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CHAPTER 1: INTRODUCTION TO
QUANTITATIVE ANALYSIS

Quantitative analysis uses a scientific approach
to decision making
 Both quantitative and qualitative factors must be
considered
 Quantitative factors:
 Determine how much our investment will be
worth in the future when deposited at a bank
at a given interest rate for a certain number of
years
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 Computing
financial ratios whose stock we
are considering
 Analyse cash flow and rates of return for
investment property
 Qualitative factors:
 Weather
 State and federal legislation
 New technological breakthroughs
 Outcome of an election
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
Quantitative analysis approach
Defining a
Problem
Testing the
Solution
Developing a
Model
Analysing the
Result
Acquiring
Input Data
Implementing
the Result
Developing a
Solution
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1. Defining the Problem
 Defining the problem can be the most
important step
 Concentrate on only a few problems
2. Acquiring Input Data
 The types of models include physical, scale,
schematic and mathematical models
 Most models contain one or more variables
and parameters
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3. Acquiring Input data
 Obtaining accurate data for a model is
essential
 Sources of data can be company reports or
any data sources
4. Developing a Solution
 Involves manipulating the model to arrive at
the best (optimal) solution to the problem
 A series of steps or procedures that are
repeated is called algorithm
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5. Testing the Solution
 Includes determining the accuracy and
completeness of the data used by the model,
done before the solution can be analysed and
implemented
6. Analysing the Result
 Starts with determining the implications of the
solution
 Sensitivity analysis or post optimality analysis
determines how the solutions will change with
a different model or input data
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7. Implementing the Result
 Incorporating the solution in the company
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
Models of Cost revenue and Profit
 Profit = Revenue – Expenses
 Revenue = Selling Price per Unit (P) ×
Number of units sold (x)
 Expenses = Fixed Cost (F) + Variable Cost
per unit (V) × Number of Units Sold (x)
 Breakeven Point = When Total Revenue
Equals Total Expenses
F
BEPx 
P V
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Breakeven Chart
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Example 1.1:
Swatch Time Plc buys, sells and repairs clock
and clock parts. It sells rebuilt spring for a price
of RM10. The fixed cost of the equipment to
build the spring is RM1, 000. The variable cost
per unit is RM5. What is the Breakeven Point?
What is the profit earned if the company sells
500 units of springs?
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CHAPTER 2: DECISION THEORY

