Solve the following problem : A dairy plant has five milk tankers, I, II, III, IV and V. These milk tankers are to be used on five delivery routes A, B, C, D and E. The distances (in kms) between the - Mathematics and Statistics

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Solve the following problem :

A dairy plant has five milk tankers, I, II, III, IV and V. These milk tankers are to be used on five delivery routes A, B, C, D and E. The distances (in kms) between the dairy plant and the delivery routes are given in the following distance matrix.

  I II III IV V
A 150 120 175 180 200
B 125 110 120 150 165
C 130 100 145 160 175
D 40 40 70 70 100
E 45 25 60 70 95

How should the milk tankers be assigned to the chilling center so as to minimize the distance travelled?

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Solution

Step 1: Row minimum

Subtract the smallest element in each row from every element in its row.

The matrix obtained is given below:

  I II III IV V
A 30 0 55 60 80
B 15 0 10 40 55
C 30 0 45 60 75
D 0 0 30 30 60
E 20 0 35 45 70

Step 2: Column minimum

Subtract the smallest element in each column of assignment matrix obtained in step 1 from every element in its column.

  I II III IV V
A 30 0 45 30 25
B 15 0 0 10 0
C 30 0 35 30 20
D 0 0 20 0 5
E 20 0 25 15 15

Step 3:

Draw minimum number of vertical and horizontal lines to cover all zeros. 

First cover all rows and columns which have maximum number of zeros.

  I II III IV V
A 30 0 45 30 25
B 15 0 0 10 0
C 30 0 35 30 20
D 0 0 20 0 5
E 20 0 25 15 15

Step 4:

From step 3, minimum number of lines covering all the zeros are 3, which is less than order of matrix, i.e., 5.

∴  Select smallest element from all the uncovered elements, i.e., 15 and subtract it from all the uncovered elements and add it to the elements which lie at the intersection of two lines.

  I II III IV V
A 15 0 30 15 10
B 15 15 0 10 0
C 15 0 20 15 5
D 0 15 20 0 5
E 5 0 10 0 0

Step 5:

Draw minimum number of vertical and horizontal lines to cover all zeros.

  I II III IV V
A 15 0 30 15 10
B 15 15 0 10 0
C 15 0 20 15 5
D 0 15 20 0 5
E 5 0 10 0 0

Step 6:

From step 5, minimum number of lines covering all the zeros are 4, which is less than order of matrix, i.e., 5.

∴ Select smallest element from all the uncovered elements, i.e., 5 and subtract it from all the uncovered elements and add it to the elements which lie at the intersection of two lines.

  I II III IV V
A 10 0 25 10 5
B 15 20 0 10 0
C 10 0 15 10 0
D 0 20 20 0 5
E 5 5 10 0 0

Step 7:

Draw minimum number of vertical and horizontal lines to cover all zeros.

  I II III IV V
A 10 0 25 10 5
B 15 20 0 10 0
C 10 0 15 10 0
D 0 20 20 0 5
E 5 5 10 0 0

Step 8:

From step 7, minimum number of lines covering all the zeros are 5, which is equal to order of the matrix, i.e., 5.

∴ Select a row with exactly one zero, enclose that zero in () and cross out all zeros in its respective column.

Similarly, examine each row and column and mark the assignment ().

The matrix obtained is as follows:

  I II III IV V
A 10 0 25 10 5
B 15 20 0 10 0
C 10 0 15 10 0
D 0 20 20 0 5
E 5 5 10 0 0

Step 9:

The matrix obtained in step 8 contains exactly one assignment for each row and column.

Optimal assignment schedule is as follows:

Routes Dairy Plant Distance (kms)
A II 120
B III 120
C V 175
D I 40
E IV 70
    525

∴ Minimum distance travelled

= 120 + 120 + 175 + 40 + 70

= 525 kms.

  Is there an error in this question or solution?
Chapter 7: Assignment Problem and Sequencing - Part I [Page 128]

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Balbharati Mathematics and Statistics 2 (Commerce) 12th Standard HSC Maharashtra State Board
Chapter 7 Assignment Problem and Sequencing
Part I | Q 2 | Page 128

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Step 1: Subtract the smallest element in each row from every element of it. New assignment matrix is obtained as follows :

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Step 2: Subtract the smallest element in each column from every element of it. New assignment matrix is obtained as above, because each column in it contains one zero.

Step 3: Draw minimum number of vertical and horizontal lines to cover all zeros:

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(Processing cost in ₹)
I II III IV
P 6 0 8 4
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Step 4: From step 3, as the minimum number of straight lines required to cover all zeros in the assignment matrix equals the number of rows/columns. Optimal solution has reached.

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(Processing cost in ₹)
I II III IV
P 6 0 8 4
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P II `square`
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Step I: Subtract the smallest element of each row from every element of that row:

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Step II: Since all column minimums are zero, no need to subtract anything from columns.

Step III: Draw the minimum number of lines to cover all zeros.

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Since minimum number of lines = order of matrix, optimal solution has been reached

Optimal assignment is A →`square`  B →`square`

C →IV  D →`square`

Total minimum time = `square` hours.


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