Zuraidah Derasit –FSKM-
CHAPTER 3
LINEAR PROGRAMMING
What is Linear Programming (LP)
•
The main issue of many managers and decision makers of company are to find the
optimal solution to a problem that requires decision about how best to use a set of
limited resources to achieve a state goal of objective.
•
Objectives of the business decision frequently involve maximizing profit or minimizing
costs.
•
Resources typically include labor, money, time and raw materials.
•
LP is a technique to help managers in planning and decision making relative to resource
allocation.
•
LP is a model consisting of linear relationships representing a firm’s decisions given an
objective and resource constraints.
Requirements of a LP Problem
1)
One objective function - maximize or minimize objective
•
The major objective of a typical manufacturer is to maximize profit.
•
The major objective of trucking or railroad distribution system might be to
minimize shipping costs.
2)
One or more constraints
•
For example , deciding how many units of each product to product is restricted by
available personnel and machinery.
3)
Alternative courses of action
•
For example, if a company produce 3 different products, management may use
LP to decide how to allocate among them its limited production resources.
4)
Objective function and constraints are linear.
•
Linear just mean that all terms used in the objective function and constraints are
of the first degree.
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Zuraidah Derasit –FSKM-
LP Model Components
•
Decision variables: mathematical symbols representing levels of activity of a firm.
•
Objective function: a linear mathematical relationship describing an objective of the firm,
in terms of decision variables, that is maximized or minimized
•
Constraints: restrictions placed on the firm by the operating environment stated in linear
relationships of the decision variables.
•
Parameters: numerical coefficients and constants used in the objective function and
constraint equations.
In QMT 425, you will learn:
 How to formulate the LP Model (Maximization & Minization Model)
 Solve the LP Model (Maximization & Minization Model)
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Zuraidah Derasit –FSKM-
Steps in Formulating The LP Model
1.
Identify the objective and the constraints.
2.
Clearly define the decision variables.
3.
Construct the objective function.
4.
Formulate the constraints.
Example 3.1 (Maximization Model)
The Beaver Creek Pottery Company is a small crafts company. The company employs skilled
artisans to produce clay bowls and mugs. The two primary resources used by the company are
special pottery clay and skilled labor. Given these limited resources the company desires to
know how many bowls and mugs to produce each day in order to maximize profit.
The two products have the following resource requirements for production and profit per item
produced.
Product
Labor
(hr/ unit)
Resource Requirements
Clay
(lb/ unit)
Profit
(RM/ unit)
Bowl
Mug
1
2
4
3
40
50
There are 40 hours of labor and 120 pounds of clay available each day for production.
Product
Labor
(hr/ unit)
Resource Requirements
Clay
(lb/ unit)
Bowl
Mug
Available
1
2
40
4
3
120
Profit
(RM/ unit)
40
50
Decision Variables:
x1= number of bowls to produce/day
x2= number of mugs to produce/day
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Zuraidah Derasit –FSKM-
Objective function
Maximize Z = 40x1 + 50x2
where Z= profit per day
Resource Constraints:
x1 + 2x2  40 hours of labor
4x1 + 3x2  120 pounds of clay
Non-negativity Constraints:
x1  0; x2  0
Therefore, Complete Linear Programming Model
x1= number of bowls to produce/day
x2= number of mugs to produce/day
Maximize Z = 40x1 + 50x2
subject to
x1 + 2x2  40
4x1 + 3x2  120
x1 , x2
0
Example 3.2 (Minimization Model)
A farmer is preparing to plant a crop and needs to fertilize a field. There are two brands of
fertilizer to choose from. Super-gro and Crop-quick. Each brand yields a specific amount of
nitrogen and phosphate, as follows:
Brand
Super-gro
Crop-quick
Chemical contribution
Nitrogen
Phosphate
(lb/ bag)
(lb/ bag)
2
4
4
3
The farmer’s requires at least 16 pounds of nitrogen and 24 pounds of phosphate. Super-gro
costs RM6 per bag and Crop-quick costs RM3. The farmer wants to know how many bags of
each brand to purchase in order to minimize the total cost of fertilizing.
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Zuraidah Derasit –FSKM-
Brand
Super-gro
Crop-quick
Requires at least
Chemical contribution
Nitrogen
Phosphate
(lb/ bag)
(lb/ bag)
2
4
4
3
16
24
Cost
(RM/ bag)
6
3
Decision variables
x1 = number bags of Super-gro
x2 = number bags of Crop-quick
Objective function
Minimize Z = 6x1 + 3x2
where 6x1 = cost of bags of Super-gro
3x2 = cost of bags of Crop-quick
Model constraints
2x1 + 4x2  16 lb (nitrogen constraint)
4x1 + 3x2  24 lb (phosphate constraint)
x1, x2  0 (non-negativity constraint)
Therefore, Complete Linear Programming Model
x1= number bags of Super-gro
x2= number bags of Crop-quick
Minimize Z = 6x1 + 3x2
subject to
2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
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Zuraidah Derasit –FSKM-
Exercise 3.1
The Flair Furniture Company produces inexpensive tables and chairs. 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 RM7, each chair produced is sold for a RM5 profit. Formulate
LP model to determine the best possible combination of tables and chairs to manufacture in
order to reach the maximum profit.
Exercise 3.2
The Kalo Fertilizer Company makes a fertilizer using two chemicals that provide nitrogen,
phosphate and potassium. A pound of ingredient 1 contributes 10 ounces of nitrogen and 6
ounces of phosphate, while a pound of ingredient 2 contributes 2 ounces of nitrogen, 6 ounces
