# EM506 Systems Engineering and Operations Research by tak20026

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```									IZMIR INSTITUTE OF TECHNOLOGY     Engineering Management   EM506 Summer2010

EM506
Systems Engineering
and
Operations Research
Summer 2010

07.09.2010                         Dr. Engin Aktaş                        1
IZMIR INSTITUTE OF TECHNOLOGY   Engineering Management   EM506 Summer2010

2.0 The Simplex Method
and Sensitivity Analysis

07.09.2010                       Dr. Engin Aktaş                        2
IZMIR INSTITUTE OF TECHNOLOGY          Engineering Management        EM506 Summer2010

The simplex method provides a complete analysis of a linear program.

When applied to a linear program
•The simplex method determines whether or not the linear
program has a feasible solution.
•If the linear program does have a feasible solution, the
simplex method determines whether or not its objective is
bounded
•If the linear program is feasible and bounded, the simplex
method finds an optimal solution to it.

The simplex method is developed by George B. Dantzig in 1947.

07.09.2010                              Dr. Engin Aktaş                             3
IZMIR INSTITUTE OF TECHNOLOGY            Engineering Management               EM506 Summer2010

Standard LP Form
1. All the constraints (with the exception of the nonnegativity
restrictions on the variables) are equations with nonnegative
right-hand side.
2. All the variables are nonnegative.
3. The objective function may be of the maximization or the
minimization problem
slack
Conversion of inequalities into equations
surplus
For example      x1  2 x2  3                      x1  2 x2  s1  3

3x1  x2  5                       3x1  x2  S1  5

Conversion of unrestricted variable into nonnegative variables

x j  x  x ,
j    j      x , x  0
j    j

07.09.2010                                Dr. Engin Aktaş                                         4
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Algorithm for Simplex Method

Step1: Using slack variables, convert the LP problem to a system of linear equations.
Step 2: Set up the initial tableau.
Step 3: Select the pivot column.
The rule for the selecting a pivot column is this: Look at all the numbers in the bottom row, excluding the
Answer column. From these, choose the negative number with the largest magnitude. Its column is the pivot
column. (If there are two candidates, choose either one.) If all the numbers in the bottom row are zero or
positive, then you are done, and the basic solution is the optimal solution.

Step 4: Select the pivot in the pivot column.
1) The pivot must always be a positive number. (This rules out zeros and negative numbers)
2) For each positive entry b in the pivot column, compute the ratio a/b, where a is the number in the rightmost
column in that row. We call this a test ratio.
3) Of these ratios, choose the smallest one. The corresponding number b is the pivot.

Step 5: Use the pivot to clear the pivot column in the normal manner. This gives the next
tableau.

Step 6: Repeat Steps 3-5 until there are no more negative numbers in the bottom row
(with the possible exception of the Answer column).
The solution for the LP problem is then the basic solution associated with the final tableau.

07.09.2010                                          Dr. Engin Aktaş                                                5
IZMIR INSTITUTE OF TECHNOLOGY           Engineering Management            EM506 Summer2010

Let’s use an example to provide simple interpretation of the simplex method.

Maximize     2A+3B
Subject to
A        ≤ 6,
A+     B ≤ 7,
2 B ≤ 9,
- A + 3 B ≤ 9,
A       ≥ 0,
B ≥ 0.
First step is to change program to a format that facilitates pivoting called Form 1.
A linear program is said to be written in Form 1 if
• Its objective is to maximize the quantity z or to minimize z.
• Each decision variable other than z is constrained to be
nonnegative.
•All other constraints are equations

07.09.2010                               Dr. Engin Aktaş                                     6
IZMIR INSTITUTE OF TECHNOLOGY             Engineering Management         EM506 Summer2010
Maximize z
Subject to
-2 A - 3 B                      +z =0
1A          + s1                    = 6,
1A+ 1 B           + s2               = 7,
2 B              + s3         = 9,
-1 A + 3 B                     + s4 = 9,
A ≥ 0, B ≥ 0, si ≥0 for i=1, 2, 3, 4.
A canonical form is the format that each linear program can be converted
Form 1 is a canonical form, which we can verify through the following observations;
•Form 1 encompasses maximization problems and minimization problems
•An equation can be included that equates z to the value of the objective
•Each inequality constraint can be converted into an equation by insertion of a
nonnegative variable
•Each variable that is unconstrained in sign can be replaced by the difference of
two nonnegative variables

07.09.2010                                 Dr. Engin Aktaş                                  7
IZMIR INSTITUTE OF TECHNOLOGY           Engineering Management            EM506 Summer2010

smallest of the
Largest magnitude               test ratio gives
Pivot column                          negative value                   the pivot

To obtain the basic solution in a tableau, look for the columns that are cleared (all
zeros except for one entry)

Active Variables   s1=6   s2=7 s3=9           s4=9   z=0

Inactive Variables             A=0      B=0

Notice that the current value of z (z = 0) hardly seems like a maximum value. In
the simplex method, we obtain larger and larger values of p by pivoting and then
looking at the new basic solution.

07.09.2010                               Dr. Engin Aktaş                                   8
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A      B        s1      s2         s3            s4       z   RHS    test ratio
1     0         1       0          0             0       0    6
1     1         0       1          0             0       0    7         7
0     2         0       0          1             0       0    9        4.5
-1     3         0       0          0             1       0    9         3
-2     -3        0       0          0             0       1    0

First Row: 1*R1
Second Row: 3*R2-R4
Third Row : 3*R3-2*R4

Pivot              Fourth Row: 1*R4
Column              Fifth Row: 1*R5+1*R4

A      B        s1      s2         s3            s4    z       RHS   test ratio
1     0         1       0          0             0    0        6        6
4     0         0       3          0            -1    0        12       3
2     0         0       0          3            -2    0        9       4.5
-1     3         0       0          0             1    0        9
-3     0         0       0          0             1    1        9

07.09.2010                             Dr. Engin Aktaş                                     9
IZMIR INSTITUTE OF TECHNOLOGY               Engineering Management              EM506 Summer2010
A           B        s1       s2         s3           s4       z      RHS    test ratio
1          0         1        0          0            0       0       6         6
4          0         0        3          0           -1       0       12        3
2          0         0        0          3           -2       0       9        4.5
-1          3         0        0          0            1       0       9
-3          0         0        0          0            1       1       9
First Row: 4*R1-1*R2
Second Row: 1R2                       No more negative values in
Third Row : 2*R3-1*R2                 the last row

