Introduction to Operations Research/ Management Science

Introduction to Manufacturing Systems /
Operations Research
What is Operations Research? Management Science?
Operations research is concerned with scientifically deciding
how to best design and operate man-machine systems, usually
under conditions requiring the allocation of scarce resources.
(Operations Research Society of America, 1976).

What is Operations Research? Management Science?
Management Science is the application of a scientific approach
to solving management problems in order to help managers
make better decisions. (Introduction to Management Science,
B.W.Taylor III, 1999).
OR is the abstraction of real world problems into mathematical
models, and the development of techniques to optimally (or
attempt to optimally) solve these problems. The application of
OR is using the models and methods to aid decision makers
(Schultz, 2002).

History of MS/OR
Origins in World War II
- radar deployment policies
- anti-aircraft fire control
- fleet convoy sizing
- detection of enemy submarine
Early MS/OR Efforts - Large Industries
- automotive
- oil&gas
- airlines
Additional MS/OR sectors
- telecommunications
- financial planning
- health care
- public service

Nature of OR / Approach to Problem Solving
- observe situation
- define / formulate the problem
- construct a model that attempts to sufficiently
represent the situation
- solve / validate the model
- apply the model (aid decision maker)

Types of models
- Math programming (Linear, Integer, non-Linear)
- Network
- Queueing
- MDP (Markov Decision Processes)
- Regression
- Forecasting
- Inventory
- Dynamic programming
- Simulation

Introduction to Optimization
and Linear Programming
Chapter 2

Introduction
 We all face decision about how to use limited
resources such as:
 Oil in the earth
 Land for dumps
 Time
 Money
 Workers

Mathematical Programming...
 MP is a field of management science that finds the
optimal, or most efficient, way of using limited
resources to achieve the objectives of an individual of a
business.
 Optimization

Applications of Optimization
 Determining Product Mix
 Manufacturing
 Routing and Logistics
 Financial Planning

Characteristics of Optimization
Problems
 Decisions
 Constraints
 Objectives

General Form of an Optimization Problem
MAX (or MIN): f0(X1, X2, …, Xn)
Subject to: f1(X1, X2, …, Xn)<=b1
:
fk(X1, X2, …, Xn)>=bk
:
fm(X1, X2, …, Xn)=bm
Note: If all the functions in an optimization are
linear, the problem is a Linear Programming (LP)
problem

Example LP Problem
Blue Ridge Hot Tubs produces two types of hot
tubs: Aqua-Spas & Hydro-Luxes.
There are 200 pumps, 1566 hours of labor, and
2880 feet of tubing available.
Aqua-Spa Hydro-Lux
Pumps 1 1
Labor 9 hours 6 hours
Tubing 12 feet 16 feet
Unit Profit $350 $300

5 Steps In Formulating LP Models:
1. Understand the problem.
2. Identify the decision variables.
X1=number of Aqua-Spas to produce
X2=number of Hydro-Luxes to produce
3. State the objective function as a linear
combination of the decision variables.
MAX: 350X1 + 300X2

5 Steps In Formulating LP Models
(continued)
4. State the constraints as linear combinations of
the decision variables.
1X1 + 1X2 <= 200 } pumps
9X1 + 6X2 <= 1566 } labor
12X1 + 16X2 <= 2880 } tubing
5. Identify any upper or lower bounds on the
decision variables.
X1 >= 0
X2 >= 0

MAX: 350X1 + 300X2
S.T.: 1X1 + 1X2 <= 200
9X1 + 6X2 <= 1566
12X1 + 16X2 <= 2880
X1 >= 0
X2 >= 0
Example-01

