(a) Find the initial basic feasible solution of the following transportation problem.

| | I | II | III | IV | Supply |
| --- | --- | --- | --- | --- | --- |
| A | 5 | 1 | 3 | 3 | 34 |
| B | 3 | 3 | 5 | 4 | 15 |
| C | 6 | 4 | 4 | 3 | 12 |
| D | 4 | 1 | 4 | 5 | 19 |
| Demand | 21

Question

(a) Find the initial basic feasible solution of the following transportation problem.

|  | I | II | III | IV | Supply |
| --- | --- | --- | --- | --- | --- |
| A | 5 | 1 | 3 | 3 | 34 |
| B | 3 | 3 | 5 | 4 | 15 |
| C | 6 | 4 | 4 | 3 | 12 |
| D | 4 | 1 | 4 | 5 | 19 |
| Demand | 21 | 25 | 17 | 17 |  |

Using
(i) North west corner rule
(ii) Least cost method
(iii) Vogel's approximation method

OR

(b) Wages of the factory workers are assumed to be normally distributed with variance $25$. A random sample of $50$ workers gives the total wages equal to $	ext{₹ } 2{,}550$. Test the hypothesis $\mu = 52$, against the alternative hypothesis $\mu = 49$ at $1\%$ level of significance.

Accepted Answer

>

> [!TLDR]
> (a) Balanced TP: NWCR $= 310$, LCM $= 214$, VAM $= 202$. (b) $z = -1.414$, within the acceptance region at $1\%$, so accept $H_0: \mu = 52$.
>
> ## Part (a) — Transportation problem
>
> Supply $= 34+15+12+19 = 80 =$ Demand $= 21+25+17+17 = 80$ (balanced). Cost matrix:
>
> |  | I | II | III | IV | Supply |
> | --- | --- | --- | --- | --- | --- |
> | A | 5 | 1 | 3 | 3 | 34 |
> | B | 3 | 3 | 5 | 4 | 15 |
> | C | 6 | 4 | 4 | 3 | 12 |
> | D | 4 | 1 | 4 | 5 | 19 |
> | Demand | 21 | 25 | 17 | 17 |  |
>
> **(i) North-West Corner Rule.** Allocations: A-I=21, A-II=13, B-II=12, B-III=3, C-III=12, D-III=2, D-IV=17.
> $$	ext{Cost} = 21(5)+13(1)+12(3)+3(5)+12(4)+2(4)+17(5) = 105+13+36+15+48+8+85 = 	ext{₹}310.$$
>
> **(ii) Least Cost Method.** Allocations: A-II=25, A-III=9, B-I=15, C-IV=12, D-I=6, D-III=8, D-IV=5.
> $$	ext{Cost} = 25(1)+9(3)+15(3)+12(3)+6(4)+8(4)+5(5) = 25+27+45+36+24+32+25 = 	ext{₹}214.$$
>
> **(iii) Vogel's Approximation Method.** Using row/column penalties: D-II=19, A-II=6, B-I=15, C-IV=12, A-I=6, A-III=17, A-IV=5.
> $$	ext{Cost} = 19(1)+6(1)+15(3)+12(3)+6(5)+17(3)+5(3) = 19+6+45+36+30+51+15 = 	ext{₹}202.$$
>
> All three have $m+n-1 = 4+4-1 = 7$ occupied cells (non-degenerate). VAM gives the lowest starting cost ($	ext{₹}202$), being closest to optimal.
>
> ## Part (b) — Test of hypothesis (known variance)
> …

Part (a) — Transportation problem

Part (b) — Test of hypothesis (known variance)

🎓Unlock everything free for 14 days

Unlock everything free for 14 days