Transportation and Transshipment Problemsvkostogl/en/files/Educational...

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TRANSPORTATION AND TRANSSHIPMENT PROBLEMS - V. Kostoglou

Transportation and Transshipment Problems

Vassilis Kostoglou

E-mail: [email protected] URL: www.it.teithe.gr/~vkostogl/en

mailto:[email protected]

http://www.it.teithe.gr/~vkostogl

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Description A transportation problem basically deals with finding the best way to fulfill the demand of n demand points using the capacities of m supply points. While trying to find this optimal combination, 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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Transportation model Companies or manufacturers produce products at locations called sources and

ship these products to customer locations called destinations.

Each source has a limited quantity that can ship and each customer-destination has to receive a required quantity of the product.

The only possible shipments are those directly from a source to a destination. A problem with the above characteristics is generally called transportation problems. These problems involve the shipment of a homogeneous product from a number of

supply locations to a number of demand locations.

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A typical transportation problem requires three sets of information:

Capacities (or supplies)

Indicate the maximum amount most each plant can supply in a given time period.

Demands (or requirements)

They are typically estimated from some type of forecasting model. The demand

estimations are often based on historical customer demand data.

Unit shipping (or possibly production) cost

It is calculated through a transportation cost analysis.

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The transportation problem involves the determination of the amount of goods that

have to be transported from a number of sources to a number of destinations.

The most usual objective is to minimize the total shipping costs or the total

distances that have to be traveled for the transportation of the goods.

The transportation problem is a specific case of Linear Programming problems and

a special algorithm has been developed to solve it.

The problem "Given the needs at the demand locations, how should we take the limited supply at supply locations and move the goods. The objective is to minimize the total transportation cost."

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Basic concept Objective: Minimize cost Variables: Quantity of goods shipped from each supply point to each demand point Restrictions:

- Non negative shipments - Supply availability in each supply location - Demand need in each demand location

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Transportation problem's main variables

Symbol Variable

m

n

ai

bj

cij

xij

C

Sources (supply or production locations)

Destinations (consumption locations)

Capacity (supply or production) of source i

Need (demand or consumption) of destination j

Unit transportation cost from source i to destination j

Quantity shipped from source i to destination j

Total transportation cost

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Mathematic model

min C = c11x11 + c12x12 + ... + c1nx1n + c21x22 + c22x22 + ... + c2nx2n +

........................................... + cm1xm1 + cm2xm2+ ... +cmnxmn

with the capacity constraints:

x11 + x12 + ... + x1n = a1

x21 + x22 + ... + x2n = a2

...........................................

xm1 + xm2 + ... +xmn = am

the needs constraints:

x11 + x21 + ... + xm1 = b1

x12 + x22 + ... + xm2 = b2

…......................................

x1n + x2n + ... +xmn = bn

and the non negativity constraints xij 0, i,j

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Transportation problem standard model

min C =

with constraints

= ai

= bj

and xij 0,

where i = 1, 2, ..., m and j = 1, 2, ..., n

m

i 1

n

j

ijijxc1

m

i

ijx1

n

j

ijx1

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This problem has feasible solutions only if the total of sources’ capacities is equal to the

total of destinations’ needs, that is:

= = xij

m

i

ia1

n

j

ib1

n

j 1

m

i 1

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Assumptions

Unit transportation cost is independent of transported quantity.

Supply and demand are known and independent of the price charged for the product.

The transportation capacity to ship across any transportation route is considered to be unlimited.

A single commodity is transported.

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Standard form of transportation tableau

i

j

(1) (2) … (n) ai

(1) x11

c11

x12

c12

...

...

x1n

c1n a1

(2) x21

c21

x22

c22

...

...

x2n

c2n a2

...

...

...

...

...

...

...

...

... ...

(m) xm1

cm1

xm2

cm2

...

...

xmn

cmn am

bj b1 b2 bn Σai = Σ bj

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Transportation problem example

A mining company extracts gravel, its basic product, from three quarries, L1, L2 and L3.

The weekly production of each quarry is 75, 150 and 75 tones of gravel respectively.

