# Transportation_ Assignment and Transshipment problems

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```					              Chapter 7
Transportation, Assignment and
Transshipment Problems

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Applications of Network
Optimization
Physical analog    Physical analog
Applications                                                 Flow
of nodes            of arcs
phone exchanges,
Cables, fiber optic Voice messages,
Communication     computers,
systems       transmission
facilities, satellites
Pumping stations                        Water, Gas, Oil,
Hydraulic systems                         Pipelines
Reservoirs, Lakes                       Hydraulic fluids
Integrated      Gates, registers,
Wires        Electrical current
computer circuits    processors
Rods, Beams,
Mechanical systems           Joints                            Heat, Energy
Springs
Passengers,
Intersections,      Highways,
Transportation                                                freight,
Airports,        Airline routes
systems                                                   vehicles,
Rail yards         Railbeds
2                                                                operators
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Description
A transportation problem basically deals with the
problem, which aims to find the best way to fulfill
the demand of n demand points using the
capacities of m supply points. While trying to find
the best way, generally a variable cost of shipping
the product from one supply point to a demand
point or a similar constraint should be taken into
consideration.

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7.1 Formulating Transportation
Problems
Example 1: Powerco has three electric
power plants that supply the electric needs
of four cities.
•The associated supply of each plant and
demand of each city is given in the table 1.
•The cost of sending 1 million kwh of
electricity from a plant to a city depends on
the distance the electricity must travel.
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Transportation tableau

A transportation problem is specified by
the supply, the demand, and the shipping
costs. So the relevant data can be
summarized in a transportation tableau.
The transportation tableau implicitly
expresses the supply and demand
constraints and the shipping cost between
each demand and supply point.

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Table 1. Shipping costs, Supply, and Demand
for Powerco Example

From                                  To
City 1 City 2     City 3 City 4       Supply
(Million kwh)
Plant 1       \$8         \$6     \$10        \$9         35
Plant 2        \$9        \$12    \$13        \$7        50
Plant 3       \$14         \$9    \$16        \$5        40
Demand         45         20     30         30
(Million kwh)
Transportation Tableau
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Solution
1. Decision Variable:
Since we have to determine how much electricity
is sent from each plant to each city;

Xij = Amount of electricity produced at plant i
and sent to city j

X14 = Amount of electricity produced at plant 1
and sent to city 4
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2. Objective function

Since we want to minimize the total cost of shipping
from plants to cities;

Minimize Z = 8X11+6X12+10X13+9X14
+9X21+12X22+13X23+7X24
+14X31+9X32+16X33+5X34

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3. Supply Constraints
Since each supply point has a limited production
capacity;

X11+X12+X13+X14 <= 35
X21+X22+X23+X24 <= 50
X31+X32+X33+X34 <= 40

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4. Demand Constraints
Since each supply point has a limited production
capacity;

X11+X21+X31 >= 45
X12+X22+X32 >= 20
X13+X23+X33 >= 30
X14+X24+X34 >= 30

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5. Sign Constraints
Since a negative amount of electricity can not be
shipped all Xij’s must be non negative;

Xij >= 0 (i= 1,2,3; j= 1,2,3,4)

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LP Formulation of Powerco’s Problem
Min Z = 8X11+6X12+10X13+9X14+9X21+12X22+13X23+7X24
+14X31+9X32+16X33+5X34

S.T.:    X11+X12+X13+X14 <= 35           (Supply Constraints)
X21+X22+X23+X24 <= 50
X31+X32+X33+X34 <= 40
X11+X21+X31 >= 45               (Demand Constraints)
X12+X22+X32 >= 20
X13+X23+X33 >= 30
X14+X24+X34 >= 30
Xij >= 0 (i= 1,2,3; j= 1,2,3,4)
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General Description of a Transportation
Problem
1. A set of m supply points from which a good is
shipped. Supply point i can supply at most si
units.
2. A set of n demand points to which the good is
shipped. Demand point j must receive at least di
units of the shipped good.
3. Each unit produced at supply point i and shipped
to demand point j incurs a variable cost of cij.

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Xij = number of units shipped from supply point i to
demand point j
i m j n
min  cijXij
i 1 j 1
j n
s.t. Xij  si (i  1,2,...,m)
j 1
i m

X
i 1
ij    dj ( j  1,2,...,n)

Xij  0(i  1,2,...,m; j  1,2,...,n)

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Balanced Transportation Problem
If Total supply equals to total demand, the
problem is said to be a balanced
transportation problem:

i m           j n

s  d
i 1
i
j 1
j

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Balancing a TP if total supply exceeds total
demand

If total supply exceeds total demand, we
can balance the problem by adding dummy
demand point. Since shipments to the
dummy demand point are not real, they are
assigned a cost of zero.

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Balancing a transportation problem if total
supply is less than total demand

If a transportation problem has a total
supply that is strictly less than total
demand the problem has no feasible
solution. There is no doubt that in such a
case one or more of the demand will be left
unmet. Generally in such situations a
penalty cost is often associated with unmet
demand and as one can guess this time the
total penalty cost is desired to be minimum
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7.2 Finding Basic Feasible
Solution for TP
Unlike other Linear Programming
problems, a balanced TP with m supply
points and n demand points is easier to
solve, although it has m + n equality
constraints. The reason for that is, if a set
of decision variables (xij’s) satisfy all but
one constraint, the values for xij’s will
satisfy that remaining constraint
automatically.

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Methods to find the bfs for a balanced TP

There are three basic methods:

1. Northwest Corner Method

2. Minimum Cost Method

3. Vogel’s Method

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1. Northwest Corner Method
To find the bfs by the NWC method:
Begin in the upper left (northwest) corner of the
transportation tableau and set x11 as large as
possible (here the limitations for setting x11 to a
larger number, will be the demand of demand
point 1 and the supply of supply point 1. Your
x11 value can not be greater than minimum of
this 2 values).

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According to the explanations in the previous slide
we can set x11=3 (meaning demand of demand
point 1 is satisfied by supply point 1).
5

6

2

3          5     2       3

3                              2

6

2

X          5     2       3
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After we check the east and south cells, we saw that
we can go east (meaning supply point 1 still has
capacity to fulfill some demand).
3           2                  X

6

2

X           3    2       3

3           2                  X

3                  3

2

X           X    2       3
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After applying the same procedure, we saw that we
can go south this time (meaning demand point 2
needs more supply by supply point 2).
3          2                 X

3    2            1

2

X          X    X       3

3          2                 X

3    2       1    X

2

23               X          X    X       2
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Finally, we will have the following bfs, which is:
x11=3, x12=2, x22=3, x23=2, x24=1, x34=2

3        2                      X

3       2        1     X

2     X

X        X       X        X

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2. Minimum Cost Method
The Northwest Corner Method dos not utilize shipping
costs. It can yield an initial bfs easily but the total
shipping cost may be very high. The minimum cost
method uses shipping costs in order come up with a
bfs that has a lower cost. To begin the minimum cost
method, first we find the decision variable with the
smallest shipping cost (Xij). Then assign Xij its largest
possible value, which is the minimum of si and dj

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After that, as in the Northwest Corner Method we
should cross out row i and column j and reduce the
supply or demand of the noncrossed-out row or
column by the value of Xij. Then we will choose the
cell with the minimum cost of shipping from the
cells that do not lie in a crossed-out row or column
and we will repeat the procedure.

