transportation problems

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TRANSPORTATION PROBLEM

Decision Making

Transportation Problem

2

Assumptions

1. Items to be shipped are homogeneous

2. Shipping cost per unit is the same (No economy/diseconomy of scale)

3. Only one route exists between an origin-destination pair

Transportation Model

A special class of linear programming

Need to know

1. The origin points and the capacity or supply per period at each

2. The destination points and the demand per period at each

3. The cost of shipping one unit from each origin to each destination

3


Ahmedabad (300 units required)

Faridkot (300 units capacity)

Erode (300 units capacity)

Davangere (100 units capacity)

Calcutta (200 units required)

Bhopal (200 units required)

4


To

From Ahmedabad Bhopal Calcutta

Davangere $5 $4 $3

Erode $8 $4 $3

Faridkot $9 $7 $5

5

Transportation Matrix

From

To

Ahmedabad Bhopal Calcutta

Davangere

Erode

Faridkot

Factory capacity

Warehouse requirement

300

300

300 200 200

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

Cost of shipping 1 unit from Faridkot factory to Bhopal warehouse

Davangere capacity constraint

Cell representing a possible source-to-destination shipping assignment (Erode to Calcutta)

Total demand and total supply

Calcutta warehouse demand

6

Transportation Problem: LP Model

7

To


Davangere $5 $4 $3

Erode $8 $4 $3

Faridkot $9 $7 $5 𝑥𝑖𝑗: 𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠ℎ𝑖𝑝𝑚𝑒𝑛𝑡 𝑓𝑟𝑜𝑚 𝑠𝑜𝑢𝑟𝑐𝑒 𝑖 𝑡𝑜 𝑑𝑒𝑠𝑡𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑗

Variables:

𝑀𝑖𝑛 5𝑥11+ 4𝑥12+ 3𝑥13+ 8𝑥21+ 4𝑥22+ 3𝑥23+ 9𝑥31+ 7𝑥32+ 5𝑥33

𝑠. 𝑡. 𝑥11+ 𝑥12+ 𝑥13 <= 100

𝑥21+ 𝑥22+ 𝑥23 <= 300

𝑥31+ 𝑥32+ 𝑥33 <= 300

𝑥11+ 𝑥21+ 𝑥31 >= 300

𝑥12+ 𝑥22+ 𝑥32 >= 200

𝑥13+ 𝑥23+ 𝑥33 >= 200

Capacity constraint for Devangere

Capacity constraint for Erode

Capacity constraint for Faridkot

Demand constraint for Ahmedabad

Demand constraint for Bhopal

Demand constraint for Calcutta

𝑥11, 𝑥12, 𝑥13, 𝑥21, 𝑥22, 𝑥23, 𝑥31, 𝑥32, 𝑥33 ≥ 0

Transportation Problem: LP Model

8

To


Davangere $5 $4 $3

Erode $8 $4 $3

Faridkot $9 $7 $5

Cost Ahmedabad Bhopal Calcutta

Devangere 5 4 3

Erode 8 4 3

Faridkot 9 7 5

Flows Ahmedabad Bhopal Calcutta Supply Constraints

Devangere 100 0 0 100 <= 100

Erode 0 200 100 300 <= 300

Faridkot 200 0 100 300 <= 300

Total Cost 3900

Demand Constraints Ahmedabad Bhopal Calcutta

300 200 200

>= >= >=

300 200 200

Transportation Problem: Generic LP Model

9

𝑐𝑖𝑗: 𝐶𝑜𝑠𝑡 𝑝𝑒𝑟 𝑢𝑛𝑖𝑡 𝑜𝑓 𝑠ℎ𝑖𝑝𝑚𝑒𝑛𝑡 𝑓𝑟𝑜𝑚 𝑠𝑜𝑢𝑟𝑐𝑒 𝑖 𝑡𝑜 𝑑𝑒𝑠𝑡𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑗

