# The Transport Problems In Europe Biology Essay

The transportation problem is concerned with transporting a single commodity from a number of supply sources to a number of demand destinations. The objective of the model is to determine the amount to be shipped from each source to each destination so as to maintain the supply and demand requirements at the lowest transportation cost. Transportation problem is a special linear problem.

Definitions

The following terms are to be defined with reference to the transportation problems:

(A) Feasible Solution (F.S.)

A set of non-negative allocations xij≥ 0 which satisfies the row and column restrictions is known as feasible solution.

(B) Basic Feasible Solution (B.F.S.)

A feasible solution to an m-origin and n-destination problem is said to be basic feasible solution if the number of positive allocations are (m+n–1). If the number of allocations in a basic feasible solutions are less than (m+n–1), it is called degenerate basic feasible solution (DBFS) (otherwise non-degenerate).

(C) Optimal Solution

A feasible solution (not necessarily basic) is said to be optimal if it minimizes the total transportation cost.

Mathematical Formulation of Transportation Problem

The transportation problem is concerned with transporting a single commodity from a number of supply sources (e.g. factories) to a number of demand destinations (e.g. warehouses).

Suppose that a manufacturer has three factories S1, S2 and S3 producing the same product. These factories, in turn, ship the product to four warehouses D1, D2, D3 and D4. The factories can supply s1, s2 and s3 units respectively. The warehouses have respective demands for d1, d2, d3 and d4 units. Each factory can supply the product to each warehouse, but the transportation costs vary for different combinations. The problem is to determine the quantity to be shipped from each factory to each warehouse so as to maintain the supply and demand requirements at the lowest transportation cost.

SUPPLY

SOURCES(factories)

DESTINATION(warehouses)

DEMAND

s1

S1

D1

d1

s2

S2

D2

d2

s3

S3

D3

D4

d3

d4

Let there be three units, producing scooter, say A1, A2 and A3 from where the scooters are to be supplied to four depots say B1, B2, B3 and B4.

Let the number of scooters produced at A1, A2 and A3 be a1,a2 and a3 respectively and the demands at the depots be b2, b1,b3 and b4 respectively.

We assume the condition

a1+a2+a3 = b1+b2+b3+b4

i.e., all scooters produced are supplied to the different depots. Let the cost of transportation of one scooter from A1 to B1 be c11.

Let out of a1scooters available at A1, x11 be taken at B1 depot, x12 be taken at B2 depot and to other depots as well.

Similarly, from A2 and A3 the scooters transported are equal to a2 and a3 respectively.

∴ x21+x22+x23+x24 = a2 (2)

And x31+x32+x33+x34 = a3 (3)

On the other hand it should be kept in mind that the total number of scooters delivered to B1

from all units must be equal to b1, i.e.,

x11+x21+x31= b1 (4)

Similarly, x12+x22+x32= b2 (5)

x13+x23+x33 = b3 (6)

x14+x24+x34 = b4 (7)

With the help of the above information we can construct the following table:

To B1

To B2

To B3

To B4

STOCK

From A1

x11(c11)

x12(c12)

x13(c13)

x14(c14)

a1

From A2

x21(c21)

x22(c22)

x23(c23)

x24(c24)

a2

From A3

x31(c31)

X32(c32)

x33(c33)

x34(c34)

a3

REQUIREMENTS

b1

b2

b3

b4

The cost of transportation from Ai (i=1,2,3) to Bj (j=1,2,3,4) will be equal to

S = ∑ cij xij , (8)

where the symbol i j ,∑ put before cij xij signifies that the quantities cij xij must be summed over all i = 1,2,3 and all j = 1,2,3,4.

Thus we come across a linear programming problem given by equations (1) to (7) and a linear function (8). Thus we come across a linear programming problem given by equations (1) to (7) and a linear function (8).

METHODS TO FIND OUT BASIC FEASIBLE SOLUTION

We shall discuss the following methods which can be used to find an initial solution;

NorthWest- Corner method

Least Cost Method

Vogel’s approximation method(VAM)

Northwest corner Method

In this method we distribute the available units in rows and column in such a way that the sum will remain the same.

UNITSS

DEPOT

To B1

To B2

To B3

To B4

STOCK

From A1

2

3

5

1

8

From A2

7

3

4

6

10

From A3

4

1

7

2

20

REQUIREMENTS

6

8

9

15

38

We have to follow the steps given below.

(a) Start allocations from north-west corner, i.e., from (1, 1) position. Here min (a1, b1), i.e., min (8, 6) =6 units. Therefore, the maximum possible units that can be allocated to this position is 6, and write it as 6(2) in the (1, 1) position of the table. This completes the allocation in the first column and cross the other positions, i.e., (2, 1) and (3, 1) in the column.

UNITSS

DEPOT

To B1

To B2

To B3

To B4

STOCK

From A1

6(2)

8-6

From A2

10

From A3

20

REQUIREMENTS

6-6=0

8

9

15

32

(b) After completion of step (a), come across the position (1, 2). Here min (8–6, 8) =2 units can be allocated to this position and write it as 2(3). This completes the allocations in the first row and cross the other positions, i.e., (1, 3) and (1, 4) in this row.