of phosphate and 1 ounce of potassium. Ingredient 1 costs RM3 per pound and ingredient 2
costs RM5 per pound. The company wants to know how many pounds of each chemical
ingredient to put into a bag of fertilizer to meet minimum requirements of 20 ounces of nitrogen,
36 ounces of phosphate and 2 ounces of potassium. Formulate LP model in order to minimize
cost.
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Zuraidah Derasit –FSKM-
Problems
3.1
A manufacturer has two versions of a toy house. The version X requires 23 sq cm of
plywood, 2 pounds of plastic and 15 minutes of assembling. The version Y requires 31 sq cm of
plywood, 3 pounds of plastic and 25 minutes of assembling. There are 9000 sq cm of plywood,
100 pounds of plastic and 42 hours of assembling time available. Each unit of version X results
in RM4 of profit and for version Y the unit profit is RM5. The manufacturer will not produce more
than 25 units of version Y and there must be a minimum of 5 units of each version produced to
cater for previous orders.
Formulate (do not solve) a linear programming model with the objective of maximizing profit.
3.2
A farmer wants to determine the number of acres of land to be planted with bananas and
starfruits. He has 80 acres of land and has allocated a capital of RM10,000 to develop the land.
The expenditures needed to plant the bananas and starfruits per acre are RM100 and RM150
respectively. The average yields per acre are 500 kg of bananas and 300kg of starfruits.
The fruits are kept in the backyard store before they are sold. The store can only hold 6,000 kg
of fruits at one time. The net profit of each kg of bananas and starfruits are RM0.60 and RM0.90
respectively.
Formulate the above problem as a linear programming model to maximize the net profit.
3.3
Rahman the successful farmer has a 50 acres farm on which to plant tomatoes and
cucumbers. He has 300 hours of labor per week and 800 tons of fertilizer available and has
contracted for shipping space for a maximum of 26 acres worth of tomatoes and 37 acres worth
of cucumbers. An acre of tomatoes requires 10 hours of labor and 8 tons of fertilizer, whereas
an acre of cucumbers requires 3 hours of labor and 20 tons of fertilizer. The profit from an acre
of tomatoes is RM400 and the profit from an acre of cucumbers is RM300. Formulate the linear
programming model for this problem to maximize profit.
3.4
A marketing company is doing research in order to maximize their profit of selling
products X, Y and Z. These products will be sold from door to door. X is sold at RM7 per unit. It
takes a salesperson 10 minutes to sell one unit of X and it costs RM1 to deliver the goods to the
customer. Y is sold at RM5 each. It takes a salesperson 15 minutes to sell one unit of Y. The
goods are left with the customer at the time of sale. Z is sold at RM12 each. It takes a
salesperson 12 minutes to sell one unit of Z and it costs RM0.80 to deliver the goods. During
any week, a salesperson is only allowed delivery expenses of not more than RM75 and selling
time is expected not to exceed 30 hours. If the unit costs of X, Y and Z are respectively RM2.20,
RM 1.80 and RM4.25, how many units of X, Y and Z must be sold to maximize the total sales?
Formulate the problem as a linear programming model (Do not solve the problem).
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Zuraidah Derasit –FSKM-
3.5
The Mighty Silver Ball Company manufactures three kinds of pinball machines, each
requiring a different manufacturing technique. Each Super Deluxe Machine (SDM) requires 17
hours of labor, 8 hours of testing and yields a profit of RM300. Each Silver Ball Special (SBS)
requires 10 hours of labor, 4 hours of testing and yields a profit of RM200. Each Bumper King
(BK) requires 2 hours of labor, 2 hours of testing and yields a profit of RM100. There are 1,000
hours of labor and 500 hours of testing time available. The company wants to produce at least
200 pinball machines. In addition, a marketing forecast has shown that the demand for the
Super Deluxe is no more than 50 machines, demand for the Silver Ball Special is no more than
80 and demand for the Bumper King is no more than 150. The manufacturer wants to determine
the optimal production schedule that will maximize his total profit. Formulate this as a linear
programming problem.
3.6
The dean from one of the private university in Selangor must plan the school’s course
offerings for the first semester next year. Student demands make it necessary to offer at least
30 undergraduate and 20 postgraduate courses in the term. Faculty contracts also dictate that
at least 60 courses be offered in total. Each undergraduate course taught costs the college an
average of RM2500 in faculty wages, and each postgraduate course costs RM3000. The dean
wants to make sure so that total faculty salaries are kept to a minimum.
Formulate the linear programming model for the above problem.
3.7
C-Bizz Mart is a retail catalog store specializing in cosmetics. Phone orders are taken
each day by a large pool of computer operators, some of whom are permanent and some
temporary. A permanent operator can process an average of 75 orders per day, whereas
temporary operator can process an average of 52 orders per day. The company receives an
average of at least 600 orders per day. The store has 10 computer workstations. A permanent
operator will process about 1.3 orders with errors each day, whereas a temporary operator will
average 4.2 orders with errors daily. The store wants to limit errors to 24 per day. A permanent
operator is paid RM60 per day including benefits and temporary operator is paid RM40 per day.
The company wants to know the number of permanent and temporary operators to hire to
minimize costs.
Formulate a linear programming model for C-Bizz Mart problem.
3.8
Maxi Designs Sdn. Bhd. has been awarded a contract to design a label for a new