Solving LP Problems:An Intuitive Approach
 Idea: Each Aqua-Spa (X1) generates the highest unit
profit ($350), so let’s make as many of them as
possible!
 How many would that be?
 Let X2 = 0
 1st constraint: 1X1 <= 200
 2nd constraint: 9X1 <=1566 or X1 <=174
 3rd constraint: 12X1 <= 2880 or X1 <= 240
 If X2=0, the maximum value of X1 is 174 and the total
profit is $350*174 + $300*0 = $60,900
 This solution is feasible, but is it optimal?
 No!
MAX: 350X1 + 300X2
S.T.: 1X1 + 1X2 <= 200
9X1 + 6X2 <= 1566
12X1 + 16X2 <= 2880
X1 >= 0
X2 >= 0

Solving LP Problems:An Intuitive Approach
 Idea: Hydro-Lux (X2) generates the highest unit profit
($300), so let’s make as many of them as possible!
 How many would that be?
 Let X1 = 0
 1st constraint: 1X2 <= 200
 2nd constraint: 6X2 <=1566 X2 <=261
 3rd constraint: 16X2 <= 2880 or X2 <= 180
 If X1=0, the maximum value of X2 is 261 and the total
profit is $300*180 + $350*0 = $54,000
 This solution is feasible, but is it optimal?
 No!
MAX: 350X1 + 300X2
S.T.: 1X1 + 1X2 <= 200
9X1 + 6X2 <= 1566
12X1 + 16X2 <= 2880
X1 >= 0
X2 >= 0

Solving LP Problems: A Graphical Approach
 The constraints of an LP problem defines its
feasible region.
 The best point in the feasible region is the optimal
solution to the problem.
 For LP problems with 2 variables, it is easy to plot
the feasible region and find the optimal solution.

X2
X1
250
200
150
100
50
0
0 50 100 150 200 250
(0, 200)
(200, 0)
boundary line of pump constraint
X1 + X2 = 200
Plotting the First Constraint

X2
X1
250
200
150
100
50
0
0 50 100 150 200 250
(0, 261)
(174, 0)
boundary line of labor constraint
9X1 + 6X2 = 1566
Plotting the Second Constraint

X2
X1
250
200
150
100
50
0
0 50 100 150 200 250
(0, 180)
(240, 0)
boundary line of tubing constraint
12X1 + 16X2 = 2880
Feasible Region
Plotting the Third Constraint

X2
Plotting A Level Curve of the
Objective Function
X1
250
200
150
100
50
0
0 50 100 150 200 250
(0, 116.67);100/300*350
(100, 0)
objective function
350X1 + 300X2 = 35000

A Second Level Curve of the Objective FunctionX2
X1
250
200
150
100
50
0
0 50 100 150 200 250
(0, 175);=150/300*350
(150, 0)
objective function
350X1 + 300X2 = 35000 objective function
350X1 + 300X2 = 52500

Using A Level Curve to Locate
the Optimal SolutionX2
X1
250
200
150
100
50
0
0 50 100 150 200 250
objective function
350X1 + 300X2 = 35000
objective function
350X1 + 300X2 = 52500
optimal solution

Calculating the Optimal Solution
• The optimal solution occurs where the “pumps” and
“labor” constraints intersect.
• This occurs where:
X1 + X2 = 200 (1)
and 9X1 + 6X2 = 1566 (2)
• From (1) we have, X2 = 200 -X1 (3)
• Substituting (3) for X2 in (2) we have,
9X1 + 6 (200 -X1) = 1566
which reduces to X1 = 122
• So the optimal solution is,
X1=122, X2=200-X1=78
Total Profit = $350*122 + $300*78 = $66,100

Enumerating The Corner Points
X2
X1
250
200
150
100
50
0
0 50 100 150 200 250
(0, 180)
(174, 0)
(122, 78)
(80, 120)
(0, 0)
obj. value = $54,000
obj. value = $0
Note: This technique will not
work if the solution is
unbounded.