Fourth Row: 4*R4+R2                   So the process is completed

Fifth Row: 4*R5+3*R2
A          B        s1        s2         s3           s4       z      RHS   test ratio
0          0         4        -3          0            1       0       12
4          0         0         3          0           -1       0       12
0          0         0        -3          6           -3       0       6
0          12        0         3          0            3       0       48
0          0         0         9          0            1       4       72
The optimal solution is then
A           B        s1       s2         s3           s4       z
3           4        3         0         1            0        18

07.09.2010                                   Dr. Engin Aktaş                                       10
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Graphical LP Solution

07.09.2010                       Dr. Engin Aktaş                       11
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Non Basic Variables                   Basic Variables
x1            x2        s1         s2          s3            s4      z      RHS      test ratio
6           4          1          0           0             0      0       24          4
1           2          0          1           0             0      0       6           6
-1           1          0          0           1             0      0       1           -1
0           1          0          0           0             1      0       2        #DIV/0!
-5           -4         0          0           0             0      1       0
Entering Variable                      Exiting Variable

x1            x2         s1        s2           s3            s4      z      RHS     test ratio
6.000          4.000    1.000     0.000        0.000         0.000   0.000   24.000       6
0.000          1.333   -0.167     1.000        0.000         0.000   0.000    2.000      1.5
0.000          1.667    0.167     0.000        1.000         0.000   0.000    5.000       3
0.000          1.000    0.000     0.000        0.000         1.000   0.000    2.000       2
0.000         -0.667    0.833     0.000        0.000         0.000   1.000   20.000

Basic Variables       Non Basic Variables    Basic Variables
x1            x2         s1        s2          s3            s4       z     RHS      test ratio
6.00          0.00       1.50     -3.00        0.00          0.00    0.00    18.00
0.00          1.33      -0.17      1.00        0.00          0.00    0.00    2.00
0.00          0.00       1.50     -5.00        4.00          0.00    0.00    10.00
0.00          0.00       0.13     -0.75        0.00          1.00    0.00    0.50
0.00          0.00       1.50      1.00        0.00          0.00    2.00    42.00

07.09.2010                                         Dr. Engin Aktaş                                         12
IZMIR INSTITUTE OF TECHNOLOGY   Engineering Management   EM506 Summer2010

Graphical LP Solution

07.09.2010                       Dr. Engin Aktaş                       13
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x1     x2        s1      s2       s3       s4   z      RHS
1       0      0.25     -0.5       0        0   0        3
0       1     -0.125    0.75       0        0   0       1.5
0       0      0.375   -1.25       1        0   0       2.5
0       0      0.125   -0.75       0        1   0       0.5
0       0      0.75      0.5       0        0   1       21

x1 = 3
x2 = 1.5
s3 = 2.5
s4 = 0.5

07.09.2010                           Dr. Engin Aktaş                           14
IZMIR INSTITUTE OF TECHNOLOGY           Engineering Management             EM506 Summer2010

Guidelines to Formulation

To formulate a linear program, one must answer these two questions
•What are the decision variables?
•What are the constraints?

Neither answer is obvious. Here are some guidelines to help construct a
linear program.

Identifying the Decision Variables

• A quantity is a natural candidate for a decision variable if it can
vary and if a cost or revenue varies with it.
•Try to organize the information into diagram or table

07.09.2010                               Dr. Engin Aktaş                                 15
IZMIR INSTITUTE OF TECHNOLOGY               Engineering Management          EM506 Summer2010

Identifying the Constraints
First lets identify several types of constraints
• A conservation constraint reflects the fact that something is
conserved.
• A bounding constraint places either a lower bound or an
upper bound on some quantity. This bound may represent
capacity.
• A ratio constraint is the linearization of a constraint on the
ratio of two linear expressions.

When building a linear program, look for
• Conservation – of matter, cash, people, flow, and so forth
• Bounds – including nonnegativity, capacities, and minimum
requirements
• Ratio – on the proportion of some quantity

07.09.2010                                   Dr. Engin Aktaş                              16
IZMIR INSTITUTE OF TECHNOLOGY         Engineering Management         EM506 Summer2010

Common Errors and Tips Avoid Them

Tips that avoid common errors:
• To make the constraints easy to read, use mnemonics
(memory aids) when you name the decision variables.
• Remember that each decision variable is a quantity. Define it
precisely, including its unit of measure.
• Check for consistent units of measure. Each term in a
constraint must have the same unit of measure; each term in the
objective must have the same unit of measure .
• Check that the linear program includes all of the relevant
information.

Tips on Inequalities

• Check that the capacity constraints and the other upper and
lower bounds are written as inequalities, not as equations.
• Remember to include the nonnegativity constraints

07.09.2010                             Dr. Engin Aktaş                               17
IZMIR INSTITUTE OF TECHNOLOGY            Engineering Management         EM506 Summer2010

Cracking and Blending Problem
A small refinery uses heat to crack crude oil into its constituents and then blends
them into two products, which are Regular and Premium grades of unleaded
gasoline.
A Barrel =42 gallons
Cracking facilities in this refinery
• Method A: Converts each barrel of crude oil into one barrel of 93 octane gasoline
• Method B: Converts each barrel of crude oil into one barrel of 85 octane gasoline
Variable cost
Method A = \$4.80      Method B = \$2.30
These cracking technologies operate independently of each other
The refinery has the capacity to process 4000 barrels per month using Method A
and 9000 barrels per month using Method B.

Regular gasoline must have an octane rating of at least 87, whereas Premium
gasoline must have an octane rating of at least 91. The octane blend of a blend
is proportional to the fractions of its constituents. For instance, blending one
barrel of 93 octane gasoline with three barrels of 85 octane gasoline produces
four barrels of 87 octane gasoline.