computer produced by KC Computer Company. Maxi Designs estimates that at least 160 hours
will be required to complete the project. Three of the firm's graphic designers are available for
assignment to this project: Aida, a senior designer and team leader; Bakri, a senior designer;
and Carol, a junior designer. Since Aida has worked on several projects for KC, management
has specified that Aida must be assigned at least 45% of the total number of hours that are
assigned to the two senior designers. To provide label-designing experience for Carol, Carol
must be assigned at least 20% of the total project time. However, the number of hours assigned
to Carol must not exceed 25% of the total number of hours that are assigned to the two senior
designers. Due to other project commitments, Aida has a maximum of 50 hours available to
work on this project. Hourly wage rates are RM40 for Aida, RM30 for Bakri and RM20 for Carol.
Formulate a linear programming model to determine the number of hours each graphic designer
should be assigned to the project in order to minimize total cost of wages.
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Zuraidah Derasit –FSKM-
Solve The LP Model
1)
2)
1)
Graphical Method
Simplex Method
Graphical Method
•
Graphical solution is limited to LP problems with only two decision variables.
•
Graphical methods provide visualization of how a solution for a linear programming
problem is obtained.
Steps in Graphical Method
1.
Plot constraints on the graph. Convert inequality into equation.
2.
Shade the feasible solution area. The feasible region is the overlapping area of problem’s
constraints. The shaded region represents the area of solution.
3.
Find the optimal solution to the problem. There are 2 approaches that can be taken either:
i)
Isoprofit/ Isocost line solution method or
ii)
Corner point solution method
Example 3.3 (Maximization model)
From example 3.1, how many bowls (x1) and mugs (x2) to produce daily, given limited amounts
of labor and clay.
Maximize Z = 40x1 + 50x2
subject to
x1 + 2x2  40
4x1 + 3x2  120
x1 , x2  0
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Zuraidah Derasit –FSKM-
Solution
Step 1: Plot the constraints on the graph.
x1 + 2x2 = 40
x1 = 0, x2 = 20
x2 = 0 , x1 = 40
4x1 + 3x2 = 120
x1 = 0, x2 = 40
x2 = 0 , x1 = 30
Step 2 : Shaded the feasible region
Shade the feasible solution area (area on the graph bounded by the constraint equation).
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Zuraidah Derasit –FSKM-
Step 3: Find the optimal solution : Use Isoprofit solution method or corner point solution method.
i)
Isoprofit solution method
Let Z = 800
800 = 40x1 + 50x2
x1 = 0, x2 = 16
x2 = 0 , x1 = 20
Optimal solution:
x1 = 24
x2 = 8
Z = 40 (24) + 50 (8) = RM1360 (highest profit)
Conclusion : The company should produce 24 units of bowls and 8 units of mugs to maximize
the profit. The maximum profit is RM1360.
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Zuraidah Derasit –FSKM-
ii)
Corner point solution method
A
B
C
D
Find the coordinates of each corner point
Point A (x1 = 0, x2 = 20)
Point C (x1 = 30, x2 = 0)
Point D (x1 = 0, x2 = 0)
Apply the simultaneous equations method to solve the corner point B, we obtain :
Point B (x1 = 24, x2 = 8)
Test the profit level at each corner point : Pick the highest profit
Corner Point
A (0,20)
B (24,8)
C (30,0)
D (0,0)
Z = 40x1 + 50x2
Z = 40(0) + 50(20) = 1000
Z = 40(24) + 50(8) = 1360 *
Z = 40(30) + 50(0) = 1200
Z = 40(0) + 50(0) = 0
Optimal solution
x1 = 24
x2 = 8
Z = RM1360 (highest profit)
Conclusion : The company should produce 24 units of bowls and 8 units of mugs to maximize
the profit. The maximum profit is RM1360.
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Zuraidah Derasit –FSKM-
Example 3.4 (Minimization model)
From example 3.2, how many bags of Super-gro (x1) and Crop-quick (x2) to purchase in order to
minimize the total cost of fertilizing.
Minimize Z = 6x1 + 3x2
subject to
2x1 + 4x2  16
4x1 + 3x2  24
x1, x2  0
Solution
Step 1: Plot the constraints on the graph.
2x1 + 4x2 = 16
x1 = 0, x2 = 4
x2 = 0 , x1 = 8
4x1 + 3x2 = 24
x1 = 0, x2 = 8
x2 = 0 , x1 = 6
Step 2 : Shaded the feasible region
Shade the feasible solution area (area on the graph bounded by the constraint equation).
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Zuraidah Derasit –FSKM-
Step 3: Find the optimal solution : Use Isocost solution method or corner point solution method.
i)
Isocost solution method
Let Z = 18
18 = 6x1 + 3x2
x1 = 0, x2 = 6
x2 = 0 , x1 = 3
Optimal solution
x1 = 0
x2 = 8
Z = 6(0) + 3(8) = RM24 (lowest cost)
Conclusion : The company should not purchase Super –pro and purchase 8 bags of Crop-quick
to minimize the total cost. The minimum total cost should be RM24.
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Zuraidah Derasit –FSKM-
ii)
Corner point solution method
A
B
C
Find the coordinates of each corner point
Point A (x1 = 0, x2 = 8)
Point C (x1 = 8, x2 = 0)
Apply the simultaneous equations method to solve the corner point B, we obtain :
Point B (x1 = 4.8, x2 = 1.6)
Test the cost level at each corner point : Pick the lowest cost
Corner Point
A (0,8)
B (4.8,1.6)
C (8,0)
Z = 6x1 + 3x2
Z = 6(0) + 3(8) = 24*
Z = 6(4.8) + 3(1.6) = 33.6
Z = 6(8) + 3(0) = 48
Optimal solution
x1 = 0
x2 = 8
Z = RM24
Conclusion : The company should not purchase Super –pro and purchase 8 bags of Crop-quick
to minimize the total cost. The minimum total cost should be RM24.
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Zuraidah Derasit –FSKM-
Exercise 3.3
Solve the following linear programming problem using the graphical method.
Maximize Profit 3x + 5y
subject to
4x + 3y  48
x + 2y  20
x, y  0
Exercise 3.4
Use the graphical method to solve the following linear programming model.
Minimize Cost 8x + 6y
subject to
3x + 2y
x + y
y
x, y