Decision theory is an analytical and systematic
way to tackle problems
 A good decision is based on logic
 Six steps in Decision Making:
1. Clearly defined the problem at hand
 Whether to expand manufacturing line or
market new product
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2. List the possible alternatives
 Alternative is defined as a course of action or
a strategy that the decision maker can choose
3. Identify the possible outcomes or states of
nature
 States of nature are those outcomes which
the decision maker has little or no control on
it, e.g. favourable market or unfavourable
market
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4. List the payoff or profit of each combination of
alternatives and outcomes
Example 2.1:
Alternative
Construct a
large plant
Construct a
small plant
Do nothing
State of Nature
Favourable
Unfavourable
Market (RM)
Market (RM)
200, 000
- 180, 000
100, 000
- 20, 000
0
0
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5. Select one of the mathematical decision theory
models
6. Apply the model and make the decision
Types of Decision Making Environments
1. Decision making under certainty
 Decision makers know with certainty the
consequences of every alternative or
decision choice. E.g. You have RM1, 000 to
invest. You have two options: Saving
account with 6% interest or Government
bond with 10% interest. Which is your
recommendation?
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2. Decision making under risk
 Decision maker knows the probability of
occurrence of each outcome. Decision theory
models normally employ two equivalent
criteria: maximisation of expected monetary
value and minimisation of expected loss
3. Decision making under uncertainty
 Here the probability are not known to the
decision maker
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Decision Making Under Risk Models
(1) Expected Monetary Value (EMV)
 EMV is the weighted sum of possible
payoffs for each alternative
EMV (alternative i) = (payoff)1 × (probability)1
(payoff)2 × (probability)2 + ………….. +
(payoff)i × (probability)i
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Example 2.2:
John of 1st Silicon has the options of the following:
Alternative
Construct a large
plant
Construct a small
plant
Do nothing
Probabilities
State of Nature
Favourable
Market (RM)
200, 000
Unfavourable
Market (RM)
- 180, 000
100, 000
- 20, 000
0
0.5
0
0.5
Which alternative would you recommend to John?
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(2) Expected Value of Perfect Information (EVPI)
 EVPI places an upper bound on what to pay
for information
EVPI = Expected value with Perfect Information
(EVwPI) – Maximum EMV
EVwPI = (best payoff)1 × (probability)1 +
(best payoff)2 × (probability)2 + ……..+
(best payoff)i × (probability)i
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Example 2.3:
John has been approached by Xcel Marketing to
help John to make decision. It will change his
environment from one decision-making under
risk to one decision-making under certainty. Xcel
Marketing would charge John RM65, 000 for the
information. What would you recommend?
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(3) Expected Opportunity Loss (EOL)
 EOL is the cost of not picking the best solution
 An alternative approach to maximise EMV is
to minimise EOL
 Opportunity loss, sometimes called regret,
refers to the difference between optimal profit
for a given state of nature and the actual
payoff received
 Solving steps: (i) Construct the opportunity
loss table, (ii) Compute the EOL
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 EOL
will always result in the same decision as
the maximum EMV
 Minimum EOL = Maximum EMV
Example 2.4:
Construct the EOL for John of Example 2.2.
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Decision Making Under Uncertainty
Models
(1) Maximax
 It finds the alternative that maximises the
maximum payoff for every alternative
 Maximax is an optimistic approach
(2) Maximin
 It finds the alternative that maximises the
minimum payoff for every alternative
 Maximin is a pessimistic approach
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Example 2.5:
Alternative
State of Nature
Favourable Unfavourable
Market (RM) Market (RM)
Construct a
200, 000
- 180, 000
large plant
Construct a
small plant
Do Nothing
100, 000
- 20, 000
0
0
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Maximum
in Row
(RM)
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Example 2.6:
Alternative
State of Nature
Favourable Unfavourable
Market (RM) Market (RM)
a
200, 000
- 180, 000
Construct
large plant
Construct a
small plant
Do Nothing
100, 000
- 20, 000
0
0
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Minimum
in Row
(RM)
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(3) Criterion of Realism (Hurwicz Criterion)
 Uses the weighted average approach
 It is a compromise between an optimistic and
a pessimistic decision
 A coefficient of realism, , is selected
 Note that when  = 1, this is the same as the
optimistic criterion, when  = 0, this is the
same as the pessimistic criterion
Weighted average =  (Maximum in row) +
(1 - )(Minimum in row)
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Example 2.7:
Assuming  for John is 0.8. What is your
recommendation?
(4) Equally Likely (Laplace)
 Computes the highest average outcome
 Select the alternative with the maximum
average payoff
Example 2.8:
Compute the Equally Likely criterion for John in
Example 2.2
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(5) Minimax Regret
 Based on Opportunity Loss
 It finds the alternative that minimises the
maximum opportunity loss within each
alternative
Example 2.9:
Develop an opportunity loss table for John in
Example 2.2. What is the minimax regret
decision?
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Decision Tree
 Any problem that can be presented in a decision
table can also be graphically illustrated in a
decision tree
 All decision trees are similar in that they contain
decision points or nodes and state of nature or
nodes:
A decision node from which one of
several alternatives may be chosen
A state of nature node out of which
one state of nature will occur
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
1.
2.
3.
4.
5.
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Five steps of Decision Tree Analysis:
Define the problem
Structure or draw the decision tree
Assign probabilities to the state of nature
Estimate payoffs for each possible combination of
alternatives and states of nature
Solve the problem by computing expected
monetary values (EMV) for each state of nature.
This is done by working backward that is, starting
at the right of the tree and working back to
decision nodes on the left. At each decision node,
the alternative with the best EMV is selected
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Example 2.10:
John of 1st Silicon has the options of the following:
Alternative
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State of Nature
Favourable
Unfavourable
Market (RM)
Market (RM)
200, 000
- 180, 000
Construct a
large plant
Construct a
100, 000
- 20, 000
small plant
Do nothing
0
0
Probabilities
0.5
0.5
Which alternative would you recommend to John? Draw
the decision tree.
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
When a sequence of decisions need to be
made, decision trees are much more powerful
tools than decision tables:
 All outcomes and alternatives must be
considered
 Most of the probabilities are conditional
probabilities
 The cost of the survey had to be subtracted
from the original payoffs
 Start by computing the EMV of each branch
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Example 2.11:
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
Expected Value of Sample Information (EVSI)
measures the value of sample information
EVSI = (EV with sample information + cost) –
(EV without sample information)
In John’s case,
EVSI = (RM49, 200 + RM10,000) – (RM40, 000)
= RM19, 200
This means that John could have paid up to
RM19, 200 for a market survey and still come
out ahead. Since it costs only RM10, 000, the
survey is indeed worthwhile.
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Utility Theory
 The overall value of the result of a decision is
called utility
 Rational people make decisions that maximise
the expected utility
 Utility assessment assigns the worst outcome a
utility of 0 and the best outcome utility of 1
 When you are indifferent, the expected utilities
are equal
 Utility values replace monetary value in utility as
a decision making criterion. The objective is to
maximise the expected utility
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Expected utility of alternative 2 = expected utility
of alternative 1
Utility of other outcome = (p)(utility of best
outcome, 1) + (1 – p)(utility of worst outcome, 0)
Utility of other outcome
= (p)(1) + (1 – p)(0)
=p
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Example 2.12:
Jane would like to construct a utility curve
revealing her preference for money between
RM0 and RM10, 000. A utility curve is the graph
that plots utility value versus monetary value.
She can either invest her money in a bank
savings account or she can invest the same
money in a real estate deal. If the money is
invested in the bank, in three years Jane would
have RM5, 000. Is she invested in real estate,
after three years she could either have nothing
or RM10, 000. Jane, however, is very
conservative.
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Unless there is an 80% chance of getting RM10,
000 from the real estate deal Jane would prefer
to have money in the bank, where it is safe.
Construct the utility assessment.
Construct the utility curve for the following
information:
U (RM0) = 0
U (RM3, 000) = 0.50
U (RM5, 000) = 0.80
U (RM7, 000) = 0.90
U (RM10, 000) = 1.00
Is Jane a risk avoider or a risk taker?
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CHAPTER 3: LINEAR
PROGRAMMING MODEL

