Store carton count and weight together
Synthetic internal pallet plan permits at most eight cartons and sixteen kilograms. Carton A is two kilograms and B three. A proposal of four A and three B has seven cartons but seventeen kilograms. These are invented planning inputs, not a vehicle or pallet safety certification.
| Constraint | Required record | Example decision rule |
|---|---|---|
| Carton composition | A count and B count | 4 + 3 = 7 cartons |
| Weight evidence | A2 kg; B3 kg per carton | 4 × 2 + 3 × 3 = 17 kg |
| Joint limits | Cartons ≤8 and kilograms ≤16 | Both conditions required |
Test a plan that passes only one limit
| Case | Input | Expected decision |
|---|---|---|
| A4/B3 | 7 cartons / 17 kg | Fails weight limit |
| A4/B2 | 6 cartons / 14 kg | Meets both internal limits |
- Read counts by carton type
- Use supplied weights
- Calculate both totals
- Accept only within both limits
Unknown carton weight leaves this plan unverified even if the carton count fits.
Replace one B in A4/B3 with one A. Does the resulting A5/B2 fit?
Show the worked answer
Yes within this example: seven cartons and 5 × 2 + 2 × 3 = sixteen kilograms, exactly at the inclusive weight limit.
Retain both quantities in the export rather than a single “fits” flag, so later limit changes can be reviewed.