07.09.2010                                Dr. Engin Aktaş                               18
IZMIR INSTITUTE OF TECHNOLOGY             Engineering Management         EM506 Summer2010

Selling price from refinery for this month:
• \$32.76 per barrel (Regular)
• \$38.22 per barrel (Premium)

Refinery pays \$18.60 per barrel of crude oil.

Refinery can sell all the Regular gasoline it can produce but can sell only 3000
barrels per month of Premium.

The refinery must produce at least at least 4500 barrels of Regular gasoline
each month because it has a long-term contract to this amount.

The refinery is not able to store crude or refined products. Each month, it
buys the amount of crude oil that will be cracked, blended, and sold.

The goal is to maximize contribution (profit). How shall the refinery be
operated this month?

07.09.2010                                 Dr. Engin Aktaş                              19
IZMIR INSTITUTE OF TECHNOLOGY             Engineering Management                EM506 Summer2010

Lines are called arcs.
A Network Flow Diagram for this problem
Circles are called nodes.
AR
2                 4             R       • Flow occurs along each arc in
A
the direction of its arrow
C                               AP
1                                                   • Flow is conserved along each
arc; what flows in the tail end of
BR
B                                               an arc comes out the head
3                 5                 P   end.
BP
• Flow is also conserved at
each node. The total of the
Decision variables                                            flows into a node equals the
C = the quantity of crude oil purchased                       total of the flows out of it.
A = the quantity of crude oil refined by Method A
B = the quantity of crude oil refined by Method B
AR = the quantity of the output of Method A cracking that are blended into Regular gasoline
AP = the quantity of the output of Method A cracking that are blended into Premium gasoline
BR = the quantity of the output of Method B cracking that are blended into Regular gasoline
BP = the quantity of the output of Method B cracking that are blended into Premium gasoline
R = the quantity of Regular gasoline produced
P = the quantity of Premium gasoline produced

07.09.2010                                 Dr. Engin Aktaş                                       20
IZMIR INSTITUTE OF TECHNOLOGY                  Engineering Management        EM506 Summer2010

What Does This Network Flow Diagram Omit?
• The capacities of the two cracking facilities
• The bound on the amount of Regular and Premium fuel that the refinery produces
• The octane requirement of each grade of gasoline
• The contribution

Lets add constraints to account for each feature
A ≤ 4000                           B ≤ 9000
R ≥ 4500                            P ≤ 3000
93 AR  85 BR
R          87  93 AR  85 BR  87 R
93 AP  85 BP
P          91  93 AP  85 BP  91 P
All that remains is to measure the contribution
32.76 R + 38.22 P – 18.6 C – 4.80 A – 2.30 B

The problem is to maximize this expression subject to the linear constraints
that have been listed above.

07.09.2010                                      Dr. Engin Aktaş                            21
IZMIR INSTITUTE OF TECHNOLOGY          Engineering Management         EM506 Summer2010
The Linear Program is then
Maximize32.76 R  38.22 P 18.6  4.80 A  2.30 B
subjected to :
C  A  B,
A  AR  AP,
B  BR  BP,
AR  BR  R,
AP  BP  P,
A  4000,
B  9000,
R  4500,
P  3000,
93 AR  85BR  87 R ,
93 AP  85BP  91 P,
C  0, A  0, B  0, AR  0, AP  0, BR  0, BP  0, R  0, P  0.

07.09.2010                              Dr. Engin Aktaş                             22
IZMIR INSTITUTE OF TECHNOLOGY   Engineering Management   EM506 Summer2010

07.09.2010                       Dr. Engin Aktaş                       23
IZMIR INSTITUTE OF TECHNOLOGY   Engineering Management   EM506 Summer2010

07.09.2010                       Dr. Engin Aktaş                       24
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07.09.2010                       Dr. Engin Aktaş                       25
IZMIR INSTITUTE OF TECHNOLOGY          Engineering Management          EM506 Summer2010

Shipping
This area of application of linear programming was pioneered by Tjalling C.
Koopmans, who formulated trans-Atlantic shipping problems as
“transportation” problems. This produces insights helped him to propel
toward a Nobel Prize in Economics.

The Flow of Material
Examples
• The flow of petroleum from wells through
refineries and markets
• The flow of commercial airplanes between cities
•The flow of messages through telecommunication
system
• etc.

07.09.2010                              Dr. Engin Aktaş                                26
IZMIR INSTITUTE OF TECHNOLOGY          Engineering Management           EM506 Summer2010

Sample Problem (Production and Distribution)
A forest products company manufactures plywood in three plants and ships
it to four depots (warehouses), from which it sold to retails. Each plant lies
within its own timber zone, and each plant has a production capacity for the
current month. Check the table below for capacities. For example, plant 1
has the capacity to produce as many as 2500 units of plywood this month.
Each depot serves its own market, which has a demand for the current
month. These demands must be satisfied exactly and are specified in the
table. For example, exactly 2000 units must be shipped this month to depot
1. Table also specifies the unit shipping costs. For example, the cost of
shipping each unit from plant 1 to depot 1 is \$4. The cost of producing
plywood has been omitted because it is the same in each plant.

07.09.2010                              Dr. Engin Aktaş                                  27
IZMIR INSTITUTE OF TECHNOLOGY        Engineering Management        EM506 Summer2010
Depots
(demand nodes)
Plants                                      =2000
(supply node)                            1

≤2500
1
=3000
2

≤4000
2
=2500
3

≤3500       3
=1500
4

07.09.2010                            Dr. Engin Aktaş                            28
IZMIR INSTITUTE OF TECHNOLOGY         Engineering Management          EM506 Summer2010

A linear program modeled like in the previous slide known as a transportation
problem. In a transportation problem
• The nodes form into two groups: supply nodes and demand nodes.
• Each shipment occurs from one of the supply nodes to one of the
demand nodes.

The definition of he transportation problem contains an element of ambiguity.
In some formulations, the demands must be satisfied exactly, whereas in
other formulations, the demands are lower bounds, These two variants of the
transportation are genuinely different because it can be cheaper to ship more
than the minimum requirements

Transshipment and Assignment Problems
Many shipping problems fail to be transportation problems because they have
transshipment points, these are nodes that receive and send shipments. 
Transshipment problem
In a special case of transportation problem, exactly one unit is available at each
supply node and exactly one unit is required at each demand node.  Assignment
problem

07.09.2010                             Dr. Engin Aktaş                                  29
IZMIR INSTITUTE OF TECHNOLOGY   Engineering Management   EM506 Summer2010

07.09.2010                       Dr. Engin Aktaş                       30
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07.09.2010                       Dr. Engin Aktaş                       31
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Tight
constraints

Plant 1 and 3 are fully utilized, the plant 2 is underutilized.

