Most economical conductor, five options (Rs 7.2)

The most economical conductor is the one with the minimum total annual cost per km: the annual charge on capital (towers plus conductor) plus the annual cost of energy lost (Kelvin's law idea).

Capital recovery factor

$$CRF = \frac{i(1+i)^n}{(1+i)^n-1} = \frac{0.1(1.1)^{20}}{(1.1)^{20}-1} = \frac{0.1\times6.7275}{5.7275} = 0.11746$$

The annual capital charge is (tower cost + conductor cost) × 0.11746. The annual loss cost is the energy lost per year × the energy rate.

Reading of the data: the energy rate (Rs 7.2/unit) is given, so the loss column is taken as energy lost per km per year in 1000 kWh, as in the 2071 Magh version of this question. A cost in 1000 Rs would make the rate unnecessary. So the annual loss cost is the loss in 1000 kWh × 7.2, in 1000 Rs.

Annual cost comparison

Sample (Goat): $2144\times0.11746 = 251.83$ and $32\times7.2 = 230.40$, total 482.23 thousand Rs/km per year.

Zebra has the lowest total. Its higher capital cost compared with Goat and Sheep is more than recovered by its lower losses. Elk saves more energy, but its extra conductor cost is not repaid.

If the loss column were instead read literally as cost in 1000 Rs, the totals would be Goat 283.83, Sheep 307.26, Zebra 324.58, Deer 332.57 and Elk 351.65 thousand Rs, and Goat would be chosen. With the energy rate given, the kWh reading above is the intended one.

Answer: Zebra is the most economical conductor, with a minimum total annual cost of Rs 473.38 thousand per km.