Fuel Bill at a State Transport Depot
A State transport depot has 400 buses on its rolls. At any time 15% are off the road for servicing, and each of the remaining buses runs 240 km a day, 26 days a month. Diesel costs ₹90 a litre and the fleet average is 4.0 km per litre. Fuel is the depot's largest single expense. Two proposals are on the table. Proposal A is a one-time engine-tuning programme costing ₹1,20,00,000, after which the fleet average is expected to improve to 4.2 km per litre. Proposal B is to withdraw 15 km from the daily run of 40 buses on unremunerative routes; this saves fuel, but the depot's own survey shows that each of those 40 buses brings in ₹2,000 a day in passenger revenue that would be lost. A clerk has calculated that Proposal A saves 5% of the fuel bill - about ₹91,800 a day - and therefore pays for itself in about 13 days. Both figures are wrong.
1. Recalculate the daily fuel bill and the true daily saving from Proposal A, and explain why a 5% gain in mileage is not a 5% saving in fuel.
Read the worked answer
Present position
- Running buses = 400 × 0.85 = 340.
- Distance = 340 × 240 = 81,600 km a day.
- Fuel = 81,600 ÷ 4.0 = 20,400 litres; cost = 20,400 × 90 = ₹18,36,000 a day.
After Proposal A
- Fuel = 81,600 ÷ 4.2 = 19,428.57 litres.
- Saving = 971.43 litres × ₹90 = ₹87,428.57 a day.
The clerk's error is treating mileage as if it sat in the numerator. Litres consumed are distance divided by mileage, so the saving is 1 - (4.0 ÷ 4.2) = 1 - 0.95238 = 4.762%, not 5%. On a bill of ₹18,36,000 that is ₹87,428, not ₹91,800. The gap looks small per day - about ₹4,370 - but it compounds: over 312 running days it is roughly ₹13.6 lakh a year, which is the size of the error a depot can least afford to make in its largest budget line.
2. Compute the payback period for Proposal A and recommend whether the depot should go ahead, using the 26-days-a-month operating pattern.
Read the worked answer
Payback = capital cost ÷ daily saving.
- 1,20,00,000 ÷ 87,428.57 = 137.3 running days.
- At 26 running days a month, that is 137.3 ÷ 26 = about 5.3 months.
The clerk's 13 days is off by a factor of ten because he divided by ₹91,800 as though it were the daily saving and then misplaced a zero; the correct order of magnitude is a little over five months, not a fortnight.
Recommendation: proceed. The tuning cost is recovered inside the first six months, and the annual saving is 87,428.57 × 26 × 12 = ₹2,72,77,714, which is more than twice the capital cost in the very first year of full operation. Two conditions should be attached: the 4.2 km per litre figure must be verified on a sample of, say, 20 buses before the full amount is released, and the ₹1,20,00,000 figure must be a firm quotation that includes labour and spares. If the verified gain is only 4.1 km per litre, the saving falls to about ₹43,700 a day and payback stretches to roughly nine months - still acceptable, which tells you the decision is not fragile.
3. Evaluate Proposal B on the depot's own numbers and state whether it is a saving at all.
Read the worked answer
Compute both sides of the ledger, not just the fuel side.
Fuel saved
- Distance removed = 40 buses × 15 km = 600 km a day.
- Fuel saved = 600 ÷ 4.0 = 150 litres.
- Money saved = 150 × ₹90 = ₹13,500 a day.
Revenue lost
- 40 buses × ₹2,000 = ₹80,000 a day.
Net effect = 13,500 - 80,000 = -₹66,500 a day, or about ₹17.3 lakh a month across 26 running days.
So Proposal B is not a saving at all; it is a service cut that costs money. The arithmetic is the easiest kind - a ₹13,500 saving cannot justify an ₹80,000 loss - and the reason the proposal looked attractive is that it was presented only on the cost side. Proposal B should be rejected unless the ₹2,000 revenue figure is itself wrong. If the depot still wants to trim those routes, it should be argued as a service decision with a stated public cost, not as a fuel-efficiency measure. Proposal A remains the only option here that genuinely reduces expenditure.