Linear programming is a technique for solving
problems for profit maximisation or cost
minimisation and resource allocation
 The word programming is simply used to denote
a series of events
 The various aspects of the problem are
expressed as linear equations
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Setting up a linear programming model:
Step 1: Define the variables
 What are the quantities that the company can
vary?
Step 2: Establish constraints
 Having defined the variables we can now
translate the two constraints into inequalities
involving the variables
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Step 3: Establish objective function
 An objective function is the mathematical
expression of the aim of a linear programming
exercise
There are two methods in linear programming:
 Graphical methods – for problems with two
variables only
 Simplex method – for problems with more than
two variables
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Graphical Method
 In the graphing model, the area on a graph of a
linear programming problem within which all of
the inequalities are satisfied in known as a
feasible area
y
y
y=6
4x + 3y = 24
x
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Example 3.1:
1.Draw the feasible area for the following inequalities:
2x + 3y
< 12
y
> 2x
2.Yung Kong Limited manufactures plastic covered steel
fencing in two qualities, standard and heavy gauge. Both
product pass through the same processes, involving
steel-forming and plastic bonding. Standard gauge
fencing sells at RM18 a roll and heavy gauge fencing at
RM24 a roll. Variable costs per roll are RM16 and RM21
respectively.
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There is an unlimited market for the standard gauge,
but demand for the heavy gauge is limited to 1, 300
rolls a year. Factory operations are limited to 2, 400
hours in each of the two production processes.
Determine the production mix, which will maximize
total contribution. Calculate the total contribution.
Processing hours per roll
Gauge
Steel-forming
Plasticbonding
Standard
0.6
0.4
Heavy
0.8
1.2
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3. Dell Computer has undertaken a contract to
supply a customer with at least 260 units in
total of two products, X and Y, during the next
month. At least 50% of the total output must be
in units of X. The products are each made by
two grades of labour, as follows.
X
Y
Hours
Hours
Grade A labour
4
6
Grade B labour
4
2
Total
8
8
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Although additional labour can be made
available at short notice, the company wishes to
make use of 1, 200 hours of Grade A labour and
800 hours of Grade B labour which has already
assigned to working on the contract next month.
The total variable cost per unit is RM120 for X
and RM100 for Y. Dell Computer Ltd. Wishes to
minimise expenditure on the contract next
month. Determine the production mix.
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Simplex Method
 Recall that the theory of LP states that the
optimal solution will lie at a corner point of the
feasible region
 In large LP problems, the feasible region cannot
be graphed because it has many dimensions,
but the concept is the same
 The simplex method systematically examines
corner points, using algebraic steps, until an
optimal solution is found
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In setting up the initial simplex solution:
 Slack variables are added to each less-than-orequal-to constraint
 Each slack variable represents an unused
resource
 Slack considers only corner points as it seeks
the best solution
 Then construct the initial simplex table
 See attached notes for steps of Simplex Method
for solving Maximisation Problems and
Minimisation Problems
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Example 3.2:
The Flair Furniture Company produces inexpensive
tables and chairs. The production process for each is
similar in that both require a certain number of hours of
carpentry work and a certain number of labour hours in
the painting and varnishing department. 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 and solve the problem using simplex
method.
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To handle > and = constraints, the simples
method makes a conversion like it made to <
constraints
 When dealing with a > constraint, we subtract a
surplus variable to form an equality
 The surplus variable tells how much the solution
exceeds the constraints resource
 Because of its analogy to a slack variable,
surplus is sometimes simply called negative
slack