07.09.2010                               Dr. Engin Aktaş                                   32
IZMIR INSTITUTE OF TECHNOLOGY        Engineering Management         EM506 Summer2010

Depots
(demand nodes)
Plants                                        =2000
(supply node)                              1

≤2500
1
=3000
2

≤4000
2
=2500
3

≤3500       3
=1500
4

07.09.2010                            Dr. Engin Aktaş                             33
IZMIR INSTITUTE OF TECHNOLOGY              Engineering Management         EM506 Summer2010
The Network Flow Model
• A network flow model arises frequently
• A network flow model is easy to spot once you learn what to look for
• Some network flow models can be guaranteed to have optimal
solutions that are integer-valued.
arc                   arc
i

node
Each arc fails into one of two classes

1)                     i                or             i

2)                     i                           j

Direction of movement
or of flow

07.09.2010                                  Dr. Engin Aktaş                             34
IZMIR INSTITUTE OF TECHNOLOGY              Engineering Management               EM506 Summer2010

A network flow model is a linear program whose decision variables
are the flows on the arcs of a directed network.

Flow in                                        Flow out
Flow in = Flow out

Flow pointing into node                      Flow pointing out of node
i

node
Flow pointing into node = Flow pointing out of node

Each arc in a network flow model has three elements of data
• A lower bound of its flow: may be zero or positive
• An upper bound of flow: at least as large as the lower bound, and may be as
large as +∞
• In minimization prob., each arc has a unit cost. In a maximization prob. has a
unit profit. The cost (or profit) of a flow on an arc equals the product of the
arc’s unit cost (profit) times the amount of the flow.

07.09.2010                                  Dr. Engin Aktaş                                   35
IZMIR INSTITUTE OF TECHNOLOGY           Engineering Management           EM506 Summer2010
Integer-valued Variables
In many applications, the flows in a network flow model are
quantities whose values must be integers.
There is a simple hypothesis under which at least one optimal solution to a
network flow model is guaranteed to be integer-valued. A network flow model is
said to have integer bounds if each arc’s lower and upper bound is an integer
or is plus infinity.
Integrality Theorem: In a network flow model that has integer
bounds, every extreme point assigns to each arc a flow whose
value is an integer.
How Can You Identify a Network Flow Model?
First write each node’s conservation of flow equation as “flow out – flow in = zero”
Flow on a arc can appear only in these ways
• In the arc’s lower bound constraint
• In the arc’s upper bound constraint
• In the flow conservation constraint (flow out of a node)
• In the flow conservation constraint (flow into a node)
• In the objective function
In any variable appears in any other way, it is not a network flow model.

07.09.2010                               Dr. Engin Aktaş                                  36
IZMIR INSTITUTE OF TECHNOLOGY          Engineering Management          EM506 Summer2010

Workforce Planning
The workforce planning problem is that of matching staffing levels to the
company’s needs.
Example
A priority mail company operates a distribution office around the clock. Its minimum
staffing needs appear in Table below. For instance, 15 employee are needed
between 6 A.M. and 10 A.M. Each employee is assigned to one of six shifts, which
start at 2 A.M. and every four hours thereafter. The second shift starts at 6 A.M.,
the third shift starts at 10 A.M., and so forth. Each shift lasts eight hours.
Employees earn 20% overtime pay for each hour worked between 6 P.M. and 6
A.M. The company has asked you to set staffing levels on each shift that meet or
exceed the staffing needs in each period and, subject to that requirement,
minimizes the payroll cost.

Minimum staffing level in each period

07.09.2010                              Dr. Engin Aktaş                                37
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07.09.2010                       Dr. Engin Aktaş                       38
IZMIR INSTITUTE OF TECHNOLOGY         Engineering Management   EM506 Summer2010

If we forget about the variables be
integer-valued, the result would not
differ.
The problem is a network flow problem
with integer bounds.
The Integrality theorem guarantee that the
simplex method finds an optimal solution
that is integer-valued. The requirement
that the variables be integer-valued is
unnecessary

07.09.2010                             Dr. Engin Aktaş                       39
IZMIR INSTITUTE OF TECHNOLOGY          Engineering Management           EM506 Summer2010
Maximizing the Flow
A max-flow problem is a network flow model that has two distinguished nodes,
which are called the source and sink.
The object of a max-flow problem is to send the maximum flow from the
source node to sink node, while satisfying the capacity constraint on each arc
and the flow conservation at each intermediary node
Example
Because of an electrical breakdown, a city is in need of electrical power. There is
ample reserve power at a nearby hydroelectric station. Figure in the next slide depicts
the portion of the power grid that connects the hydroelectric station to the city. Each
power link is represented by an arc or by a pair of arcs. The hydroelectric station is
connected by a power line (arc) to node 1 of this figure. This city is connected by a
power line (arc) to node 8. Each node depicts the junction of at least three power lines.
Adjacent to each intermediary arc is the spare capacity of its power line in the arc’s
direction of flow. For instance, the power line between nodes 3 and 6 has eight units of
spare capacity in the 3-to-6 direction and three units of spare capacity in the 6-to03
direction . Transmission losses along the power lines are negligible, so flow is
conserved along each arc. Flow is also conserved at each node of the power grid.
How much added power can be supplied to this city?
How can it be routed to the city?

07.09.2010                              Dr. Engin Aktaş                                 40
IZMIR INSTITUTE OF TECHNOLOGY       Engineering Management                      EM506 Summer2010
8
3
3                                       6
18                                                        12
5                       4
1
4                   2       9        8
9                       11
18
19
2           3           8               7
5
2
5

07.09.2010                           Dr. Engin Aktaş                                          41
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07.09.2010                       Dr. Engin Aktaş                       42
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Problem on Recreational Vehicles

In a plant 3 different models of
hours per week                                recreational       vehicle    is
produced.