18
5
6
0
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Zuraidah Derasit –FSKM-
Problems
Solve the following linear programming problems using the graphical method.
3.9
Maximize Z = 3x + 6y
subject to
3x + 2y
x+ y
x
x, y
 18
5
≤4
 0
3.11 Maximize Z = 4x1 + 4x2
subject to
3x1 + 5x2  150
x1 - 2x2 ≤10
5x1 + 3x2 ≤150
x1, x2  0
3.13 Minimize Z = x +y
subject to
x + 2y  7
2x + y  5
x + 6y  11
x, y  0
3.10 Maximize Z = x1 + 5x2
subject to
5x1 + 5x2  25
2x1 + 4x2 ≤16
x1
≤5
x1, x2  0
3.12 Maximize Z = 2x1 + 3x2
subject to
x1 + 2x2 ≥ 160
2x1 + x2 ≤ 200
4x2 ≤ 200
x1, x2  0
3.14 Minimize Z = 10x1 + 20x2
subject to
6x1 + 2x2 ≥ 36
2x1 + 4x2 ≥ 32
x2 ≤ 20
x1, x2  0
3.15 Minimize Z = X + 2Y
subject to
2X + 3Y = 600
Y ≥ 2X
X
≤ 150
X, Y
≥0
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Zuraidah Derasit –FSKMStandard Form LP Model (To handle  constraint)
Slack Variables



