1. Modeling with Linear Programming and Graphical Solution

2. Steps in formulating Linear Programming (LP) Models
Understand the problem.
Identify the decision variables
State the objective function as a linear combination of the decision variables.
State the constraints as linear combinations of the decision variables.
Identify any upper or lower bounds on the decision variables.

3. LP Model Formulation: A Maximization Example 1 – Product mix problem - Beaver Creek Pottery Company (1 of 4)
Beaver Creek Pottery Company employs skilled artisans to produce clay bowls and mugs. The two resources used by the company are special pottery clay and skilled labor. Given these limited resources, the company desires to know how many bowls and mugs to produce each day in order to maximize profit.
This is generally referred to as a product mix problem.

4. Resource Requirements
LP Model Formulation
A Maximization Example 1 (2 of 4)
Resource Requirements
Product
Labor
(Hr./Unit)
Clay
(Lb./Unit)
Profit
($/Unit)
Bowl 1 4 40
Mug 2 3 50
Product mix problem - Beaver Creek Pottery Company
How many bowls and mugs should be produced to maximize profits given labor and materials constraints?
Resource Availability: hrs of labor per day (labor constraint)
120 lbs of clay (material constraint)

5. A Maximization Example 1 (3 of 4)
LP Model Formulation
A Maximization Example 1 (3 of 4)
Resource hrs of labor per day
Availability: 120 lbs of clay
Decision x1 = number of bowls to produce per day
Variables: 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:

6. A Maximization Example 1 (4 of 4)
LP Model Formulation
A Maximization Example 1 (4 of 4)
Complete Linear Programming Model:
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0

7. Feasible Solutions
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
Clay constraint check: 4(5) + 3(10) = 70 ≤ 120 pounds

8. Infeasible Solutions
An infeasible solution violates at least one of the constraints:
Example: x1 = 10 bowls
x2 = 20 mugs
Z = $40x1 + $50x2 = $1400
Labor constraint check: 1(10) + 2(20) = 50 > 40 hours

9. Example 2 of LP Problem - Blue Ridge Hot Tubs
Blue Ridge Hot Tubs produces two types of hot tubs: Aqua-Spas & Hydro-Luxes.
Aqua-Spa Hydro-Lux
Pumps 1 1
Labor 9 hours 6 hours
Tubing 12 feet 16 feet
Unit Profit $350 $300
There are 200 pumps, 1566 hours of labor, and 2880 feet of tubing available.

10. Formulating LP Model: Blue Ridge Hot Tubs
1. Understand the problem
2. Identify the decision variables
X1=number of Aqua-Spas to produce
X2=number of Hydro-Luxes to produce
3. State the objective function as a linear combination of the decision variables.
MAX: 350X X2
4. State the constraints as linear combinations of the decision variables
1X1 + 1X2 <= 200 } pumps
9X1 + 6X2 <= 1566 } labor
12X1 + 16X2 <= 2880 } tubing
5. Identify any upper or lower bounds on the decision variables.
X1 >= 0 , X2 >= 0

11. Summary of the LP Model for Blue Ridge Hot Tubs
MAX Z = 350X X2
S.T.
1X X <= 200
9X X2 <= 1566
12X1 + 16X2 <= 2880
X1 >= 0
X2 >= 0

12. Example 3:Wershon Suit Company
Jackets Slacks Available
Profit, $/unit
Material,
Square yards
Person Hours
How many jackets and slacks should be produced ? 25

13. Example 3:Wershon Suit Company
Type of Objective Function
Maximize Profit
Variable Definition
J = number of jackets produced / week
S = number of slacks produced / week 25

14. Example 3:Wershon Suit Company
Max Z = $34 J + $40 S
Such that
2 J + 5 S  50 (Material)
4 J + 2 S  36 (sewing)
J , S  (Nonnegativity) 25

15. Example 4: Product Mix Problem
TJ’s, Inc., makes 2 nut mixes for sale to grocery chains in the states. The 2 mixes, referred to as the Regular Mix and the Deluxe Mix, are made by mixing different percentages of 3 types of nuts. In preparation for the fall season, TJ’s has just purchased 6000 pounds of almonds, 6000 pounds of pecans, and 7500 pounds of walnuts. The Regular Mix consists of 30% almonds, 20% pecans, and 50% walnuts. The Deluxe Mix consists of 35% of almonds, 30% pecans, and 35% walnuts. TJ’s accountant has analyzed the cost of packaging materials, sales price per pound, and other factors and has determined that the profit contribution per pound is $1.65 for the Regular Mix and $2.00 for the Deluxe Mix. TJ’s is committed to using the available nuts to maximize total profit contribution over the fall season. 20

16. Example 4: Product Mix Problem
Let R = amount of regular mix (pounds)
D = amount of deluxe mix (pounds)
Max z = 1.65R D (Total profit)
s.t. 0.3R D  6000 (Availability with
0.2R + 0.3D  ingredient
0.5R D  specifications)
R, D  (Nonnegativity) 21

17. Example 5: Transportation Problem
A product is to be shipped in the amounts al, a2, ..., am from m shipping origins and received in amounts bl, b2, ..., bn at each of n shipping destinations.
The cost of shipping a unit from the ith origin to the jth destination is known for all combinations of origins and destinations.
The problem is to determine the amount to be shipped from each origin to each destination such that the total cost of transportation is a minimum.

