The Wyndor Glass Company
Problem (Hillier and Liberman)
The Wyndor Glass Company is planning to launch two new
products. Product 1 is an 8-foot glass door with aluminum
framing and Product 2 a 4x6 foot double-hung wood-framed
window
Aluminum frames are made in Plant 1, wood frames are made
in Plant 2, and Plant 3 produces the glass and assembles the
products. Product 1 requires some of the production capacity
in Plants 1 and 3, but none in Plant 2. Product 2 needs only
Plants 2 and 3. The marketing division has concluded that
the company could sell as much of either product as could be
processed by these plants. The management of the company
wants to determine what mixture of both products would be
the most profitable. The following table provides the
information available.
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The Wyndor Glass Company
Problem (Hillier and Liberman)
Plant
1
2
3
Profit per batch
Production time per batch, hours
Product
1
2
1
0
3
$3,000
0
2
2
$5,000
Production
time available
per week,
hours
4
12
18
2© 2003 by Prentice Hall, Inc.
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The Wyndor Glass Company
Problem Formulation
Formulation as a linear programming problem
Decision variables:
x1 = Number of batches of product 1 produced per week
x2 = Number of batches of product 2 produced per week
Objective function:
Maximize z = 3 X1 + 5X2 (in thousands of dollars)
Subject to:
x1
2x2
3x1+ 2x2
x1, x2 0



4
12
18
(Production Available in Plant 1)
(Production Available in Plant 2)
(Production Available in Plant 3)
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Montana Wood Products
Problem
Montana Wood Products manufacturers two-high
quality products, tables and chairs. Its profit is
$15 per chair and $21 per table. Weekly
production is constrained by available labor and
wood. Each chair requires 4 labor hours and 8
board feet of wood while each table requires 3
labor hours and 12 board feet of wood. Available
wood is 2400 board feet and available labor is 920
hours. Management also requires at least 40
tables and at least 4 chairs be produced for every
table produced. To maximize profits, how many
chairs and tables should be produced?
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Montana Wood Products
Problem Formulation
Objective Function
Max Z = 15x1 + 21x2
Subject To
4x1 + 3x2 < 920 ( labor constraint)
8x1 + 12x2 < 2400 ( wood constraint)
x2 - 40 > 0 (make at least 40 tables)
x1 - 4 x2 > 0 (at least 4 chairs for every table)
x1 , x 2 > 0
(non-negativity constraints)
5© 2003 by Prentice Hall, Inc.
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The Sureset Concrete
Company Problem
• The Sureset Concrete Company produces
concrete. Two ingredients in concrete are sand
(costs $6 per ton) and gravel (costs $8 per ton).
Sand and gravel together must make up exactly
75% of the weight of the concrete. Also, no more
than 40% of the concrete can be sand and at least
30% of the concrete be gravel. Each day 2000
tons of concrete are produced. To minimize costs,
how many tons of gravel and sand should be
purchased each day?
6© 2003 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
The Sureset Concrete Company
Problem Formulation
Objective Function
Min Z = 6x1 + 8x2
Subject To
x1 + x2 = 1500 ( mix constraint)
x1 < 800 ( mix constraint)
x2 > 600 ( mix constraint )
x1 , x2 > 0
(non-negativity constraints)
7© 2003 by Prentice Hall, Inc.
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Widgets Problem
Suppose a company produces two types of widgets, manual and
electric. Each requires in its manufacture the use of three
machines; A, B, and C. A manual widget requires the use of the
machine A for 2 hours, machine B for 1 hour, and machine C for
1 hour. An electric widget requires 1 hour on A, 2 hours on B, and
1 hour on C. Furthermore, suppose the maximum numbers of
hours available per month for the use of machines A, B, and C are
180, 160, and 100, respectively. The profit on a manual widget is
$4 and on electric widget it is $6. See the table below for a
summary of data. If the company can sell all the widgets it can
produce, how many of each type should it make in order to
maximize the monthly profit?
Manual Electric
A
B
C
profit
2
1
1
$4
1
2
1
$6
Hours
available
180
160
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by Prentice Hall, Inc.
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Widgets Problem Formulation
X1 = number of manual widgets
X2 = number of electric widgets
Objective Function
max Z = 4X1 + 6X2
ٍٍ
Subject To
2X1 + X2  180
X1+ 2X2  160
X1+ X2  100
X1  0, X2  0
9© 2003 by Prentice Hall, Inc.
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Product-Mix Problem
The Handy-Dandy Company wishes to schedule the production of
a kitchen appliance which requires two resources – labor and
material. The company is considering three different models of
this appliance and its engineering department has furnished the
following data:
Resources required Product/Model
to produce 1 unit
A
B
C
Labour (hours/unit)
7
3
6
Material (lbs./unit)
4
4
5
Profit ($/unit)
4
2
3
Resources
Available
150
200
The supply of raw materials is restricted to 200 pounds per day.