𝑥𝑖𝑗: 𝐴𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑠ℎ𝑖𝑝𝑚𝑒𝑛𝑡 𝑓𝑟𝑜𝑚 𝑠𝑜𝑢𝑟𝑐𝑒 𝑖 𝑡𝑜 𝑑𝑒𝑠𝑡𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑗

Parameters:

𝐾𝑖: 𝐶𝑎𝑝𝑎𝑐𝑖𝑡𝑦 𝑜𝑓 𝑠𝑜𝑢𝑟𝑐𝑒 𝑖

𝐷𝑗 ∶ 𝐷𝑒𝑚𝑎𝑛𝑑 𝑎𝑡 𝑑𝑒𝑠𝑡𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑗

Variables:

𝑀𝑖𝑛 𝑐11𝑥11+ 𝑐12𝑥12+ 𝑐13𝑥13+ 𝑐21𝑥21+ 𝑐22𝑥22+ 𝑐23𝑥23+ 𝑐31𝑥31+ 𝑐32𝑥32+ 𝑐33𝑥33

𝑠. 𝑡. 𝑥11+ 𝑥12+ 𝑥13 <= 𝐾1 Capacity constraint for Source 1

𝑥21+ 𝑥22+ 𝑥23 <= 𝐾2 Capacity constraint for Source 2


𝑥11+ 𝑥21+ 𝑥31 >= 𝐷1 Demand constraint for Destination 1



𝑥11, 𝑥12, 𝑥13, 𝑥21, 𝑥22, 𝑥23, 𝑥31, 𝑥32, 𝑥33 ≥ 0

LP Model for Balanced Transportation Problem

10

Balanced Transportation Problem: Total Demand = Total Supply


𝑠. 𝑡. 𝑥11+ 𝑥12+ 𝑥13 = 𝐾1 Capacity constraint for Source 1

𝑥21+ 𝑥22+ 𝑥23 = 𝐾2 Capacity constraint for Source 2

𝑥31+ 𝑥32+ 𝑥33 = 𝐾3 Capacity constraint for Source 3

𝑥11+ 𝑥21+ 𝑥31 = 𝐷1 Demand constraint for Destination 1



𝑥11, 𝑥12, 𝑥13, 𝑥21, 𝑥22, 𝑥23, 𝑥31, 𝑥32, 𝑥33 ≥ 0

Initial Solution for a Balanced Transportation Problem

Start in the upper left-hand cell (or northwest corner) of the

table and allocate units to shipping routes as follows:

1. Exhaust the supply (factory capacity) of each row before moving down to the next row

2. Exhaust the (warehouse) requirements of each column before moving to the next column

3. Check to ensure that all supplies and demands are met

11

Northwest-Corner Rule


1. Assign 100 units from Davangere to Ahmedabad (exhausting Davangere’s supply)

2. Assign 200 units from Erode to Ahmedabad (exhausting Ahmeadbad’s demand)

3. Assign 100 units from Erode to Bhopal (exhausting Erode’s supply)

4. Assign 100 units from Faridkot to Bhopal (exhausting Bhopal’s demand)

5. Assign 200 units from Faridkot to Calcutta (exhausting Calcutta’s demand and Faridkot’s

supply)

12

To (A) Ahmedabad

(B) Bhopal

(C) Calcutta

(D) Davangere

(E) Erode

(F) Faridkot

Warehouse requirement 300 200 200

Factory capacity

300

300

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

From


100

100

100

200

200

Means that the firm is shipping 100 units from Faridkot to Bhopal

13

Northwest-Corner Rule Computed Shipping Cost

Route

From To Tubs Shipped Cost per Unit Total Cost

D A 100 $5 $ 500 E A 200 8 1,600 E B 100 4 400 F B 100 7 700 F C 200 5 $1,000

Total: $4,200

This is a feasible solution but not necessarily the

lowest cost alternative

14

Intuitive Lowest-Cost Method

1. Identify the cell with the lowest cost

2. Allocate as many units as possible to that cell without exceeding supply or demand;

then cross out the row or column (or both) that is exhausted by this assignment

3. Find the cell with the lowest cost from the remaining cells

4. Repeat steps 2 and 3 until all units have been allocated

15



To (A) Ahmedabad

(B) Bhopal

(C) Calcutta

(D) Davangere

(E) Erode

(F) Faridkot


Factory capacity

300

300

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

From

100

First, $3 is the lowest cost cell so ship 100 units from Davangere to Calcutta and cross off

the first row as Davangere’s capacity is exhausted.