The plant manager wishes to learn what
production mix maximizes the profit that
can be obtained from this plant

Some definitions                                                                     hours per car
Variable Costs: expense that is incurred if the action is taken
Fixed Costs: A cost that is incurred whether or not that action is taken
Contribution: Revenue that an action creates less its variable cost

07.09.2010                                          Dr. Engin Aktaş                                     43
IZMIR INSTITUTE OF TECHNOLOGY        Engineering Management   EM506 Summer2010

Problem on Recreational Vehicles

07.09.2010                            Dr. Engin Aktaş                       44
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The decision variables are

S = # of Standard model vehicles made per week
F = # of Fancy model vehicles made per week
L = # of Luxury model vehicles made per week

The linear program is
Maximize 840 S + 1120 F + 1200 L
subjected to
Engine Shop:           3S + 2F + 1L           ≤     120,
Body Shop:             1S + 2F + 3L           ≤      80,
Standard Finishing:     2S                    ≤      96,
Fancy Finishing:              3F              ≤     102,
Luxury Finishing:                   2L        ≤      40,
S ≥ 0, F ≥ 0, L ≥ 0.

07.09.2010                               Dr. Engin Aktaş                               45
IZMIR INSTITUTE OF TECHNOLOGY           Engineering Management          EM506 Summer2010

Slack and Tight Constraints
An inequality constraint is said to be tight if it is satisfied as an
equation and to be slack if it is satisfied as a strict inequality.
Shaded constraints are tight constraints. They will help us to relate
the optimal solution to the situation that the linear program models.

07.09.2010                               Dr. Engin Aktaş                              46
IZMIR INSTITUTE OF TECHNOLOGY             Engineering Management             EM506 Summer2010

The Perturbation Theorem

If the data of a linear program are perturbed by small
amounts, its optimal solution can change, but its tight
constraints stay tight, and its slack constraints stay slack.
Optimal Solution?
For example considering the recreational vehicle problem
• Make Standard model vehicles 20 per week, make Fancy
model 30 per week, make no Luxury model vehicles
•Keep the Engine and Body shops busy making Standard
and Fancy model vehicles and make no Luxury model
vehicles
In case the model is exact, both are identical.

But when data is a little off, first one may not be optimal
The second way is still valid when the data are not exact. It is the
perturbation theorem that assures us it stays feasible and optimal for a
range of data.

07.09.2010                                 Dr. Engin Aktaş                                 47
IZMIR INSTITUTE OF TECHNOLOGY          Engineering Management            EM506 Summer2010
How a Linear Program Approximates Reality
The reasons why the model is an approximation
• Uncertainty
•Aggregation
•Linearization
Are the data we used in recreational vehicle model certain?
No, the data are UNCERTAIN!!!
For example: Engine shop capacity taken to be 120 machine in a representative week.
But in the cases of
•Routine maintenance
•Machine breakdowns
•Shortages of vital parts
•Absence of key workers
•Power failures and other unforeseen events
As it is clear, the number of hours in a particular week could be larger or smaller. This
is also true for other constraints. Because of that uncertainty in the data is one reason
that models are approximate.

07.09.2010                              Dr. Engin Aktaş                                  48
IZMIR INSTITUTE OF TECHNOLOGY         Engineering Management          EM506 Summer2010

In the models we lump together several activities an a single entity. For
example Engine Shop, it is actually a system consisting of people , tools
and machines. Because of the aggregation the model is approximate.

Finally, linearization makes the model an approximation. When we express
the capacity constraints, we assumed a linear interaction between the
vehicle types. Unfortunately real interactions are more complicated and
nonlinear in nature.

We simplified the problem and constructed a model by aggregation and
linearization. That is what we should do, otherwise solution would be
intractable.

IMPORTANT!!!!! We have to check the influence of the
simplifications on optimal solution.

07.09.2010                             Dr. Engin Aktaş                               49
IZMIR INSTITUTE OF TECHNOLOGY             Engineering Management         EM506 Summer2010

Is the optimal solution to the linear program robust?

Sensitivity Analysis is required to answer that.

Change the data susceptible, rerun the linear program and
measure the changes in optimal solution.

Linear programming allows to do that without rerunning by the
help of Perturbation Theorem.

07.09.2010                                 Dr. Engin Aktaş                               50
IZMIR INSTITUTE OF TECHNOLOGY           Engineering Management         EM506 Summer2010

A Shadow Price and its Range
Due to Perturbation Theorem, optimal solution should keep the
tight constraints tight
3S + 2F + 1L            =     120 + d ,
1S + 2F + 3L            =      80,
L       =       0.
Solving the above system gives
S = 20 + 0.5 d,
F = 30 - 0.25 d,
L =     0.
And optimal value becomes
Optimal value = 840 x (20 + 0.5 d) + 1120 x (30 – 0.25 d) + 1200 x 0
= 50,400 + 140 d.
A change in capacity of the Engine Shop produces a a change of 140 d in the
optimal values.
Each constraint's shadow price equals the change in the optimal
value per unit change in that constraint’s right-hand-side value.

07.09.2010                               Dr. Engin Aktaş                               51
IZMIR INSTITUTE OF TECHNOLOGY               Engineering Management      EM506 Summer2010

Unit of Measure

The unit of measure of the shadow price is equal to the unit of measure
of objective function divided by unit of measure of the constraint

For example in recreational vehicle problem
Objective function      \$/week
Constraints             hours/week

So unit of measure of shadow price for this example is

\$/week
= \$ / hour
hours/week

07.09.2010                                   Dr. Engin Aktaş                           52
IZMIR INSTITUTE OF TECHNOLOGY          Engineering Management         EM506 Summer2010

Breakeven Prices
The shadow prices are breakeven prices.

The manager of the recreational vehicle plant should:

•Augment the Engine shop capacity if the cost is below \$140 per hour
•Not augment the Engine shop capacity if the price is above \$140/hr
•Decrement (rent out) some Engine shop capacity if she can receive a
price above \$140/hr
•Not decrement the Engine shop capacity at a price below \$140/hr.