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Example 3.3:
Constraint 1:
Rewritten as:
5X1 + 10X2 + 8X3 > 210
5X1 + 10X2 + 8X3 – S1 = 210
In setting up an initial simplex table, since all real
variables such as X1, X2 and X3 are set to “0”, S1
takes a negative value
 All
variables in LP problems must be
nonnegative at all times

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Hence, we need to introduce another type of
variable called artificial variable
 Artificial variables are needed in > and =
constraints

Constraint 1 completed:
5X1 + 10X2 + 8X3 – S1 + A1 = 210
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Example 3.4:
Constraint 2:
Rewritten as:
25X1 + 30X2 = 900
25X1 + 30X2 + A2 = 900
Artificial variables have no physical meaning and
drop out of the solution mix before the final table
 To make sure that an artificial variable is forced
out before the final solution is reached, it is
assigned a very high cost

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Example 3.5:
Minimise cost:
Constraints:
Z = $5X1 +$9X2 + $7X3
5X1 +10X2 + 8X3 > 210
25X1 +30X2 = 900
What would the completed objective function
and constraints be?
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Example 3.6:
The Tetra Chemical Corporation must produce
exactly 1, 000kg of a special mixture of
phosphate and potassium for a customer.
Phosphate costs $5 per kg and potassium costs
$6 per kg. No more than 300kg of phosphate
can be used, and at least 150kg of potassium
must be used. The problem is to determine the
least-cost blend of the two ingredients.
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
Four special cases in LP:
1. Infeasibility
2. Unboundedness
3. Redundancy
4. Alternate optimal solutions
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CHAPTER 4: TRANSPORTATION
AND ASSIGNMENT MODELS
Transportation problem:
 Distribution of items from several sources to
several destinations
 Supply
capacities
and
destination
requirements are known
 Assignment problem:
 One-to-one assignment of people to jobs

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Importance of special purpose algorithms:
 Fewer, less complicated, computations than
with simplex
 Less computer memory required
 Produce integer solutions
 Methods for solving transportation problem:
 Northwest corner rule
 Stepping-stone method
 MODI method
 Vogel’s approximation

