transportation assignment & transshipment problems-solving
TRANSCRIPT
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Transportation, Assignment and
Transshipment ProblemsMohon disarikan & di-email dalam bentuk e-
paper dg syarat: pilih satu topik:
1. Masalah Transportatsi dg iterasi solusi North-west
Corner, di halaman 20
2. Masalah Transportatsi dg iterasi solusi Minimum
cost, di halaman 25
3. Masalah Transportatsi dg iterasi solusi dg solusiVogel, di halaman 34
4. Masalah assignememt di halaman 51
5. Masalah transhipment di halaman 55
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Formulating Transportation
ProblemsExample 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 ofelectricity 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
(M il lion kwh)
Plant 1 $8 $6 $10 $9 35
Plant 2 $9 $12 $13 $7 50
Plant 3$14 $9 $16 $5 40
Demand
(Mil lion kwh)
45 20 30 30
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 productioncapacity;
X11+X12+X13+X14
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5. Sign Constraints
Since a negative amount of electricity can not be
shipped all Xijs must be non negative;
Xij >= 0 (i= 1,2,3; j= 1,2,3,4)
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LP Formulation of Powercos Problem
Min Z = 8X11+6X12+10X13+9X14+9X21+12X22+13X23+7X24
+14X31+9X32+16X33+5X34
S.T.: X11+X12+X13+X14 = 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 pointsfrom which a good is
shipped. Supply point ican supply at mostsi
units.
2. A set of n demand pointsto which the good is
shipped. Demand pointjmust receive at least di
units of the shipped good.
3. Each unit produced at supply point iand shippedto demand pointjincurs a variable costof cij.
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Xij = number of units shipped fromsupply point ito
demand point j
),...,2,1;,...,2,1(0
),...,2,1(
),...,2,1(..
min
1
1
1 1
njmiX
njdX
misXts
Xc
ij
mi
i
jij
nj
j
iij
mi
i
nj
j
ijij
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Balanced Transportation Problem
If Total supply equals to total demand, theproblem is said to be a balanced
transportation problem:
nj
j
j
mi
i
i ds11
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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 thedummy 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 feasiblesolution. 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 unmetdemand and as one can guess this time the
total penalty cost is desired to be minimum
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Finding Basic Feasible Solution
for Transportation ProblemUnlike other Linear Programming
problems, a balancedTP 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 (xijs) satisfy all but
one constraint, the values for xijs willsatisfy 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. Vogels 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 x11as large as
possible (here the limitations for setting x11to a
larger number, will be the demand of demandpoint 1 and the supply of supply point 1. Your
x11value 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
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 shippingcosts. 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 costmethod, first we find the decision variable with the
smallest shipping cost (Xij). Then assignXijits largest
possible value, which is the minimum ofsiand 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 orcolumn 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
2 1 3 5
3 8 4 6
5
10
15
12 8 4 6
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Step 2: Cross-out column 2
2 3 5 6
2 1 3 5
8
3 8 4 6
12 X 4 6
5
2
15
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Step 3: Find the new cell with minimum shipping
cost and cross-out row 2
2 3 5 6
2 1 3 5
2 8
3 8 4 6
5
X
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
5
2 1 3 5
2 8
3 8 4 6
X
X
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
5
2 1 3 5
2 8
3 8 4 6
5
X
X
10
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
5
2 1 3 5
2 8
3 8 4 6
5 4
X
X
6
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
5
2 1 3 5
2 8
3 8 4 6
5 4 6
X
X
X
X X X X
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3. Vogels 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 smallestshipping 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 Vogels Method
Step 1: Compute the penalties.
Supply Row Penalty
6 7 8
15 80 78
Demand
Column Penalty 15-6=9 80-7=73 78-8=70
7-6=1
78-15=63
15 5 5
10
15
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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
15 80 78
Demand
Column Penalty 15-6=9 _ 78-8=70
8-6=2
78-15=63
15 X 5
5
15
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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
5 5
15 80 78
Demand
Column Penalty 15-6=9 _ _
_
_
15 X X
0
15
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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
0 5 5
15 80 78
Demand
Column Penalty _ _ _
_
_
15 X X
X
15
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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
0 5 5
15 80 78
15
Demand
Column Penalty _ _ _
_
_
X X X
X
X
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The Transportation Proble as a
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 followingsteps 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 . The variable
corresponding to this odd cell will leave the basis. To
perform the pivot, decrease the value of each odd cell
by and increase the value of each even cell by . The
variables that are not in the loop remain unchanged.The pivot is now complete. If =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 thepivot. If more than one odd cell in the loop equals ,
you may arbitrarily choose one of these odd cells to
leave the basis; again a degenerate bfs will result
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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 tominimize 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 Machincos problem can be
formulated as follows (for i,j=1,2,3,4):
10
1
1
1
1
1
1
1
1..
10629387
5612278514min
44342414
43332313
42322212
41312111
44434241
34333231
24232221
14131211
4443424134333231
2423222114131211
ijij orXX
XXXX
XXXX
XXXX
XXXX
XXXX
XXXX
XXXXXXXXts
XXXXXXXX
XXXXXXXXZ
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For the model on the previous page note that:
Xij=1 if machine iis assigned to meet the demands of
jobj
Xij=0 if machine iis not assigned to meet the demands
of jobj
In general an assignment problem is balancedtransportation 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
Minimize 1 1
1
1
1 1
1 1
0
, , ,
, , ,
or 1,
n n
i j i j
i j
n
i j
j
n
i j
i
i j
c x
x i n
x j n
x i j
Each supply is 1
Each demand is 1
Comments on the Assignment
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Comments on the Assignment
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 hour.
Al h h h i i l b
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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 thatreason 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 mxmcost 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 eachcolumn. 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 optimalsolution 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 eachuncovered element of the reduced cost matrix and add k
to each element that is covered by two lines. Return to
step2.
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Transshipment Problems
A transportation problem allows only shipments that godirectly 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 whichgoods 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.
Transshipment Problem
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Transshipment Problem
An extension of a transportation problem
More general than the transportation problem in that in thisproblem 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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Each supply point will have a supply equal to its
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 (points original supply)+s and a demand equal
to (points original demand)+s. This ensures that anytransshipment point that is a net supplier will have a net
outflow equal to points original supply and a net
demander will have a net inflow equal to points original
demand. Although we dont know how much will beshipped through each transshipment point, we can be
sure that the total amount will not exceed s.
Transshipment Example
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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 theDenver factory can produce as many as 200 widgetsper day. Widgets are shipped by air to customers inLA and Boston. The customers in each city require130 widgets per day. Because of the deregulation of
airfares, Widgetco believes that it may be cheaperfirst fly some widgets to NY or Chicago and then flythem to their final destinations. The cost of flying awidget are shown next. Widgetco wants to minimizethe total cost of shipping the required widgets tocustomers.
Transportation Tableau Associated
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Transportation Tableau Associatedwith 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
Limitations of Transportation Problem
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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: (UNLESStransporting any one of the multiple commodities iscompletely independent of transporting any othercommodity 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 acapacity constraint
Example: where economy of scale can be achieved bytransporting the two products on the same link at a largertotal volume and at a lower unit cost of transportation
Limitations of Transportation Problem
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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-linearprogramming
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