Question

Q2. A firm has three stores with a total of 80 computers. An order is received from the Local Authority for 70 units to be delivered to 4 schools. The transport costs in $`s from stores to schools are shown below together with the availabilities and requirements. (15mks)
schools
Stores
I
II
III A B C D supply
3 5 2 6 35
4 3 3 1 25
2 1 5 2 20
Demand 18 27 17 8

a) It is required to make the most economic deliveries.
Set up the initial tableau and make the initial feasible deliveries.
b) find the transportation schedule that minimizes cost.

Answer 1

Answer on Question #65631 – Economics – Economics of Enterprise

Here Total Demand = 70 is less than Total Supply = 80. So we add a dummy demand constraint (b5) with 0 unit cost and with allocation 10.

a) Initial tableau

The rim values for a1 =35 and b1 =18 are compared.

The smaller of the two i.e. min(35,18) = 18 is assigned to for a1b1

This meets the complete demand of b1 and leaves 35 - 18 = 17 units with a1

The rim values for a2 =25 and b2=10 are compared.

The smaller of the two i.e. min(25,10) = 10 is assigned to a2b2

This meets the complete demand of b2 and leaves 25 - 10 = 15 units with a2

Total transportation cost $= 3 \times 18 + 5 \times 17 + 3 \times 10 + 3 \times 15 + 5 \times 2 + 2 \times 8 + 0 \times 10 = 240$

b) Transportation schedule that minimize cost

$\mathrm{Ui} + \mathrm{Vj} = \mathrm{Pij}$

1) $V5 = 0$ ;

2) $\mathrm{U}3 = \mathrm{P}3,5 - \mathrm{V}5$

3) V3 = P3,3 - U3;

4) V4 = P3,4 - U3;

5) $\mathrm{U}2 = \mathrm{P}2,3 - \mathrm{V}3$

6) $\mathrm{V}2 = \mathrm{P}2,2 - \mathrm{U}2$

7) $\mathrm{U}1 = \mathrm{P}1,2 - \mathrm{V}2$

8) $\mathrm{V}1 = \mathrm{P}1,1 - \mathrm{U}1$

Now, for all non-filled cells of the matrix calculate the Sij, according to the formula: $\mathrm{Sij} = \mathrm{Pij} - \mathrm{Ui} - \mathrm{Vj}$ (green). If there are negative valuations it means that the plan can be improved

M = 10

M = 7

M = 8

M = 10

The table does not contain negative assessments (the plan can not be improved), therefore the optimal solution is reached.

Minimized transportation cost = 137

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