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An example for Minimum Cost Method
Step 1: Select the cell with minimum cost.

2          3          5           6
5

2          1          3           5
10

3          8          4           6
15

12          8          4           6

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Step 2: Cross-out column 2

2            3         5             6
5

2            1         3             5
2
8

3            8         4             6
15

12            X         4             6

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Step 3: Find the new cell with minimum shipping
cost and cross-out row 2

2           3       5          6
5

2           1       3          5
X
2            8

3           8       4          6
15

10           X       4          6

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Step 4: Find the new cell with minimum shipping
cost and cross-out row 1

2           3        5          6
X
5

2           1        3          5
X
2           8

3           8        4          6
15

5           X        4          6

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Step 5: Find the new cell with minimum shipping
cost and cross-out column 1

2           3        5          6
X
5

2           1        3          5
X
2           8

3           8        4          6
10
5

X           X        4          6

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Step 6: Find the new cell with minimum shipping
cost and cross-out column 3

2           3           5       6
X
5

2           1           3       5
X
2           8

3           8           4       6
6
5                       4

X           X           X       6

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Step 7: Finally assign 6 to last cell. The bfs is found
as: X11=5, X21=2, X22=8, X31=5, X33=4 and X34=6

2           3           5           6
X
5

2           1           3           5
X
2           8

3           8           4           6
X
5                       4           6

X           X           X           X

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3. Vogel’s Method
Begin with computing each row and column a penalty.
The penalty will be equal to the difference between
the two smallest shipping costs in the row or column.
Identify the row or column with the largest penalty.
Find the first basic variable which has the smallest
shipping cost in that row or column. Then assign the
highest possible value to that variable, and cross-out
the row or column as in the previous methods.
Compute new penalties and use the same procedure.

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An example for Vogel’s Method
Step 1: Compute the penalties.

Supply   Row Penalty

6            7          8
10         7-6=1

15           80         78
15       78-15=63

Demand              15            5          5

Column Penalty     15-6=9     80-7=73      78-8=70

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Step 2: Identify the largest penalty and assign the
highest possible value to the variable.

Supply   Row Penalty

6            7          8
5         8-6=2
5

15           80         78
15       78-15=63

Demand            15            X          5

Column Penalty   15-6=9         _        78-8=70

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Step 3: Identify the largest penalty and assign the
highest possible value to the variable.

Supply   Row Penalty

6            7            8
0           _
5            5

15           80           78
15           _

Demand            15            X            X

Column Penalty   15-6=9         _            _

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Step 4: Identify the largest penalty and assign the
highest possible value to the variable.

Supply   Row Penalty

6            7            8
X           _
0             5            5

15           80           78
15           _

Demand               15            X            X

Column Penalty       _             _            _

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Step 5: Finally the bfs is found as X11=0, X12=5,
X13=5, and X21=15

Supply   Row Penalty

6            7            8
X           _
0             5            5

15           80           78
X           _
15

Demand                X            X            X

Column Penalty        _            _            _

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7.3 The Transportation Simplex
Method

In this section we will explain how the simplex
algorithm is used to solve a transportation problem.

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How to Pivot a Transportation Problem

Based on the transportation tableau, the following
steps should be performed.
Step 1. Determine (by a criterion to be developed
shortly, for example northwest corner method) the
variable that should enter the basis.
Step 2. Find the loop (it can be shown that there is
only one loop) involving the entering variable and
some of the basic variables.
Step 3. Counting the cells in the loop, label them as
even cells or odd cells.
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Step 4. Find the odd cells whose variable assumes the
smallest value. Call this value θ. The variable
corresponding to this odd cell will leave the basis. To
perform the pivot, decrease the value of each odd cell
by θ and increase the value of each even cell by θ. The
variables that are not in the loop remain unchanged.
The pivot is now complete. If θ=0, the entering
variable will equal 0, and an odd variable that has a
current value of 0 will leave the basis. In this case a
degenerate bfs existed before and will result after the
pivot. If more than one odd cell in the loop equals θ,
you may arbitrarily choose one of these odd cells to
leave the basis; again a degenerate bfs will result
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7.5. Assignment Problems
Example: Machineco has four jobs to be completed.
Each machine must be assigned to complete one job.
The time required to setup each machine for completing
each job is shown in the table below. Machinco wants to
minimize the total setup time needed to complete the
four jobs.

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Setup times
(Also called the cost matrix)
Time (Hours)
Job1     Job2   Job3    Job4
Machine 1       14     5      8      7
Machine 2       2     12      6      5
Machine 3       7      8      3      9
Machine 4       2      4      6     10

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The Model
According to the setup table Machinco’s problem can be
formulated as follows (for i,j=1,2,3,4):
min Z  14 X 11  5 X 12  8 X 13  7 X 14  2 X 21  12 X 22  6 X 23  5 X 24
 7 X 31  8 X 32  3 X 33  9 X 34  2 X 41  X 42  6 X 43  10 X 44
s.t. X 11  X 12  X 13  X 14  1
X 21  X 22  X 23  X 24  1
X 31  X 32  X 33  X 34  1
X 41  X 42  X 43  X 44  1
X 11  X 21  X 31  X 41  1
X 12  X 22  X 32  X 42  1
X 13  X 23  X 33  X 43  1
X 14  X 24  X 34  X 44  1
Xij  0orXij  1
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For the model on the previous page note that:

Xij=1 if machine i is assigned to meet the demands of
job j

Xij=0 if machine i is not assigned to meet the demands
of job j

In general an assignment problem is balanced
transportation problem in which all supplies and
demands are equal to 1.
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The Assignment Problem
In general the LP formulation is given as
n     n

Minimize    c
i 1 j 1
ij   xij

n

x
j 1
ij    1, i  1,   ,n   Each supply is 1

n

 xij  1, j  1,
i 1
,n   Each demand is 1

xij  0 or 1, ij
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Problem
• The Air Force has used this for assigning
thousands of people to jobs.

• This is a classical problem. Research on the
assignment problem predates research on LPs.

• Very efficient special purpose solution techniques
exist.
– 10 years ago, Yusin Lee and J. Orlin solved a problem
with 2 million nodes and 40 million arcs in ½ hour.

