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
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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:
Product
Bowl
Mug
Resource Requirements
Labor
Clay
Profit
(hr/unit)
(lb/unit)
($/unit)
1
4
40
2
3
50
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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:
x10; x2  0
Complete Linear Programming Model:
maximize Z=$40x1 + 50x2
subject to
1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
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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
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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.
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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, x 2  0
Coordinates for graphical analysis
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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, x 2  0
Graph of the labor constraint line
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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
x 1, x2  0
The labor constraint area
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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
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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, x 2  0
Graph of both model Constraints
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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
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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, x 2  0
Objective function line for Z 5 $800
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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, x 2  0
Alternative objective function lines for profits, Z, of $800, $1,200, and $1,600
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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, x 2  0
Identification of optimal solution point
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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
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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, x 2  0
Solutions at all corner points
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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, x 2  0
The optimal solution with Z 5 70x1 1 20x2
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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
Nitrogen
(lb/bag)
Phosphate
(lb/bag)
Super-gro
2
4
Crop-quick
4
3
Brand
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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)
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A Minimization Model Example
Complete 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, x 2  0
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A Minimization Model Example
Feasible 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
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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
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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
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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
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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
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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
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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).
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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.
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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
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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
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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
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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
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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
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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
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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.
- $25,000 available for a production run.
- 500 dozen blank T-shirts and sweatshirts in stock.
- How many dozens (boxes) of each type of shirt to produce?
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Product Mix Example
Data
Processing
Time (hr)
Per dozen
Cost
($)
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
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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
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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
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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  $25,000
x1 + x2  500 dozed sweatshirts
x3 + x4  500 dozen T-shirts
x1, x2, x3, x4  0
QP5013 – LINEAR PRORAMMING
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Diet Example
Data and Problem Definition
Breakfast
Food
1. Bran cereal (cup)
2. Dry cereal (cup)
3. Oatmeal (cup)
4. Oat bran (cup)
5. Egg
6. Bacon (slice)
7. Orange
8. Milk-2% (cup)
9. Orange juice (cup)
10.Wheat toast (slice)
Fat
Calories (g)
90 0
110 2
100 2
90 2
75 5
35 3
65 0
100 4
120 0
65 1
Cholesterol
(mg)
0
0
0
0
270
8
0
12
0
0
Iron Calcium Protein Fiber
(mg)
(mg)
(g)
(g)
6
20
3
5
4
48
4
2
2
12
5
3
3
8
6
4
1
30
7
0
0
0
2
0
1
52
1
1
0
250
9
0
0
3
1
0
1
26
3
3
Cost
($)
0.18
0.22
0.10
0.12
0.10
0.09
0.40
0.16
0.50
0.07
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.
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Diet Example
Model 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
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Diet Example
Model 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
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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($)
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A Marketing Example
Data and Problem Definition
Exposure
(people/ad or
commercial)
Cost
Television
commercial
Radio commercial
20,000
$15,000
12,000
6,000
Newspaper ad
9,000
4,000
- 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.
QP5013 – LINEAR PRORAMMING
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Transportation Example
Problem 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
From
Warehouse
700
total
600
To Store
A
B
C
1
2
$16
14
$18
12
$11
13
3
13
15
17
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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
At least 60% of 1
At least 10% of 2
18
Extra
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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
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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
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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
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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
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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
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Example Problem Solution
Problem 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?
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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,
Can content constraint:
xhb- xfb - xcb  0
xhm + xfm + xcm + xhb + xfb+ xcb  36,000 ounces
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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
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Additional Exercises
•
•
•
•
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.
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