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Add a dummy job $0$. Give the arc $0\to i$ cost equal to job $i$'s initial setup, the arc $i\to j$ its changeover cost, and every arc $i\to 0$ zero cost. A Hamiltonian cycle through the dummy represents a schedule, starting just after $0$ and ending just before it. The total processing time is constant on a single machine, so minimizing cycle cost minimizes completion time. Dropping subtour elimination constraints gives an assignment problem lower bound.

Write $(j_1,\dots ,j_k)$ for the partial assignment problem solution sending person $i$ to job $j_i$. The prescribed assignment lower bound from independent task minima isEvery completion must pay at least the minimum in each remaining column, so this is a valid lower bound, even if the minima use the same person several times. At $k=3$, the one remaining job must go to person 4, and the bound is the exact completed cost.

Use branch and bound with the smallest open bound expanded first. The initial bounds areFor example, $L(2)=12+11+13+22=58$. Expand $(2)$, givingExpand $(2,3)$. Its children are $(2,3,1)$ with bound 64 and $(2,3,4)$ with bound 65. Completing the first gives $(2,3,1,4)$ at cost 64, our first incumbent. The second cannot improve it. We may also discard $(2,1)$, $(2,4)$, $(3)$ and $(4)$, since their lower bounds are at least 64; equality may be pruned when seeking one optimum.

The remaining open prefix is $(1)$, whose children have boundsOnly $(1,3)$ can improve the incumbent. Its children $(1,3,2)$ and $(1,3,4)$ have exact completed costs 69 and 61, respectively. The latter completes to $(1,3,4,2)$ and replaces the incumbent. Every other branch has already been bounded above this value. Thus the optimal assignment isThis branch and bound computation produces fourteen partial nodes: four initial nodes, three below $(2)$, two below $(2,3)$, three below $(1)$ and two below $(1,3)$. The last person is assigned automatically at a depth-three node.

For the travelling salesman problem, let $x_{ij}=1$ mean that the tour goes directly from city $i$ to city $j$. Retain only the constraintsThis is an assignment problem between cities as departure points and cities as arrival points, solvable by the Hungarian algorithm. Its optimum is a cycle cover of a directed graph, possibly with several disjoint cycles. Since every tour is such a cover, its cost is a lower bound on the best tour cost. This is the assignment relaxation of the travelling salesman problem.

If the minimizing cover is a single Hamiltonian cycle, it supplies a tour and can update the incumbent. Otherwise choose one proper subtour $C$. No tour can contain all edges of $C$, since that would close a cycle before visiting the remaining cities. Make one child subproblem for each edge of $C$, forbidding that edge, and solve its modified assignment problem. These branches may overlap but together retain every possible tour. Prune a child if it is infeasible or its assignment bound is at least the incumbent cost. Further subtours cause further branching. Each branch adds a new forbidden edge, so the search is finite, and the covering property plus the valid lower bounds proves optimality when no open branch remains. This enforces the effect of subtour elimination constraints through branch and bound.

Drop the subtour elimination constraints, retaining the one-incoming/one-outgoing equations and any branch decisions. This is an assignment problem, solvable in polynomial time. Its optimum is a lower bound on the best schedule in that branch because every admissible schedule satisfies the relaxed constraints. Add the constant processing-time sum to compare completion-time bounds.

Keep a feasible schedule as an incumbent upper bound. At each branch and bound node, solve the restricted assignment problem. If it is infeasible or its bound is at least the incumbent, discard the node. If its optimum is a single tour, it is a feasible schedule attaining the bound, so update the incumbent and close the node. If it has a proper subtour $C$ with arcs $e_1,\dots ,e_r$, every valid Hamiltonian tour must omit at least one of those arcs.

For a disjoint exhaustive branching rule, child $h$ forces $e_1,\dots ,e_{h-1}$ and forbids $e_h$, for $h=1,\dots ,r$. Every valid tour falls into the child indexed by the first omitted arc. Re-solve the corresponding assignment relaxations, prioritize small bounds, and repeat. Finitely many binary choices give an exact terminating search; its worst-case tree can be exponential. Assignment optima provide lower bounds, while complete tours provide upper bounds. Patching subtours or local improvement can improve the incumbent without changing the exact pruning argument.

The total processing time ${\sum }_{i=1}^{40}{t}_i$ 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 $0$, and setLet the binary variable $x_{ij}$ indicate that $j$ follows $i$ in the augmented tour. A Hamiltonian cycle through $0$ specifies exactly one schedule, read starting after $0$. 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 ${\pi }_r$. For a nonempty list definean empty machine has $C_r=0$. Minimize $C_{\max }$ subject to $C_r\le C_{\max }$ 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 ${\max }_r{C}_r$. 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.