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Although the transportation simplex appears to be very
efficient, there is a certain class of transportation
problems, called assignment problems, for which the
transportation simplex is often very inefficient. For that
reason there is an other method called The Hungarian
Method. The steps of The Hungarian Method are as
listed below:
Step1. Find a bfs. Find the minimum element in each row
of the mxm cost matrix. Construct a new matrix by
subtracting from each cost the minimum cost in its row.
For this new matrix, find the minimum cost in each
column. Construct a new matrix (reduced cost matrix) by
subtracting from each cost the minimum cost in its
column.
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Step2. Draw the minimum number of lines (horizontal
and/or vertical) that are needed to cover all zeros in the
reduced cost matrix. If m lines are required , an optimal
solution is available among the covered zeros in the
matrix. If fewer than m lines are required, proceed to step
3.

Step3. Find the smallest nonzero element (call its value
k) in the reduced cost matrix that is uncovered by the
lines drawn in step 2. Now subtract k from each
uncovered element of the reduced cost matrix and add k
50step2.
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7.6 Transshipment Problems
A transportation problem allows only shipments that go
directly from supply points to demand points. In many
situations, shipments are allowed between supply points
or between demand points. Sometimes there may also
be points (called transshipment points) through which
goods can be transshipped on their journey from a
supply point to a demand point. Fortunately, the optimal
solution to a transshipment problem can be found by
solving a transportation problem.

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Transshipment Problem
• An extension of a transportation problem
– More general than the transportation problem in that in this
problem there are intermediate “transshipment points”. In
addition, shipments may be allowed between supply points
and/or between demand points
• LP Formulation
– Supply point: it can send goods to another point but cannot
receive goods from any other point
– Demand point It can receive goods from other points but
cannot send goods to any other point
– Transshipment point: It can both receive goods from other
points send goods to other points

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The following steps describe how the optimal solution to
a transshipment problem can be found by solving a
transportation problem.
Step1. If necessary, add a dummy demand point (with a
supply of 0 and a demand equal to the problem’s excess
supply) to balance the problem. Shipments to the dummy
and from a point to itself will be zero. Let s= total
available supply.
Step2. Construct a transportation tableau as follows: A
row in the tableau will be needed for each supply point
and transshipment point, and a column will be needed for
each demand point and transshipment point.
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Each supply point will have a supply equal to it’s
original supply, and each demand point will have a
demand to its original demand. Let s= total available
supply. Then each transshipment point will have a supply
equal to (point’s original supply)+s and a demand equal
to (point’s original demand)+s. This ensures that any
transshipment point that is a net supplier will have a net
outflow equal to point’s original supply and a net
demander will have a net inflow equal to point’s original
demand. Although we don’t know how much will be
shipped through each transshipment point, we can be
sure that the total amount will not exceed s.
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Transshipment Example
• Example 5: Widgetco manufactures widgets at two
factories, one in Memphis and one in Denver. The
Memphis factory can produce as 150 widgets, and the
Denver factory can produce as many as 200 widgets
per day. Widgets are shipped by air to customers in
LA and Boston. The customers in each city require
130 widgets per day. Because of the deregulation of
airfares, Widgetco believes that it may be cheaper
first fly some widgets to NY or Chicago and then fly
them to their final destinations. The cost of flying a
widget are shown next. Widgetco wants to minimize
the total cost of shipping the required widgets to
customers.

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Transportation Tableau Associated
with the Transshipment Example
•             NY Chicago LA        Boston Dummy Supply
•   Memphis \$8       \$13    \$25 \$28          \$0      150
•   Denver \$15 \$12 \$26 \$25                   \$0      200
•   NY         \$0    \$6      \$16 \$17          \$0      350
•   Chicago \$6       \$0     \$14 \$16          \$0      350
•   Demand 350      350     130    130       90
•   Supply points: Memphis, Denver
•   Demand Points: LA Boston
•   Transshipment Points: NY, Chicago
•   The problem can be solved using the transportation simplex
method
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Limitations of Transportation Problem
• One commodity ONLY: any one product supplied
and demanded at multiple locations
– Merchandise
– Electricity, water
• Invalid for multiple commodities: (UNLESS
transporting any one of the multiple commodities is
completely independent of transporting any other
commodity and hence can be treated by itself alone)
– Example: transporting product 1 and product 2 from the
supply points to the demand points where the total amount
(of the two products) transported on a link is subject to a
capacity constraint
– Example: where economy of scale can be achieved by
transporting the two products on the same link at a larger
total volume and at a lower unit cost of transportation
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Limitations of Transportation Problem
– Difficult to generalize the technique to accommodate
(these are generic difficulty for “mathematical
programming,” including linear and non-linear
programming
• Economy of scale the per-unit cost of transportation on a link
decreasing with the volume (nonlinear and concave; there is a trick
to convert a “non-linear program with a piecewise linear but
convex objective function to a linear program; no such tricks exists
for a piecewise linear but concave objective function)
• Fixed-cost: transportation usually involves fixed charges. For
example, the cost of truck rental (or cost of trucking in general)
consists of a fixed charge that is independent of the mileage and a
mileage charge that is proportional to the total mileage driven.
Such fixed charges render the objective function NON-LINEAR
and CONCAVE and make the problem much more difficult to
solve
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Chapter 8
Network Models

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Networks are Everywhere
• Physical Networks
–   Railway Networks
–   Airline traffic Networks
–   Electrical networks, e.g., the power grid
• Abstract networks
– organizational charts
– precedence relationships in projects
• Others?
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Overview:
• Networks and graphs are powerful
modeling tools.
• Most OR models have networks or graphs
as a major aspect
• Each representation has its advantages
– Major purpose of a representation
• efficiency in algorithms
• ease of use

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Description
Many important optimization problems can be analyzed
by means of graphical or network representation. In this
chapter the following network models will be discussed:

1. Shortest path problems
2. Maximum flow problems
3. CPM-PERT project scheduling models
4. Minimum Cost Network Flow Problems
5. Minimum spanning tree problems

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WHAT IS CPM/PERT FOR?
CPM/PERT are fundamental tools of project
management and are used for one of a kind, often
large and expensive, decisions such as building
docks, airports and starting a new factory. Such
decisions can be described via mathematical
models, but this is not essential. Some would
argue that CPM/PERT is not a pure OR topic.
CPM/PERT really falls into gray area that can be
claimed by fields other than OR also.

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General Comments on CPM/PERT vs. ALB
Assembly Line Balancing (ALB) are naturally not
discussed in this text, but it is important to be aware of the
huge difference between the ALB and CPM/PERT
concepts because the precedence diagrams look so similar.
Activity on node (AON) method of network precedence
diagram drawing and the ALB diagram are identical
looking at first. The ALB deals with small repetitive
items such as TV’s while CPM/PERT deals with large one
of a kind projects.

