Water for a Twelve-Day Camp
You are the logistics officer for a 12-day combined training camp for 240 personnel at a site with no borewell. Water is brought by one 4,000-litre tanker from a filling point 32 km away, one way, over a road on which vehicles average 20 km/h. Loading takes 40 minutes and unloading 30 minutes, and the tanker may operate only between 0600 and 1800. The site has 2,000 litres of usable storage. The approved daily scale is 4 litres per person for drinking, 3 for cooking, and 12 for washing and sanitation, plus 1,000 litres a day for vehicle washing, kitchens and the aid post. Your commander wants a 10 per cent reserve held at all times. On the morning of day 5 the tanker's gearbox fails and the fitter forecasts 36 hours before it is mobile again. A civilian tanker can be hired at short notice - capacity 3,000 litres, no time restriction - at a cost the camp budget can absorb once, but not twice.
1. Work out the daily and total requirement, the reserve, and the number of tanker trips needed.
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Daily demand:
- drinking: 240 x 4 = 960 L
- cooking: 240 x 3 = 720 L
- washing and sanitation: 240 x 12 = 2,880 L
- vehicles, kitchens, aid post: 1,000 L
- total: 5,560 L per day
Over 12 days: 5,560 x 12 = 66,720 L. Add the 10 per cent reserve: 66,720 x 1.10 = 73,392 L. Subtract usable storage already on site: 73,392 - 2,000 = 71,392 L to be delivered. At 4,000 L per trip: 71,392 / 4,000 = 17.85, so 18 trips.
Eighteen trips deliver 72,000 L; with storage that is 74,000 L against 66,720 L consumed, a surplus of 7,280 L - a little under 11 per cent, which satisfies the commander's reserve with nothing to spare. Note the useful check: the answer 17.85 must be rounded up, not down; rounding down leaves the camp 608 L short of the reserve on the last day, and the error would only appear on day 12.
2. Show whether one tanker can physically deliver those trips, and identify the binding constraint.
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One trip:
- round trip 64 km at 20 km/h = 3.2 h = 192 min
- loading 40 min + unloading 30 min = 70 min
- cycle = 262 min = 4 h 22 min
In the 12-hour window, three trips would need 786 min (13 h 06 min) and do not fit. Two trips per day (8 h 44 min) is the maximum, leaving about 3 h 16 min of slack for a late start, traffic or a queue at the filling point.
At 8,000 L delivered per day against 5,560 L consumed, net storage grows by 2,440 L daily. Eighteen trips therefore need 9 operating days out of the 12 available.
The binding constraint is the daylight driving window combined with a single vehicle - not the pump, not the driver and not the road. That matters for planning: the camp has three spare days built in, so a short interruption is a scheduling problem, not a supply crisis. It also means the reserve is real only if the tanker runs on at least 9 of the 12 days.
3. The tanker is down for 36 hours from the morning of day 5. Quantify the effect and decide what, if anything, you cut and whether you hire the civilian tanker.
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Quantify first. Days 1-4 delivered 8 trips = 32,000 L; consumption was 4 x 5,560 = 22,240 L. Storage on the morning of day 5 is 2,000 + 32,000 - 22,240 = 11,760 L. Days 5 and 6 lose both trips each, so consumption of 11,120 L comes entirely from storage, leaving about 640 L on the morning of day 7.
That is the real problem: not the total, but the storage floor. Drinking water would sit unprotected at 640 L, one contamination incident away from a health emergency.
Decision. Re-sequence rather than hire. Twelve operating days remain after repair on day 7; the camp needs 9. But for days 5 and 6, suspend the 1,000 L daily vehicle-and-kitchen washing allocation, which saves 2,000 L and holds storage near 2,640 L. Never cut the drinking scale of 4 L or the cooking scale of 3 L: dehydration and food-borne illness cost more water, more manpower and more medical risk than dirty vehicles. Do not hire the civilian tanker yet; set the trigger at the tanker still being immobile at 0600 on day 8, when the remaining operating days fall below 9, and report that trigger to the commander in writing.