The Shared Studio Deadline
Three students share a studio and a single drying rack with 12 shelves. Each student needs to complete a multi-part work involving wet plaster over armature, which must dry undisturbed for 48 hours before any further work. The assessment is in five days. Student A has four plaster components ready to dry; Student B has five; Student C has three, but C's components are the largest and each occupies two shelves. A also needs to apply a final surface treatment to two already-dried pieces, which requires 24 hours of drying after application. The studio is only accessible during school hours, and the rack cannot be moved or loaded beyond 12 shelves. All three students want their work finished and cannot agree on priority.
1. Calculate the shelf demand and determine whether all components can dry simultaneously. Show your reasoning.
Read the worked answer
Shelf demand: A needs 4 shelves for wet components. B needs 5 shelves. C's 3 components each take two shelves, so 6 shelves. Total simultaneous demand is 4 + 5 + 6 = 15 shelves, against a capacity of 12. So all components cannot dry at once; there is a shortfall of 3 shelves. Additionally, A's two already-dried pieces need a 24-hour drying period after surface treatment, which requires 2 more shelves at some point, raising peak demand further if scheduled concurrently. The rack cannot be overloaded, so the group must sequence the work. The critical path is the 48-hour drying time: with five days available and school-hours access, there is time for at least two drying cycles, but only if the first cycle is loaded immediately and unloaded promptly.
2. Propose a fair scheduling plan that meets the deadline, and justify how it balances the competing claims.
Read the worked answer
A defensible plan: cycle one loads C's three large components (6 shelves), B's five components (5 shelves), and one of A's four (1 shelf), totalling 12 shelves exactly. This clears C entirely in one cycle, which is fair because C's components are indivisible and occupy the most space per item, so delaying C would block the most capacity later. Cycle two, after 48 hours, loads A's remaining three components plus A's two surface-treated pieces (5 shelves) and B's... note B is already complete, leaving spare capacity. A's surface treatment should be applied at the start of cycle two so its 24-hour dry overlaps the 48-hour plaster dry. This completes all work with time to spare before assessment. Justification: the plan respects the hard 12-shelf limit, sequences around the 48-hour constraint, front-loads the largest indivisible items, and gives A a clear slot for both remaining plaster and surface treatment. Fairness is achieved through transparent rotation rather than equal simultaneous loading, which is physically impossible.