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Network Analysis and Their LP
Connections
• Several Major Basic Classes of Network Problems
– How to recognize and formulate them? What are the
features they can be used to model? What are their
limitations?
– All can be formulated as an LP
– Several important “ad-hoc” algorithms and their rationales
will be provided
• Most efficient transportation of goods, information
etc. through a network
– Transshipment problem

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Network Analysis and Their LP
Connections
• Most efficient way to go from one point to another in a
distance network or networks representing non-distance
phenomenon, e.g., the “cost network” representing
production, inventory, and other costs
– Shortest path problem:
• Find the shortest path between two points in a network
• Dijkstra algorithm
• Limitations: Breakdown of the Dijkstra’s algorithm when “side
• LP formulation
• LP formulation to accommodate some side constraints, e.g.,
disallowing use of two particular links in the shortest path
– An example application to non-distance contexts:
Minimum production and inventory cost in the context of
dynamic programming.
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Network Analysis and Their LP
Connections
– Maximum amount of flow from one point to
another in a capacitated network
• Maximum flow problem
• The flow-augmenting algorithm
• Limitations: breakdown of the flow-augmenting
• LP formulation
• LP formulation to accommodate some “side-
constraints,” e.g., no link carrying more than 30% of the
whole flow so as to avoid drastic reduction of flow after
failure of any one flow-carrying link

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8.1 Basic Definitions
A graph or network is defined by two sets of symbols:
• Nodes: A set of points or vertices(call it V) are called
nodes of a graph or network.
Nodes
1                        2

• Arcs: An arc consists of an ordered pair of vertices and
represents a possible direction of motion that may occur
between vertices.                   Arc

1                        2

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• Chain: A sequence of arcs such that every arc has
exactly one vertex in common with the previous arc is
called a chain.
Common vertex
between two arcs

1           2

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• Path: A path is a chain in which the terminal node of
each arc is identical to the initial node of next arc.
For example in the figure below (1,2)-(2,3)-(4,3) is a
chain but not a path; (1,2)-(2,3)-(3,4) is a chain and a
path, which represents a way to travel from node 1 to
node 4.
1               4

2               3
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Essence of Dijkstra’s Shortest- Path
Algorithm
• Key Points regarding the nature of the
algorithm
– In each iteration, the shortest path from the origin
to one of the rest of the nodes is found. That is, we
obtain one new “solved” node in each iteration.
(More than one such path and node may be found
in one iteration when there is a tie. There may also
exist multiple shortest paths from the origin to
some nodes.)
– The algorithm stops when the shortest path to the
destination is found

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Essence of Dijkstra’s Shortest- Path
Algorithm
• General thought process involved in each
iteration
– Let S be the current set of “solved nodes” (the set
of nodes whose shortest paths from the origin been
found), N be the set of all nodes, and N – S be the
set of “unsolved nodes
• 1. The next “solved” node should be reachable directly
from one of the solved nodes via one direct link or arc
(these nodes can be called neighboring nodes of the
current solved nodes). Therefore, we consider only such
nodes and all the links providing the access from the
current solved nodes to these neighboring nodes (but no
72
Essence of Dijkstra’s Shortest- Path
Algorithm
– 2. For each of these neighboring nodes, find the
shortest path from the origin via only current
solved nodes and the corresponding distance from
the origin
– 3. In general, there exist multiple such neighboring
nodes.The shortest path to one of these nodes is
claimed to have been found. This node is the one
that has the shortest distance from the origin
among these neighboring nodes being considered.
Call this new node “solved node.”

73
.
Algorithm for the Shortest Path Problem
• Objective of the nth iteration: Find the nth nearest node to the
origin (to be repeated for n = 1, 2, … until the nth nearest node is
the destination)
• Input for the nth Iteration: (n – 1) nearest nodes to the origin
(solved for at the previous iterations), including their shortest
path and distance from the origin. (These nodes plus the origin
will be called solved nodes; the others are unsolved nodes)
• Candidates for the nth nearest node: Each solved node that is
directly connected by a link to one or more unsolved nodes
provides one candidate  the unsolved node with the shortest
• Calculation of nth nearest node: For each solved node and its
candidate, add the distance between them and the distance of the
shortest path from the origin to this solved node. The candidate
with the smallest such total distance is the nth nearest node (ties
provide additional solved nodes), and its shortest path is the one
generating this distance.

74
.
• Cars are not allowed into the park
• There is a narrow winding road system for trams and
for jeeps driven by the park rangers
–   The road system is shown without curves in the next slide
–   Location O is the entrance into the park
–   Other letters designate the locations of the ranger stations
–   The scenic wonder is at location T
–   The numbers give the distance of these winding roads in
miles
• The park management wishes to determine which
route from the park entrance to station T has the
smallest total distance for the operation of the trams

75
.

7
A
D       5
2
2
4                           T
1
5
O                       B                                   7
3
4               1
E
4
C

76
.
Dijkstra’s Algorithm for Shortest Path on a
Network with Positive Arc Lengths
• Oth iteration: Shortest distance from node O to
Node O. S = {O}.
• Ist iteration:
– Step 1: Neighboring Nodes = {A, B, C}
– Step 2: Shortest path from O to neighboring nodes
that traverse through the current set of solved
nodes S. Min {2, 5, 4} = 2 (corresponding to node
A).
– Step 3: The shortest path from O to A has been
found with a distance of 2. S = {O, A}

77
.
Dijkstra’s Algorithm for Shortest Path on a
Network with Positive Arc Lengths

A
Solved Nodes
2

O                      B
5

4

C
78
.
Dijkstra’s Algorithm for Shortest Path on a
Network with Positive Arc Lengths
• 2nd Iteration:
– Step 1: Neighboring nodes = {B, C, D}
– Step 2: Min (Min (2 + 2, 5), 4, (2 + 7)) = 4.
– Step 3: Shortest path from B and C has been
found. S = {O, A, B, C}
•                                                7
Current Solved Nodes         A                           D
(2)               2

(0)                                 B
5
C
O
79                                        4
.
Dijkstra’s Algorithm for Shortest Path on a
Network with Positive Arc Lengths
• 3rd Iteration:
– Step 1: Neighboring nodes = {D, E}. Only AD, BD, BE,
and CE
– Step 2: Min(Min(2 + 7, 4+4), Min(4 + 3, 4+4)) = 7
– Step 3: The shortest path to E has been found S = {O, A, B,
C, E}

