Fair Bandobast: Calculating the Shortfall Before the Crowd Arrives
A one-day district fair opens at 6 a.m. and closes at 6 p.m. and is expected to draw 48,000 visitors. Special Branch estimates the largest simultaneous crowd at 9,600 people, between 4 p.m. and 6 p.m. Standing norms for the fair are: one constable for every 200 persons of the peak crowd; eighteen fixed posts, each needing two personnel throughout the day; and an armed reserve equal to 25 per cent of the total personnel deployed, to be found only from police constables, never from home guards. The ground has three exit gates, each clearing about 40 persons a minute in a controlled flow. The district has sanctioned 78 constables and 20 home guards for this fair. The Additional SP wants the shortfall, the number of gates needed to clear the ground in 45 minutes, and the effect of a 25 per cent rise in the peak crowd.
1. Compute the total personnel required under the stated norms and the shortfall against the sanctioned strength.
Read the worked answer
Crowd duty: 9,600 ÷ 200 = 48 constables.
Fixed posts: 18 posts × 2 = 36 personnel.
Sub-total to be deployed: 48 + 36 = 84.
Armed reserve: 25 % of 84 = 84 × 0.25 = 21.
Total requirement: 84 + 21 = 105.
Sanctioned strength: 78 constables + 20 home guards = 98.
Shortfall: 105 − 98 = 7 personnel.
Check the reserve rule before concluding: the 21 reserve personnel must come only from constables, and 21 ≤ 78, so that condition is satisfiable. But it also means the home guards can never be counted as reserve - they may hold fixed posts and crowd lines, and the 21 reserve must be exclusively police. Practical reading: 7 additional personnel must be requested from the Police Lines or reserves of neighbouring stations, and the Additional SP should be told in writing that without them either the crowd norm of 1:200 or the 25 % reserve will be breached, and which one he is choosing to breach.
2. How many gates does the ground need to clear in 45 minutes, and what does that imply for the existing three gates?
Read the worked answer
Required rate: 9,600 persons ÷ 45 minutes = 213.3 persons per minute.
Capacity of one gate: 40 persons per minute.
Gates needed: 213.3 ÷ 40 = 5.33, and gates cannot be fractional, so round up to 6 gates.
Implication: the ground has only 3 gates giving 3 × 40 = 120 persons per minute, which clears 9,600 people in 9,600 ÷ 120 = 80 minutes - 35 minutes longer than the target. Three additional gates must be opened (or the flow rate at each gate raised, which is unsafe and unreliable in a crowd).
Two supporting points worth recording in the plan:
- Real crowd flow falls below 40 per minute per gate once people push; plan for the slower figure, not the ideal one.
- The gates must be manned and lit as well; six gates need more than six men if each is to be supervised, and this must be added to the 84 personnel computed earlier - an omission that quietly breaks a good calculation.
3. The peak crowd rises by 25 per cent. Recompute the requirement and state the new shortfall.
Read the worked answer
New peak crowd: 9,600 × 1.25 = 12,000.
Crowd duty: 12,000 ÷ 200 = 60.
Fixed posts: unchanged at 36.
Sub-total: 60 + 36 = 96.
Reserve: 25 % of 96 = 24.
Total requirement: 96 + 24 = 120.
Added against the earlier figure: 120 − 105 = 15 more personnel.
New shortfall: 120 − 98 = 22 personnel (against 7 before).
Notice the multiplier effect: a 25 % rise in the crowd raises the total requirement by only about 14 %, because the 36 fixed-post men do not change - but it nearly triples the shortfall, because the sanctioned strength is fixed. That is the number the Additional SP actually needs: 22, not 15. Gates must also be revisited: 12,000 ÷ (45 × 40) = 6.67, so 7 gates would be needed for the 45-minute target.