In graphical solution, the model constraints are considered as equations (=), rather than 
or  inequalities.
There is a standard procedure for transforming inequality constraints into equations.
The complete LP model with slack variables referred as standard form LP model.
Slack variables
A slack variable is added to a  constraint to convert it to an equation (=).


Slack variable represents the amount of unused resources.
Slack variable contribute nothing to the objective function value.
Example 3.5 (Maximization model)
From example 3.1,
Maximize Z = 40x1 + 50x2
subject to
x1 + 2x2  40 (labor)
4x1 + 3x2  120 (clay)
x1 , x2  0
Convert to standard form of LP model
Maximize Z = 40x1 + 50x2 + 0s1 + 0s2
subject to
x1 + 2x2 + s1
= 40 (labor)
4x1 + 3x2
+ s2 = 120 (clay)
x1 , x2 , s1 , s2
0
Consider a hypothetical solution x1 = 5, x2 = 10. Substituting these values into the equation
yields.
Profit: Z = 40(5) + 50(10) = 700
Constraint (labor): (5) + 2(10) + s1 = 40, then s1 = 15
Constraint (clay): 4(5) + 3(10) + s2 = 120, then s2 = 70



In this example, x1 = 5 bowls, x2 = 10 mugs.
s1 = 15 hours represents the amount of unused labor
s2 = 70 lb represents the amount of unused clay
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Zuraidah Derasit –FSKM-
2)
Simplex Method
Setting up the first simplex tableau
1. Write the problem in standard form.
Consider the maximization problem:
Maximize Z = 40x1 + 50x2
subject to
x1 + 2x2  40
4x1 + 3x2  120
x1 , x2  0
In standard form, the problem written as:
Maximize Z = 40x1 + 50x2 + 0s1 + 0s2
subject to
x1 + 2x2 + s1
= 40
4x1 + 3x2
+ s2 = 120
x1 , x2 , s1 , s2  0
2. Basic feasible solution.
All real variables are set equal to zero:
x1 = 0 ,
x2 = 0
First simplex tableau:
Cj
0
0
Solution mix
s1
s2
Zj
Cj-Zj
40
x1
1
4
0
40
50
x2
2
3
0
50
0
s1
1
0
0
0
0
s2
0
1
0
0
Quantity
40
120
0
Calculations for the values of Zj are as follows:
Z (for column x1) = 0(1) + 0(4) = 0
Z (for column x2) = 0(2) + 0(3) = 0
Z (for column s1) = 0(1) + 0(0) = 0
Z (for column x1) = 0(0) + 0(1) = 0
Cj
Zj
Cj-Zj
40
0
40
50
0
50
0
0
0
0
0
0
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Zuraidah Derasit –FSKM-
Explanation about the values/variables inside the simplex tableau

Variables in solution mix are referred as basic variables, in this example are s1 and s2.

Variables not in the solution mix are called non-basic variables (x1 and x2).

Cj : Profit contribution per unit of each variable.

Zj : In the quantity column, provides total contribution.
The basic feasible solution:
x1 = 0
x2 = 0
s1 = 40
s2 = 120
Zj = 0
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Zuraidah Derasit –FSKM-
Simplex solution procedure (After setting up the first simplex tableau)
1. Determine which variable to enter into the next solution mix. Identify the column with the
largest positive number in the Cj-Zj row. The column identified is called the pivot
column.
2. Determine which variable to replace. Divide each amount in the quantity column with
amount in pivot column. The row with the smallest nonnegative number will be
replaced in the next tableau. This row referred as pivot row.
3. Compute new values for the pivot row. Divide every number in the row by pivot number.
4. Compute new values for each remaining row. All remaining rows are calculated as
follows:
New row
number
=
Old row
number
-
Its pivot column
coefficient number
New pivot row
number
x
5. Compute the Zj and Cj-Zj rows. If all numbers in the Cj-Zj row are zero or negative, we
have found the optimal solution.
Procedure 1 : Determine which variable to enter.
Iteration 1
Cj
0
0
Basic
variables
s1
s2
Zj
Cj-Zj
x1
40
1
4
0
40
x2
50
2
3
0
50
s1
0
1
0
0
0
s2
0
0
1
0
0
Quantity
s1
0
1
0
0
0
s2
0
0
1
0
0
Quantity
40
120
0
Procedure 2 : Determine which variable to replace.
Cj
0
0
Basic
variables
s1
s2
Zj
Cj-Zj
x1
40
1
4
0
40
x2
50
2
3
0
50
40/2=20
120/3=40
0
Identify the pivot number (Pivot number = the number at the intersection of the pivot row and
pivot column) = 2
The variable to enter is X2
The variable to remove is s1
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Zuraidah Derasit –FSKM-
Procedure 3 : Compute new values for the pivot row.
Iteration 2
Cj
50
0
Basic
variables
x2
s2
Zj
Cj-Zj
x1
40
1/2
4
0
40
x2
50
1
3
0
50
s1
0
1/2
0
0
0
s2
0
0
1
0
0
Quantity
x2
50
1
0
0
50
s1
0
1/2
-3/2
0
0
s2
0
0
1
0
0
Quantity
x2
50
1
0
50
0
s1
0
1/2
-3/2
25
-25
s2
0
0
1
0
0
Quantity
40/20
120
0
Procedure 4 : Compute new values for each remaining row.
Cj
50
0
Basic
variables
x2
s2
Zj
Cj-Zj
x1
40
1/2
5/2
0
40
20
60
0
Procedure 5 : Compute the Zj and Cj-Zj rows
Cj
50
0
Basic
variables
x2
s2
Zj
Cj-Zj
x1
40
1/2
5/2
25
15
20
60
1000
* Since not all number in Cj-Zj row are 0 or negative, the second simplex tableau is not optimal,
and we must repeat the five simplex step.
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Zuraidah Derasit –FSKM-
The result after repeat all five steps:
Iteration 3
Cj
50
40
Basic
variables
x2
x1
Zj
Cj-Zj
x1
40
0
1
40
0
x2
50
1
0
50
0
s1
0
4/5
-3/5
16
-16
s2
0
-1/5
2/5
6
-6
Quantity
8
24
1360
Therefore, the optimal solution:
x1 = 24 bowls
x2 = 8 mugs
s1 = 0 (unused labor)
s2 = 0 (unused clay)
Zj = RM1360
52
Zuraidah Derasit –FSKM-
Exercise
3.5
Solve the following LP model using simplex method.
Maximize Z = 4x1 + 5x2
subject to
2x1 + 2x2  20
3x1 + 7x2  42
x1 , x2  0
3.6
Write down the standard form and construct the initial simplex tableau for the following
linear programming problem:
Maximize Z = 3x1 + 2x2 + x3
subject to
x1 + 2x2 + 2x3  30
2x1 + 3x2 + x3  50
x1 - 2x2  0
x1 , x2 , x3  0
3.7
Consider a linear programming model;
Maximize Z = x1 + 3/2x2 + 2x3
subject to
2x1 + x2 + 2x3  30
x1 + 3x2 + 2x3  45
x1 , x2 , x3  0
The second simplex tableau for the above problem is given below:
Cj
Basis
x1
x2
x3
s1
3/2
2
0
1
2
x3
1
½
1
½
0
s2
-1
2
0
-1
Zj
2
1
2
1
Cj-Zj
-1
1/2
0
-1
s2
0
0
1
0
0
RHS
15
15
30
Is the solution optimal? Explain. If not, obtain the optimal solution.
53
Zuraidah Derasit –FSKMStandard Form LP Model (To handle  ,  and = constraints)
Slack Variables
A slack variable is added to a  constraint to convert it to an equation (=).
Surplus Variables
A surplus variable is subtracted from a  constraint to convert it to an equation (=).