16


To (A) Ahmedabad

(B) Bhopal

(C) Calcutta

(D) Davangere

(E) Erode

(F) Faridkot


Factory capacity

300

300

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

From

100

100

Second, $3 is again the lowest cost cell so ship 100 units from Erode to Calcutta and cross

off column C as demand for Calcutta is satisfied 17


To (A) Ahmedabad

(B) Bhopal

(C) Calcutta

(D) Davangere

(E) Erodee

(F) Faridkot


Factory capacity

300

300

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

From

100

100

200

Third, $4 is the lowest cost cell so ship 200 units from Erode to Bhopal and cross off column

B and row E as both Erode’s capacity and Bhopal’s demand are satisfied

18


To (A) Ahmedabad

(B) Bhopal

(C) Calcutta

(D) Davangere

(E) Erode

(F) Faridkot


Factory capacity

300

300

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

From

100

100

200

300

Finally, ship 300 units from Ahmedabad to Faridkot as this is the only remaining cell to

complete the allocations 19


To (A) Ahmedabad

(B) Bhopal

(C) Calcutta

(D) Davangere

(E) Erode

(F) Faridkot


Factory capacity

300

300

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

From

100

100

200

300

Total Cost = $3(100) + $3(100) + $4(200) + $9(300)

= $4,100 20


To (A) Ahmedabad

(B) Bhopal

(C) Calcutta

(D) Dvangere

(E) Erode

(F) Faridkot


Factory capacity

300

300

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

From

100

100

200

300

Total Cost = $3(100) + $3(100) + $4(200) + $9(300)

= $4,100

This is a feasible solution, and an

improvement over the previous solution, but

not necessarily the lowest cost alternative 21

Vogal’s Approximation method

1. Identify the cell with minimum unit cost and the next minimum unit cost in each row; write

the difference as penalty by the side of corresponding row.

2. Identify the cell with minimum unit cost and the next minimum unit cost in each column;

write the difference as penalty by the side of corresponding column.

3. In the row or column with the largest penalty, select the cell with the minimum unit cost.

Allocate maximum resources possible to the demand in that cell.

4. If possible, eliminate that row / column.

5. Repeat steps 1-4 on the reduced table until all demands are met.

(NB: Ties in penalty can be broken arbitrarily)

22



From

To


Davangere

Erode

Faridkot

Factory capacity


300

300

300 200 200

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

23

1

1

2

3 0 0 3


From

To


Davangere

Erode

Faridkot

Factory capacity


300

300

300 200 200

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

24

1

1

2

3 0 0 3

$5


From

To


Davangere

Erode

Faridkot

Factory capacity


300

300

200 200 200

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

25

1

2

1 3 2

100


From

To


Davangere

Erode

Faridkot

Factory capacity


300

300

200 200 200

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

26

1

2

1 3 2

100

3

$4


From

To


Davangere

Erode

Faridkot

Factory capacity


100

300

200 200 200

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

27

100

200


From

To


Davangere

Erode

Faridkot

Factory capacity


100

300

200 200 200

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

28

5

4

1 2

100

200


From

To


Davangere

Erode

Faridkot

Factory capacity


100

300

200 200 200

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

29

5

4

1 2

100

200 5 $3


From

To


Davangere

Erode

Faridkot

Factory capacity


100

300

200 200 100

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

30

100

200 100


From

To


Davangere

Erode

Faridkot

Factory capacity


300

300

300 200 200

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

31

100

200 100

200 100

Total Cost = $5(100) + $4(200) + $3(100) + $9(200)+$5(100)

= $3,900

Stepping-Stone Method

1. Select any unused cell to evaluate.

2. Beginning at this cell, trace a closed path back to the original cell via cells

that are currently being used.