The gravel has to be transported to five main consumers, K1, K2, K3, K4 and K5

requiring for their needs 100, 60, 40, 75 and 25 tones of gravel per week respectively.

The problem that concerns company's management is the minimization of the required

cost for the transportation of the product to the consumers. For this purpose a detailed

cost analysis was carried out which gave the results of the following table (the numbers

denote the transportation cost in € per ton of gravel).

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Cost table of gravel transportation

Consumers

Κ1 Κ2 Κ3 Κ4 Κ5

L1 3 2 3 4 1

Quarries L2 4 1 2 4 2

L3 1 0 5 3 2

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Methods for finding an initial basic feasible solution

1. Northwest corner method

2. Minimum cost method

3. Vogel method

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Northwest corner method

1. The maximum possible quantity is assigned to the northwest (up left) cell according

to the supply and demand of the corresponding row or column. The supply of the

row and the demand of the column are adjusted appropriately.

2. Either the row of which the supply has been exhausted or the column of which the

demand has been satisfied is crossed out.

3. If all supplies have been exhausted and all demands have been satisfied then:

END OF THE METHOD, otherwise: transfer to step 1.

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In more detail:

Starting from cell (1, 1) the maximum value is given to variable xij which either satisfies

the needs of destination j or exhausts the remaining capacity of source i; obviously the

smaller of these two quantities. Then a value is given to variable xij+1 in the former case

or to variable xi+1j in the latter. Due to the main property of the transportation problem

(Σai = Σbj), the value of the last variable xmn is such that the capacity of source m and

the need of destination n are satisfied simultaneously.

The Northwest Corner Method is simple and fast in its use; however does not take into

account at all the shipping costs. It can yield an initial basic feasible solution easily, but

the corresponding total shipping cost may be quite high.

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Initial feasible solution with the Northwest corner method

Κ1 Κ2 Κ3 Κ4 Κ5

L1 75

3

2

3

4

1 75

L2 25

4

60

1

40

2

25

4

2 150

L3

1

0

5

50

3

25

2 75

100 60 40 75 25

Kmin = € 765

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Minimum cost method

The minimum cost method uses the shipping costs in order to come up with a basic

feasible solution that has lower total cost.

To begin the method, at first the variable xij with the smallest shipping cost is located.

The largest possible value is assigned to variable xij; this value is the minimum of the

corresponding ai and bj.

After that (as in the Northwest Corner Method) row i or column j is crossed out and the

supply or the demand of the non-crossed out row or column is reduced by the value of

xij. The next cell with the minimum shipping cost is chosen among the ones which do

not belong to the crossed-out row or column. This procedure is repeated until all

capacities are exhausted and all demands are satisfied.

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Initial feasible solution with the minimum cost method

Κ1 Κ2 Κ3 Κ4 Κ5

L1 50

3

2

3

4

25

1 75

L2 35

4

1

40

2

75

4

2 150

L3 15

1

60

0

5

3

2 75

100 60 40 75 25

Kmin = € 710

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Vogel method

Methodology steps

1. Addition - below and right of the transportation tableau – of a new row and a new column with elements the difference of the two smaller cost elements of each row and each column respectively.

2. Selection of the largest element of the added two new lines.

3. Location of the minimum element of row i or column j in which belongs the element identified in step 2.

4. Assignment of the value xij = min (ai, bj) to the route corresponding to the position of the smallest element in order to meet the capacity of a source or the demand of a destination.

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5. If the capacity of a source is exhausted, then the demand bj of the corresponding destination is reduced by ai. In contrary, if the demand of a destination is satisfied, then the capacity ai of the corresponding source is reduced by bj. The source (row) or destination (column) that was satisfied is crossed-out and is not taken further into account.

Each time the above procedure is repeated the capacity of a source is exhausted or the

needs of a destination are satisfied. The implementation of the method is completed

when the capacity of the last row and the needs of the last column are simultaneously

satisfied. The solution yielded is feasible because it meets all capacities and all needs.