7
A
Current Solved                                              D
(2)            4
Nodes
3
B

O                     (4)                             E
80                                           C
.
4
(4)
Dijkstra’s Algorithm for Shortest Path on a
Network with Positive Arc Lengths
• Iteration 4
– Step 1: Include (only) Nodes D and T. Include
only arcs AD, BD, ED, & ET
– Step 2: Min((min(2+7, 4+4, 7+1), (7+7))) = 8
– Step 3: Shortest path from node O to Node D has
been found. S = {O, A, B, C, D, E}
7
A
(2)                              D
4
O                                          1
(4)   B
E             T
Current solved nodes                             7
81                  (4)                   (7)
C   .
Dijkstra’s Algorithm for Shortest Path on a
Network with Positive Arc Lengths
• Iteration 5
– Step 1: Include only node T and include arcs DT and ET
– Step 2: Min(8+5, 7+7) = 13 (no other competing nodes)
– Step 3: The shortest path from the origin to T, the
destination, has been found, with a distance of 13

A   (2)
Current solved nodes           D           5
(8)
B   (4)                           T
(0)
O
7
82                                           E
(4)             (7)
C          .
Dijkstra’s Algorithm for Shortest Path on a
Network with Positive Arc Lengths
• Final Solution
• Incidentally, we have also found the nth
nearest node from the origin sequentially
A       (2)
(8)
2
2                                          5
D
4
T
O
B                             1
(0)       4                         3                         (13)
(4)
E
(7)
83                           C
(4)
.
Shortest-Path Algorithm Applied to
n        Solved Nodes       Closest      Total     Nth      Minimum       Last
Directly Connected   Connected   Distance   Nearest   Distance   Connection
to Unsolved Nodes    Unsolved    Involved    Node
Node
1            O                A           2         A          2          OA

2, 3          O                C          4          C          4          OC
A                B        2+2=4        B          4          AB
4            A                D        2+7=9
B                E        4+3=7        E          7          BE
C                E        4+4=8
5            A                D        2+7=9
B                D        4+4=8        D          8          BD
E                D        7+1=8        D          8          ED
6            D                T        8+5=13       T         13          DT
E                T        7+7=14
84
.
LP Formulation of the Shortest Path
Problem
• Consider the following shortest path problem
from node 1 to node 6
3
2                   4           2
4
3
6
1                   2                       6
1
4
3                                       2

3
2           3                       5           7

5

85
• Send one unit of flow from node 1 to node 6
.
LP Formulation of the Shortest Path
Problem
• Use flow conservation constraints
– (Outflow from any node – inflow to that node) = 0
– For origin = 1
– For destination = -1
– For all other nodes = 0
– Let xj denote the flow along link j, j = 1, 2, .., 7, xj
= 0 or 1
– It turns out that this 0-1 constraints can be replaced
by 0  xj  1, which can in turn be replaced by xj 
0

86
.
LP Formulation of the Shortest Path
Problem
• Min 4x1 + 3x2 + 3x3 + 2x4 + 3x5 + 2x6 + 2x7
• S.t. x1 + x2                          =1
•     -x1       + x3 + x4               =0
•         - x2           + x5           =0
•              - x3           + x6      =0
•                   - x4 – x5      + x7 = 0
•                             - x6 – x7 = -1
• Xj  0, j = 1, 2, …, 7 (xj integers)
87
.
Maximum Flow Problem
• Maximum flow problem description
– All flow through a directed and connected network
originates at one node (source) and terminates at one
another node (sink)
– All the remaining nodes are transshipment nodes
– Flow through an arc is allowed only in the direction
indicated by the arrowhead, where the maximum amount of
flow is given by the capacity of that arc. At the source, all
arcs point away from the node. At the sink, all arcs point
into the node
– The objective is to maximize the total amount of flow from
the source to the sink (measured as the amount leaving the
source or the amount entering the sink)
88
.
Maximum Flow Problem
• Typical applications
– Maximize the flow through a company’s
distribution network from its factories to its
customers
– Maximize the flow through a company’s supply
network from its vendors to its factories
– Maximize the flow of oil through a system of
pipelines
– Maximize the flow of water through a system of
aqueducts
– Maximize the flow of vehicles through a
transportation network
89
.
Maximum Flow Algorithm
• Some Terminology
– The residual network shows the remaining arc
capacities for assigning additional flows after some
flows have been assigned to the arcs

The residual capacity for assigning some flow from node B to node O

5
2
O                                               B

The residual capacity for flow from node O to Node B
90
.
Maximum Flow Algorithm
– An augmenting path is a directed path from the source to
the sink in the residual network such that every arc on this
path has strictly positive residual capacity
– The residual capacity of the augmenting path is the
minimum of these residual capacities (the amount of flow
that can feasibly be added to the entire path)
• Basic idea
– Repeatedly select some augmenting path and add a flow
equal to its residual capacity to that path in the original
network. This process continues until there are no more
augmenting paths, so that the flow from the source to the
sink cannot be increased further

91
.
Maximum Flow Algorithm
• The Augmenting Path Algorithm
– Assume that the arc capacities are either integers or rational
numbers
• 1. identify an augmenting path by finding some
directed path from the source to the sink in the
residual network such that every arc on this path has
strictly positive residual capacity. If no such path
exists, the net flows already assigned constitute an
optimal flow pattern
• 2. Identify the residual capacity c* of this augmenting
path by finding the minimum of the residual
capacities of the arcs on this path. Increase the flow in
92
this path by c*
.
Maximum Flow Example
• During the peak season the park management of the
Seervada park would like to determine how to route
the various tram trips from the park entrance (Station
O) to the scenic (Station T) to maximize the number
of trips per day. Each tram will return by the same
route it took on the outgoing trip so the analysis
focuses on outgoing trips only. To avoid unduly
disturbing the ecology and wildlife of the region,
strict upper limits have been imposed on the number
of outgoing trips allowed per day in the outbound
direction of travel for outgoing trips is indicated by
an arrow in the next slide. The number at the base of
the arrow gives the upper limit on the number of
outgoing trips allowed per day.
93
.
Maximum Flow Example
• Consider the problem of sending as many units from node O to
node T for the following network (current flow, capacity):
(0,3)
A
D           (0,9)
(0,5)
(0,1)
(0,4)                                  T
(0,1)
(0,7)
B                                          (0,6)
O
(0,5)
(0,4)          (0,2)
(0,4)          E

C

94
.
Maximum Flow Example
•   Iteration 1: one of the several augmenting paths is OBET, which has a
residual capacity of min{7, 5, 6} = 5. By assigning the flow of 5 to this path, the
resulting network is shown above
(0,3)
A
D           (0,9)
(0,5)
(0,1)
(0,4)                                       T
(0,1)
(5,7)
B                                          (5,6)
O
(5,5)
(0,4)          (0,2)
(0,4)          E

