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Linear Programming
Chapter 13 Supplement
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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.
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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
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Objective Function
• Profit• $40 per bowl (x)• $50 per mug (y)
• Maximize Z = 40x + 50y
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Constraint Table
Constrained Quantity
Bowls x
Mugsy Type
Max. orMinimum
Labor 1
4
2
3
<
<
40
120Clay
Constraints: x + 2y < 40 4x + 3y < 120
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Nonnegativity Constraints
• We cannot make a negative amount of either product
• Add nonnegativity constraints
x > 0
y > 0
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Problem Statement
Maximize Z = 40x + 50y
subject to
x + 2y < 40
4x + 3y < 120
x > 0
y > 0
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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.
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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.
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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.
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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.
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Problem Statement
Maximize Z = 40x + 50y
subject to
x + 2y < 40
4x + 3y < 120
x > 0
y > 0
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Feasible Region for This Problem
This region is bounded by the lines
x + 2y = 40 (Labor)
4x + 3y = 120 (Clay)
x = 0
y = 0
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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)
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Bowls (x)10 20 30
10
20
30
40
Labor
Mu
gs
(y) Clay
FeasibleRegion
Points in feasibleregion satisfyall constraints.
40
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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)
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Bowls10 20 30
10
20
30
40
Labor
Mu
gs
Clay
40
Iso-profit Line
Profit = $800
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Bowls10 20 30
10
20
30
40
Labor
Mu
gs
Clay
40
Iso-profit Linesfor profits of $800 and $1,000
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Bowls10 20 30
10
20
30
40
Labor
Mu
gs
Clay
40
Iso-profit Lines
Optimal Point=(24,8)
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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
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Maximum Profit
• Optimal solution is (24,8)
• Objective function is 40x + 50y
$40(24) + $50(8) = $1,360
• Maximum profit is $1,360
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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.
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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.
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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.
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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 >.
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Linear programming maximizes or minimizes the objective function
subject to constraints.
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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
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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
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Uses of Linear Programming (3)
• Distribution• Minimize cost
• Facility location• Minimize transportation cost.
• Emergency response systems• Minimize average response time.
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Conditions to Use Linear Programming
• Objective function must be linear
• Constraints must be linear inequalities or linear equations
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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
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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.
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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.
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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
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Bowls10 20 30
10
20
30
40
Labor
Mu
gs
Clay
40 80
NewLabor
Optimal point = (0,40). Profit = $2,000