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Successive steps of Vogel method

1

st step

Κ1 Κ2 Κ3 Κ4 Κ5 Difference Κ1 Κ2 Κ3 Κ4 Κ5

L1 3 2 3 4 1 1 L1 75

L2 4 1 2 4 2 1 L2 150

L3 1 0 5 3 2 1 L3 75 75 0

Difference 2 1 1 1 1 100

25

60 40 75 25

2

nd step

Κ1 Κ2 Κ3 Κ4 Κ5 Differencee Κ1 Κ2 Κ3 Κ4 Κ5

L1 3 2 3 4 1 1 L1 25 75 50

L2 4 1 2 4 2 1 L2 150

L3 L3 75 0

DDifference 1 1 1 0 1 25

60 40 75 25

0

3rd

step Κ1 Κ2 Κ3 Κ4 Κ5 Difference Κ1 Κ2 Κ3 Κ4 Κ5

L1 3 2 3 4 1 L1 25 25 50 25

L2 4 1 2 4 1 L2 150

L3 L3 75 0

Difference 1 1 1 0 25

0

60 40 75 0

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4th


L1 2 3 4 1 L1 25 25 25 25 0

L2 1 2 4 1 L2 150

L3 L3 75 0

Difference 1 1 0 0

60

35

40 75 0

5th


L1 L1 25 25 25 25

L2 1 2 4 1 L2 35 150 115

L3 L3 75 0

Difference - - - 0 35

0

40 75 0

6th


L1 L1 25 25 25 25

L2 2 4 1 L2 35 40 115 75

L3 L3 75 0

Difference - - 0 0

40

0

75 0

25


Initial basic feasible solution (Vogel method)

Κ1 Κ2 Κ3 Κ4 Κ5

L1 25

3

2

3

4

25

1 75

L2

4

35

1

40

2

75

4

2 150

L3 75

1

0

5

3

2 75

100 60 40 75 25

Kmin = € 640

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Degenerate initial solution

Κ1 Κ2 Κ3 Κ4 Κ5

L1 100

3

2

3

4

1 100

L2

4

60

1

40

2

25

4

2 125

L3

1

0

5

50

3

25

2 75

100 60 40 75 25

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Initial solution that is no longer degenerate

Κ1 Κ2 Κ3 Κ4 Κ5

L1 100

3

ε

2

3

4

1

100+ε

L2

4

60

1

40

2

25

4

2

125

L3

1

0

5

50

3

25

2

75

100 60+ε 40 75 25

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Transportation problems solution methodology

Initial step

Creation of an initial basic feasible solution using one of the three relevant methods.

Transfer to the termination rule.

Repeating step

1. Determination of the incoming variable that will be introduced to the base:

Selection of the non basic variable xij with maximum negative difference cij - ui - vj

2. Determination of the outgoing variable that will be removed from the base:

Identification of the (unique) loop, which’s all vertices are basic variables.

Assignment of the maximum possible value to the incoming variable.

For its determination is selected among the donor routes the one with the

smallest value. The corresponding variable is removed from the basis.

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3. Determination of the new basic feasible solution:

Addition of the quantity θ to each recipient route and deduction from each donor

route, so that all constraints of sources and destinations are met.

Termination rule

Calculation of the elements ui and vj.

[It is suggested to select the line (row or column) with the largest number of basic

variables, giving to the corresponding ui (vj) the zero value and solving the simple

system of equations cij = ui + vj for each basic route (i, j)]

Solution optimality control

If for each non basic route (i,j) ui + vj <= cij then the solution is optimal End.

In not, go to the repetitive step.

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Successive steps of finding the optimal solution

1st

repetition

Κ1 Κ2 Κ3 Κ4 Κ5

vj

ui

4 1 2 4 3

L1

-1

75

3

2

3

4

*

1

75

L2

-0

25-θ

4

60

1

40

2

25+θ

4

*

2

150

L3

-1

θ *

1

0

5

50-θ

3

25

2

75

100 60 40 75 25

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The largest quantity that can be transferred from route L2 - Κ1 (outgoing variable) to

the route L3 - Κ1 (incoming variable as c31 = u3 - v1 = 1-(-1)-4 = -2 < 0) is θmax = 25.

