Peak Crowd, Thin Blue Line: Sizing a Festival Deployment
The control room of a municipal police station in central Kerala must clear a deployment plan for a one-day temple procession along a 3 km route. Intelligence estimates a peak crowd of 96,000 people spread over four hours, with about 20 per cent held in barricaded grounds while the procession passes. District norms: one policeperson per 250 persons in the crowd; one policeperson for every 50 m of barricaded route edge, the route being barricaded on both sides; and a 15 per cent reserve over the computed total. The event, including one hour before and one hour after, runs six hours, and one shift is six hours. Barricade panels are 2.5 m long, and a 4 m emergency gap must be left every 100 m on each barricaded edge. Holding ground A measures 120 m by 80 m, is only 75 per cent usable because of stalls and a tree line, and is planned at 2.5 persons per square metre. The station has 210 personnel, of whom 30 are committed to other law-and-order calls.
1. Compute the total number of personnel the plan requires and state whether this station can meet it. Show every step.
Read the worked answer
Crowd component: 96,000 ÷ 250 = 384.
Route component: 3 km = 3,000 m per edge; both edges = 6,000 m; 6,000 ÷ 50 = 120.
Sub-total = 384 + 120 = 504.
Reserve: 15% of 504 = 75.6, so 504 + 75.6 = 579.6. Round up to 580, because a fraction of a person cannot be posted.
Availability: 210 − 30 committed elsewhere = 180.
Shortfall = 580 − 180 = 400.
Conclusion: the station cannot staff this plan alone. The wrong response is to spread 180 people thinly along 6 km of edge, which creates gaps exactly where a surge will happen. The professional response is a written request for 400 personnel from the district armed reserve and neighbouring stations, with this computation attached, submitted at least 24 hours before the event, plus a fallback plan (shorten the barricaded stretch, stagger the procession) if sanction is not received in time.
2. Compute the number of barricade panels needed for the route, allowing for the emergency gaps. Note any additional quantities an officer should check before indenting.
Read the worked answer
Total barricaded edge = 3,000 m × 2 = 6,000 m.
Gaps: one every 100 m per edge → 3,000 ÷ 100 = 30 gaps per edge; × 2 edges = 60 gaps.
Length given to gaps = 60 × 4 m = 240 m.
Panelled length = 6,000 − 240 = 5,760 m.
Panels = 5,760 ÷ 2.5 = 2,304.
Checks a good answer adds:
- Round up if the last panel does not divide evenly - here it does, so no extra is needed for arithmetic.
- The 4 m gaps are relief points and must not sit on a bend, a bridge or a narrow stretch, so a handful of extra panels may be needed to shift a gap a few metres.
- Holding-ground perimeter barricading is outside this figure. For a 120 × 80 m rectangle the perimeter is 2(120 + 80) = 400 m, i.e. 160 panels plus its own gaps - compute it separately.
3. Test whether holding ground A can absorb the 20 per cent diverted crowd. If it cannot, what are the options, and what must not be done?
Read the worked answer
Diverted crowd = 20% of 96,000 = 19,200 persons.
Ground A area = 120 × 80 = 9,600 m².
Usable area = 75% of 9,600 = 7,200 m².
Capacity at 2.5 persons/m² = 7,200 × 2.5 = 18,000.
Shortfall = 19,200 − 18,000 = 1,200 persons.
Options: (a) identify a second ground with at least 1,200 ÷ 2.5 = 480 m² of usable area; (b) reduce the diverted share to 18,000 ÷ 96,000 = 18.75 per cent and hold the rest on the route; (c) stagger the waves so no ground is ever at its ceiling.
What must not be done: raise the density to absorb the overflow. 2.5 persons/m² is already the planning ceiling - at 4 persons/m² the crowd cannot move, a single stumble cascades, and the barricade becomes a trap rather than a barrier. Also check capacity at the peak of each wave, not against the day's average, because grounds fill and empty unpredictably.