1. Part I: Linear Programming Model Formulation and Graphical Solution
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.
QP5013 – LINEAR PRORAMMING

3. 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.
QP5013 – LINEAR PRORAMMING

4. 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:
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:
x10; 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
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. A slack variable represents unused resources.
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.
QP5013 – LINEAR PRORAMMING

21. 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
QP5013 – LINEAR PRORAMMING

22. 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 ?
QP5013 – LINEAR PRORAMMING

23. 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. Complete model formulation:
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, x2  0
QP5013 – LINEAR PRORAMMING

25. 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
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)
QP5013 – LINEAR PRORAMMING

28. 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
QP5013 – LINEAR PRORAMMING

30. 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.
QP5013 – LINEAR PRORAMMING

34. 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  50
x2  200
x1 - 2x2  0
QP5013 – LINEAR PRORAMMING

38. Solve the following model graphically:
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
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
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
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 = $90x x2 + 45x3 + 65x4
Model constraints:
0.10x x x x4  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 = $90x x2 + 45x3 + 65x4
subject to:
0.10x x x x4  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 = $90x x2 + 45x3 + 65x4
subject to:
0.10x x x x4  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 Example Data 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.
QP5013 – LINEAR PRORAMMING

48. 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
QP5013 – LINEAR PRORAMMING

49. Diet Example Model Summary
minimize Z =0.18x x x x x x x x x9 0.07x10
subject to
90x x x3 + 90x4 + 75x5 + 35x6 + 65x x x9 + 65x10  420
2x x3 + 2x4 + 5x5 + 3x6 + 4x8 + x10  20
270x5 + 8x6 + 12x8  30
6x1 + 4x2 + 2x3 + 3x4+ x5 + x7 + x10  5
20x1 + 48x x3 + 8x4+ 30x5 + 52x x8 + 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 = $0.085x x x x4
subject to
x1  14,000
x2 - x1 - x3- x4  0
x2 + x3  21,000
-1.2x1 + x2 + x x4  0
x1 + x2 + x3 + x4 = 70,000
x1, x2, x3, x4  0
where
x1 = amount invested in municipal bonds ($)
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.
QP5013 – LINEAR PRORAMMING

52. Transportation Example Problem Definition and Data
Warehouse supply of televisions sets: Retail store demand for television sets:
1- Cinncinnati A. - New York 150
2- Atlanta B. - Dallas 250
3- Pittsburgh C. - Detroit 200
total total 600
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.
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 x2s x3s  0
0.70x2s x1s x3s  0
0.60x1p x2p x3p  0
0.75x3p x1p x2p  0
0.40x1e- 0.60x2e x3e  0
0.90x2e x1e x3e  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
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
QP5013 – LINEAR PRORAMMING

58. A Data Envelopment Analysis (DEA) Example Decision Variables and Model Summary
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 y y y3 = 1
86x x2 + 71x3 .06y y y3
82x x2 + 67x3  .05y y y3
81x x2 + 80x3  .08y y y3
81x x2 + 69x3  .06y y y3
xi, yi  0
QP5013 – LINEAR PRORAMMING

59. 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?
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) (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 xfm xcm xhb xfb 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  xhb- xfb - xcb  xhm + xfm + xcm + xhb + xfb+ xcb  36,000 ounces xij  0
QP5013 – LINEAR PRORAMMING

62. Additional Exercises Chap 2 - No 36 & 38; page 62;
Chap 3 – No 8, 9, 10, 17, 19; pg 92;
Chap , 21; pg 146
Group assignment: Case Problem “Summer Sports Camp at State University”
Please try these questions & we will discuss it during our class.
QP5013 – LINEAR PRORAMMING