3. Beginning with a plus (+) sign at the unused corner, place alternate minus

and plus signs at each corner of the path just traced

32

Optimal Solution for a Balanced Transportation Problem


4. Calculate an improvement index by first adding the unit-cost figures found in

each cell containing a plus sign and subtracting the unit costs in each cell

containing a minus sign

5. Repeat steps 1 though 4 until you have calculated an improvement index for all

unused cells. If all indices are ≥ 0, you have reached an optimal solution.

33

To (A) Ahmedabad

(B) Bhopal

(C) Calcutta

(D) Davangere

(E) Erode

(F) Faridkot


Factory capacity

300

300

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

From

Initial Solution (Using Northwest-Corner Rule)

100

100

100

200

200

34

$5

$8 $4

$4

+ -

+ -


To (A) Ahmedabad

(B) Bhopal

(C) Calcutta

(D) Davangere

(E) Erode

(F) Faridkot


Factory capacity

300

300

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

From

100

100

100

200

200

+ -

- +

1 100

201 99

99

100 200

Davangere – Bhopal index

= $ 4 - $ 5 + $ 8 - $ 4

= +$ 3

35


To (A) Ahmedabad

(B) Bhopal

(C) Calcutta

(D) Davangere

(E) Erode

(F) Faridkot


Factory capacity

300

300

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

From

100

100

100

200

200

Start

+ -

+

- +

-

Davangere - Calcutta = $ 3 - $ 5 + $ 8 - $ 4 + $ 7 - $ 5 = +$ 4

36


To (A) Ahmedabad

(B) Bhopal

(C) Calcutta

(D) Davangere

(E) Erode

(F) Faridkot


Factory capacity

300

300

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

From

100

100

100

200

200

Erode - Calcutta index

= $3 - $4 + $7 - $5 = +$1

(Closed path = EC - EB + FB - FC)

Faridkot - Ahmedabad index

= $9 - $7 + $4 - $8 = -$2

(Closed path = FA - FB + EB - EA) 37


1. If an improvement is possible, choose the route (unused cell) with the most negative

improvement index (or any unused cell with a negative improvement index)

2. On the closed path for that route, select the smallest number found in the cells containing

minus signs

3. Add this number to all cells on the closed path with plus signs and subtract it from all cells

with a minus sign

38


To (A) Ahmedabad

(B) Bhopal

(C) Calcutta

(D) Davangere

(E) Erode

(F) Faridkot


Factory capacity

300

300

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

From

100

100

100

200

200 +

+ -

-

Add 100 units on route FA

Subtract 100 from routes FB

Add 100 to route EB

Subtract 100 from route EA 39


To (A) Ahmedabad

(B) Bhopal

(C) Calcutta

(D) Davangere

(E) Erode

(F) Faridkot


Factory capacity

300

300

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

From

100

200

100

100

200

Total Cost = $ 5(100) + $ 8(100) + $ 4(200) + $ 9(100) + $ 5(200)

= $ 4,000 40


41

To (A) Ahmedabad

(B) Bhopal

(C) Calcutta

(D) Davangere

(E) Erode

(F) Faridkot


Factory capacity

300

300

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

From

100

200

100

100

200 +

+ -

-

Erode - Calcutta index

= $3 - $8 + $9 - $5

= - $1

Add 100 units on route EC

Subtract 100 from routes EA

Add 100 to route FA

Subtract 100 from route FC


42

To (A) Ahmedabad

(B) Bhopal

(C) Calcutta

(D) Davangere

(E) Erode

(F) Faridkot


Factory capacity

300

300

100

700

$5

$5

$4

$4

$3

$3

$9

$8

$7

From

200

200

100

100 +

+ -

-

Total Cost = $ 5(100) + $ 4(200) + $ 3(100) + $ 9(200) + $ 5(100)

= $ 3,900

Is this optimal? Why?