Surplus variable represents an excess above a constraint requirement level.

Surplus variable contribute nothing to the objective function value.
Artificial Variables

An artificial variable is a variable that has no physical meaning in terms of a real world LP
problem. It simply allows us to create a basic feasible solution to start the simplex algorithm.
An artificial variable is not allowed to appear in the final solution to the problem.

Whenever an artificial or surplus variable is added to one of the constraints, it must also be
included in the other equations and in the objective function, just as what we have done for
slack variables.

Each artificial variable is assigned an extremely high cost to ensure it does not appear in the
final solution.
Standard form of LP model




To handle  constraint, slack variable is added.
To handle  constraint, surplus variable is subtracted and then artificial variable is added.
To convert an equality (=), we add an artificial variable to the equation.
In the objective function, surplus and artificial variable must be included , just as we done for
slack variable and
Maximization model  put -MA to represent the artificial variable
Minimization model  put +MA to represent the artificial variable
54
Zuraidah Derasit –FSKM-
Convert the following LP model into standard form by adding appropriate slack variable(s)
surplus variable(s) and artificial variable(s) to be used in the initial simplex tableau.
Example 3.6
Maximize Z = x1 + 2x2 + 2x3
subject to
x1 + x2 + 2x3  12
2x1 + x2 + 5x3 = 20
x1 + x2 - x3  8
x1, x2, x3  0
Maximize Z = x1 + 2x2 + 2x3 + 0s1 + 0s2 - MA1 - MA2
subject to
x1 + x2 + 2x3 + s1
= 12
2x1 + x2 + 5x3
+ A1 = 20
x1 + x2 - x3 - s2 + A2 = 8
x1 , x2 , x3, s1 , s2 , A1 , A2  0
Example 3.7
Minimize Z = 6x1 + 3x2
subject to
2x1 + 4x2  16
4x1 + 3x2 = 24
x1, x2  0
Minimize Z = 6x1 + 3x2 + 0s1 + MA1 + MA2
subject to
2x1 + 4x2 - s1 + A1 = 16
4x1 + 3x2
+ A2 = 24
x1 , x2 , s1 , A1 , A2  0
Exercise
Convert the following LP models to standard form.
3.8
3.9
Maximize Z = 30x1 + 40x2 + 35x3
subject to
3x1 + 4x2 + 2x3  90
2x1 + x2 + x3 ≤ 54
x1 + 3x2 + 2x3 = 93
x1, x2, x3  0
Minimize Z = 84x1 + 4x2 + 30x3
subject to
8x1 + x2 + 3x3  240
16x1 + x2 + 7x3 ≥ 480
8x1 - x2 + 4x3 ≥ 160
x1, x2, x3  0
55
Zuraidah Derasit –FSKM-
2)
Simplex Method (Slack Variable, Surplus Variable and Artificial Variable)
Solve this LP model using simplex method
Maximize Z = 4x1 + 3x2
Subject to
2x1 + 2x2 ≤ 20
x1 + x2 ≥ 2
x1 , x2 ≥ 0
Standard form of LP model:
Maximize Z = 4x1 + 3x2 + 0S1 + 0S2 – MA1
Subject to
2x1 + 2x2 + s1
= 20
x1 + x2 - s2 + A1 = 2
x1 , x2 , s1 , s2 , A1 ≥ 0
Solution:
Iteration 1
Cj
0
-M
Iteration 2
Cj
0
4
Iteration 3
Cj
0
4
Basic
Var
s1
A1
Zj
Cj - Zj
4
x1
2
1
-M
4+M
3
x2
2
1
-M
3+M
0
s1
1
0
0
0
0
s2
0
-1
M
-M
-M
A1
0
1
-M
0
Quantity
20
2
-2M
Basic
Var
s1
x1
Zj
Cj - Zj
4
x1
0
1
4
0
3
x2
0
1
4
-1
0
s1
1
0
0
0
0
s2
2
-1
-4
4
Quantity
Basic
Var
s2
x1
Zj
Cj - Zj
4
x1
0
1
4
0
3
x2
0
1
4
-1
0
s1
1/2
1/2
2
-2
0
s2
1
0
0
0
Quantity
16
2
8
8
10
40
Optimal solution:
x1 = 10
x2 = 0
s1 = 0
s2 = 8
Z = 40
56
Zuraidah Derasit –FSKM-
Special Cases in LP
Four special cases that may occur in solving LP problems:
1.
2.
3.
4.
An infeasible problem
An unbounded problem
Alternate optimal solutions
Degeneracy
1.
An Infeasible Problem – LP problem with no feasible solution.
o
Infeasibility is a condition that arises when there is no solution to a LP problem
that satisfies all of the constraints given.
Example 3.8
Maximize Z= 2x1 + 6x2
subject to
4x1 + 3x2  12
2x1 + x2  8
x1, x2  0
 No feasible solution
57
Zuraidah Derasit –FSKM-
o
Simplex tableau : An infeasible problem is detected when an artificial variable
still in the final solution mix.
2
Basic
Var
x1
x1
2
1
x2
6
3/4
s1
0
1/4
s2
0
0
-M
A2
0
-1/2
-1/2
-1
2
Zj
2
3/2+1/2M
1/2+1/2M
M
6-2M
Cj-Zj
0
-1/2M+9/2
-1/2M-1/2
-M
Cj
2.
RHS
3
An Unbounded Problem – the objective function value (for maximization problem) gets
very large without bound
o
o
The objective function can increase indefinitely without reaching a maximum
value.
The solution space is not completely closed in.
Example 3.9
Maximize Z= 2x1 + 6x2
subject to:
4x1 + 3x2  12
2x1 + x2  8
x1, x2  0
 The solution is unbounded
58
Zuraidah Derasit –FSKM-
o
Cj
6
0
3.
Simplex tableau : An unbounded problem is detected when all entries in the pivot
column are non positive, hence there will be no leaving variable.
Basic
Var
x2
s1
Zj
Cj-Zj
x1
2
2
2
12
-10
x2
6
1
0
6
0
s1
0
0
1
0
0
s2
0
-1
-3
-6
6
RHS
8
12
48
Ratio
8/-1 = - 8
12/-3= - 4
Alternate Optimal Solutions – LP problem has more than one optimal solution.
o
o
This is the case when the objective function’s isoprofit or isocost line runs
perfectly parallel to one of the problem’s constraint.
Provide greater flexibility to the decision maker.
Example 3.10
Maximize Z = 60x1 + 60x2
subject to
3x1 + 3x2  90
2x1 + 4x2  80
x1 , x2  0
 Alternate optimal solution
Corner points:
59
Zuraidah Derasit –FSKM-
X1
0
30
0
20
X2
0
0
20
10
Z
0
1800
1200
1800
 A corner points having the same maximum value indicate the LP problem has
alternate optimal solution.
o
Simplex tableau : Alternate optimal solutions is detected if the final tableau has
Cj – Zj value equal to 0 for a non-basic variable.
An alternate optimal solution
Cj
Basic
Var
60
x1
60
x2
Zj
Cj-Zj
4.
x1
60
1
0
60
0
x2
60
0
1
60
0
s1
0
2/3
-1/3
20
-20
s2
0
-1/2
1/2
0
0
RHS
20
10
1800
Degeneracy
o
A redundant constraint is one that does not affect the feasible solution region.
Example 3.11
Maximize Z = 80x1 + 70x2
subject to
2x1 + x2  120
x1 + x2  60
x1  70
x1 , x2  0
60
Zuraidah Derasit –FSKM-
Redundant
constraint
 Degeneracy
o
Simplex tableau : Degeneracy is detected in simplex tableau when more than
one row have similar smallest nonnegative numbers (ratio).
Iteration 1
Basic
Cj
Var
0
s1
x1
80
2
x2
70
1
s1
0
1
s2
0
0
s3
0
0
RHS
Ratio
120
120/2 = 60
0
s2
1
0
0
1
0
70
70/1 = 70
0
s3
1
1
0
0
1
60
60/1 = 60
Zj
0
0
0
0
0
0
0
Cj-Zj
80
70
0
0
0
0
Choosing row s1 as the pivot row produces the following tableau:
61
Zuraidah Derasit –FSKM-
Iteration 2
Basic
Cj
Var
80
x1
x1
80
1
x2
70
1/2
s1
0
1/2
s2
0
0
s3
0
0
RHS
Ratio
60
60/1/2 =120
0
s2
0
-1/2
-1/2
1
0
10
10/-1/2 = x
0
s3
0
1/2
1/2
0
1
0
0/1/2 = 0
Zj
80
40
40
0
0
4800
Cj-Zj
0
30
-40
0
0
 Degeneracy causes a basic variable to have zero value. In the second iteration above,
the basic variable s3 = 0.
Iteration 3
80
Basic
Var
x1
x1
80
1
x2
70
0
s1
0
1
s2
0
0
s3
0
-1
0
s2
0
0
-1
1
1
10
70
x2
0
1
-1
0
2
0
Zj
80
70
10
0
60
4800
Cj-Zj
0
0
-10
0
-60
Cj
RHS
60
In the third iteration, the basic variable x2 = 0.
62
Zuraidah Derasit –FSKM-
Exercises
Solve the following linear programming problems using the graphical method.
3.10
3.11
Maximize Z = x +y
Maximize Z = 2x1 + x2
subject to
subject to
x+y  1
-3x + y  3
x, y  0
3.12
3x1 + x2  3
4x1 + 3x2 ≥ 6
x1 + 2x2 ≥ 2
x1, x2  0
3.13
Maximize Z = 2x1 + 4x2
Maximize Z = 4x1 + 5x2
subject to
subject to
x1 + 2x2  5
x1 + x2 ≤ 4
x1, x2  0
2x1 + 2x2 ≤ 100
x2 ≤ 25
x1
≤ 60
x1, x2  0
63
Zuraidah Derasit –FSKM-
Sensitivity Analysis with the simplex tableau