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Northwest Corner Rule:
Start in the upper left-hand cell and allocate
units to shipping routes as follows:
1. Exhaust the supply (factory capacity) of each
row before moving down to the next row
2. Exhaust
the
demand
(warehouse)
requirements of each column before moving to
the next column to the right
3. Check
that all supply and demand
requirements are met
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Example 4.1:
Transportation costs, factory capacity and
warehouse requirement are as follows.
What is the optimal distribution method?
From
(Sources)
To (destinations)
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Factory
Capacity
Everise
Upwell
Choice
Daily
Jacobs
$5
$4
$3
100
Danone
$8
$4
$3
300
Julie’s
$9
$7
$5
300
Warehouse
Requirement
300
200
200
700
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Stepping-Stone Method:
1. Select any unused square to evaluate
2. Begin at this square. Trace a closed path back
to the original square via squares that are
already being used (only horizontal or vertical
moves allowed)
3. Place + in unused square: alternate – and + on
each corner square of the closed path
4. Calculate improvement index: add together the
unit cost figures found in each square
containing a +; subtract the unit cost figure in
each containing a –
5. Repeat steps 1 ~ 4 for each unused square
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MODI Method:
1. Compute the values for each row and column:
set Ri + Kj = Cij for those squares currently
used or occupied
2. After writing all equations, set R1 = 0
3. Solve the system of equations for Ri and Kj
values
4. Compute the improvement index for each
unused square by the formula improvement
index: Cij – Ri + Kj
5. Select the largest negative index and proceed
to solve the problem as you did using the
stepping stone method
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Vogel’s Approximation Method:
1. For each row/column of table, find difference
between two lowest cost (opportunity cost)
2. Find the greatest opportunity cost
3. Assign as many units as possible to lowest cost
square in row/column with greatest opportunity
cost
4. Eliminate row or column, which has been
completely satisfied
5. Begin again, omitting eliminated rows/columns
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The Hungarian Method (Assignment Model):
 Subtract the smallest number in each row from
every number in that row
 Subtract the smallest number in each column
from every number in that column
 Draw the minimum number of vertical and
horizontal straight lines necessary to cover zeros
in the table
 If the number of lines equals the number of
rows or columns, than one can make an
optimal assignment
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If the number of lines does not equal the number
of rows and columns:
 Subtract the smallest number not covered by
a line from every other uncovered number
 Add the same number to any number lying at
the intersection of any two lines
 Return to Step 2
 Make optimal assignments at locations of zeros
within the table

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Example 4.2:
What is the optimal assignment?
Person
Project No.
1
2
3
Adams
11
14
6
Brown
8
10
11
Cooper
9
12
7
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CHAPTER 5: NETWORK MODELS

1.
2.
3.
4.
The presentation will cover three network
models that can be used to solve a variety of
problems:
the minimal-spanning tree technique,
the maximal-flow technique,
the shortest-route technique, and
PERT / CPM
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Minimal-Spanning Tree Technique
Definition:
The minimal-spanning tree technique
determines the path through the network
that connects all the points while
minimizing total distance.
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For example:
If the points represent houses in a subdivision,
the minimal spanning tree technique can be used
to determine the best way to connect all of the
houses to electrical power, water systems, etc.
in a way that minimizes the total distance or length
of power lines or water pipes.
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The Maximum Flow Technique
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Definition:
The maximal-flow technique finds the maximum
flow of any quantity or substance through a
network.
For example:
This technique can determine the maximum
number of vehicles (cars, trucks, etc.) that can
go through a network of roads from one location
to another.
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Shortest Route Technique
Definition:
Shortest route technique can find the shortest
path through a network.
For example:
This technique can find the shortest route from
one city to another through a network of roads.
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1.
2.
3.
4.
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Minimal-Spanning Tree
Steps
Selecting any node in the network.
Connecting this node to the nearest node
minimizing the total distance.
Finding and connecting the nearest unconnected
node.
 If there is a tie for the nearest node, one can
be selected arbitrarily.
 A tie suggests that there may be more than
one optimal solution.
Repeating the third step until all nodes are
connected.
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Example 5.1: Network for Lauderdale Construction
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




Solving the network for Melvin Lauderdale
construction
Start by arbitrarily selecting node 1.
Since the nearest node is the third node at a
distance of 2 (200 feet), connect node 1 to node
3.
Considering nodes 1 and 3, look for the nextnearest node.
 This is node 4, which is the closest to node 3
with a distance of 2 (200 feet).
Once again, connect these nodes
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Maximal-Flow Technique

The maximal-flow technique allows the maximum
amount of a material that can flow through a
network to be determined.
 For example:
 It has been used to find the maximum number
of automobiles that can flow through a state
highway system.
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Example 5.2:
 Waukesha is in the process of developing a
road system for downtown.
 City planners would like to determine the
maximum number of cars that can flow
through the town from west to east.
 The road network is shown in next slide
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Road Network for Waukesha
 Traffic can flow in both directions
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Maximal-Flow Technique (continued)
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The Four Maximal-Flow Technique Steps:
1.
Pick any path from the start (source) to the finish (sink) with some
flow.
 If no path with flow exists, then the optimal solution has been
found.
2. Find the arc on this path with the smallest flow capacity available.
 Call this capacity C.
 This represents the maximum additional capacity that can be
allocated to this route.
3. For each node on this path, decrease the flow capacity in the
direction of flow by the amount C.