If the capacity is altered, then according to Perturbation Theorem the
production levels should be adjusted to keep the tight constraints tight and
slack constraints slack.
How much capacity alteration can be made??????

07.09.2010                              Dr. Engin Aktaş                                 53
IZMIR INSTITUTE OF TECHNOLOGY              Engineering Management           EM506 Summer2010

The Range for Which a Shadow Price Applies
For each shadow price there is a range where the Perturbation Theorem
holds. At each end of a range, a slack constraint becomes tight.
The shadow price of the Engine shop (\$140/hr) was found by solving
the following system
S = 20 + 0.5 d,
F = 30 - 0.25 d,
L = 0.
•The perturbed solution must stay feasible
•The production levels must stay nonnegative
•The capacity constraint on each finishing shop must be satisfied
•   0≤   S = 20 + 0.5 d       d = -20 / 0.5 = -40
Largest
•   0≤   F = 30 - 0.25 d     d = -30 /(- 0.25) = 120
d = 56
• 96 ≥ 2 S = 40 + d           d =56
Smallest
•102 ≥ 3 F = 90 – 0,75 d      d =12/(- 0.75) = -16
d = -16
• 40 ≥ 2 L = 0               all the d values

07.09.2010                                  Dr. Engin Aktaş                               54
IZMIR INSTITUTE OF TECHNOLOGY           Engineering Management             EM506 Summer2010

So, for d > 56 or d < -16, this perturbed solution is not feasible.
Why?
Because the constraints are violated.

In general
Each shadow price applies to a range of changes in the right-
hand-side value of the constraint
•The end-points of this range are called the allowable
increase and allowable decrease.
•At each end-point, a slack constraint becomes tight.

07.09.2010                               Dr. Engin Aktaş                                 55
IZMIR INSTITUTE OF TECHNOLOGY          Engineering Management          EM506 Summer2010

Sensitivity to the Right-Hand-Side Values

These are slack constraints
The capacities are underutilized
Perturbing their right-hand-side value do not change optimal value.
If the optimal value satisfies a constraint as a strict inequality, that
constraint’s shadow price equal to zero.
The shadow prices apply to simultaneous changes in the right-hand-side
values if and only if these changes are small enough that the solution that
keeps the tight constraints stays feasible.

07.09.2010                              Dr. Engin Aktaş                               56
IZMIR INSTITUTE OF TECHNOLOGY          Engineering Management          EM506 Summer2010
Opportunity Cost and Marginal Profit
The opportunity cost of doing something equals the reduction in profit that occurs if
you set aside (free up) the resources that are needed to do a new thing.

=B\$8-B\$10                                =SUMPRODUCT(D3:D7,\$H3:\$H7)
The marginal profit for doing something equals its contribution less the opportunity cost
of the resources that must be freed up to accomplish that thing.

07.09.2010                              Dr. Engin Aktaş                                57
IZMIR INSTITUTE OF TECHNOLOGY             Engineering Management           EM506 Summer2010

Suppose that the optimal solution to a linear program equates a variable to a
positive value. Then
•This variable’s contribution equals the opportunity cost of the resources
that would need to be diverted to make one unit if the commodity that it
represents.
•Equivalently, this variable’s marginal profit equals zero.

07.09.2010                                 Dr. Engin Aktaş                                58
IZMIR INSTITUTE OF TECHNOLOGY              Engineering Management           EM506 Summer2010

Sensitivity to Objective Function Coefficients
The optimal solution is unchanged for a range of
increases and decreases in an objective coefficient
Final Reduced Objective       Allowable  Allowable
Cell          Name             Value   Cost   Coefficient    Increase   Decrease
\$B\$9 value of variable S           20        0         840           200       280
\$C\$9 value of variable F           30        0        1120           560       100
\$D\$9 value of variable L             0    -200        1200           200     1E+30

Optimum solution does not change when a type vehicle contribution lies between
Obj. Coeff. – Allow. Dec. and Obj. Coeff. + Allow. Inc.

Beyond that range the optimal solution changes.

In a maximization problem, the reduced cost of each variable equals its
marginal profit, namely, its contribution less the opportunity cost of the
resources that are needed to make one unit of the commodity that it
represents
Actually it is marginal profit.

07.09.2010                                  Dr. Engin Aktaş                                  59
IZMIR INSTITUTE OF TECHNOLOGY           Engineering Management              EM506 Summer2010

Computing the Shadow Prices
The information we have
•If a constraint is slack, its shadow price is zero.
•If a variable is positive, its contribution equals its opportunity
cost, which means that its reduced cost equals zero.
Shadow Prices: E, B, SF, LF and FF
Last three are slack variables
SF = 0     FF = 0           LF = 0

The optimal solution gives S and F as positive values and
reduced cost should be zero
0 = 840 – [ 3 E + 1 B + 2 SF]
0 = 1120 – [2 E + 2 B + 3 FF]
Solving this system gives
E = 140
B = 420

07.09.2010                               Dr. Engin Aktaş                                     60
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07.09.2010                       Dr. Engin Aktaş                       61
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07.09.2010                       Dr. Engin Aktaş                       62
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An Alternate Use of Capacity?
If someone offers to rent five hours of Engine shop capacity for one
week for \$1000. Is it profitable to accept this offer?

In analysis we found out that the shadow price is \$140 per hour
and valid for decreases up to 16 hours.

Net profit = \$1000 – 5 x \$140 = \$300

So, accepting that offer will increase contribution by \$300.
A New Tool?
An engineer proposes a new tool to cut time spent for Luxury model in
Body shop from 3 hrs to 2.6 hrs. Should this tool be bought?

The opportunity cost of one Luxury model vehicle would
change.

O.C. of one Luxury vehicle = 1 x 140 + 2.6 x 420 + 2 x 0 = \$1232
So, the marginal profit by making one Luxury model vehicle would
equal 1200 – 1232 = -32. It is not worthwhile, independent of its cost.

07.09.2010                             Dr. Engin Aktaş                                  63
IZMIR INSTITUTE OF TECHNOLOGY           Engineering Management            EM506 Summer2010
An Improved Tool?
An engineer proposes a new tool to cut time spent for Luxury model in
Body shop from 3 hrs to 2 hrs. Should this tool be bought?