Graphical LP Solution Procedure
1) Formulate the LP problem
2) Plot the constraints on a graph
3) Identify the feasible solution region
4) Plot two objective function lines
5) Determine the direction of improvement
6) Find the most attractive corner
7) Determine the coordinates of the MAC
8) Find the value of the objective function

A farmer can plant up to 8 acres of land with wheat and barley. He can earn
$5,000 for every acre he plants with wheat and $3,000 for every acre he plants
with barley. His use of a necessary pesticide is limited by federal regulations
to 10 gallons for his entire 8 acres. Wheat requires 2 gallons of pesticide for
every acre planted and barley requires just 1 gallon per acre.
What is the maximum profit he can make?
Exercise

let x = the number of acres of wheat ; let y = the number of acres of barley.
since the farmer earns $5,000 for each acre of wheat and $3,000 for each acre of barley, then the total profit the
farmer can earn is 5000*x + 3000*y.
let p = total profit that can be earned. your equation for profit becomes: p = 5000x + 3000y ; that's your objective
function. it's what you want to maximize
the constraints are:
number of acres has to be greater than or equal to 0, number of acres has to be less than or equal to 8.; amount of
pesticide has to be less than or equal to 10.; your constraint equations are: x >= 0, y >= 0;
x + y <= 8 ; 2x + y <= 10
to graph these equations, solve for y in those equations that have y in them and then graph the equality portion of
those equations. ; x >= 0; y >= 0 ; y <= 8-x; y <= 10 - 2x
x = 0 is a vertical line that is the same line as the y-axis. ; y = 0 is a horizontal line that is the same line as the x-
axis.
the area of the graph that satisfies all the constraints is the region of feasibility.
the maximum or minimum solutions to the problem will be at the intersection points of the lines that bound the
region of feasibility.
the graph of your equations looks like this:

y = 8 – x ; y = 10 - 2x
subtract the first equation from the second equation and you get: 0 = 2
– x add x to both sides of this equation and you get: x = 2 ; substitute 2
for x in either equation to get y = 6.
that makes the intersection point (x,y) = (2,6).
the objective equation is: p = 5000x + 3000y
profit will be maximum at the intersection points of the region of
feasibility on the graph.
the profit equation is evaluated at each of these points as shown in the
following table.
intersection point of (x,y) p
(0,0) $0
(0,8) $24,000
(2,6) $28,000 *****
(5,0) $25,000

A painter has exactly 32 units of yellow dye and 54 units of green
dye. He plans to mix as many gallons as possible of color A and
color B. Each gallon of color A requires 4 units of yellow dye and
1 unit of green dye. Each gallon of color B requires 1 unit of
yellow dye and 6 units of green dye.
Find the maximum number of gallons he can mix.
Exercise-02

The Bead Store sells material for customers to make their
own jewelry. Customer can select beads from various bins.
Grace wants to design her own Halloween necklace from
orange and black beads. She wants to make a necklace that
is at least 12 inches long, but no more than 24 inches long.
Grace also wants her necklace to contain black beads that
are at least twice the length of orange beads. Finally, she
wants her necklace to have at least 5 inches of black beads.
Find the constraints, sketch the problem and find the
vertices (intersection points)
Exercise-03

x >= 0 is there because the number of inches of black beads can't be negative.
y >= 0 is there because the number of inches of orange beads can't be negative.
x + y >= 12 is there because the total length of the necklace has to be greater than or
equal to 12 inches.
x + y <= 24 is there because the total length of the necklace has to be less than or
equal to 24 inches.
x >= 2y is there because the length of the black beads has to be greater than or equal
to twice the length of the orange beads.
x >= 5 is there because the number of inches of black beads has to be greater than or
equal to 5.
to graph these equations, we have to solve for y in each equation that has y in it and
then graph the equality portion of each of them.
your equations for graphing are:
x >= 0; y >= 0 ; y >= 12 – x ; y <= 24 – x ; y <= x/2 ; x >= 5