The daily availability of manpower is 150 hours. Formulate a
linear programming model to determine the daily production rate
of the various models of appliances in order
to maximize the total
10© 2003 by Prentice Hall, Inc.
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profit.
Product-Mix Problem
Formulation
XA= daily production of model A
XB= daily production of model B
XC= daily production of model C
Objective Function
Maximize Z = 4XA + 2XB + 3XC
subject to the constrains
7XA + 3XB + 6XC  150
4XA + 4XB + 5XC  200
XA  0, XB  0, XC  0
11© 2003 by Prentice Hall, Inc.
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An Electric Company Problem
• An electric company manufacturers two radio models,
each on a separate rate production line. The daily capacity
of the first line is 60 radios and that of the second is 75
radios. Each unit of the first model uses 10 pieces of a
certain electronic component, whereas each unit of the
second model requires 8 pieces of the same component.
The maximum daily availability of the special component
is 800 pieces. The profit per unit of models 1 and 2 is $30
and $20, respectively. Determine the optimum daily
production of each model.
12© 2003 by Prentice Hall, Inc.
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An Electric Company Problem
Formulation
X1 = number of radios of model 1
X2 = number of radios of model 2
Objective Function
max Z =30X1 + 20X2
ٍٍ
Subject To
X1 60
X2 75
10 X1+8X2  800
X1  0, X2  0
13© 2003 by Prentice Hall, Inc.
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Furniture Factory Problem
• A small furniture factory manufacturers tables and chairs.
It takes 2 hours to assemble a table and 30 minutes to
assemble a chair. Assembly is carried out by four workers
on the basis of a single 8-hour shift per day. Customers
usually buy at least four chairs with each table, meaning
that the factory must produce at least four times as many
chairs as tables. The sale price is $150 per table and $50
per chair. Determine the daily production mix of chairs and
tables that would maximize the total daily revenue to the
factory.
14© 2003 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Furniture Factory Problem
Formulation
X1 = number of tables
X2 = number of chairs
Objective Function
max Z =150X1 + 50X2
ٍٍ
Subject To
X2 - 4X1  0
120 X1+30X2  4x8x60
X1  0, X2  0
15© 2003 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458
Giapetto’s Woodcarving
Problem
• Giapetto’s, Inc., manufactures wooden soldiers and trains.
Each soldier built:
Each train built
•Sell for $27 and uses $10 worth of
raw materials.
•Giapetto’s variable labor/overhead
costs by $14.
•Requires 2 hours of finishing labor.
•Requires 1 hour of carpentry labor
•Sell for $21 and used $9 worth of
raw materials.
•Giapetto’s variable labor/overhead
costs by $10.
•Requires 1 hour of finishing labor.
•Requires 1 hour of carpentry labor.
Each week Giapetto can obtain: all needed raw material, only 100 finishing
hours & only 80 carpentry hours.
Demand for the trains is unlimited, at most 40 soldiers are bought each
week.
Giapetto wants to maximize weekly profit. Formulate a mathematical model
of Giapetto’s situation that can be used to maximize
16© 2003weekly
by Prentice Hall,profit.
Inc.
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Giapetto’s Woodcarving
Problem Formulation
x1 = number of soldiers produced each week
x2 = number of trains produced each week
Weekly profit = weekly revenue – weekly raw material costs – the weekly variable
costs = 3x1 + 2x2
• objective function
Maximize z = 3x1 + 2x2
• Subject to
•
2 x1 + x2 ≤ 100
•
x1 + x2 ≤ 80
•
x1
≤ 40
•
x1 ≥ 0, x2 ≥ 0
(finishing constraint)
(carpentry constraint)
(constraint on demand for soldiers)
17© 2003 by Prentice Hall, Inc.
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Graphical Solution of an LP
Problem
• Used to solve LP problems with two (and sometimes three)
decision variables
• Consists of two phases
• Finding the values of the decision variables for which
all the constraints are met (feasible region of the
solution space)
• Determining the optimal solution from all the points in
the feasible region (from our knowledge of the nature
of the optimal solution)
18© 2003 by Prentice Hall, Inc.
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Finding the Feasible Region
(2D)
• Steps
• Use the axis in a 2-dimensional graph to represent the
values that the decision variables can take
• For each constraint, replace the inequalities with
equations and graph the resulting straight line on the 2dimensional graph
• For the inequality constraints, find the side (half-space) of
the graph meeting the original conditions (evaluate
whether the inequality is satisfied at the origin)
• Find the intersection of all feasible regions defined by all
the constraints. The resulting region is the (overall)
feasible region.