UNITSS

DEPOT

To B1

To B2

To B3

To B4

STOCK

From A1

6(2)

2(3)

2-2=0

From A2

10

From A3

20

REQUIREMENTS

0

8-2=6

9

15

30

(c) Now come to second row, here the position (2, 1) is already been struck off, so consider the position (2, 2). Here min (10, 8–2) =6 units can be allocated to this position and write it as 6(3). This completes the allocations in second column so strike off the position (3, 2) (see Table 5)

UNITSS

DEPOT

To B1

To B2

To B3

To B4

STOCK

From A1

6(2)

2(3)

0

From A2

6(3)

10-6=4

From A3

20

REQUIREMENTS

0

0

9

15

24

(d) Again consider the position (2, 3). Here, min (10–6, 9) =4 units can be allocated to this position and write it as 4(4). This completes the allocations in second row so struck off the position (2, 4) (see Table 6).

UNITSS

DEPOT

To B1

To B2

To B3

To B4

STOCK

From A1

6(2)

2(3)

0

From A2

6(3)

4(4)

4-4=0

From A3

20

REQUIREMENTS

0

0

9-4=5

15

20

(e) In the third row, positions (3, 1) and (3, 2) are already been struck off so consider the position (3, 3) and allocate it the maximum possible units, i.e., min (20, 9–4) =5 units and write it as 5(7). Finally, allocate the remaining units to the position (3, 4), i.e., 15 units to this position and write it as 15(2).

UNITSS

DEPOT

To B1

To B2

To B3

To B4

STOCK

From A1

6(2)

2(3)

## -

## -

8

From A2

## -

6(3)

4(4)

## -

10

From A3

## -

## -

5(7)

15(2)

20

REQUIREMENTS

6

8

9

15

38

From the above table calculate the cost of transportation as

6×2 + 2×3 + 6×3 + 4×4 + 5×7 + 15×2

= 12 + 6 + 18 + 16 + 35 + 30

= 117

i.e. Rs. 11700.

Lowest cost entry method

In this method we start with the lowest cost position. Here it is (1,4) and (3,2) positions, allocate the maximum possible units to these positions, i.e., 8 units to the position (1,4) and 8 units to position (3,2), write them as 8(1) and 8(1) respectively, then strike off the other positions in row 1 and also in column 2, since all the available units are distributed to these positions.

UNITSS

DEPOT

To B1

To B2

To B3

To B4

STOCK

From A1

x

x

x

8(1)

0

From A2

x

10

From A3

## -

8(1)

12

REQUIREMENTS

6

0

9

7

22

Consider the next higher cost positions, i.e., (1, 1) and (3, 4) positions, but the position (1, 1) is already been struck off so we can’t allocate any units to this position. Now allocate the maximum possible units to position (3, 4), i.e., 7 units as required by the place and write it as 7(2). Hence the allocations in the column 4 is complete, so strike off the (2, 4) position.

UNITSS

DEPOT

To B1

To B2

To B3

To B4

STOCK

From A1

x

x

x

8(1)

0

From A2

x

x

10

From A3

## -

8(1)

7(2)

5

REQUIREMENTS

6

0

9

0

15

(c) Again consider the next higher cost position, i.e., (1, 2) and (2, 2) positions, but these positions are already been struck off so we cannot allocate any units to these positions.

(d) Consider the next higher positions, i.e., (2, 3) and (3, 1) positions, allocate the maximum possible units to these positions, i.e., 9 units to position (2, 3) and 5 units to position (3, 1), write them as 9(4) and 5(4) respectively. In this way allocation in column 3 is complete so strike off the (3, 3) position.

UNITSS

DEPOT

To B1

To B2

To B3

To B4

STOCK

From A1

x

x

x

8(1)

0

From A2

x

9(4)

x

1

From A3

5(4)

8(1)

x

7(2)

0

REQUIREMENTS

1

0

0

0

1

(e) Now only the position (2, 1) remains and it automatically takes the allocation 1 to complete the total in this row, therefore, write it as 1(7).

With the help of the above facts complete the allocation table as given below.

UNITSS

DEPOT

To B1

To B2

To B3

To B4

STOCK

From A1

x

x

x

8(1)

8

From A2

1(7)

x

9(4)

x

10

From A3

5(4)

8(1)

x

7(2)

20

REQUIREMENTS

6

8

9

15

38

From the above facts, calculate the cost of transportation as

8×1 + 1×7 + 9×4 + 5×4 + 8×1 + 7×2

= 8 + 7 + 36 + 20 + 8 + 14

= 93

i.e., Rs. 9300.

Vogel’s approximation method

(a1) Write the difference of minimum cost and next to minimum cost against each row in the penalty column. (This difference is known as penalty).

(a2) Write the difference of minimum cost and next to minimum cost against each column in the penalty row. (This difference is known as penalty).

We obtain the table as given below.

