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Solving Linear Programming ModelsPowerPoint Presentation

Solving Linear Programming Models

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Solving Linear Programming Models

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Topics

- Computer Solution
- Sensitivity Analysis

Product mix problem - Beaver Creek Pottery Example (1 of 2)

- 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: 40 hrs of labor per day (labor constraint)
120 lbs of clay (material constraint)

Product mix problem - Beaver Creek Pottery Example (2 of 2)

Complete Linear Programming Model:

x1 = number of bowls to produce per day

x2 = number of mugs to produce per day

MaximizeZ = $40x1 + $50x2

subject to:1x1 + 2x2 40

4x2 + 3x2 120

x1, x2 0

Excel Spreadsheet – Data Screen (1 of 5)

Click on “Data” tab to invoke “Solver.”

=C6*B10+D6*B11

=G6-E6

Decision variable—bowls (x1)=B10; mugs (x2)=B11

=G7-E7

=C7*B10+D7*B11

Objective function =C4*B10+D4*B11

“Solver” Parameter Screen(2 of 5)

Objective function

Decision variables

Click on “Add” to add model contraints.

=C7*B10+D7*B11<20

=C6*B10+D6*B11<40

x1, x2>0

Select “Simplex LP” method

Solver parameters

“Solver” Settings (4 of 5)

Slack—S1=0 and S2=0

Slack S1 = 0

and S2 = 0

Solution screen

Graphical Solution

Maximize Z = $40x1 + $50x2

subject to: x1 + 2x2 40

4x1 + 3x2 120

x1, x2 0

Optimal solution point

Sensitivity analysis determines the effect on the optimal solution of changes in parameter values of the objective function and constraint equations.

Changes may be reactions to anticipated uncertainties in the parameters or to new or changed information concerning the model.

Objective Function Coefficient

Sensitivity Range

objective function Z = $40x1 + $50x2 sensitivity range for:

x1: 25 c1 66.67 x2: 30 c2 80

The sensitivity range for an objective function coefficient is the range of values over which the current optimal solution point will remain optimal.

Objective Function Coefficient Ranges

Beaver Creek Example Sensitivity Report

Sensitivity ranges for objective function coefficients

sensitivity range for:

x1: 25 c1 66.67 x2: 30 c2 80

Changes in Constraint Quantity Values

Sensitivity Range

- The sensitivity range for a right-hand-side value is the range of values over which the quantity’s value can change without changing the solution variable mix, including the slack variables.
- Recall the Beaver Creek Pottery example.

Maximize Z = $40x1 + $50x2 subject to:

x1 + 2x2 40 hr of labor

4x1 + 3x2 120 lb of clay

x1, x2 0

Constraint Quantity Value Ranges by Computer

Excel Sensitivity Range for Constraints

Sensitivity ranges for constraint quantity values

the sensitivity range for clay quantity q2 is 60≤ q2 ≤160 lb.

the sensitivity range for the labor hours q1 is 30 ≤q1 ≤80 hr.

Excel Sensitivity Report for Beaver Creek Pottery

Shadow Prices Example

Maximize Z = $40x1 + $50x2 subject to:

x1 + 2x2 40 hr of labor

4x1 + 3x2 120 lb of clay

x1, x2 0

Shadow prices (dual values)

The shadow price (or marginal value) for labor is $16 per hour, and the shadow price for clay is $6 per pound. This means that for every additional hour of labor that can be obtained, profit will increase by $16 and for every additional lb of clay the profit increases by $6. The upper limit of the sensitivity range for the labor & clay are 80 hours & 160 lb and the lower limits are 30 hours & 60 lb, before the optimal solution mix changes.

- With every linear programming problem, there is associated another linear programming problem which is called the dualof the original (or the primal) problem.

Shadow price is also called as the marginal value of one additional unit of resource.

The sensitivity range for a constraint quantity value is also the range over which the shadow price is valid.

Primal and Dual problems for Beaver Creek Pottery Example

Primal Problem

Maximize Z = $40x1 + $50x2

subject to:

x1 + 2x2 40

4x1 + 3x2 120

x1, x2 0

Dual Problem

Minimize P = 40y1 + 120y2

subject to:

y1 + 4y2 ≥ 40

2y1 + 3y2 ≥ 50

y1, y2 0

Flair Furniture Company

The Flair Furniture Company produces tables and chairs. The production process for each is similar in that both require a certain number of hours of carpentry work and a certain number of labor hours in the painting and varnishing department. Each table takes 4 hours of carpentry and 2 hours in the painting and varnishing shop. Each chair requires 3 hours in carpentry and 1 hour in painting and varnishing. During the current production period, 240 hours of carpentry time are available and 100 hours in painting and varnishing time are available. Each table sold yields a profit of $70; each chair produced is sold for a $50 profit.

Formulate the LP Model.

Solve the model graphically.

Solve this model by using Excel.

Flair Furniture Company

T = number of tables to be produced per week

C = number of chairs to be produced per week

Maximize profit Z= $70T + $50C

subject to the constraints

4T + 3C ≤ 240 (carpentry constraint)

2T + 1C ≤ 100 (painting and varnishing constraint)

T, C ≥0 (non-negativity constraints)

sensitivity range for T: 66.7 c1 100 C: 35 c2 52.5

The sensitivity range for the carpentry hours q1 is 200 ≤q1 ≤ 300

The sensitivity range for painting hours q2 is 80≤ q2 ≤120

The shadow price (or marginal value) for carpentry is $15 per hour, and the shadow price for painting is $5 per hour. This means that for every additional hour of carpentry that can be obtained, profit will increase by $15 and for every additional hour of painting the profit increases by $5.

Transportation Problem – Example

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.

Demand & Supply

Each warehouse has the following supply of televisions available for shipment each month:

Warehouse Supply (sets)

1. Cincinnati 300

2. Atlanta 200

3. Pittsburgh 200

700

Each retail store has the following monthly demand for television sets:

Store Demand (sets)

A. New York 150

B. Dallas 250

C. Detroit 200

600

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

2 14 12 13

3 13 15 17

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

Computer Solution with Excel

The solution is

x1C = 200 TVs shipped from Cincinnati to Detroit

x2B = 200 TVs shipped from Atlanta to Dallas

x3A = 150 TVs shipped from Pittsburgh to New York

x3B = 50 TVs shipped from Pittsburgh to Dallas

Z = $7,300 shipping cost