19© 2003 by Prentice Hall, Inc.
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Equation Form
Basic data
Tons of raw material per ton of
Exterior paint
Raw material, M1
Raw material, M2
Profit per ton ($1000)
6
1
5
Interior paint
4
2
4
Maximum daily
availability (tons)
24
6
*Decision variables: Need to determine the amounts to be produced of exterior and interior
paints.
x1 = tons produced daily of exterior paint
x2 = tons produced daily of interior paint
*Objective (goal) aims to optimize: The company wants to increase its profit as much as
possible.
z represents the total daily profit (in thousands of dollars)
Maximize z = 5x1 + 4x2
*Constraints: Restrict raw materials usage and demand
(Usage of a raw material) ≤ (Maximum raw material)
by both paints
availability
6x1 + 4x2 ≤ 24 (Raw material M1)
x1 + 2x2 ≤ 6 (Raw material M2)
-x1 + x2 ≤ 1 (Demand Limit)
x2 ≤ 2 (Demand Limit)
20© 2003 by Prentice Hall, Inc.
Nonnegativity restrictions: x1, x2 ≥
0
Upper Saddle River, NJ 07458
Graphical LP Solution
Graphical procedure includes 2 steps
Determination: 1) The solution space that defines all feasible solutions of the model
2) The optimum solution from among all the feasible points in the solution space
Nonnegativity constraints
Step 1: Determination of the Feasible Solution Space
x1 ≥ 0
x2
x2 ≥ 0
first quadrant
6
Replace each inequality (≤) with equations (=)
6x1 + 4x2 ≤ 24
6x1 + 4x2 = 24
5
4
x1 + 2x2 ≤ 6
x1 + 2x2 = 6
-x1 + x2 ≤ 1
x2 ≤ 2
-x1 + x2 = 1
x2 = 2
3
Determine 2 points
6x1 + 4x2 = 24
let x1 = 0, x2 = 6
let x2 = 0, x1 = 4
2 points: (0,6) and (4,0)
2
1
0
2
3
4
1
2
3
4
5
6
21© 2003 by Prentice Hall, Inc.
Graph of equations
Upper Saddle River, NJ 07458
x1
1
Feasible Space
Reference point
Convenient computationally: (0,0)
Note: select another point if the line passes
through the origin, can’t use (0,0)
Ex: 6x1 + 4x2 ≤ 24
Feasible half-space
(0,0)
6 x 0 + 4 x 0 = 0 (0 ≤ 24)
Not a feasible side
(6,0)
6 x 6 + 4 x 0 = 36 (36 ≥ 24)
x2
6
5
4
Constraints:
3
x1 + 2x2 ≤
2
1
0
1
6 2
1 3
24
0 5
0 6
6x1 + 4x2 ≤ 24
Solution
space
1
2
-x1 + x2 ≤
x2 ≤
x1
4
5
6
3
Feasible space of the Reddy Mikks
model
22© 2003 by Prentice Hall, Inc.
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x1 ≥
x2 ≥
Optimum Solution
Step 2: Determination of the Optimum Solution
The feasible space consists of an infinite number of points
Optimum solution: identify the piont in which the profit function z = 5x1 + 4x2 increases
(maximize z)..
Values of x1 and x2 associated with the optimum point
6x1 + 4x2 = 24 solve
x1 + 2x2 = 6
simultaneously
x1 = 3 and x2 = 1.5
z = 21
x2
3
x1 + 2x2 ≤ 6
Optimum: x1 = 3 tons
x2 = 1.5 tons
z = $21,000
2
6x1 + 4x2 ≤ 24
1
0
1
2
3
4
x1
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Reddy Mikks Problem
Formulation
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Reddy Mikks Problem
Solution
Constraint 2: 2XE + XI  8

Constraint 3: -XE + XI  1
I
Constraint 4: XI  2
Constraint 1: XE + 2XI  6
Feasible
Region
1
2
3
4
5
6
7
0 + 2(0)  6
25© 2003 by Prentice Hall, Inc.
Upper Saddle River, NJ 07458

E
Example(2)
• Objective Function
Minimize z = 3x – y
• Subject to
x– y1
x+ y5
x  0, and y  0.
26© 2003 by Prentice Hall, Inc.
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Example(2)\solution
vertex
value of z at vertex
(0, 0)
(1, 0)
(3, 2)
(0, 5)
z = 3(0) – (0) = 0
z = 3(1) – (0) = 3
z = 3(3) – (2) = 7
z = 3(0) – (5) = –5
x  y 1
y
(0, 5)
(3, 2)
x
(1, 0)
x y 5
The minimum value of z is –5 and this occurs at (0, 5).
27© 2003 by Prentice Hall, Inc.
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