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Linear Programming
Chapter 13 Supplement
Pottery Example
Beaver Creek Pottery Company is located on a Native American reservation. Each day, the company has available 40 hours of labor and 120 pounds of clay. The firm makes two products, bowls and mugs. A bowl requires 1 hour of labor and 4 pounds of clay. A mug requires 2 hours of labor and 3 pounds of clay. The firm's profit is $40 per bowl and $50 per mug. The company wants to maximize profit.
Pottery Example
Business Objective: Determine
• Number of bowls to make
• Number of mugs to make
to maximize profit.
Decision variables:
x = number of bowls to make
y = number of mugs to make
Objective Function
• Profit• $40 per bowl (x)• $50 per mug (y)
• Maximize Z = 40x + 50y
Constraint Table
Constrained Quantity
Bowls x
Mugsy Type
Max. orMinimum
Labor 1
4
2
3
<
<
40
120Clay
Constraints: x + 2y < 40 4x + 3y < 120
Nonnegativity Constraints
• We cannot make a negative amount of either product
• Add nonnegativity constraints
x > 0
y > 0
Problem Statement
Maximize Z = 40x + 50y
subject to
x + 2y < 40
4x + 3y < 120
x > 0
y > 0
Linear Programming Terminology
• Feasible solution: Any solution which satisfies all constraints.
• Feasible region: Set of points which satisfy all constraints.
• Optimal solution(s): A point which • Satisfies all constraints• Maximizes or minimizes the value of the
objective function.
Solving a Linear ProgrammingProblem by the Graphical Method
(1) Set up the problem:• Decision variables and their definitions.• Write objective function and state whether it
should be maximized or minimized.• Write the constraints as mathematical
inequalities or equalities.
(2) Draw the feasible region.
Solving a Linear ProgrammingProblem by the Graphical Method(2)
(3) Determine which point(s) in the feasible
region give an optimal solution• Point(s) which maximize or minimize the
objective function.
• Note: To use the graphical method, the problem must have only 2 variables.
To Draw the Feasible Region
• Convert each constraint into an equation.
• Draw the corresponding line.
• The set of points bounded by these lines is the feasible region.
Problem Statement
Maximize Z = 40x + 50y
subject to
x + 2y < 40
4x + 3y < 120
x > 0
y > 0
Feasible Region for This Problem
This region is bounded by the lines
x + 2y = 40 (Labor)
4x + 3y = 120 (Clay)
x = 0
y = 0
Plot x + 2y = 40
• x-intercept: Set y = 0.
x + 0 =40 ==> x = 40.
Point is (40,0)
• y-intercept: Set x=0.
(0)+2y=40 ==> y = 20.
Point is (0,20)
Bowls (x)10 20 30
10
20
30
40
Labor
Mu
gs
(y) Clay
FeasibleRegion
Points in feasibleregion satisfyall constraints.
40
Iso-Profit Lines
A set of points on which the objective function is constant
Example: The set of points satisfying
40x + 50y = 800
x-intercept: (20,0)
y-intercept: (0,16)
Bowls10 20 30
10
20
30
40
Labor
Mu
gs
Clay
40
Iso-profit Line
Profit = $800
Bowls10 20 30
10
20
30
40
Labor
Mu
gs
Clay
40
Iso-profit Linesfor profits of $800 and $1,000
Bowls10 20 30
10
20
30
40
Labor
Mu
gs
Clay
40
Iso-profit Lines
Optimal Point=(24,8)
Optimal Solution
• Satisfies x + 2y = 40 and 4x + 3y = 120
• Solution is (24,8): 24 bowls and 8 mugs
• Verification:
24 + 2(8) = 40
4(24) + 3(8) = 120
Maximum Profit
• Optimal solution is (24,8)
• Objective function is 40x + 50y
$40(24) + $50(8) = $1,360
• Maximum profit is $1,360
Fundamental Theorem of Linear Programming
In a linear programming problem, the optimal solution occurs at an extreme point (vertex or corner point) of the
feasible region.
Objective of Linear Programming
• Maximize the use of resources (or minimize costs) to achieve competitive priorities.
• Optimum solution is a base case which must be adjusted to reflect business realities.
Ch 11 Supp - 2© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Linear Programming Model
• Decision variables are mathematical symbols representing activity levels.
• Objective function is a mathematical, linear function which represents the organization’s objectives.• Used to compare alternative courses of
action.
Ch 11 Supp - 2© 1998 by Prentice-Hall IncRussell/Taylor Oper Mgt 2/e
Linear Programming Model (2)
• Constraints are mathematical, linear relationships representing restrictions on decision making• Resource constraints.• Policy or legal constraints.• Sales constraints.
• Constraints may be <, =, or >.
Linear programming maximizes or minimizes the objective function
subject to constraints.
Uses of Linear Programming
• Production Scheduling• Maximize profit• Minimize cost
Different objectives yield different schedules.
• Determine product or service mix• Maximize profit• Maximize revenue• Minimize costs
Uses of Linear Programming (2)
• Scheduling labor in services• Minimize cost
• Production-location problem• Allocate products and customers to plants
to minimize total cost of production and distribution
Uses of Linear Programming (3)
• Distribution• Minimize cost
• Facility location• Minimize transportation cost.
• Emergency response systems• Minimize average response time.
Conditions to Use Linear Programming
• Objective function must be linear
• Constraints must be linear inequalities or linear equations
Methods for Solving LP Problems
• Graphical method: limited to 2 variables
• Any linear programming problem• Simplex method• Karmarkar’s algorithm
• Transportation method:• Facility location problems• Production-location problem: minimize total
cost of production and shipment to customers
Slack Variables
• Slack = unused amount of any resource or constrained quantity
• S1 = Amount of unused labor
= 40 - (x + 2y) = 40 - x - 2y = 40 - 24 - 8(2)
= 0. Optimum solution uses all labor.
• S2 = Amount of unused clay
= 120 - (4x + 3y) = 120 - 4x - 3y
= 120 - 4(24) - 3(8) = 0.
Optimum solution uses all clay.
Sensitivity Analysis (Ranging)
• Dual value or shadow price• Incremental increase or decrease in the
objective function if one more unit of a resource is added.
• The amount we would pay to get one more unit of a resource.
• Valid only for resources which have no slack.
Shadow Price for Labor
• Dual value = shadow price = $16.
• If labor hours increase from 40 to 41, profit increases from $1360 to $1376.
• Amount to buy
= upper bound - original value
= 80 hours - 40 hours = 40 hours
Bowls10 20 30
10
20
30
40
Labor
Mu
gs
Clay
40 80
NewLabor
Optimal point = (0,40). Profit = $2,000
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