For each node on this path, increase the flow capacity in the
reverse direction by the amount C.
4. Repeat these steps until an increase in flow is no longer possible.
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Solving the Waukesha
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
Start by arbitrarily picking the path 1–2–6, at the
top of the network
 What is the maximum flow from west to east? It
is 2 because only 2 units (200 cars) can flow
from node 2 to node 6
 Now we adjust the flow capacities. As you can
see, we subtracted the maximum flow of 2 along
the path 1–2–6 in the direction of the flow (west
to east) and added 2 to the path in the direction
against the flow (east to west)
 The result is the new path in next slide
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Capacity Adjustment
Iteration 1
Add 2
2
1
2
2
3
West
Point
1
Subtract 2
3
2
6
East
Point
6
East
Point
0
4
1
West
Point
New path
1
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Final Network Flow
The maximum flow of 500 cars per hour is
summarized in the following table:
(Cars per Hour)
PATH
1-2-6
1-2-4-6
1-3-5-6
Total
FLOW
200
100
200
=500
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The Shortest-Route Technique
 The shortest-route technique minimizes the


distance through a network.
The shortest-route technique finds how a person
or item can travel from one location to another
while minimizing the total distance traveled.
The shortest-route technique finds the shortest
route to a series of destinations.
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Example 5.3: From Ray’s Plant to Warehouse
For example,
 Every day, Ray Design, Inc., must transport beds,
chairs, and other furniture items from the factory to
the warehouse.
 This involves going through several cities.
 Ray would like to find the route with the shortest
distance.
 The road network is shown on the next slide.
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Shortest-Route Technique (continued)
Roads from Ray’s Plant to Warehouse:
Plant
200
2
1
4
150
50
3
40
6
5
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83
Steps of the Shortest-Route Technique
1.
2.
3.
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Find the nearest node to the origin (plant). Put
the distance in a box by the node.
Find the next-nearest node to the origin (plant),
and put the distance in a box by the node. In
some cases, several paths will have to be
checked to find the nearest node.
Repeat this process until you have gone through
the entire network.
The last distance at the ending node will be the
distance of the shortest route.
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Shortest-Route Technique (continued)
Ray Design: 1st Iteration
Plant
100
2
1
200
4
150
50
3
40
6
5
Warehouse
 The nearest node to the plant is node 2, with
a distance of 100 miles.
 Thus, connect these two nodes.
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PROJECT MANAGEMENT: PERT / CPM
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Project management can be used to manage
complex projects.
 The first step in planning and scheduling a
project is to develop the work breakdown
structure.
 This involves identifying the activities that must
be performed in the project. Each detail and
each activity may be broken into its most basic
components.
 The
time, cost, resource requirements,
predecessors, and person(s) responsible are
identified.

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One of the earliest techniques was the Gantt
chart (Used by US Navy).
 This type of chart shows the start and finish
times of one or more activities, as shown below:

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The program evaluation and review technique
(PERT) and the critical path method (CPM) are
two popular quantitative analysis techniques that
help managers plan, schedule, monitor, and
control large and complex projects.
 They were developed because there was a
critical need for a better way to manage.

