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 Bowls
Quantity
x
Labor
1
Clay
4
Mugs
y
Max. or
Type Minimum
2
<
40
3
<
120
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 Programming
Problem 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 Programming
Problem 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)
Mugs (y)
40
Clay
Points in feasible
region satisfy
all constraints.
30
20
10
Feasible
Region
10
20
Bowls (x)
Labor
30
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)
Iso-profit Line
Mugs
40
Clay
30
20
10
Labor
10
20
Bowls
30
40
Mugs
40
Clay
Iso-profit Lines
for profits of
$800 and $1,000
30
20
10
Labor
10
20
Bowls
30
40
Iso-profit Lines
Mugs
40
Clay
30
20
Optimal Point=(24,8)
10
Labor
10
20
Bowls
30
40
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.
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.
© 1998 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 2/e
Ch 11 Supp - 2
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 >.
© 1998 by Prentice-Hall Inc
Russell/Taylor Oper Mgt 2/e
Ch 11 Supp - 2
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
Mugs
Optimal point = (0,40). Profit = $2,000
40
30
Clay
New
Labor
20
10
Labor
10
20
Bowls
30
40
80