18. Transportation Problem – Example 5
The Zephyr Television Company ships televisions from three warehouses to three retail stores on a monthly basis. Each warehouse has a fixed supply per month, and each store has a fixed demand per month. The manufacturer wants to know the number of television sets to ship from each warehouse to each store in order to minimize the total cost of transportation.

19. Demand & Supply
Each warehouse has the following supply of televisions available for shipment each month:
Warehouse Supply (sets)
1. Cincinnati
2. Atlanta
3. Pittsburgh
700
Each retail store has the following monthly demand for television sets:
Store Demand (sets)
A. New York
B. Dallas
C. Detroit
600

20. Cost Matrix
Costs of transporting television sets from the warehouses to the retail stores vary as a result of differences in modes of transportation and distances. The shipping cost per television set for each route is as follows:
To Store
From
Warehouse A B C
1 $16 $18 $11

21. Model Development
The model for this problem consists of nine decision variables, representing the number of television sets transported from each of the three warehouses to each of the three stores:
xij = number of television sets shipped from warehouse i to store j
where i = 1, 2, 3, and j = A, B, C
The objective function of the television manufacturer is to minimize the total transportation costs for all shipments. Thus, the objective function is the sum of the individual shipping costs from each warehouse to each store:
minimize Z = $16x1A + 18x1B + 11x1C + 14x2A + 12x2B + 13x2C + 13x3A + 15x3B + 17x3C
In a "balanced" transportation model, supply equals demand such that all constraints are equalities; in an "unbalanced" transportation model, supply does not equal demand, and one set of constraints is ≤.

22. Constraints Supply Constraints x1A + x1B + x1C ≤ 300
Demand Constraints
x1A + x2A + x3A = 150
x1B + x2B + x3B = 250
x1C + x2C + x3C = 200

23. Model Summary
minimize Z = $16x1A + 18x1B + 11x1C + 14x2A + 12x2B + 13x2C + 13x3A + 15x3B + 17x3C
subject to
The transportation model can also be optimally solved by Linear Programming

24. Example 6: Scheduling Problem
The personnel manager must schedule a security force in order to satisfy staffing requirements shown below. Each worker has an eight hour shift and there are six such shifts each day. The starting and ending time for each of the 6 shifts is also given below. The personnel manager wants to determine how many people need to work each shift in order to minimize the total number of officers employed while satisfying the staffing requirements.
Time
# Officers Required
Shift
Shift Time
12am-4am 5 1
12am-8am 7 15 12 9
4am-8am
8am-noon
noon-4pm
4pm-8pm
8pm-12am
4am-noon
8am-4pm
noon-8pm
4pm-12am
8pm-4am 2 3 4 6 26

25. Example 6: Scheduling Problem
Let xi = number of officers who work on shift i, i = 1, ..., 6
Min z = x1 + x2 + x3 + x4 + x5 + x6
(Total number of officers employed)
s.t. x6 + x1  5 (12am-4am)
x1 + x2  7 (4am-8am)
x2 + x3  15 (8am-noon)
x3 + x4  7 (noon-4pm)
x4 + x5  12 (4pm-8pm)
x5 + x6  9 (8pm-12am)
xi  0, i = 1, ..., 6 (Nonnegativity) 27

26. Example 7: An Investment Example
Kathleen Allen, an individual investor, has $70,000 to divide among several investments. The alternative investments are municipal bonds with an 8.5% annual return, certificates of deposit with a 5% return, treasury bills with a 6.5% return, and a growth stock fund with a 13% annual return. The investments are all evaluated after 1 year. Kathleen wants to know how much to invest in each alternative in order to maximize the return.
The following guidelines have been established for diversifying the investments and lessening the risk perceived by the investor:
1. No more than 20% of the total investment should be in municipal bonds.
2. The amount invested in certificates of deposit should not exceed the amount invested in the other three alternatives.
3. At least 30% of the investment should be in treasury bills and certificates of deposit.
4. To be safe, more should be invested in CDs and treasury bills than in municipal bonds and the growth stock fund, by a ratio of at least 1.2 to 1.
Kathleen wants to invest the entire $70,000.

27. An Investment Example - Model
Decision Variables:
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
The Objective Function:
Maximize Z = $0.085x x x x4
Constraints:
x1 ≤$14,000
Model Summary
x2 ≤x1 + x3 + x4
x2 + x3 ≥$21,000
[(x2 + x3)/(x1 + x4)] ≥1.2
x1 + x2 + x3 + x4 = $70,000

28. Linear Programming (LP)
General Description
Problem: to determine decision variables
Objective: to maximize or minimize an objective function
Restrictions: represented by constraints
Solution methods: graphical, simplex, computer 14

29. General Form of a Linear Programming (LP) Problem
MAX (or MIN): c1X1 + c2X2 + … + cnXn
Subject to a11X1 + a12X2 + … + a1nXn  b1 :
ak1X1 + ak2X2 + … + aknXn ≥ bk
am1X1 + am2X2 + … + amnXn = bm
X1 , X2 …… Xn ≥ 0
Note: If all the functions in the model are linear, the problem is a Linear Programming (LP) problem

30. Graphical Solution of LP Models
Graphical solution is limited to linear programming models containing only two decision variables.
Graphical methods provide visualization of how a solution for a linear programming problem is obtained.