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






There are six steps common to both PERT and
CPM.
Define the project and all of its significant
activities or tasks.
Develop the relationships among the activities.
Decide which activities must precede others.
Draw the network connecting all of the
activities.
Assign time and/or cost estimates to each
activity.
Compute the longest time path through the
network; this is called the critical path.
Use the network to help plan, schedule,
monitor, and control the project.
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
The critical path is important because
activities on the critical path can delay the
entire project.
 PERT is probabilistic, whereas CPM is
deterministic.
 Almost any large project can be subdivided
into a series of smaller activities or tasks
that can be analyzed with PERT.
 Projects can have thousands of specific
activities and it is important to be able to
answer many associated questions.
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Example 5.4:
Background:
 General Foundry, Inc., a metal works plant has long
been trying to avoid the expense of installing air
pollution control equipment.
 The local environmental protection group has recently
given the foundry 16 weeks to install a complex air
filter system on its main smokestack.
 General Foundry was warned that it will be forced to
close unless the device is installed in the allotted
period.
 They want to make ensure that installation of the
filtering system progresses smoothly and on time.
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
Define the project and all project activities.
 Immediate predecessors are determined in
the second step.
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
Activities and events are drawn and connected.
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
The fourth step is to assign activity times.
 Time is usually given in units of weeks.
 Without solid historical data, managers are
often uncertain as to activity times.
 The developers of PERT thus employed a
probability distribution based on three time
estimates for each activity:
 Optimistic time
 Pessimistic time
 Most likely time
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
Optimistic time (a) = time an activity will take if
everything goes as well as possible. There
should be only a small probability (say, 1/1000)
of this occurring.

Pessimistic time (b) = time an activity would
take assuming very unfavorable conditions.
There should also be only a small probability
that the activity will really take this long.

Most likely time (m) = most realistic time
estimate to complete the activity.
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
Time Estimates (weeks) for General Foundry,
Inc.
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
The fifth step is to compute the longest path
through the network— the critical path.
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




To find the critical path, need to determine the
following quantities for each activity in the
network:
Earliest start time (ES): the earliest time an
activity can begin without violation of immediate
predecessor requirements.
Earliest finish time (EF): the earliest time at
which an activity can end.
Latest start time (LS): the latest time an activity
can begin without delaying the entire project.
Latest finish time (LF): the latest time an activity
can end without delaying the entire project.
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The earliest times are found by beginning at the
start of the project and making a forward pass
through the network
 The latest times are found by beginning at the
finish of the project and making a backward pass
through the network
 Slack is the length of time an activity can be
delayed without delaying the whole project.

Mathematically, Slack = LS-ES or LF-EF
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The Network Diagram:
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Slack Time:
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CHAPTER 6: FORECASTING

Forecasting techniques:
 Qualitative techniques
 Time series methods: moving average,
exponential smoothing, trend projections
 Causal methods: regression analysis, multiple
regression
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
Time series can be decomposed into:
 Trend (T): gradual up or down movement over
time
 Seasonality (S): pattern of fluctuations above
or below trend line that occurs every year
 Cycles (C): patterns in data that occur every
several years
 Random variations (R): ‘blips’ in the data
caused by change and unusual situations
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Demand
Components of Decomposition
650
550
450
350
250
150
50
-50
-150
0
1
2
3
4
5
Time (Years)
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Annual Sales
Scatter Diagram
450
400
350
300
250
200
150
100
50
0
0
2
4
6
8
10
12
Time (Years)
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Multiplicative model:
Demand = T × S × C × R
 Additive model:
Demand = T + S + C + R
 Moving Average
=  Demand in previous n periods
n
 Exponential Smoothing:
New forecast = Previous forecast
+  (Previous actual – Previous forecast)

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Forecast errors allow one to see how well the
forecast model works and compare that model
with other forecast models.
Forecast error = actual value – forecast value
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

Measures of forecast accuracy include:
Mean Absolute Deviation (MAD)
=  |forecast errors|
n

Mean Squared Error (MSE)
=  (errors)
n
2

Mean Absolute Percent Error (MAPE)
error
=  actual
n
× 100%
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Trend Projection
The formula for the trend projection is:
Y = b 0 + b1 X
where:
Y = predicted value
b1 = slope of the trend line
b0 = intercept
X = time period (1,2,3…n)
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X  Y 


XY 
 X  X Y  Y   XY  nXY 



b
 X  X   X  n X
X
2
1
2
2
n
2  X

n
2
Y
X


b Y b X  n b n
0
1
1
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Correlation Coefficient

The correlation coefficient (r) measures the
strength of the linear relationship.
r
nXYXY
[nX X ][nY (Y Y ]
2
2
2
2
2
Note: -1 < r < 1
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
Tracking signals measure how well predictions fit
actual data.
RSFE
TrackingSi gnal 
MAD
Σ(actual demand in period i  forecast demand in period i)

MAD
where
Σ | forecast errors |
MAD 
n
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