Percentages, Ratios and Averages in a Traffic-Control Deployment
A traffic sergeant has 48 constables for a four-hour evening deployment across three intersections, A, B and C. Rules from the deployment order: every intersection must have at least 8 constables; the number at B must be 1.5 times the number at A; no more than 40 percent of the total may be at any one intersection. Two constables must be held back at the control room and cannot be deployed. During the deployment, the sergeant learns that intersection C is handling 25 percent more vehicles per hour than planned, while A is handling 20 percent fewer. He can move constables only in whole numbers and only once, and he must keep every rule satisfied after the move.
1. How many constables are available for the three intersections, and what is the maximum number allowed at any single intersection? Show the arithmetic.
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Available for deployment: 48 − 2 held back = 46 constables.
Maximum at any one intersection: 40 percent of the total. The question is which total the rule means. The deployment order caps a share of the constables being deployed, so use 46:
0.40 × 46 = 18.4, so at most 18 constables at any one intersection (whole people, and 19 would exceed 18.4).
Check the alternative reading: if the cap were 40 percent of all 48, that is 19.2, allowing 19. The safer operational reading is the stricter one, 18, because it satisfies both interpretations at once. A constable should state the interpretation used, since deployment orders are often challenged on exactly this point.
2. Find a valid starting allocation across A, B and C that satisfies all rules, and show why it is valid.
Read the worked answer
Let A = a. Then B = 1.5a, and C = 46 − a − 1.5a = 46 − 2.5a.
Constraints:
- a ≥ 8, 1.5a ≥ 8 (so a ≥ 5.34, weaker), C ≥ 8.
- Each ≤ 18.
- All whole numbers, so 1.5a must be whole, meaning a is even.
Try a = 12: B = 18, C = 46 − 30 = 16. Check: A = 12 ≥ 8; B = 18 ≥ 8; C = 16 ≥ 8. Maximum is 18, which equals the cap of 18 - allowed, since the rule says no more than 40 percent.
So A = 12, B = 18, C = 16 is valid. Total 12 + 18 + 16 = 46, all rules met.
A second valid option is a = 14: B = 21, which breaks the cap, so it fails. a = 10: B = 15, C = 21, which breaks the cap. This shows the cap, not the minimum, is the binding constraint on the upper side, and the ratio plus the minimum bind on the lower side.
3. C now handles 25 percent more traffic and A handles 20 percent fewer. Using the starting allocation, propose one whole-number move that respects all rules, and explain the limit on how far the sergeant can rebalance.
Read the worked answer
Relative load per constable, using the starting allocation:
- A: load index 0.80 (20 percent fewer vehicles) with 12 constables → 0.80 / 12 ≈ 0.067 per constable.
- C: load index 1.25 (25 percent more) with 16 constables → 1.25 / 16 ≈ 0.078 per constable.
C is more burdened per constable than A, so move one constable from A to C: A = 11, C = 17, B = 18.
Check every rule after the move:
- Minimum 8: 11, 18, 17 all ≥ 8. Pass.
- Cap 18: maximum is 18 at B. Pass (equal to cap).
- Ratio B = 1.5 × A: 1.5 × 11 = 16.5, but B = 18. Fail.
So that move breaks the ratio rule. The ratio is the hidden handcuff: B must always be 1.5 times A, so A can only be even, and once B sits at the cap of 18, A is locked at 12. Any move away from A = 12 either breaks the ratio or pushes B over 18.
Corrected proposal: keep A = 12, B = 18 and take the relief for C from nowhere - it cannot come from A without breaking the ratio. The sergeant's real options are to (a) accept C = 16 and manage congestion by tasking, or (b) request a rules waiver to relax the 1.5 ratio for this evening. The lesson: when one rule couples two variables, rebalancing must be checked against the coupling rule, not just the minimum and cap.