O.C. of one Luxury vehicle = 1 x 140 + 2 x 420 + 2 x 0 = \$980
1200 – 980 = 220 > 0, Luxury model vehicles would become profitable
The linear program would be rerun, and judged whether that
amount justifies the cost of tool.
General        If the optimal solution equates a variable to zero, you can use
Principle      the shadow prices to see how to reduce its opportunity cost
below its contribution, thereby making it profitable.
A New Activity?
After the linear program run, we might want to see whether adding
another activity would be profitable?
Computing the marginal profit of such an activity is known as pricing out that
activity.
A new vehicle, Nifty is proposed; it requires 2 hrs in the E.S., 2.5 hrs in the B.S., and
1.5 hrs in the S.F.S., each Nifty would be sold with a contribution of \$1300.
O.C. of one Nifty = 2 x 140 + 2.5 x 420 + 1.5 x 0 = \$1330
Marginal profit = 1300 – 1330 = -30        The Nifty model is not profitable.

07.09.2010                               Dr. Engin Aktaş                                  64
IZMIR INSTITUTE OF TECHNOLOGY         Engineering Management            EM506 Summer2010

Insight

•The Perturbation theorem describes the optimal
solution in a way that can implemented when the
model is inexact.

•The ranges of the right-hand-side values and on the
objective coefficients give a feel for the extent t which
varying these data keeps the solution optimal.

•The shadow prices indicate why this solution take
the form that it does.

07.09.2010                             Dr. Engin Aktaş                                65
IZMIR INSTITUTE OF TECHNOLOGY             Engineering Management          EM506 Summer2010

Minimizing Cost
Up to now we have worked on maximization problems. Sometimes an
optimization problem may require to find the minimum.
For example
• A commuter may wish to find the quickest way to destination
•An engineer may wish to complete a project as inexpensively as
possible and in the time allotted.

Switching the Sense of Optimization
It is easy to convert minimization and maximization problems into each
other, by simply multiplying each objective coefficient by -1 and reverse the
sense of optimization.
Net revenue = Income – Expenditure
Net cost = Expenditure - Income
But you don’t need to convert minimization problem to maximization one
to solve. Solver is capable of solving the minimization problem as well.

07.09.2010                                 Dr. Engin Aktaş                                   66
IZMIR INSTITUTE OF TECHNOLOGY             Engineering Management            EM506 Summer2010

Solver uses the definitions
In any linear program – maximization or minimization –
Each constraint’s shadow price equals the change in the optimal value
per unit change in that constraint’s right-hand-side value.
Each variable’s reduced cost has this meaning;
• If the variable is positive, its reduced cost equals zero.
• If the variables is zero, its reduced cost equals the change in the
objective value if the optimal solution is perturbed by equating that
variable 1 and adjusting the values of other variables of the other
variables so as to keep the tight constraints tight.

Signs of Reduced Costs
• Maximization problem  if variable is zero, its reduced cost
cannot be positive
• Minimization problem  if variable is zero, its reduced cost
cannot be negative.

07.09.2010                                 Dr. Engin Aktaş                                67
IZMIR INSTITUTE OF TECHNOLOGY        Engineering Management           EM506 Summer2010

Reduced                                 Its contribution less
Maximization                                  =    Marginal Profit   its opportunity cost
Problem                      Cost

The coefficient of a variable in an objective function is
Minimization        direct cost instead of contribution
Problem
Opportunity cost of doing something equals the
increase in cost that occurs if you set aside the resources
that are needed to do that thing

Marginal cost of doing something equals its direct cost
plus its opportunity cost.

07.09.2010                            Dr. Engin Aktaş                                        68
IZMIR INSTITUTE OF TECHNOLOGY            Engineering Management                 EM506 Summer2010

Diseconomy of Scale
For a function that measures total profit
Decreasing Marginal Return
contribution from
quantity that is produced
the marginal unit

For a function that measures total cost
Increasing Marginal Cost
cost of the
quantity that is produced                  marginal unit

Decreasing marginal return and increasing marginal cost
manifest a diseconomy of scale because the marginal profit
(or cost) can only get better as the scale (quantity) increases.

07.09.2010                                Dr. Engin Aktaş                                     69
IZMIR INSTITUTE OF TECHNOLOGY                                     Engineering Management                  EM506 Summer2010
Each of the first 12 units of the Standard model vehicle has \$840
Example                               as its contribution, and any units in excess of 12 have a contribution
of \$500 each.
16000

14000

12000
Slope equals
\$840
10000
Total Contribution

Slope equals
8000
\$500
6000

4000
Define two new decision variables
S1 = Standard vehicles sold at \$840
2000
S2 = Standard vehicle sold at \$500

0
0       2     4    6     8      10      12     14     16   18      20
Quantity

07.09.2010                                                         Dr. Engin Aktaş                                      70
IZMIR INSTITUTE OF TECHNOLOGY         Engineering Management                  EM506 Summer2010

The linear program is then
Maximize 0 S + 840 S1 + 500 S2 + 1120 F + 1200 L
subjected to
Engine Shop:           3S +                                  2F +     1L     ≤   120,
Body Shop:             1S +                                  2F +     3L     ≤    80,
Standard Finishing:     2S                                                   ≤    96,
Fancy Finishing:                                             3F              ≤   102,
Luxury Finishing:                                                      2L    ≤    40,
Counting:                S-     S1 -   S2                                    =     0,
Sales limit                     S1                                           ≤    12,
S ≥ 0,   S1 ≥ 0, S2 ≥ 0,          F ≥ 0,   L ≥ 0.

Solving using Solver gives

S1 = 12, S2 = 0,    S = 12,       F = 34,      L = 0.

A linear program readily accommodates decreasing marginal return,
increasing marginal cost, and other diseconomies of scale. Its
formulation introduces unintended options, but optimization rules them
out.

07.09.2010                             Dr. Engin Aktaş                                       71
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No Economy of Scale
For a function that measures total profit
Increasing Marginal Return
contribution from
quantity that is produced
the marginal unit

For a function that measures total cost
Decreasing Marginal Cost
cost of the
quantity that is produced                  marginal unit

Increasing marginal return and decreasing marginal cost
manifest an economy of scale because the marginal profit (or
cost) can only get worse as the scale (quantity) increases.