Sensitivity analysis trace the changes that we make to our parameter.
» Changes in Resources/RHS Values
-
Making changes in the resources/RHS values result in changes in the feasible region
and often the optimal point. This would also affect the value of the objective function.
-
Making changes in the resources/RHS values also leads us to important topic of
shadow price.
-
Shadow price is:

The changes in value of the objective function for every additional unit of scarce
resource.

It is also refer to maximum amount the firm should pay for every additional unit of
resource to make available.
Example 3.12
High Note Sound is a company that makes compact disks CD players (x1) and stereo receivers
(x2). The company wants to maximize profit.
LP model:
Maximize Profit = 50x1 + 120x2
subject to
2x1 + 4x2  80 (hours of electricians’ time available)
3x1 + x2  60 (hours of audio technicians’ time available)
x1 , x2  0
In the High Note Sound company; there are two resources available;
1. hours of electrician’s time
2. hours of technician’s time
For electrician’s time, there are 80 hours available; and 60 hours of technician’s time available.
64
Zuraidah Derasit –FSKM-
The graphical solution of the above problem is as follows:
A
B
D
C
Corner point solution:
A (x1 = 0, x2 = 20, Z = 2,400)*
B (x1 = 16, x2 = 12, Z = 2,240)
C (x1 = 20, x2 = 0, Z = 1,000)
D (x1 = 0, x2 = 0, Z = 0)
Optimal solution:
x1 = 0 CD player
x2 = 20 stereo receivers
Z = RM2400
With 80 hours and 60 hours of electrician and technician’s time available, we are able to
produce RM2400 profit.
Now, if we add one hour of technician’s time from 80 hours to 81 hours; the following result is
obtained:
65
Zuraidah Derasit –FSKM-
Change RHS value from 80 to 81 hours of electricians’ time available. Solve LP model.
Maximize Profit = 50x1 + 120x2
subject to
2x1 + 4x2  81
3x1 + x2  60
x1 , x2  0
A
B
D
C
Corner point solution:
A (x1 = 0, x2 = 20.25, Z = 2,430)*
B (x1 = 15.9, x2 = 12.3, Z = 2,271)
C (x1 = 20, x2 = 0, Z = 1,000)
D (x1 = 0, x2 = 0, Z = 0)
Noticed that; 1 additional unit of the electrician’s hour, assuming no cost incurred, causes
the total profit to increase to RM2430. This is an increase of RM30.
Hence, the 1 additional hour of electrician’s time is worth RM30. This is known as shadow price
for electrician’s hour.
Shadow prices can be simply obtained from an optimal solution simplex tableau. These are the
numbers in the z-row of slack variable columns.
66
Zuraidah Derasit –FSKM-
The simplex solution for High Note Sound model:
Final simplex tableau:
x2
x1
50
½
x2
120
1
s1
0
¼
s2
0
0
s2
5/2
0
-1/4
1
40
Zj
60
120
0
2400
Cj-Zj
-10
0
30
-30
Cj
Basic
Var
120
0
Optimal solution:
x1 = 0 (CD players)
x2 = 20 (stereo receivers)
s1 = 0 (hour of unused time of electrician’s)
s2 = 40 (hours of unused time of audio technician’s)
Z = RM2400
Quantity
20
0
Shadow price (ignore negative)
RM30 for one additional hour of
electrician’s time
RM0 for one additional hour of
technician’s time
One additional hour of electrician’s time is worth RM30.
If the company increases 1 hour of electrician’s time, the new profit would be: RM2400+RM30.
One additional hour of technician’s time is worth nothing.
If the company increases 1 hour of technician’s time, the new profit would be: RM2400+RM0.
67
Zuraidah Derasit –FSKM-
Example 3.13
The management of a company has formulated the following linear programming problem.
Let
x1 = number of product 1 to produce
x2 = number of product 2 to produce
x3 = number of product 3 to produce
Maximize Profit, Z = 7 x 1  3 x 2  9 x 3
subject to
4x1 + 5x2 + 6x3  360 hours (labor hour)
2x1 + 4x2 + 6x3  300 hours (machine hour)
9x1 + 5x2 + 6x3  600 kg (raw material)
x1,x2,x3  0
The following is the final simplex tableau for the above problem. Use this to answer the
questions below.
Cj
Basic
x1
7
x2
3
x3
9
s1
0
s2
0
s3
0
RHS
7
9
0
x1
x3
s2
Zj
Cj – Zj
1
0
0
7
0
0
0.83
-1
7.5
-4.5
0
1
0
9
0
-0.2
0.3
-1.4
1.3
-1.3
0
0
1
0
0
0.2
-0.13
0.4
0.2
-0.2
48
28
36
a
1. Why we said that the above tableau is the optimal tableau?
All Cj – Zj row values are less and equal to zero.
2. State and interpret the optimal solution.
x1 = 48; produce 48 units of product 1
x2 = 0; produce 0 unit of product 2
x3 = 28; produce 28 units of product 3
s1 = 0; raw material is fully utilized
s2 = 36; 36 hours machine time is unused
s3 = 0; workers’ time is fully utilized
68
Zuraidah Derasit –FSKM-
3. Determine the value of a.
Z = a = 7(48) + 9(28) = RM588
4. Which resource(s) is/are not fully utilized? State the value.
Machine time is not fully utilized, hours left is 36 hours.
5. How many hours of labor and machine and amount of raw material are used to reach the
optimal solution?
Used:
Labor hour = 360 hours
Machine hour = 300 – 36 = 264 hours
Raw material = 600kg
6. What are the shadow prices of the resources?
The shadow prices of labor hour
= RM1.30
machine hour = RM0
raw material = RM0.20
7. What does a zero shadow price mean? How could this happen?
Shadow price RM0 means that 1 additional unit of the resource is worth nothing.
The resource is not fully utilized. e.g. optimal solution s2 = 36
8. Is it worthwhile to pay RM2 for 1 additional hour of the labor? Explain.
Not worthwhile because the maximum amount to pay is RM1.30.
9. Should the company pay RM3 for 3 more hours of the labor? Explain.
RM3
 RM1/hour ;
3hours
Worth, because the maximum amount to pay is RM1.30.
10. What will be the new Z value if the company increase 2 hours of labor? Increase 3 hours
of machine time? Decrease 3 kilograms of raw material?
Z = RM588 + 2(1.30) = RM590.60
Z = RM588 + 3(0)
= RM588
Z = RM588 - 3(RM0.20) = RM587.40
11. Is there any alternative optimal solution? Why?
No. Because in Cj – Zj row, there is no non-basic variable equal to zero.
69
Zuraidah Derasit –FSKM-
Exercises
3.14 IMB Sdn. Bhd. manufactures three products, a, b and c. The products must go through
three production processes: Machining (Resource 1), Assembly (Resource 2) and Packaging
(Resource 3). The company wants to maximize profit. Profits for the products a, b and c are
RM50, RM20 and RM30 respectively. The simplex tableau for this problem is given below.
Cj
Basic
Variables
A
a
b
c
s1
s2
s3
RHS
1
-1
0
1
-1
0
100
C
0
3
1
-1
2
0
40
s3
0
-1
0
0
1
1
25
Zj
Cj – Zj
a)
Complete the above tableau.
b)
Is the above tableau optimal? Explain.
c)
State the optimal solution.
d)
Which resource(s) is/are fully utilized.
e)
Which resource(s) is/are not fully utilized and state the value(s).
f)
State the shadow price for each resource and interpret.
g)
What would be the effect on total profit if resource 1 was increased by 10 units and
resource 3 was increased by 5 units?
70
Zuraidah Derasit –FSKM-
3.15 Beautiful Cotton Manufacturer produces two types of cotton cloths: denim and corduroy.
This company wants to maximize profit. Given below is the LP model in order to maximize profit.
Maximize profit = 6X1 + 4.5X2
subject to
3.5 X1  2.5 X 2  3250 (cotton usage, meters)
3.2 X1  3 X 2  3000 (processing time, hours)
X 1  510 (demand for corduroy, meters)
X 1, X 2  0
The final simplex tableau for the above problem is shown below:
Cj
0
4.5
6
Basic
Variable
S1
X2
X1
Zj
Cj – Zj
6
X1
0
0
1
6
0
4.5
X2
0
1
0
4.5
0
0
S1
1
0
0
0
0
0
S2
-0.83
0.33
0
1.5
-1.5
0
S3
-0.83
-1.07
1
1.2
-1.2
Quantity
325
456
510
5112
X1 = number of meters of corduroy produced,
X2 = number of meters of denim produced,
S1, S2, and S3 are the slack variables for cotton usage, processing time and demand for
corduroy respectively.
a)
Determine the optimal production level and the maximum profit obtained.
b)
How much cotton and processing time are left over in the optimal solution?
c)
What is the maximum amount the manufacturer is willing to spend for an additional
processing time?
d)
If the manufacturer increase one meter of cotton at cost RM3 will it be worthwhile? Why?
e)
Does this problem have alternate optimal solution? Explain.
71
Zuraidah Derasit –FSKM-
The Dual