2nd

repetition

Κ1 Κ2 Κ3 Κ4 Κ5

vj

ui

2 1 2 4 3

L1

1

75-θ

3

2

3

*

4

θ *

1

75

L2

-0

4

60

1

40

2

50

4

*

2

150

L3

-1

25+θ

1

0

5

25

3

25-θ

2

75

100 60 40 75 25

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Total transportation cost: 715

Incoming variable : x15 (c15-u1-v5 = -1)

Outgoing variable : x35

Transferred quantity : θmax = 25

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3rd

repetition

Κ1 Κ2 Κ3 Κ4 Κ5

vj

ui

2 1 2 4 3

L1

1

50-θ

3

2

3

θ *

4

25

1

75

L2

-0

4

60

1

40

2

50

4

2

150

L3

-1

50+θ

1

0

5

25-θ

3

2

75

100 60 40 75 25

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Total transportation cost : 640 (equal to Vogel method's initial solution)



Transferred quantity : θmax = 25

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4th

repetition

Κ1 Κ2 Κ3 Κ4 Κ5

vj

ui

2 1 2 4 3

L1

1

25

3

2

3

25

4

25

1

75

L2

-0

4

60

1

40

2

50

4

2

150

L3

-1

75

1

0

5

3

2

75

100 60 40 75 25

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This solution is optimal as for each non basic variable xij the condition ui + vj cij is

valid. The minimum total transportation cost is € 615.

Thus, the optimal transportation and distribution program of the 300 tones of gravel is

the following:

25 tones to consumer Κ1

from quarry L1 25 - " - Κ4

25 - " - Κ5

60 tones to consumer Κ1

from quarry L2 40 - " - Κ3

50 - " - Κ4

from quarry L3 75 tones to consumer Κ1

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Degenerate solution

Κ1 Κ2 Κ3 Κ4 Κ5

L1

100

3

Ε

2

3

4

1

100+ε

L2

4

60

1

40

2

25

4

2

125

L3

θ

1

0

5

50

3

25

2

75

100 60+ε 40 75 25

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1st

repetition

Κ1 Κ2 Κ3 Κ4 Κ5

vj

ui

2 1 2 4 3

L1

1

100

3

ε-θ

2

3

*

4

θ*

1

100+ε

L2

0

4

60-θ

1

40

2

25-θ

4

*

2

125

L3

-1

1

0

5

50-θ

3

25-θ

2

75

100 60+ε 40 75 25

With asterisk (*) are marked all the routes with negative difference cij-ui-vj

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Total transportation cost

:

740



Transferring quantity : θmax = ε

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2nd

repetition

Κ1 Κ2 Κ3 Κ4 Κ5

vj

ui

5 1 2 4 3

L1

1

100-θ

3

2

3

4

ε+θ

1

100+ε

L2

0

*

4

60+ε

1

40

2

50-ε

4

*

2

125

L3

-1

θ*

1

0

5

50+ε

3

25-ε-θ

2

75

100 60+ε 40 75 25

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Total transportation cost : 740



Transferring quantity

: θmax = 25-ε

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3rd

repetition

Κ1 Κ2 Κ3 Κ4 Κ5

vj

ui

2 1 2 4 3

L1

1

75-θ

3

2

3

θ*

4

25

1

100+

ε

L2

0

4

60+ε

1

40

2

25-ε

4

2

125

L3

-1

25-ε+θ

1

0

5

50+ε-θ

3

2

75

100 60+ε 40 75 25

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Total transportation cost : 665



Transferring quantity

: θmax = 50+ε

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4th

repetition

Κ1 Κ2 Κ3 Κ4 Κ5

vj

ui

3 1 2 4 1

L1

0

25

3

2

3

50+ε

4

25

1

100+ε

L2

0

4

60+ε

1

40

2

25-ε

4

2

150

L3

-1

75

1

0

5

3

2

75

100 60+ε 40 75 25

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For all non basic routes of the last table the condition ui + vj cij is valid, so the current

solution is optimal. In order to identify this solution the auxiliary variable ε is deleted,

having completed its contribution.

The total transportation cost (€ 615) and the analytical program of transportation and

distribution of 300 tones of gravel can now be determined.

