Past exam of the mathematics course of the University of Cambridge 2004 iii Paper 33 5 Solution 2026-10-14
The total processing time is constant for a single machine, so only the initial setup and changeover sum depends on the order. Use the dummy-job reduction for sequence-dependent setup times. Add vertex , and setLet the binary variable indicate that follows in the augmented tour. A Hamiltonian cycle through specifies exactly one schedule, read starting after . An exact integer programming formulation isThe degree equations alone describe a cycle cover of a directed graph, possibly several separate graph cycles. The subtour elimination constraints exclude every proper directed graph cycle; the cover is therefore a single tour through all jobs. Conversely every schedule supplies a feasible tour with exactly its completion time as objective. Exponentially many written inequalities are permitted in this exact formulation; an implementation may add them only when violated.
For three parallel machines, the relevant objective is the parallel-machine makespan with sequence-dependent setups, not simply total setup cost. Partition jobs into three ordered lists . For a nonempty list definean empty machine has . Minimize subject to for each machine and each job assigned once. This can be written as a mixed-integer extension with assignment variables, one dummy tour per machine and conditional subtour cuts.
A practical strategy starts with a balanced assignment, sequences each machine using the single-machine methods, and improves by relocating or swapping jobs between machines together with within-machine 2-opt moves. Score each move by its change in . Branch and bound can combine machine-assignment decisions with within-machine assignment bounds. For nonnegative times, the total processing time divided by three and the largest individual processing time are valid lower bounds, supplemented by setup-dependent route bounds. Balancing processing and changeovers jointly is essential; independently minimizing the three tour costs does not by itself minimize completion time.