A garden shop wishes to prepare a supply of special fertilizer at a
minimal cost by mixing two fertilizers, A and B.
The mixture is to contain:
at least 45 units of phosphate
at least 36 units of nitrate
at least 40 units of ammonium
Fertilizer A costs the shop $.97 per pound.
Fertilizer B costs the shop $1.89 per pound.
fertilizer A contains 5 units of phosphate and 2 units of nitrate and 2
units of ammonium.
fertilizer B contains 3 units of phosphate and 3 units of nitrate and 5
units of ammonium.
how many pounds of each fertilizer should the shop use in order to
minimize their cost.
Exercise-04

the objective function becomes:
c = .97x + 1.89y
the constraint equations are:
x >= 0
y >= 0
5x + 3y >= 45 ; 2x + 3y >= 36 ; 2x + 5y >= 40
the equations to be graphed are:
x >= 0 ; y >= 0 ; y >= (45-5x)/3 ; y >= (36 - 2x)/3 ; y >= (40-2x)/5

intersection points (x,y) c = .97x + 1.89y minimum solution
(0,15) 28.35
(3,10) 21.81
(15,2) 18.33 *****
(20,0) 19.40
the table suggests that we have a minimum cost solution when the value of x is equal
to 15 and the value of y is equal 2.
when x = 15 and y = 2, the number of pounds of potassium, nitrates, and ammonium
are:
phosphate = 5x + 3y = 5*15 + 3*2 = 75 + 6 = 81
nitrate = 2x + 3y = 2*15 + 3*2 = 30 + 6 = 36
ammonium = 2x + 5y = 2*15 + 5*2 = 30 + 10 = 40
all the constraints associated with the minimum cost objective have been met.

An elementary school wants to send children on a field trip to a
museum. The museum staff has informed the school that tours can
be scheduled for no more than 50 total people and the school must
provide at least one adult chaperone for every 9 students.
Make a list of constraints.
graph the feasible region.
calculate and label the vertices.
Exercise-05

A group of artists has decided to produce hand-drawn cards for
Valentine's Day and donate the money generated to charity.
The artists will produce ink drawings and watercolors.
They have volunteered to spend at most 120 hours for preparation of
the cards and a maximum of 60 hours for packaging the cards.
The preparation of an ink drawing takes 0.3 hours and the
preparation of a watercolor takes 0.5 hours.
The packaging of each requires 0.2 hours.
Make a list of the constraints and sketch the feasible region and label
the vertices.
Exercise-06

A company produces 2 portable CD players.
They are called the Shuffle Man and the Walk On.
The profits per unit are $20 for the Shuffle Man and $15 for the Walk On.
The product time (in hours) for one unit of each product is given in the chart.
The company has at most 750 worker-hours of manufacturing time available and
200 worker-hours of shipping time available each day.
At least 300 Shuffle Man players and 500 Walk On players must be produced each
day.
Write the profit function to be maximized.
How many of each product should be produced each day?
What will the profit be?
If the profits for each Shuffle Man falls to $16 per unit, how many of each product
should be produced?
What will the new total profit be?
Exercise-07

Simplex Method
Blue Ridge Hot Tubs produces two types of hot
tubs: Aqua-Spas & Hydro-Luxes.
There are 200 pumps, 1566 hours of labor, and
2880 feet of tubing available.
Aqua-Spa Hydro-Lux
Pumps 1 1
Labor 9 hours 6 hours
Tubing 12 feet 16 feet
Unit Profit $350 $300

Exercise
Resource Unit Requirements Amount
Available
Table Chair
Wood 30 20 300
Labor 5 10 110
Unit
Profit
6 8

Solution
Let: XT = number of tables produced
XC = number of chairs produced
MAX Z = 6 XT + 8 XC
s.t. 30 XT + 20 XC < 300
5 XT + 10 XC < 110
where: XT, XC > 0

Exercise
9-65
Maximize:Objective: 21 57 XX +
Hours Required to Produce One Unit
Department
X1
Tables
X2
Chairs
Available
Hours This
Week
Carpentry
Painting/Varnishing
4
2
3
1
240
100
Profit/unit
Constraints:
$7 $5
( painting & Varnishing)10012 21 + XX
)(carpentry2403 21 + XX