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Optimal solution

Κ1 Κ2 Κ3 Κ4 Κ5

L1

25

3

2

3

50

4

25

1

100

L2

4

60

1

40

2

25

3

2

125

L3

75

1

0

5

3

2

75

100 60 40 75 25

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Lack of balance

1. Surplus production

(where i = 1,2,...,m)

(where j = 1,2,...,n)

Creation of a fictitious destination (demand location)

The values of the additional column’s cells depend on the type and characteristics of

the specific problem.

i

m

j

ij ax 1

i

n

i

ij bx 1

n

ij

i

m

ij

i ba

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2. Deficient production

(where i = 1,2,...,m)

(where j = 1,2,...,n)

Creation of a fictitious source (supply location)

i

n

ii

ij bx

n

ij

i

m

ij

i ba

i

m

ij

ij ax

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Regarding the elements of the additional row:

If the shortage cost of a quantity that has to be transported to a destination is zero,

the corresponding cost element is equal to zero.

If the inability to meet the demand implies some economic consequences

(penalties, discounts, cost of good reputation, etc), then the cost of each cell

of the additional row is put equal to the corresponding unit shortage cost.

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3. Satisfaction obligation

In these cases the transportation tableau has to be formed in such a way that

the corresponding additional route of the fictitious source or destination will not

participate in any case in the final solution.

For this purpose, a very large positive value (M) is given to the cost element of this

route. This ensures that this route will not participate in any case in the final solution.

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4. Surplus or deficient production

It is possible that in some transportation problems it is not known in advance whether

the production will be surplus or deficient. This can happen if the elements cij of the

objective function represent results (profit, loss or another efficiency criterion) from the

satisfaction of a demand location. Some of these elements may be negative. Therefore,

it may be preferable not to satisfy at all a demand if the corresponding economic results

to a loss.

In order to address such a situation a fictitious source (row) and a fictitious destination

(column) are added to the initial transportation table. The capacity and demand of the

two additional lines should be such as neither the satisfaction of all demands or the

exhaustion of all capacities is compulsory.

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Solution of maximization problems

In such a case the target is the maximization of the objective function. The

methodology of finding the optimal solution of such a problem is nearly identical to that

of minimization problems. The successive steps after the determination of an initial

feasible solution remain unchanged. The only difference refers to the termination

rule examining the optimality of the solution:

The current solution is optimal if for any non-basic route is valid: ui + vj cij

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Differences in the methods of finding an initial solution

Northwest corner method: No difference.

Minimum cost method: In practice it is renamed to “maximum profit method”.

So, keeping the same logic, the priority is the assignment of the maximum possible

quantity to the route with the highest unit profit.

Vogel method: The differences, due to its complexity, are more:

- Step 1: The elements of the additional column and row are the differences

between the two larger profit elements of each row and column respectively.

- Step 3: The largest element of the appropriate column or row is searched out in

the transportation table.

.

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TRANSSHIPMENT PROBLEMS

One of the main requirements of the transportation problems is the a priori knowledge

of the way the units will be transported from every source i to every destination j, and

the corresponding unit cost cij can be determined.

Nevertheless, sometimes the best way of transportation is not clear due to the

possibility of transshipments. In this case the products are sent through intermediate

transfer points (that can also be other sources or destinations). Such cases should be

examined from scratch in order to identify the most economical route from each source

to each destination. If there are many possible intermediate transfer points, the

determination of this cheapest route can become extremely complex and time

consuming.

For this purpose a special algorithm determining the transferred quantities for each

source to each destination, as well as the route that will follow each load in order to

minimize the total shipping cost.

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The extension of the transportation problem that includes the decisions about the

intermediate routings is called transshipment problem.

A simple way of transshipment problem formulation has been developed so that it can

be adapted to the transportation problem standard model. So, the Simplex method

used for transportation problems can also be used for solving transshipment problems.

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Transshipment problem example

The canned industry "Canned legumes" has ascertained that the cost of transporting its

products (measured in truck loads) can be reduced by interrupting the use of its trucks

and using transportation companies (through subcontracting) for the transport of its

products.

