (j = 1, 2, 3, 4, 5, 6)
oj = overtime production of computers per week j (j = 1, 2, 3, 4, 5, 6)
ij = extra computers carried over as inventory in week j (j = 1, 2, 3, 4, 5)
Model summary:
minimize Z = $190(r1 + r2 + r3 + r4 + r5 + r6) + $260(o1 + o2 + o3 + o4 + o5 +o6) + 10(i1, + i2 + i3 + i4 + i5)
subject to rj 160 (j = 1, 2, 3, 4, 5, 6)
oj 150 (j = 1, 2, 3, 4, 5, 6)
r1 + o1 - i1 105
r2 + o2 + i1 - i2 170
r3 + o3 + i2 - i3 230
r4 + o4 + i3 - i4 180
r5 + o5 + i4 - i5 150
r6 + o6 + i5 250
rj, oj, ij 0
QP5013 – LINEAR PRORAMMING 57
A Data Envelopment Analysis (DEA) Example Problem Definition and Data
DEA compares a number of service units of the same type based on their inputs (resources) and outputs. The result indicates if a particular unit is less productive, or efficient, than other units.
Elementary school comparison:
input 1 = teacher to student ratio output 1 = average reading SOL score
input 2 = supplementary $/student output 2 = average math SOL score
input 3 = parent education level output 3 = average history SOL score
Inputs
Outputs
School
1
2
3
1
2
3
Alton
.06
$260
11.3
86
75
71
Beeks
.05
320
10.5
82
72
67
Carey
.08
340
12.0
81
79
80
Delancey
.06
460
13.1
81
73
69
QP5013 – LINEAR PRORAMMING 58
A Data Envelopment Analysis (DEA) Example Decision Variables and Model Summary
Decision variables:
xi = a price per unit of each output where i = 1, 2, 3
yi = a price per unit of each input where i = 1, 2, 3
Model summary:
maximize Z = 81x1 + 73x2 + 69x3
subject to
.06 y1 + 460y2 + 13.1y3 = 1
86x1 + 75x2 + 71x3 .06y1 + 260y2 + 11.3y3
82x1 + 72x2 + 67x3 .05y1 + 320y2 + 10.5y3
81x1 + 79x2 + 80x3 .08y1 + 340y2 + 12.0y3
81x1 + 73x2 + 69x3 .06y1 + 460y2 + 13.1y3
xi, yi 0
QP5013 – LINEAR PRORAMMING 59
Example Problem SolutionProblem Statement and Data
- Canned catfood, Meow Chow; dogfood, Bow Chow.
- Ingredients/week: 600lb horse meat; 800 lb fish; 1000 lb cereal.
- Recipe requirement: Meow Chow at least half fish; Bow Chow at least half horse meat.
- 2,250 sixteen-ounce cans available each week.
-Profit /can: Meow Chow $0.80; Bow Chow$0.96.
- How many cans of Bow Chow and Meow Chow should be produced each week in order to maximize profit?
QP5013 – LINEAR PRORAMMING 60
Example Problem Solution Model Formulation
Step 1: Define the Decision Variables
xij = ounces of ingredient i in pet food j per week, where i = h (horse meat), f (fish) and
c (cereal), and j = m (Meow chow) and b (Bow Chow).
Step 2: Formulate the Objective Function
maximize Z = $0.05(xhm + xfm + xcm) + 0.06(xhb + xfb + xcb)
Step 3: Formulate the Model Constraints
Amount of each ingredient available each week:
xhm + xhb 9,600 ounces of horse meat
xfm + xfb 12,800 ounces of fish
xcm + xcb 16,000 ounces of cereal additive
Recipe requirements:
Meow Chow xfm/(xhm + xfm + xcm) 1/2, or, - xhm + xfm- xcm 0
Bow Chow xhb/(xhb + xfb + xcb) 1/2, or, xhb- xfb - xcb 0
Can content constraint: xhm + xfm + xcm + xhb + xfb+ xcb 36,000 ounces
QP5013 – LINEAR PRORAMMING 61
Example Problem Solution Model Summary and Solution with QM for Windows
Step 4: Model Summary
maximize Z = $0.05 xhm + 0.05 xfm + 0.05 xcm + 0.06 xhb + 0.06 xfb + 0.06 x subject to xhm + xhb 9,600 ounces of horse meat xfm + xfb 12,800 ounces of fish xcm + xcb 16,000 ounces of cereal additive - xhm + xfm- xcm 0 xhb- xfb - xcb 0 xhm + xfm + xcm + xhb + xfb+ xcb 36,000 ounces
xij 0
QP5013 – LINEAR PRORAMMING 62
• Chap 2 - No 36 & 38; page 62;
• Chap 3 – No 8, 9, 10, 17, 19; pg 92;
• Chap 4 - 20, 21; pg 146
• Group assignment: Case Problem “Summer Sports Camp at State University”
Please try these questions & we will discuss it during our class.
Additional Exercises