Exercise
9-66
Department
X1
Tables
X2
Chairs
Available
Hours This
Week
Carpentry
Painting/Varnishing
4
2
3
1
240
100
Profit/unit
Constraints:
$7 $5

Redwood Furniture Problem
XT = 4 tables
XC = 9 chairs
P = 6(4) + 8(9) = 96
dollars

Summary of Graphical Solution
to LP Problems
1. Plot the boundary line of each constraint
2. Identify the feasible region
3. Locate the optimal solution by either:
a. Plotting level curves
b. Enumerating the extreme points

Special Conditions in LP Models
 A number of anomalies can occur in LP problems:
 Alternate Optimal Solutions
 Redundant Constraints
 Unbounded Solutions
 Infeasibility

Example-02
9-70
Department
X1
Tables
X2
Chairs
Available
Hours This
Week
Carpentry
Painting/Varnishing
4
2
3
1
240
100
Profit/unit
Constraints:
$7 $5

9-71
NumberofChairs
100
80
60
40
20
0 20 40 60 80 100 X
X2
Number of Tables
B = (0,80)
C = (30,40)
D = (50,0)
Feasible
Region
24034 21 + XX
10012 11 + XX

Simplex Steps for Maximization
1. Choose the variable with the greatest positive
Cj - Zj to enter the solution.
2. Determine the row to be replaced by selecting
that one with the smallest (non-negative)
quantity-to-pivot-column ratio.
3. Calculate the new values for the pivot row.
4. Calculate the new values for the other row(s).
5. Calculate the Cj and Cj - Zj values for this
tableau. If there are any Cj - Zj values greater
than zero, return to Step 1.
9-72

Simplex Steps for Minimization
1. Choose the variable with the greatest negative Cj - Zj to enter
the solution.
2. Determine the row to be replaced by selecting that one with
the smallest (non-negative) quantity-to-pivot-column ratio.
3. Calculate the new values for the pivot row.
4. Calculate the new values for the other row(s).
5. Calculate the Cj and Cj - Zj values for this tableau. If there
are any Cj - Zj values less than zero, return to Step 1.
9-73

9-83
Department
X1
Tables
X2
Chairs
Available
Hours This
Week
Carpentry
Painting/Varnishing
4
2
3
1
240
100
Profit/unit
Constraints:
$7 $5
Example-03

9-84
(painting and Varnishing )10012 21
+ XX
(carpentry )24034 21
+ XX
Constraints:
Constraints with Slack Variables
10012
24034
221
121
=++
=++
SXX
SXX
21
57 XX +
Objective Function
Objective Function with Slack Variables
2121
0057 SSXX +++
(painting and Varnishing )
(carpentry )

9-85
Profit
per
Unit
Column Prod.
Mix
Column
Real Variables
Columns
Slack Variables
Columns Constant
Column
Cj
Solution
Mix
X1 X2 S1 S2 Quantity
$7 $5 $0 $0
Profit
per
unit row
2 1 1 0
4 3 0 1
$0 $0 $0 $0
$7 $5 $0 $0
$0
$0
S1
S2
Zj
Cj - Zj
100
240
$0
$0
Constraint
equation
rows
Gross
Profit
row
Net
Profit
row

9-86
Cj
Solution
Mix
X1 X2 S1 S2 Quantity
$7 $5 $0 $0
2 1 1 0
4 3 0 1
$0 $0 $0 $0
$7 $5 $0 $0
$0
$0
S1
S2
Zj
Cj - Zj
100
240
$0
$0
Pivot
row
Pivot number
Pivot column

9-87
Cj
Solution
Mix X1 X2 S1 S2 Quantity
$7 $5 $0 $0
1 1/2 1/2 0
0 1 -2 1
$7 $7/2 $7/2 $0
$0 $3/2 -$7/2 $0
$7
$0
X1
S2
Zj
Cj - Zj
50
40
$350