UNITSS

DEPOT

To B1

To B2

To B3

To B4

STOCK

Penalties

From A1

2

3

5

1

8

(1)

From A2

7

3

4

6

10

(1)

From A3

4

1

7

2

20

(1)

REQUIREMENTS

6

8

9

15

38

Penalties

(2)

(2)

(1)

(1)

(b)Identify the maximum penalties. In this case it is at column one and at column two. Consider any of the two columns, (here take first column) and allocate the maximum units to the place where the cost is minimum (here the position (1,1) has minimum cost so allocate the maximum possible units, i.e., 6 units to this position). Now write the remaining stock in row one. After removing the first column and then by repeating the step (a), we obtain as follows:

UNITSS

DEPOT

To B2

To B3

To B4

STOCK

Penalties

From A1

(3)

(5)

(1)

2

(2)

From A2

(3)

(4)

(6)

10

(1)

From A3

(1)

(7)

(2)

20

(1)

REQUIREMENTS

8

9

15

32

Penalties

(2)

(1)

(1)

(b)Identify the maximum penalties. In this case it is at row one and at column two. Consider any of the two (let it be first row) and allocate the maximum possible units to the place where the cost is minimum (here the position (1, 4) has minimum cost so allocate the maximum possible units, i.e., 2 units to this position). Now write the remaining stock in column four. After removing the first row and by repeating the step (a), we obtain table 14 as given below.

UNITSS

DEPOT

To B2

To B3

To B4

STOCK

Penalties

From A2

(3)

(4)

(6)

10

(1)

From A3

(1)

(7)

(2)

20

(1)

REQUIREMENTS

8

9

13

30

Penalties

(2)

(3)

(4)

(d)Identify the maximum penalties. In this case it is at column four. Now allocate the maximum possible units to the minimum cost position (here it is at (3,4) position and allocate maximum possible units, i.e., 13 to this position). Now write the remaining stock in row three. After removing the fourth column and then by repeating the step (a) we obtain table 15 as given below.

UNITSS

DEPOT

To B2

To B3

STOCK

Penalties

From A2

(3)

(4)

10

(1)

From A3

(1)

(7)

7

(6)

REQUIREMENTS

8

9

Penalties

(2)

(3)

(e) Identify the maximum penalties. In this case it is at row three. Now allocate the maximum possible units to the minimum cost position (here it is at (3,2) position and allocate maximum possible units, i.e., 7 to this position). Now in order to complete the sum, (2,2) position will take 1 unit and (2,3) position will be allocated 9 units. This completes the allocation and with the help of the above information draw table 16 as under.

UNITSS

DEPOT

To B1

To B2

To B3

To B4

STOCK

From A1

6(2)

2(1)

8

From A2

1(3)

9(4)

10

From A3

7(1)

13(2)

20

REQUIREMENTS

6

8

9

15

38

From the above facts calculate the cost of transportation as

6×2 + 2×1 + 1×3 + 9×4 + 7×1 + 13×2

= 12 + 2 + 3 + 36 + 7 + 26

= 86

i.e., Rs. 8600.

Formulation of Transportation Problem

COUNTRIES

GRAINS

Wheat

Barley

Oats

SUPPLY

England

162

121.5

82.5

## 70

France

93.6

108

75

## 110

Spain

158.4

100.8

100.8

## 80

DEMAND

## 125

## 60

## 75

## 200

Basic Feasible Solution using North- West Corner Method

TABLE 1

GRAINS

COUNTRIES

Wheat

Barley

Oats

SUPPLY

England

70

162

121.5

82.5

## 70

France

93.6

108

75

## 110

Spain

158.4

100.8

100.8

## 80

DEMAND

## 125

## 60

## 75

## 200

TABLE 2

GRAINS

COUNTRIES

Wheat

Barley

Oats

SUPPLY

England

## -

## -

## -

## 0

France

55

93.6

108

75

## 110

Spain

158.4

100.8

100.8

## 80

DEMAND

## 55

## 60

## 75

## 190

TABLE 3

GRAINS

COUNTRIES

Wheat

Barley

Oats

SUPPLY

England

## -

## -

## -

## 0

France

## -

55

108

75

## 55

Spain

## -

100.8

100.8

## 80

DEMAND

## 0

## 60

## 75

## 135

TABLE 4

GRAINS

COUNTRIES

Wheat

Barley

Oats

SUPPLY

England

## -

## -

## -

## 0

France

## -

## -

## -

## 0

Spain

## -

5

100.8

75

100.8

## 80

DEMAND

## 0

## 5

## 75

## 80

FINAL TABLE

GRAINS

COUNTRIES

Wheat

Barley

Oats

SUPPLY

England

70

162

121.5

82.5

## 70

France

55

93.6

55

108

75

## 110

Spain

158.4

5

100.8

75

100.8

## 80

DEMAND

## 125

## 60

## 75

## 200

Thus number of occupied cells = 5

m+n-1=3+3-1=5

Thus, the solution is not degenerate.

CALCUALTION OF TOTAL COST

Total cost =162*70+93.6*55+108*55+100.8*5+100.8*75

Total cost = 11340+5148+5940+504+7560

Total cost =30492

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