07.09.2010                                Dr. Engin Aktaş                                     72
IZMIR INSTITUTE OF TECHNOLOGY                                       Engineering Management                EM506 Summer2010
Each of the first 12 units of the Standard model vehicle has \$500
Example                                 as its contribution, and any units in excess of 12 have a contribution
14000
of \$800 each.

12000

10000

Slope equals
Total Contribution

8000
\$500
6000
Slope equals
\$800
4000

2000

0
0        2     4     6      8      10        12   14      16   18     20
Quantity

07.09.2010                                                           Dr. Engin Aktaş                                    73
IZMIR INSTITUTE OF TECHNOLOGY            Engineering Management                EM506 Summer2010

The linear program is then
Maximize 0 S + 500 S1 + 800 S2 + 1120 F + 1200 L
subjected to
Engine Shop:           3S +                                     2F +     1L   ≤   120,
Body Shop:             1S +                                     2F +     3L   ≤    80,
Standard Finishing:     2S                                                    ≤    96,
Fancy Finishing:                                                3F            ≤   102,
Luxury Finishing:                                                        2L   ≤    40,
Counting:                S-     S1 -   S2                                     =     0,
Sales limit                     S1                                            ≤    12,
S ≥ 0,   S1 ≥ 0, S2 ≥ 0,             F ≥ 0,   L ≥ 0.

A linear program fails to accommodate increasing marginal return,
decreasing marginal cost, and other economics of scale. Its formulation
introduces unintended options, and optimization selects them.

07.09.2010                                Dr. Engin Aktaş                                     74
IZMIR INSTITUTE OF TECHNOLOGY        Engineering Management        EM506 Summer2010

An Integer Program
Generally speaking, an economy of scale can only be handled by an
integer program.

Integer Program: one or more of its variables is
restricted to integer values.

07.09.2010                            Dr. Engin Aktaş                            75
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Linear Programs and Plane Geometry

Maximize   2A+3B
Subject to
A        ≤ 6,
A+     B ≤ 7,
2 B ≤ 9,
- A + 3 B ≤ 9,
A       ≥ 0,
B ≥ 0.

07.09.2010                             Dr. Engin Aktaş                       76
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8

7            A+ B≤ 7

6                                       - A + 3B ≤ 9
2B≤9
5
B

4

3                                                           A≤ 6

2             Feasible Region

1

0
0   1       2        3             4        5      6       7        8
A

07.09.2010                               Dr. Engin Aktaş                               77
IZMIR INSTITUTE OF TECHNOLOGY                   Engineering Management                 EM506 Summer2010

8

7                                            Objective        Points in the direction of
vector          increase of the objective
6                                                                        In this case only a
2A + 3B = 18
single    point   is
3            Optimal
5
optimal solution, if
2                     solution              two    points   are
Extreme         optimal, then so is
2A + 3B = 12                 (3,4)                               every point on the
B

4
points
line segment that
connects them.
3

2
2A + 3B = 6                                                          Isoprofit lines

1

2A + 3B = 0
0
0          1         2       3             4          5          6            7            8
A

07.09.2010                                       Dr. Engin Aktaş                                              78
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Linear Programs and Solid Geometry

07.09.2010                               Dr. Engin Aktaş                       79
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The Big M Method
If an LP has any > or = constraints, a starting bfs may not be readily apparent.
When a bfs is not readily apparent, the Big M method or the two-phase simplex
method may be used to solve the problem. The Big M method is a version of the
Simplex Algorithm that first finds a bfs by adding "artificial" variables to the problem.
The objective function of the original LP must, of course, be modified to ensure that
the artificial variables are all equal to 0 at the conclusion of the simplex algorithm.

07.09.2010                               Dr. Engin Aktaş                                     80
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Steps
1. Modify the constraints so that the RHS of each constraint is nonnegative (This requires that
each constraint with a negative RHS be multiplied by -1. Remember that if you multiply an
inequality by any negative number, the direction of the inequality is reversed!). After
modification, identify each constraint as a <, >, or = constraint.

2. Convert each inequality constraint to standard form (If constraint i is a < constraint, we add
a slack variable si; and if constraint i is a > constraint, we subtract an excess variable ei).

3. Add an artificial variable ai to the constraints identified as > or = constraints at the end of
Step 1. Also add the sign restriction ai > 0.

4. Let M denote a very large positive number. If the LP is a min problem, add (for each
artificial variable) Mai to the objective function. If the LP is a max problem, add (for each
artificial variable) -Mai to the objective function.

5. Since each artificial variable will be in the starting basis, all artificial variables must be
eliminated from row 0 before beginning the simplex. Now solve the transformed problem
by the simplex (In choosing the entering variable, remember that M is a very large
positive number!).

If all artificial variables are equal to zero in the optimal solution, we have found the
optimal solution to the original problem.
If any artificial variables are positive in the optimal solution, the original problem is
infeasible!!!
07.09.2010                                    Dr. Engin Aktaş                                           81
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Example illustrating Big M (Oranj Juice)
(Bevco manufactures an orange flavored soft drink called Oranj by
combining orange soda and orange juice. Each ounce of orange soda
contains 0.5 oz of sugar and 1 mg of vitamin C. Each ounce of orange juice
contains 0.25 oz of sugar and 3 mg of vitamin C. It costs Bevco 2¢ to
produce an ounce of orange soda and 3¢ to produce an ounce of orange
juice. Marketing department has decided that each 10 oz bottle of Oranj
must contain at least 20 mg of vitamin C and at most 4 oz of sugar. Use LP
to determine how Bevco can meet marketing dept.’s requirements at
minimum cost.
LP Model:
Let x1 and x2 be the quantity of ounces of orange soda and orange juice
(respectively) in a bottle of Oranj.

07.09.2010                             Dr. Engin Aktaş                              82
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Solving Oranj Example with Big M Method

The RHS of each constraint is nonnegative

07.09.2010                               Dr. Engin Aktaş                          83
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07.09.2010                       Dr. Engin Aktaş                       84
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07.09.2010                       Dr. Engin Aktaş                       85

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