solving linear programming models. topics computer solution sensitivity analysis
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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)
Resource Requirements
ProductLabor
(Hr./Unit)Clay
(Lb./Unit)Profit($/Unit)
Bowl 1 4 40
Mug 2 3 50
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
Maximize Z = $40x1 + $50x2
subject to: 1x1 + 2x2 404x2 + 3x2 120x1, x2 0
Beaver Creek Pottery ExampleExcel Spreadsheet – Data Screen (1 of 5) Click on “Data”
tab to invoke “Solver.”
=C6*B10+D6*B11 =G6-E6
=G7-E7
=C7*B10+D7*B11
Decision variable—bowls (x1)=B10; mugs (x2)=B11Objective
function =C4*B10+D4*B11
Beaver Creek Pottery Example“Solver” Parameter Screen (2 of 5)
Solver parameters
Objective function
Decision variables Click on “Add”
to add model contraints.
=C6*B10+D6*B11<40
=C7*B10+D7*B11<20
x1, x2 >0
Select “Simplex LP” method
Beaver Creek Pottery Example“Solver” Settings (4 of 5)
Solution screen
Slack S1 = 0and S2 = 0
Slack—S1=0 and S2=0
Maximize Z = $40x1 + $50x2
subject to: x1 + 2x2 40 4x1 + 3x2 120
x1, x2 0
Optimal solution
point
Beaver Creek Pottery ExampleGraphical Solution
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.
Sensitivity Analysis
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 CoefficientSensitivity Range
objective function Z = $40x1 + $50x2 sensitivity range for:
x1: 25 c1 66.67 x2: 30 c2 80
Objective Function Coefficient RangesBeaver 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 ValuesSensitivity 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 labor4x1 + 3x2 120 lb of clay
x1, x2 0
Constraint Quantity Value Ranges by ComputerExcel Sensitivity Range for Constraints
Sensitivity ranges for constraint quantity values
the sensitivity range for the labor hours q1 is 30 ≤q1 ≤80 hr.
the sensitivity range for clay quantity q2 is 60≤ q2 ≤160 lb.
Maximize Z = $40x1 + $50x2 subject to: x1 + 2x2 40 hr of labor4x1 + 3x2 120 lb of clay
x1, x2 0
Excel Sensitivity Report for Beaver Creek PotteryShadow Prices Example
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.
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.
Duality (Shadow Prices)
With every linear programming problem, there is associated another linear programming problem which is called the dual of the original (or the primal) problem.
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 CompanyThe 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 CompanyT = number of tables to be produced per weekC = number of chairs to be produced per week
Maximize profit Z= $70T + $50Csubject to the constraints4T + 3C ≤ 240 (carpentry constraint)2T + 1C ≤ 100 (painting and varnishing constraint)T, C ≥0 (non-negativity constraints)
Flair Furniture Company- Solver Solution
Target Cell (Max)Cell Name Original Value Final Value
$B$12 Profit= 0 4100
Adjustable CellsCell Name Original Value Final Value
$B$10 Tables= 0 30$B$11 Chairs= 0 40
ConstraintsCell Name Cell Value Formula Status Slack
$E$6 Carpentry Usage 240$E$6<=$G$6 Binding 0$E$7 Painting Usage 100$E$7<=$G$7 Binding 0
Sensitivity Report
sensitivity range for T: 66.7 c1 100 C: 35 c2 52.5The sensitivity range for the carpentry hours q1 is 200 ≤q1 ≤ 300The 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 $5.
Adjustable Cells Final Reduced Objective Allowable Allowable
Cell Name Value Cost Coefficient Increase Decrease$B$10 Tables= 30 0 70 30
3.333333333
$B$11 Chairs= 40 0 50 2.5 15
Constraints Final Shadow Constraint Allowable Allowable
Cell Name Value Price R.H. Side Increase Decrease
$E$6Carpentry Usage 240 15 240 60 40
$E$7 Painting Usage 100 5 100 20 20
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 & SupplyEach warehouse has the following supply of televisions available for shipment each month:Warehouse Supply (sets)1. Cincinnati 3002. Atlanta 2003. Pittsburgh 200
700Each retail store has the following monthly demand for television sets:Store Demand (sets)A. New York 150B. Dallas 250C. Detroit 200
600
Cost MatrixCosts 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 StoreFromWarehouse A B C
1 $16 $18 $112 14 12 133 13 15 17
Model Summaryminimize Z = $16x1A + 18x1B + 11x1C + 14x2A + 12x2B + 13x2C + 13x3A + 15x3B + 17x3Csubject to
The transportation model can also be optimally solved by Linear Programming