C

95
.
Maximum Flow Example
• Iteration 2: Assign a flow of 3 to the augmenting path
OADT. The resulting residual network is
(3,3)
A
D           (3,9)
(3,5)
(0,1)
(0,4)                                  T
(0,1)
(5,7)
B                                          (5,6)
O
(5,5)
(0,4)         (0,2)
(0,4)          E

C

96
.
Maximum Flow Example
• Iteration 3: Assign a flow of 1 to the augmenting path
OABDT. The resulting residual network is
(3,3)
A
D           (4,9)
(4,5)
(1,1)
(1,4)                                  T
(0,1)
(5,7)
B                                          (5,6)
O
(5,5)
(0,4)         (0,2)
(0,4)          E

C

97
.
Maximum Flow Example
• Iteration 4: Assign a flow of 2 to the augmenting path OBDT. The
resulting residual network is

(3,3)
A
D           (6,9)
(4,5)
(1,1)
(3,4)                                  T
(0,1)
(7,7)
B                                          (5,6)
O
(5,5)
(0,4)         (0,2)
(0,4)          E

C

98
.
Maximum Flow Example
Iteration 5: Assign a flow of 1 to the augmenting path
OCEDT. The resulting residual network is
(3,3)
A
D           (7,9)
(4,5)
(1,1)
(3,4)                                  T
(1,1)
(7,7)
B                                          (5,6)
O
(5,5)
(1,4)         (0,2)
(1,4)          E

C

99
.
Maximum Flow Example
Iteration 6: Assign a flow of 1 to the augmenting path
OCET. The resulting residual network is
(3,3)
A
D           (7,9)
(4,5)
(1,1)
(3,4)                                  T
(1,1)
(7,7)
B                                          (6,6)
O
(5,5)
(2,4)         (0,2)
(2,4)          E

C

100
.
Maximum Flow Example
• There are no more flow augmenting paths, so the current
flow pattern is optimal

3
A                                   7
D
1
4
T
13                                        3       1
O                   B                                       13
6
7
2                               5
E

C                   2
101
.
Maximum Flow Example
• Recognizing optimality
• Max-flow min-cut theorem can be useful
• A cut is defined as any set of directed arcs
containing at least one arc from every directed
path from the source to the sink
• For any particular cut, the cut value is the sum
of the arc capacities of the arcs of the cut
• The theorem states that, for any network with a
single source and sink, the maximum feasible
flow from the source to the sink equals the
minimum cut value for all cuts of the network
102
.
8.3 Maximum Flow Problems
Many situations can be modeled by a network in which
the arcs may be thought of as having a capacity that
limits the quantity of a product that may be shipped
through the arc. In these situations, it is often desired to
transport the maximum amount of flow from a starting
point (called the source) to a terminal point (called the
sink). Such problems are called maximum flow
problems.

103
.
An example for maximum flow problem
Sunco Oil wants to ship the maximum possible amount
of oil (per hour) via pipeline from node so to node si as
shown in the figure below.
Arc     Capacity
a0                            (so,1)      2
(1)3                                 (so,2)      3
so          1              2                si   (1,2)       3
(2)2          (1)3            (2)2
(1,3)       4
(1)4   3         (1)1
(3,si)      1
(2,si)      2
The various arcs represent pipelines of different diameters. The
maximum number of barrels of oil that can be pumped through
each arc is shown in the table above (also called arc capacity).
104
.
For reasons that will become clear soon, an artificial arc
called a0 is added from the sink to the source. To
determine the decision variable.

Xij = Millions of barrels of oil per hour that will pass
through arc(i,j) of pipeline.

For a flow to be feasible it needs to be in the following
range:
0 <= flow through each arc <= arc capacity
And
Flow into node i = Flow out from node i
105
.
Let X0 be the flow through the artificial arc, the conservation of
flow implies that X0 = total amount of oil entering the sink. Thus,
Sunco’s goal is to maximize X0.

Max Z= X0
S.t.   Xso,1<=2          (Arc Capacity constraints)
Xso,2<=3
X12<=3
X2,si<=2
X13<=4
X3,si<=1
X0=Xso,1+Xso,2    (Node so flow constraints)
Xso,1=X12+X13     (Node 1 flow constraints)
Xso,2+X12=X2,si   (Node 2 flow constraints)
X13=X3,si         (Node 3 flow constraints)
X3,si+X2,si=X0    (Node si flow constraints)
Xij>=0
106
.
One optimal solution to this LP is Z=3, Xso,1=2, X13=1,
X12=1, Xso,2=1, X3,si=1, X2,si=2, Xo=3.

107
.
8.6 Minimum Spanning Tree
Problems
Suppose that each arc (i,j) in a network has a length
associated with it and that arc (i,j) represents a way of
connecting node i to node j. For example, if each node
in a network represents a computer in a computer
network, arc(i,j) might represent an underground cable
that connects computer i to computer j. In many
applications, we want to determine the set of arcs in a
network that connect all nodes such that the sum of the
length of the arcs is minimized. Clearly, such a group of
arcs contain no loop.
108
.
Minimum Spanning Tree Problem
• An undirected and connected network is being
considered, where the given information includes
some measure of the positive length (distance, cost,
time, etc.) associated with each link
• Both the shortest path and minimum spanning tree
problems involve choosing a set of links that have the
shortest total length among all sets of links that
satisfy a certain property
– For the shortest-path problem this property is that the
chosen links must provide a path between the origin and
the destination
– For the minimum spanning tree problem, the required
property is that the chosen links must provide a path
between each pair of nodes

109
.
Some Applications
• Design of telecommunication networks (fiber-optic
networks, computer networks, leased-line telephone
networks, cable television networks, etc.)
• Design of lightly used transportation network to
minimize the total cost of providing the links (rail
• Design of a network of high-voltage electrical
transmission lines
• Design of a network of wiring on electrical
equipment (e.g., a digital computer system) to
minimize the total length of the wire
• Design of a network of pipelines to connect a number
of locations

110
.
For a network with n nodes, a spanning tree is a group
of n-1 arcs that connects all nodes of the network and
contains no loops.