31. Example 1 – Product mix problem - Beaver Creek Pottery Company
Complete Linear Programming Model:
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0

32. Graphical Solution of Maximization Model (1 of 11)
Coordinate Axes
Graphical Solution of Maximization Model (1 of 11)
X2 is mugs
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
X1 is bowls
Figure: Coordinates for graphical analysis

33. Graphical Solution of Maximization Model (2 of 11)
Labor Constraint
Graphical Solution of Maximization Model (2 of 11)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
Figure: Graph of labor constraint

34. Graphical Solution of Maximization Model (3 of 11)
Labor Constraint Area
Graphical Solution of Maximization Model (3 of 11)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
Figure: Labor constraint area

35. Graphical Solution of Maximization Model (4 of 11)
Clay Constraint Area
Graphical Solution of Maximization Model (4 of 11)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
Figure: The constraint area for clay

36. Graphical Solution of Maximization Model (5 of 11)
Both Constraints
Graphical Solution of Maximization Model (5 of 11)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
Figure: Graph of both model constraints

37. Feasible Solution Area
Graphical Solution of Maximization Model (6 of 11)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
Figure: The feasible solution area

38. Objective Function Solution = $800
Graphical Solution of Maximization Model (7 of 11)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
Figure: Objective function line for Z = $800

39. Alternative Objective Function Solution Lines
Graphical Solution of Maximization Model (8 of 11)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
Figure: Alternative objective function lines for profits Z of $800, $1,200, and $1,600

40. Graphical Solution of Maximization Model (9 of 11)
Optimal Solution
Graphical Solution of Maximization Model (9 of 11)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
Figure: Identification of optimal solution point

41. Optimal Solution Coordinates
Graphical Solution of Maximization Model (10 of 11)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
Figure: Optimal solution coordinates

42. Extreme (Corner) Point Solutions
Graphical Solution of Maximization Model (11 of 11)
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2  40
4x2 + 3x2  120
x1, x2  0
The optimal solution of a maximization problem is at the extreme point farthest to the origin.
Figure: Solutions at all corner points

43. Slack Variables
Standard form requires that all constraints be in the form of equations (equalities).
A slack variable is added to a  constraint to convert it to an equation (=).
A slack variable typically represents an unused resource.
A slack variable contributes nothing to the objective function value.

44. Linear Programming Model: Standard Form
Max 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
Figure: Solutions at points A, B, and C with slack

45. LP Model Formulation – Minimization (1 of 6)
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 ?

46. LP Model Formulation – Minimization (2 of 6)
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 (non-negativity constraint)

47. Constraint Graph – Minimization (3 of 6)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2  16
4x2 + 3x2  24
x1, x2  0
Figure: Constraint lines for fertilizer model

48. Feasible Region– Minimization (4 of 6)
Minimize Z = $6x1 + $3x2
subject to: 2x1 + 4x2  16
4x2 + 3x2  24
x1, x2  0
Figure: Feasible solution area

49. Graphical Solutions – Minimization (5 of 6)
Minimize Z = $6x1 + $3x2 + 0s1 + 0s2
subject to: 2x1 + 4x2 – s1 = 16
4x2 + 3x2 – s2 = 24
x1, x2, s1, s2  0
The optimal solution of a minimization problem is at the extreme point closest to the origin.
Figure: Graph of the fertilizer example

50. Surplus Variables – Minimization (6 of 6)
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.
A surplus variable contributes 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)

51. Irregular Types of Linear Programming Problems
For some linear programming models, the general rules do not apply.
Special types of problems include those with:
Multiple optimal solutions
Infeasible solutions
Unbounded solutions

52. Multiple Optimal Solutions Beaver Creek Pottery
The objective function is
parallel to a constraint line.
Maximize Z=$40x1 + 30x2
subject to: 1x1 + 2x2  40
4x1 + 3x2  120
x1, x2  0
Where:
x1 = number of bowls
x2 = number of mugs
Figure: Example with multiple optimal solutions

53. 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
Figure: Graph of an infeasible problem

54. An Unbounded Problem Value of the objective
function increases indefinitely:
Maximize Z = 4x1 + 2x2
subject to: x1  4
x2  2
x1, x2  0
Figure: Graph of an unbounded problem

55. Characteristics of Linear Programming Problems
A decision amongst alternative courses of action is required.
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.
Restrictions (represented by constraints) exist that limit the extent of achievement of the objective.
The objective and constraints must be definable by linear mathematical functional relationships.

56. 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.
Divisibility - 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).