100

Stepping-Stone is a Simplex Method, specialized for the special structure of

a balanced transportation problem

Hence, gives proven optimal solution

43


Unbalanced Transportation Problems

Total Demand not equal to Total Supply

Common situation in the real world

Resolved by introducing dummy sources or dummy destinations as necessary with

cost coefficients of zero

44

Special Issues in Modeling

New Davangere capacity

To (A) Ahmedabad

(B) Bhopal

(C) Calcutta

(D) Davangere

(E) Erode

(F) Faridkot


Factory capacity

300

300

250

850

$5

$5

$4

$4

$3

$3

$9

$8

$7

From

50 200

250

50

150

Dummy

150

0

0

0

150

Total Cost = 250($ 5) + 50($ 8) + 200($ 4) + 50($ 3) + 150($ 5) + 150(0)

= $ 3,350

45


Degeneracy

To use the stepping-stone methodology, the number of occupied cells

in any solution must be equal to the number of rows in the table plus

the number of columns minus 1

A solution not satisfying this rule is degenerate

46


To Customer 1

Customer 2

Customer 3

Warehouse 1

Warehouse 2

Warehouse 3

Customer demand 100 100 100

Warehouse supply

120

80

100

300

$8

$7

$2

$9

$6

$9

$7

$10

$10

From

0 100

100

80

20

How to find the entering variable?

47


To Customer 1

Customer 2

Customer 3

Warehouse 1

Warehouse 2

Warehouse 3

Customer demand 100 100 100

Warehouse supply

120

80

100

300

$8

$7

$2

$9

$6

$9

$7

$10

$10

From

0 100

100

80

20

Initial solution is degenerate

Place a zero quantity in an unused cell and proceed computing

improvement indices

48

Example

A washing machine manufacturer wants to establish regional warehouses in South India.

The market is geographically and organizationally split into 5 segments. The forecast

demands in the five regions are 2000, 1500, 1200, 2800, 2500. It has decided to build 4

warehouses to handle monthly demands of 2900, 2300, 3700 and 1100. Transportation

costs between each warehouse-market pair is given. Identify the most cost-effective ways

of serving markets from these four warehouses.

49

Multiple facilities location problem Transportation table (Example)

Market 1 Market 2 Market 3 Market 4 Market 5 Supply

70 40 10 0 0 Warehouse A

2800 100 2900

0 65 0 95 10 Warehouse B

2000 300 2300

55 0 35 20 0 Warehouse C

400 900 2400 3700

0 20 65 65 50 Warehouse D

1100 1100

Demand 2000 1500 1200 2800 2500 10000

Market 1 Market 2 Market 3 Market 4 Market 5 Supply

100 70 50 30 40 Warehouse A

2900

30 95 40 125 50 Warehouse B

2300

75 20 65 40 30 Warehouse C

3700

20 40 95 85 80 Warehouse D

1100

Demand 2000 1500 1200 2800 2500 10000

Problem

Find transportation costs using all

the four procedures

50

Transportation Like Problems

51

Can other non-transportation problems be solved using the efficient Stepping-Stone

method (developed for balanced transportation problem)?

Yes, if the underlying mathematical model has the structure of a transportation

model.

How can we recognize the underlying transportation model?

Remember the structure of a transportation model (see next slide).

Non-Transportation Problems

52


𝑠. 𝑡. 𝑥11+ 𝑥12+ 𝑥13 <= 𝐾1 Capacity constraint for Source 1






𝑥11, 𝑥12, 𝑥13, 𝑥21, 𝑥22, 𝑥23, 𝑥31, 𝑥32, 𝑥33 ≥ 0

transportation problems

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