Five Trips or Six: The Cash Van Deadline
A currency chest at a regional office must move sealed cash boxes to three linked branches before a 4:00 p.m. deadline. Branch P requires 24 boxes, Branch Q requires 21 boxes and Branch R requires 18 boxes. One armoured van is available and its certified capacity is 15 boxes.
Because every box is tagged to a branch, a single trip may serve only one branch; boxes for different branches cannot be mixed in the same trip. Each trip begins at the chest with 12 minutes of loading and sealing. Travel time from the chest to any of the three branches is 18 minutes, and unloading at a branch takes 10 minutes. After every trip except the last one, the van must return to the chest, which again takes 18 minutes. The van leaves the chest at 9:30 a.m.
The driver is entitled to one 20-minute break, taken after the third trip. The supervisor wants to know whether the schedule holds, and what margin exists.
1. Calculate the minimum number of trips needed and the total boxes to be moved, showing how the branch-tagging rule affects the answer.
Read the worked answer
Total demand: 24 + 21 + 18 = 63 boxes.
If mixing were allowed: 63 ÷ 15 = 4.2, so 5 trips would suffice (15 + 15 + 15 + 15 + 3).
But trips cannot be mixed. Each branch must be served by its own trips, so the count is computed branch by branch using the ceiling of demand divided by capacity:
- P: 24 ÷ 15 = 1.6 → 2 trips (15 + 9)
- Q: 21 ÷ 15 = 1.4 → 2 trips (15 + 6)
- R: 18 ÷ 15 = 1.2 → 2 trips (15 + 3)
Minimum trips = 2 + 2 + 2 = 6.
The tagging rule costs one extra trip. Note how much unused capacity the rule creates: the second trip to each branch carries only 9, 6 and 3 boxes respectively. That wasted capacity is the price of branch integrity, and it is a scheduling rule the guard cannot waive.
2. Assuming the six trips are run back to back, at what time does the van finish, and how much margin remains against the 4:00 p.m. deadline?
Read the worked answer
Time per round trip = loading 12 + outward 18 + unloading 10 + return 18 = 58 minutes.
Last trip has no return = 12 + 18 + 10 = 40 minutes.
Six trips means five round trips plus one final one-way trip:
- 5 × 58 = 290 minutes
- 1 × 40 = 40 minutes
- Subtotal = 330 minutes, that is 5 hours 30 minutes
Driver break = 20 minutes, taken after trip 3, so it falls inside the schedule.
Total elapsed = 350 minutes = 5 hours 50 minutes.
Start 9:30 a.m. + 5 h 50 min = 3:20 p.m.
Margin against 4:00 p.m. = 40 minutes.
The schedule holds, but the margin is thin enough that a second delay of more than 40 minutes would push the last delivery past the deadline - which matters because the chest cannot hold cash overnight without additional escort arrangements.
3. Branch R later confirms it needs only 16 boxes instead of 18. Does the trip count change, and what reduction in R's demand would actually save a trip?
Read the worked answer
Check R against capacity: 16 ÷ 15 = 1.07, which still rounds up to 2 trips. The trip count therefore stays at 6 and the finish time is unchanged at 3:20 p.m. The supervisor saves nothing in time, only two boxes of load.
Find the threshold instead of guessing. R needs a second trip only while its demand exceeds 15 boxes. So:
- R = 15 → 1 trip; total trips fall to 5.
- R = 16 → 2 trips; total stays at 6.
The saving from reaching 15 is one full round trip, 58 minutes, moving the finish to about 2:22 p.m. and lifting the margin to nearly 100 minutes.
The practical lesson is that in capacity problems the useful question is not 'has the number gone down?' but 'has the number crossed the capacity boundary?' A reduction that does not cross a multiple of 15 buys no time at all. The supervisor should also resist the temptation to slip R's single leftover box into a trip for P or Q: that breaks the branch-tagging rule and creates exactly the reconciliation problem the rule exists to prevent.