Feeder-route withdrawal under a retention floor
Vasudha, a transport cooperative in a Tier-2 city, runs four feeder routes from the railway station. Daily records show: Route A - 18 trips, 42 seats per bus, 78% average occupancy, fare ₹15, operating cost ₹9,800. Route B - 22 trips, 36 seats, 64% occupancy, fare ₹18, cost ₹10,600. Route C - 14 trips, 50 seats, 88% occupancy, fare ₹12, cost ₹9,100. Route D - 20 trips, 30 seats, 55% occupancy, fare ₹20, cost ₹8,900. Every route currently runs at a deficit. The board can withdraw only one route this quarter; it has promised the municipal body that at least 80% of the present total daily passenger-trips will be retained. Occupancy is uniform across trips, and a passenger-trip means one passenger on one trip. The board also wants the withdrawn route to be the one that most reduces the daily operating deficit among all feasible choices. Before the meeting, the finance officer asks for the arithmetic behind the recommendation and for a separate note on whether Route C can ever break even through occupancy alone.
1. Compute the daily deficit of each route and recommend which route to withdraw. Show the passenger-retention check.
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Passenger-trips = trips × seats × occupancy.\n\nA: 18 × 42 × 0.78 = 589.68; revenue = 589.68 × 15 = ₹8,845.20; deficit = 9,800 − 8,845.20 = ₹954.80.\nB: 22 × 36 × 0.64 = 506.88; revenue = 506.88 × 18 = ₹9,123.84; deficit = ₹1,476.16.\nC: 14 × 50 × 0.88 = 616; revenue = 616 × 12 = ₹7,392; deficit = ₹1,708.\nD: 20 × 30 × 0.55 = 330; revenue = 330 × 20 = ₹6,600; deficit = ₹2,300.\n\nTotal passenger-trips = 589.68 + 506.88 + 616 + 330 = 2,042.56. The 80% floor is 1,634.048 passenger-trips.\n\nCut A: 2,042.56 − 589.68 = 1,452.88 - fails.\nCut B: 2,042.56 − 506.88 = 1,535.68 - fails.\nCut C: 2,042.56 − 616 = 1,426.56 - fails.\nCut D: 2,042.56 − 330 = 1,712.56 - passes.\n\nD is the only feasible withdrawal and it also removes the largest deficit, so it is the recommendation. The retention floor, not the deficit ranking, is what makes the choice unique.
2. Can Route C break even by raising occupancy alone? If not, what minimum fare at current occupancy would break even?
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At ₹12, the maximum possible revenue is 14 × 50 × 12 = ₹8,400, which is below the cost of ₹9,100. Even at 100% occupancy the route loses ₹700, so occupancy alone cannot break even.\n\nBreak-even passenger-trips at ₹12 = 9,100 / 12 = 758.33, but capacity is only 700.\n\nAt the current 616 passenger-trips, the required fare is 9,100 / 616 = ₹14.7727..., i.e. about ₹14.78. If fares must be set in 10-paise steps, the minimum is ₹14.80. The conclusion is that Route C needs a fare increase or a cost reduction; occupancy improvement alone is arithmetically insufficient.
3. Suppose the board raises Route D's fare to ₹24 without changing occupancy. Recompute its deficit and decide whether the withdrawal recommendation changes under the 80% rule.
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New D revenue = 330 × 24 = ₹7,920; deficit = 8,900 − 7,920 = ₹980. D is no longer the largest deficit; C (₹1,708) and B (₹1,476.16) are larger.\n\nHowever, the feasibility test is unchanged: cutting A, B or C still leaves passenger-trips below 1,634.05, while cutting D leaves 1,712.56. So D remains the only permissible withdrawal. The recommendation does not flip; the binding constraint is the retention promise, not the deficit ranking. A useful lesson is that a route can stop being the worst performer and still be the only one the board is allowed to cut.