12
1            2

(1,2)-(2,3)-(3,1) is a loop
4
7

(1,3)-(2,3) is the minimum spanning tree
3

111
.
Minimum Spanning Tree Problem
Description
• You are given the nodes of the network but not the
positive length for each if it is inserted into the
network (alternative measures for length of a link
include distance, cost, and time)
• You wish to design the network by inserting enough
links to satisfy the requirement that there be a path
between every pair of nodes
• The objective is to satisfy this requirement in a way
that minimizes the total length of links inserted into
the network
112
.
Minimum Spanning Tree Algorithm
• Greedy Algorithm
– 1. Select any node arbitrarily, and then connect (i.e., add a
link) to the nearest distinct node
– 2. Identify the unconnected node that is closest to a
connected node, and then connect these two nodes (i.e., add
a link between them). Repeat the step until all nodes have
been connected
– 3. Tie breaking: Ties for the nearest distinct node (step 1)
or the closest unconnected node (step 2) may be broken
arbitrarily, and the algorithm will still yield an optimal
solution.
• Fastest way of executing algorithm manually is the
graphical approach illustrated next
113
.
Example: The State University campus has five
computers. The distances between computers are given
in the figure below. What is the minimum length of
cable required to interconnect the computers? Note that
if two computers are not connected this is because of
underground rock formations.
1           1
2
2
2
6
5               4
2           3
4
4
5                   3
114
.
Solution: We want to find the minimum spanning tree.
• Iteration 1: Following the MST algorithm discussed
before, we arbitrarily choose node 1 to begin. The
closest node is node 2. Now C={1,2}, Ć={3,4,5}, and
arc(1,2) will be in the minimum spanning tree.

1           1
2
2
2
6
5                4
2           3
4
4
5                   3
115
.
• Iteration 2: Node 5 is closest to C. since node 5 is two
blocks from node 1 and node 2, we may include either
arc(2,5) or arc(1,5) in the minimum spanning tree. We
arbitrarily choose to include arc(2,5). Then C={1,2,5}
and Ć={3,4}.

1           1
2
2
2
6
5                4
2           3
4
4
5                   3
116
.
• Iteration 3: Since node 3 is two blocks from node 5,
we may include arc(5,3) in the minimum spanning tree.
Now C={1,2,5,3} and Ć={4}.

1           1
2
2
2
6
5                4
2           3
4
4
5                   3
117
.
• Iteration 4: Node 5 is the closest node to node 4. Thus,
we add arc(5,4) to the minimum spanning tree.
We now have a minimum spanning tree consisting of
arcs(1,2), (2,5), (5,3), and (5,4). The length of the
minimum spanning tree is 1+2+2+4=9 blocks.

1           1
2
2
2
6
5               4
2           3
4
4
5                   3
118
.
8.4 CPM and PERT
Network models can be used as an aid in the scheduling of large
complex projects that consist of many activities.

CPM: If the duration of each activity is known with certainty, the
Critical Path Method (CPM) can be used to determine the length
of time required to complete a project.

PERT: If the duration of activities is not known with certainty,
the Program Evaluation and Review Technique (PERT) can be
used to estimate the probability that the project will be completed

119
.
CPM and PERT are used in many applications
including the following:
• Scheduling construction projects such as office
buildings, highways and swimming pools
• Developing countdown and “hold” procedure for the
launching of space crafts
• Installing new computer systems
• Designing and marketing new products
• Completing corporate mergers
• Building ships

120
.
Project Planning, Scheduling and
Control

• Planning: organized approach to accomplish the goal of
minimizing elapsed time of project
a network; estimates time and resources
• Scheduling: a time-phased commitment of resources
– identifies critical tasks which, if delayed, will delay the
project’s completion time.
• Control: means of monitoring and revising the progress
of a project
121
.
Network Representation
• Tasks (or activities) are represented by arcs
– Each task has a duration denoted by tj
– Node 0 represents the “start” and node n denotes the “finish” of
the project
• Precedence relations are shown by “arcs”
– specify what other tasks must be completed before the task in
question can begin.
• A path is a sequence of linked tasks going from beginning
to end
• Critical path is the longest path
122
.
To apply CPM and PERT, we need a list of activities
that make up the project. The project is considered to be
completed when all activities have been completed. For
each activity there is a set of activities (called the
predecessors of the activity) that must be completed
before the activity begins. A project network is used to
represent the precedence relationships between
activities. In the following discussions the activities will
be represented by arcs and the nodes will be used to
represent completion of a set of activities (Activity on
arc (AOA) type of network).
A                       B
1                        2                     3
Activity A must be completed before activity B starts
123
.
While constructing an AOA type of project diagram one
should use the following rules:
• Node 1 represents the start of the project. An arc should lead
from node 1 to represent each activity that has no predecessors.
• A node (called the finish node) representing the completion of
the project should be included in the network.
• Number the nodes in the network so that the node representing
the completion time of an activity always has a larger number than
the node representing the beginning of an activity.
• An activity should not be represented by more than one arc in the
network
• Two nodes can be connected by at most one arc.
To avoid violating rules 4 and 5, it can be sometimes necessary to
utilize a dummy activity that takes zero time.
124
.
Formulating the CPM Problem

Input Data:
Precedence relationships and durations

Decision Variable:
ESi : Earliest starting times for each of the tasks

Objective:
Minimize the elapsed time of the project
where node n is the last node in the graph
125
.
Constraints
• If tj is the earliest starting time of a task, ESi is
the earliest starting time of an immediate
predecessor and ti is the duration of the
immediate predecessor, then we have
ESj  ESi + ti for every arc (i, j)

126
.
Critical Path Definitions
• Earliest Start Time (ES) is the earliest time a task
can feasibly start
• Earliest Finish Time (EF) is the earliest time a
• Latest Start Time (LS) is the latest time a task can
feasibly start, without delaying the project at all.
• Latest Finish Time (LF) is the latest time a task
can feasibly end, without delaying the project at
all

127
.
Critical Path Method
• Forward Pass
– Go through the jobs in order
– Start each job at the earliest time while satisfying the
precedence constraints
– It finds the earliest start and finish times
– EFi = ESi + ti
– Earliest start time for an activity leaving a particular
node is equal to the largest of the earliest finish times
for all activities entering the node.

128
.
CPM: The Backward Pass
• Fix the finishing time
• Look at tasks in reverse order
• Lay out tasks one at a time on the Gantt chart
starting at the finish and working backwards to the
start
• Start the task at its latest starting time
– LSi = LFi - ti
– Latest finish time for an activity entering a particular
node is equal to the smallest of the latest start times for
all activities leaving the node.
129
.
CPM and Critical Path
• Theorem: The minimum length of the schedule is
the length of the longest path.
The longest path is called the critical path
Look for tasks whose earliest start time and latest start time are
the same. These tasks are critical, and are on a critical path.

130
.
CPM and Critical Path
• Critical path are found by identifying those tasks where
ES=LS (equivalently, EF=LF)
• No flexibility in scheduling tasks on the critical path
• The makespan of the critical path equals the LF of the
• Slack is the difference between LS and ES, or LF and EF.
An activity with a slack of zero is on the critical path

131
.
An example for CPM
Widgetco is about to introduce a new product. A list of
activities and the precedence relationships are given in
the table below. Draw a project diagram for this project.