Initial transportation tableau of the canned industry

Distribution

centers

1 2 3 4

1 464 513 654 867 75

Canneries 2 352 416 690 791 125

3 995 682 388 685 100

80 65 70 85

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As there is no transportation company that can service the whole region containing

company's canneries and distribution centers, some loads need at least once during

their route from source to final destination to get transshipped to another truck. These

transshipments can be carried out to intermediate canneries or distribution centers od

to five other locations called junctions.

The transportation cost per load among canneries, distribution centers and junctions is

presented in the following summary table.

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Data of independent transshipments for the canned industry To

From

Cannery

1 2 3

Junction

1 2 3 4 5

Distribution center

1 2 3 4

Production

1

Cannery 2

3

146 …*

146 …

324 286 … … …

373 212 570 609 …

324 … 405 419 158

452 505 … 871

335 407 688 784

… 685 359 673

75

125

100

1

2

Junction 3

4

5

322 371 656

284 210 …

… 569 403

… 608 418

… … 158

262 398 430 …

262 406 421 644

398 406 81 272

431 472 81 287

… 647 274 288

503 234 329 …

305 207 464 558

597 253 171 282

613 280 236 229

831 501 291 482

1

Distribution 2

center 3

4

453 336 …

505 407 683

… 687 357

868 781 670

505 307 599 615 831

235 208 254 281 500

329 464 171 236 290

… 282 282 229 480

359 706 587

357 362 341

705 362 457

587 340 457

Allocation 80 65 70 85

* the symbol … denotes that direct transportation is not feasible

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The most effective approach for reaching the solution of transshipment problem is its

reformulation as a transportation problem, so that it can be solved with the already

known method. The basic idea for this purpose is the consideration of each movement

of a truck as transportation form a source to a destination. Therefore all locations

(canneries, junctions and distribution centers) are considered as both; possible sources

and possible destinations for all transportations..

Thus it is considered that there are 12 sources (3 canneries + 4 distribution centers + 5

junctions) and 12 destinations. The corresponding transportation costs cij (where i = 1,

2, … 12 και j = 1, 2, … 12) are given in the following table. The big Μ (Μ >> 0) is used

for the non feasible transportations (that were denoted with ... in the previous table).

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The number of loads that are transshipped through each location has to be included to

both; the needs of this location as destination and the supply of this location as source.

As this number is not known in advance, an upper limit has to be added in the demand

and the supply of the location. Then a dummy variable has to be inserted in the

demand and supply constraints, in order to allocate this surplus amount. As it is not

economically efficient to transship a load to the same location more than once, a safe

upper limit for any location is the total number of the 300 produced loads.

The final cost table with the corresponding supplies and demands can now be compiled

as a transportation problem model of this transshipment problem.

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Costs table of the transshipment problem transformed as

transportation problem

Destination

Source

Canneries

1 2 3

Junctions

1 2 3 4 5

Distribution centers

1 2 3 4

Supply

1

Canneries 2

3

0 146 Μ

146 0 Μ

Μ Μ 0

324 286 Μ Μ Μ

373 212 570 609 Μ

658 Μ 405 419 158

452 505 Μ 871

335 407 688 784

Μ 685 359 673

375

425

400

1

2

Junctions 3

4

5

322 371 656

284 210 Μ

Μ 569 403

Μ 608 418

Μ Μ 158

0 262 398 430 Μ

262 0 406 421 644

398 406 0 81 272

431 472 81 0 287

Μ 647 274 288 0

503 234 329 Μ

305 207 464 558

597 253 171 282

613 280 236 229

831 501 291 482

300

300

300

300

300

1

Distribution 2

3

4

453 336 Μ

505 407 683

Μ 687 357

868 781 670

505 307 599 615 831

235 208 254 281 500

329 464 171 236 290

Μ 282 282 229 480

0 359 706 587

357 0 362 341

705 362 0 457

587 340 457 0

300

300

300

300

Demand 300 300 300 300 300 300 300 300 80 65 70 85

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The optimal solution of the transshipment problem of the canned industry "Canned

legumes" can now be determined with the use of the Simplex method of transportation

problems. Due to the relatively large size of the problem (m=n=12) the use of software

is essential.

Transportation and Transshipment Problemsvkostogl/en/files/Educational...

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