9-88
Cj
Solution
$7 $5 $0 $0
1 1/2 1/2 0
0 1 -2 1
$7 $7/2 $7/2 $0
$0 $3/2 -$7/2 $0
$7
$0
X1
S2
Zj
Cj - Zj
50
40
$350
(Total
Profit)
Pivot
row
Pivot number
Pivot column

9-89
= - x1
0
3/2
-1/2
30
1
1/2
1/2
0
50
(1/2)
(1/2)
(1/2)
(1/2)
(1/2)
(0)
(1)
(-2)
(1)
(40)
= - x
= - x
= - x
= - x






















































=










rowX
newin
number
ingCorrespond
number
pivot
above
Number
rowX
oldin
Number
RowX
Newin
Number
ii

9-90
Cj
Solution
$7 $5 $0 $0
1 0 3/2 -1/2
0 1 -2 1
$7 5 $1/2 $3/2
$0 $0 -$1/2 -$3/2
$7
$5
X1
X2
Zj
Cj - Zj
30
40
$410

Solve the following LP using Graphical method and Simplex
Method
Answer
Exercise-08

Method
Answer
Exercise-09

Method
Answer
Exercise-10

Method
Answer
Exercise-11

Method
Answer
Exercise-12

Simplex Method – Big M method

X2
X1
250
200
150
100
50
0
0 50 100 150 200 250
(0, 180)
(240, 0)
boundary line of tubing constraint
12X1 + 16X2 = 2880
Feasible Region
BOUNDED REGION
MAX: 350X1 + 300X2
S.T.: 1X1 + 1X2 <= 200
9X1 + 6X2 <= 1566
12X1 + 16X2 <= 2880
X1 >= 0
X2 >= 0

Simplex Method : Big ‘M’ Method
Example 06

Surplus & Artificial Variables
9-124
Constraints
Constraints-Surplus & Artificial Variables
9003025
2108105
21
321
=+
++
XX
XXX
9003025
2108105
221
11321
=++
=+++
AXX
ASXXX
Objective Function
321 795 XXX ++:Min
211321 0795 MAMASXXX +++++:Min
Objective Function-Surplus & Artificial Variables

Special Cases
Infeasibility
9-125
02M+21M-31200Cj - Zj
1800+2
M
0-21-M31-M-285Zj
201-1-1000A2M
1000-12110X28
2000-13-201X15
QtyA2A1
S2S1X2X1Sol
Mix
MM0085Cj

Special Cases
Unboundedness
9-126
Pivot Column
Cj 6 9 0 0
Sol
Mix
X1 X2 S1 S2 Qty
X1 -1 1 2 0 30
S1 -2 0 -1 1 10
Zj -9 9 18 0 270
Cj - Zj 15 0 -18 0

Special Cases
Degeneracy
9-127
Pivot Column
C j 5 8 2 0 0 0
Solution
Mix
X 1 X 2 X 3 S 1 S 2 S 3 Qty
8 X 2 1/4 1 1 -2 0 0 10
0 S 2 4 0 1/3 -1 1 0 20
0 S 3 2 0 2 2/5 0 1 10
Z j 2 8 8 16 0 0 80
C j -Z j 3 0 6 16 0 0

Special Cases
Multiple Optima
9-128
Cj 3 2 0 0
Sol
Mix
X1 X2 S1 S2 Qty
2 X1 3/2 1 1 0 6
0 S2 1 0 1/2 1 3
Zj 3 2 2 0 12
Cj - Zj 0 0 -2 0

Sensitivity Analysis
High Note Sound Company
9-129
6013
8042
12050
21
21
21
+
+
+
XX
XX
XX
:toSubject
:Max

Sensitivity Analysis
9-130

Simplex Solution
9-131
Cj 50 120 0 0
Sol
Mix
X1 X2 S1 S2 Qty
120 X2 1/2 1 1/4 0 20
0 S2 5/2 0 -1/4 1 40
Zj 60 120 30 0 2400
Cj - Zj -10 0 -30 0