Activity              Predecessors   Duration(days)
A:train workers                     -               6
B:purchase raw materials           -                9
C:produce product 1               A, B              8
D:produce product 2               A, B              7
E:test product 2                    D              10
F:assemble products 1&2            C, E            12
132
.
Project Diagram for Widgetco

C8               F 12
3                   5           6
A6
D7
1        Dummy
E 10
B9    2             4

Node 1 = starting node
Node 6 = finish node

133
.
Project Diagram for Widgetco Forward Pass (ES,EF)
(9,17)                   (26,38)
C8                   F 12
(0,6)    3                      5                   6
A6
D7
1           Dummy
E 10
(16,26)
B9              (9,16)
2                4
(0,9)
Node 1 = starting node
Node 6 = finish node

134
.
Project Diagram for Widgetco Backward Pass (LS,LF)
(18,26)                   (26,38)
C8                    F 12
(3,9)      3                       5                   6
A6
D7
1             Dummy
E 10
(16,26)
B9                (9,16)
2                4
(0,9)
Node 1 = starting node
Node 6 = finish node

135
.
For Widgetco example ES(i)’s and LS(i)’s are as
follows:

Activity         ES(i)       LS(i)
A             0            3
B             0            0
C             9            18
D             9            9
E             16           16
F             26           26
136
.
According to the table on the previous slide the slacks
are computed as follows:

Activity B:    0
Activity A:    3
Activity D:   0
Activity C:   9
Activity E:   0
Activity F:   0

137
.
Critical path

• An activity with a slack of zero is a critical activity
• A path from node 1 to the finish node that consists
entirely of critical activities is called a critical path.

For Widgetco example B-D-E-F is a critical path.

The Makespan is equal to 38

138
.
Using LP to find a critical path
Decision variable:

Xij:the time that the event corresponding to node j occurs

Since our goal is to minimize the time required to
complete the project, we use an objective function of:

Z=XF-X1

Note that for each activity (i,j), before j occurs , i must
occur and activity (i,j) must be completed.
139
.
Min Z =X6-X1
S.T. X3>=X1+6            (Arc (1,3) constraint)
X2>=X1+9            (Arc (1,2) constraint)
X5>=X3+8            (Arc (3,5) constraint)
X4>=X3+7            (Arc (3,4) constraint)
X5>=X4+10           (Arc (4,5) constraint)
X6>=X5+12           (Arc (5,6) constraint)
X3>=X2              (Arc (2,3) constraint)

The optimal solution to this LP is Z=38, X1=0, X2=9,
X3=9, X4=16, X5=26, X6=38
140
.
– each takes 1 unit of time on average
– the time it takes is uniformly distributed between 0
and 2.

CPM (uses expected value)             Modeling Randomness
141        The random schedule takes longer
.
On Incorporating Variability

• Program Evaluation and Review Technique
(PERT)
– Attempts to incorporate variability in the durations
– Assume mean, m, and variance, s2, of the durations
can be estimated
• Simulation
– Model variability using any distribution
– simulate to see how long a schedule will take

142
.
PERT
CPM assumes that the duration of each activity is
known with certainty. For many projects, this is clearly
not applicable. PERT is an attempt to correct this
shortcoming of CPM by modeling the duration of each
activity as a random variable. For each activity, PERT
requires that the project manager estimate the following
three quantities:
a : estimate of the activity’s duration under the most
favorable conditions
b : estimate of the activity’s duration under the least
favorable conditions
m : most likely value for the activity’s duration
143
.
Let Tij be the duration of activity (i,j). PERT requires
the assumption that Tij follows a beta distribution.
According to this assumption, it can be shown that the
mean and variance of Tij may be approximated by

a  4m  b
E (Tij ) 
6
(b  a) 2
var Tij 
36

144
.
PERT requires the assumption that the durations of all
activities are independent. Thus,

 E (T )
( i , j )path
ij      : expected duration of activities on any path

 var T
( i , j )path
ij : variance of duration of activities on any path

145
.
Let CP be the random variable denoting the total
duration of the activities on a critical path found by
CPM. PERT assumes that the critical path found by
CPM contains enough activities to allow us to invoke
the Central Limit Theorem and conclude that the
following is normally distributed:

CP                 T       ij
( i , j )criticalpath

146
.
a, b and m for activities in Widgetco

Activity     a        b        m
(1,2)      5        13        9
(1,3)      2        10        6
(3,5)      3        13        8
(3,4)      1        13        7
(4,5)      8        12       10
(5,6)      9        15       12
147
.
According to the table on the previous slide:
5  13  36                 (13  5) 2
E (T 12)               9, var T 12              1.78
6                          36
2  10  24                 (10  2) 2
E (T 13)               6, var T 13               1.78
6                          36
3  13  32                 (13  3) 2
E (T 35)               8, var T 35              2.78
6                          36
1  13  28                 (13  1) 2
E (T 34)               7, var T 34             4
6                          36
8  12  40                   (12  8) 2
E (T 45)               10, var T 45               0.44
6                           36
9  15  48                   (15  9) 2
E (T 56)               12, var T 56              1
148                   6                           36
.
Of course, the fact that arc (2,3) is a dummy arc yields
E(T23)=varT23=0

The critical path was B-D-E-F. Thus,
E(CP)=9+0+7+10+12=38
varCP=1.78+0+4+0.44+1=7.22

Then the standard deviation for CP is (7.22)1/2=2.69
And
CP  38 35  38
P(CP  35)  P(                )  P(Z  1.12)  0.13
2.69    2.69
149
.
PERT implies that there is a 13% chance that the
project will be completed within 35 days.

150
.
More on Project Management
• CPM
–   Advantage of ease of use
–   Lays out the Gantt chart (nicely visual)
–   Identifies the critical path
–   Used in practice on large projects
• e.g., used for the big dig

• Other issues
– Tasks take resources, which are limited
– Task times are really random variables
151     – Unpredictable things happen
.
Incorporating Resource
Constraints
• Each task can have resources that it needs
– 3 construction workers
– 1 crane
– etc
• In scheduling, do not use more resources
than are available at any time
– Makes the problem much more difficult to
solve exactly. Heuristics are used.
152
.
Dealing with the unknown
• VERY hard to model
• How does one model totally unforeseen
events?
– In the Big Dig, there is a leak in digging a
tunnel despite assurances it would not happen
– In the Hubble telescope, a firm assures that the
mirrors are ground properly. By the time the
mistake is discovered, the telescope is in outer
space.
153
.
Project Management Software
• Explosive growth in software packages using
these techniques
• Cost and capabilities vary greatly
• Yearly survey in PM Network
• Microsoft Project is most commonly used
package today
– Free 60 day trial versions:
http://www.microsoft.com/office/98/project/
154     trial/info.htm     .
Summary
• Project management
• Simple model: we use estimates of the time
for each task, and we use the precedence
constraints

155
.

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