Every LP primal model has a dual model.
The first way of stating a linear problem is called the primal of the problem. The second
way of stating the same problem is called the dual.
Step to form a dual:
Primal
Dual
Objective function maximization
Objective function minimization
Objective function minimization
Objective function maximization
RHS values
Objective function coefficients
Constraints coefficients
Transpose of primal constraints
coefficients
Objective function coefficients
RHS values
Constraint inequality sign
Inequality signs reversed
Example 3.14
Primal model
Maximize Profit = 50x1 + 120x2
subject to
2x1 + 4x2  80
3x1 + x2  60
x1 , x2  0
Dual model
Minimize Opportunity Cost = 80y1 + 60y2
subject to
2y1 + 3y2  50
4y1 + y2  120
y1, y2  0
72
Zuraidah Derasit –FSKM
In the final simplex tableau of primal problem; the absolute value of the number in the Cj-Zj
row under slack variables represent the solutions to the dual problem

It also happens that the absolute value of the Cj-Zj values of the slack variables in the dual
solution represent the optimal values of the primal X1 and X2.

The minimum opportunity cost derived in the dual must always equal to the maximum
profit derived in the primal
Example 3.15
Primal model
Maximize Profit = 50x1 + 120x2
subject to
2x1 + 4x2  80
3x1 + x2  60
x1 , x2  0
Optimal tableau of primal model:
Cj
120
0
Basic
Var
x2
s2
Zj
Cj -Zj
Primal optimal solution:
x1 = 0
x2 = 20
s1 = 0
s2 = 40
Z = RM2400
x1
50
½
5/2
60
-10
x2
120
1
0
120
0
s1
0
¼
-1/4
30
-30
s2
0
0
1
0
0
Quantity
20
40
2400
Dual optimal solution:
y1 = 30
y2 = 0
s1 = 10
s2 = 0
Z = RM2400
73
Zuraidah Derasit –FSKM-
Dual model
Minimize Opportunity Cost = 80y1 + 60y2
subject to
2y1 + 3y2  50
4y1 + y2  120
y1, y2  0
Optimal tableau of dual model:
Dual model
Cj
80
0
Basic
Var
y1
s1
y1
80
1
0
y2
60
¼
-5/2
s1
0
0
1
s2
0
-1/4
-1/2
Zj
80
20
0
-20
Zj - Cj
0
-40
0
-20
Dual optimal solution:
y1 = 30
y2 = 0
s1 = 10
s2 = 0
Z = RM2400
Quantity
30
10
2400
Primal optimal solution:
x1 = 0
x2 = 20
s1 = 0
s2 = 40
Z = RM2400
74
Zuraidah Derasit –FSKM-
Exercises
Write the dual.
3.16
3.17
Maximize Z = 3x1 + 9x2
Maximize Z = 5x1 +7x2
subject to
subject to
2x1 + 4x2  8
x1 + 7x2 ≤ 7
x1, x2  0
3.18
x1
 6
2x1 + 3x2 ≤ 19
x1 + x2 ≤ 8
x1, x2  0
3.19
Minimize Z = x 1  2x 2
Maximize Z =30x1+ 20x2 + 40x3
subject to
subject to
x1 + x2  40
2x1 + 4x2  60
x1 + x2  0
5x1 + 9x2 + 8x3 ≤ 450
3x1 – 4x2
≥ 120
x1, x2, x3 ≥ 0
75
Zuraidah Derasit –FSKM-
Problems
3.16
Consider the following linear programming problem.
Z  2000 x 1  700 x 2  1600 x 3
Minimize
Subject to
5 x 1  2x 2  2x 3  20
4 x 1  x 2  2x 3  30
x 1, x 2 , x 3  0
a)
Write the dual for the above linear programming problem in variable Y.
The following is an incomplete final simplex tableau for the dual model (a).
Cj
30
0
0
Basic
variable
Y2
S2
S3
20
Y1
1.25
0.75
-0.5
30
Y2
1
0
0
0
S1
0.25
-0.25
-0.5
0
S2
0
1
0
0
S3
0
0
1
Quantity
500
200
600
Zj
Cj-Zj
b)
c)
d)
Complete the above simplex tableau.
State the optimal solution for the dual.
State the optimal solution for the primal based on the final simplex tableau.
3.17 The following is the incomplete final simplex tableau for the primal, where S1, S2 and S3
are the slack variable.
Cj
Solution
Mix
10
X1
24
X2
50
X3
Zj
Cj - Zj
a)
b)
c)
10
X1
1
0
0
24
X2
0
1
0
50
X3
0
0
1
0
S1
4
0
0
0
S2
-2
2
0
0
S3
-1
-1
1
Quantity
4
3
2
Complete the above simplex tableau.
Explain why this tableau gives the optimal solution.
What is the optimal solution for the primal? Dual?
76
Zuraidah Derasit –FSKM-
3.18 The Toy Race Car Company manufactures three different race-cars called Racer 1,
Racer 2 and Racer 3. The resources required to make the products are technical services, labor
and administration. Table below gives the requirements for the production.
Product
Technical
services (hour)
Labor
(hour)
Administration
(hour)
Unit profit
(RM)
Racer 1
Racer 2
Racer 3
1
2
3
10
4
5
2
2
6
10
6
4
Each week there are 100 hours of technical services available, 600 hours of labor and 300
hours of administration.
a) Formulate the problem as a linear programming problem.
The optimal simplex tableau for the above problem is given below. The S1, S2, S3 are the slack
variables for technical services, labor and administration, respectively.
Cj
10
X1
0
6
X2
1
4
X3
5/6
0
S1
5/3
0
S2
-1/6
0
S3
0
Quantity
6
Basic
Variables
X2
10
0
X1
S3
1
0
0
0
1/6
4
-2/3
-2
1/6
0
0
1
100/3
100
200/3
Zj
Cj – Zj
b)
Fill in the values on Zj and Cj-Zj rows.
c)
Specify the optimal daily production levels of the three products. What is the total profit?
d)
Which resource(s) is not fully utilized? If so, how much spare capacity is left?
e)
If the company had an opportunity to purchase additional resources, which resource would
be the most valuable? How much would the company be willing to pay for this resource?
f)
Write down the dual of the problem.
g)
Give the solution of the dual problem.
77