Nonbasic Objective Function
Coefficients
9-132
Cj 50 120 0 0
Sol
Mix
X1 X2 S1 S2 Qty
120 X2 1/2 1 1/4 0 20
0 S2 5/2 0 -1/4 1 40
Zj 60 120 30 0 2400
Cj – Zj -10 0 -30 0

Basic Objective Function Coefficients
9-133
Cj 50 120 0 0
Sol
Mix
X1 X2 S1 S2 Qty
120+

X1 1/2 1 1/4 0 20
0 S2 5/2 0 -1/4 1 40
Zj 60+ /2 120+  30+ /4 0 2400+20

Cj - Zj -10- /2 0 -30- /4 0

Simplex Solution
9-134
Objective increases by 30 if 1 additional
hour of electricians time is available.
Cj 50 120 0 0
Sol
Mix
X1 X2 S1 S2 Qty
X1 ½ 1 1/4 0 20
S2 5/2 0 -1/4 1 40
Zj 60 120 30 0 40
Cj - Zj 0 0 -30 0 2400

Shadow Prices
 Shadow Price: Value of One Additional Unit of a
Scarce Resource
 Found in Final Simplex Tableau in C-Z Row
 Negatives of Numbers in Slack Variable Column
9-135

Steps to Form the Dual
To form the Dual:
 If the primal is max., the dual is min., and vice versa.
 The right-hand-side values of the primal constraints become
the objective coefficients of the dual.
 The primal objective function coefficients become the right-
hand-side of the dual constraints.
 The transpose of the primal constraint coefficients become
the dual constraint coefficients.
 Constraint inequality signs are reversed.
9-136

Ex
9-137
Dual
6013
8042
21
21
+
+
XX
XX
Subject to:
12014
5032
21
21
+
+
UU
UU
Subject to:
12050 21 + XX:Max 6080 21 + UU:Min

Ex
9-138
Primal:
6013
8042
21
21
+
+
XX
XX
Subject to:
12014
5032
21
21
+
+
UU
UU
Subject to:
12050 21 + XX:Max 6080 21 + UU:Min

Primal & Dual
9-139
Primal: Dual
6013
8042
21
21
+
+
XX
XX
Subject to:
12014
5032
21
21
+
+
UU
UU
Subject to:
12050 21 + XX:Max 6080 21 + UU:Min

Comparison of the Primal and Dual Optimal
Tableaus
9-140
Primal’sOptimalSolution
Dual’sOptimalSolution
Cj
Solution
Mix
Quantity
$7
$5
X2
S2
Zj
Cj - Zj
20
40
$2,400
X1 X2 S1 S2
$50 $120 $0 $0
1/2 1 1/4 0
5/2 0 -1/4 1
60 120 30 0
-10 0 -30 0
Cj
Solution
Mix
Quantity
$7
$5
U1
S1
Zj
Cj - Zj
30
10
$2,400
X1 X2 S1 S2
80 60 $0 $0
1 1/4 0 -1/4
0 -5/2 1 -1/2
80 20 0 -20
$0 40 0 20
A1 A2
M M
0 1/2
-1 1/2
0 40
M M-40

Maximize z = -4x1 - 2x2
subject to
3x1 + x2 ≥ 27
x1 + x2 ≥ 21
x1 + 2x2 ≥ 30
x1, x2 ≥ 0
Answer:
Maximize z = -4x1 - 2x2 + 0x3 + 0x4 + 0x5 – MA1 – MA2 – MA3
3x1 + x2 - x3 + A1 = 27
x1 + x2 - x4 + A2 = 21
x1 + 2x2 - x5 + A3 = 30
x1 ,x2, x3, x4, x5, A1, A2 ≥ 0
Where:
x and x are surplus variables

Introduction to Operations Research/ Management Science

Introduction to Operations Research/ Management Science

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