Tranportation and Transshipment Problem Ppt

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. 1 Transportation, Transportation, Assignment and Assignment and Transshipment Problems Transshipment Problems

Transcript of Tranportation and Transshipment Problem Ppt

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Transportation, Transportation, Assignment and Assignment and Transshipment ProblemsTransshipment Problems

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DescriptionDescriptionA 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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Formulating Formulating Transportation ProblemsTransportation 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 tableTransportation tableA 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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Shipping costs, Supply, and Shipping costs, Supply, and Demand for Power co ExampleDemand for Power co 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 (Million kwh)

45 20 30 30

Transportation Table

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SolutionSolution1. 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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Objective functionObjective 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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Supply ConstraintsSupply ConstraintsSince 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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. Demand Constraints. Demand ConstraintsSince 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 Constraints5. Sign ConstraintsSince 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 LP Formulation of Powerco’s ProblemProblemMin 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 General Description of a Transportation ProblemTransportation 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

),...,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 ProblemBalanced Transportation ProblemIf Total supply equals to total demand, the problem 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 Balancing a TP if total supply exceeds total demandexceeds 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 Balancing a transportation problem if total supply is less than problem if total supply is less than total demandtotal 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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Finding Basic Feasible Finding Basic Feasible Solution for TPSolution 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 Methods to find the bfs for a balanced TPbalanced TP

There are three basic methods:

1. Northwest Corner Method

2. Minimum Cost Method

3. Vogel’s Method

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Northwest Corner MethodNorthwest Corner MethodTo 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 According to the explanations in the previous slide we can set xprevious slide we can set x1111=3 =3 (meaning demand of demand point 1 is (meaning demand of demand point 1 is satisfied by supply point 1).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 After we check the east and south cells, we saw that we can go east (meaning supply saw that we can go east (meaning supply point 1 still has capacity to fulfill some point 1 still has capacity to fulfill some demand).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 After applying the same procedure, we saw that we can go south this time (meaning that we can go south this time (meaning demand point 2 needs more supply by supply demand point 2 needs more supply by supply point 2).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, Finally, we will have the following bfs, which is:which is: x x1111=3, x=3, x1212=2, x=2, x2222=3, x=3, x2323=2, x=2, x2424=1, x=1, x3434=2=2

3 2 X

3 2 1 X

2 X

X X X X

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Minimum Cost MethodMinimum 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 MethodAn example for Minimum Cost MethodStep 1: Select the cell with minimum cost.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 2Step 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 Step 3: Find the new cell with minimum shipping cost and cross-out row 2shipping 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 Step 4: Find the new cell with minimum shipping cost and cross-out row 1shipping 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 Step 5: Find the new cell with minimum shipping cost and cross-out column 1shipping 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 Step 6: Find the new cell with minimum shipping cost and cross-out column 3shipping 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 Step 7: Finally assign 6 to last cell. The bfs is found as: Xbfs is found as: X1111=5, X=5, X2121=2, X=2, X2222=8, =8, XX3131=5, X=5, X3333=4 and X=4 and X3434=6=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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Vogel’s MethodVogel’s MethodBegin 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 MethodAn example for Vogel’s MethodStep 1: Compute the penalties.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 Step 2: Identify the largest penalty and assign the highest possible value to the assign the highest possible value to the variable.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 Step 3: Identify the largest penalty and assign the highest possible value to the assign the highest possible value to the variable.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 Step 4: Identify the largest penalty and assign the highest possible value to the assign the highest possible value to the variable.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 XStep 5: Finally the bfs is found as X1111=0, =0, XX1212=5, X=5, X1313=5, and X=5, and X2121=15=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 Simplex The Transportation Simplex MethodMethod

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 How to Pivot a Transportation ProblemProblem

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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Assignment ProblemsAssignment ProblemsExample: Machine 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. Machine 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 ModelThe ModelAccording to the setup table Machinco’s 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

XXXX

XXXXts

XXXXXXXX

XXXXXXXXZ

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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 ProblemThe Assignment Problem

In general the LP formulation is given as

Minimize 1 1

1

1

1 1

1 1

0

, , ,

, , ,

or 1,

n n

ij iji j

n

ijj

n

iji

ij

c x

x i n

x j n

x ij

Each supply is 1

Each demand is 1

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Comments on the Assignment Comments on the Assignment ProblemProblemThe 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 to each element that is covered by two lines. Return to step2.

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Transshipment ProblemsTransshipment ProblemsA 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 ProblemTransshipment 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 ExampleTransshipment ExampleExample 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 Table Associated Transportation Table Associated with the Transshipment with the Transshipment ExampleExample

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 Limitations of Transportation ProblemProblemOne 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 Limitations of Transportation ProblemProblem

◦ 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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Presented by Presented by

Sumit jhaFuture innoversity