Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 35 (Mathematics of Operational Research)

Introduce nonnegative slack variables $s_i$ and write $z=c^{\mathsf{T}}x$. A maximizing simplex dictionary is dual feasible when every nonbasic coefficient in the objective is nonpositive. After inserting a constraint, the dual simplex algorithm retains this property while repairing any negative basic value. For a leaving row $x_B=\beta +{\sum }_j{d}_j{x}_j$ with $\beta <0$, only $d_j>0$ can enter; choose a column minimizing the ratio of its objective penalty to $d_j$.

For the first linear program, $x_1$ has the best profit-to-resource ratio. Pivot it into the initial slack basis to obtainAll basic values are nonnegative and all reduced costs are nonpositive, so the optimal point is $(5,0,0,0)$, with value five. Adding the second constraint givesThe same basis remains feasible and optimal: the second constraint does not change the answer.

The third constraint gives the new rowThe current dictionary is dual feasible but primal infeasible. The eligible entering variables are $x_3,x_4$, with dual ratio-test values $2$ and $3$ respectively. Enter $x_3$ and leave $s_3$; solving that row and substituting into the old rows yieldsThis is a feasible optimal simplex dictionary. Thus the new optimal point is $(2,0,1,0)$, with value three. As an independent linear programming optimality certificate, dual multipliers $(3,0,2)$ have weighted objective bound $3\cdot 5+2(-6)=3$ and weighted coefficient row $(1,4,1,2)$, which dominates the profit row.

On inserting the fourth constraint the slack isIt has a negative constant and no eligible entering column. Since all its nonbasic variables are nonnegative, this row cannot be satisfied: the fourth linear program is infeasible, and adding the fifth constraint cannot restore feasibility. The requested optimal values are therefore

The arbitrary-data selection question is a maximum feasible subsystem problem. With rational input it is in NP: guess the selected row indices, and test their linear programming feasibility in polynomial time. Equivalently, a nonempty rational polyhedron has a feasible rational point of polynomially bounded encoding length, which serves as a certificate.

For an explicit NP-hardness reduction, take an independent set instance with graph $H$ having $r$ vertices and desired independent-set size $K$. Use variables $x_v\ge 0$, one inequality $-x_v\le -1$ for each vertex, and $M=r+1$ identical copies of $x_u+x_v\le 1$ for each edge. Ask for a feasible subsystem of size $M|E(H)|+K$. An independent set of size $K$ gives such a subsystem: include all edge rows and those $K$ vertex rows, and use its indicator vector. Conversely, omitting every copy of even one edge row leaves at most $M(|E(H)|-1)+r<M|E(H)|+K$ selectable rows. Hence any qualifying subsystem retains every edge inequality and at least $K$ vertex inequalities. Their vertices must be independent, since two adjacent selected vertices would force $x_u+x_v\ge 2$. The construction has polynomial size; $c$ is irrelevant to feasibility. Thus the rational-input decision problem is NP-complete, not merely difficult because there are many subsets to enumerate.

A chain of $r$ diamond-shaped graph stages has $3r+1$ vertices and $4r$ edges, but $2^r$ distinct source-to-target paths: at each stage choose either of two branches. Each path has its own inequality. Therefore the number of displayed path constraints is not polynomially bounded in the graph size. Redundancy among some inequalities would not change this count of the given formulation.

The positive integer edge costs imply, for any feasible point,Thus the feasible set lies in the Euclidean ball of radius $100$ centered at zero. That ball lies inside $[-100,100{]}^m$, so its volume is at most $200^m$. This proves the requested $O(200^m)$ enclosing-volume bound, and provides an initial ellipsoid, for example with center zero and shape matrix $10^{4}I$.

The ellipsoid method needs a separation oracle, rather than an explicit list of every constraint. For a proposed center $z$, first scan the coordinates and the budget inequality. If some $z_e<0$, some $z_e>1$, or ${\sum }_e{c}_e{z}_e>k$, return the corresponding violated inequality as the separating half-space. These tests take $O(m)$ arithmetic operations.

Once those tests pass, assign nonnegative edge length $z_e$ to every edge and solve the shortest path problem from $s$ to $t$. The Dijkstra algorithm finds a minimum-length path and stores its predecessors in polynomial time. If its length is less than one, returnfor the minimizing path $p$. This inequality contains every feasible point and is violated by $z$. If the shortest length is at least one, every path constraint holds; if no path exists, those constraints are vacuous. In either case $z$ is feasible. This is path-constraint separation by shortest paths; checking a single shortest path replaces checking exponentially many paths.

The rational ellipsoid method feasibility theorem now gives polynomial running time in the dimension and coefficient encoding bounds when the oracle runs in polynomial time. Here every path row has coefficients zero or one, and all other coefficients and right sides are bounded integers. Although there are exponentially many rows, their individual encoding lengths are polynomial in $m$. The containing-radius bound and the rational precision bounds are also polynomially describable. For a simple graph $m=O(n^2)$, this gives a worst-case running time polynomial in $n$; in general the natural bound is polynomial in the encoded graph size $n+m$.

Use the exact rational feasibility version of the ellipsoid method, which handles lower-dimensional polyhedra by rational precision bounds or an appropriate relaxation. Nonemptiness must not be inferred from a positive-volume argument: this feasible set can have empty interior. The allowed algorithmic theorem supplies that detail, while the explicit shortest-path oracle supplies the crucial efficient separation.

For NP-completeness, use the decision version with an objective threshold. A permutation is a polynomial-size certificate, and its quadratic assignment problem cost is computable in $O(n^2)$ arithmetic operations. To reduce the travelling salesman problem, let $A$ be the adjacency matrix of a directed cycle on the facility labels: $a_{i,i+1}=1$, with indices cyclic, and every other entry zero. Let $B$ be the matrix of city-to-city distances. Thenwhich is exactly the tour length. Every tour arises from such a permutation. Hence the decision QAP is NP-complete, and finding its optimal value is NP-hard. No factor of two appears in this directed-cycle construction, even when the distance matrix is symmetric.

Fix any complete assignment $\pi$. Its restriction to the facilities other than $i$ is an admissible bijection in the definition of ${\ell }_{i,\pi (i)}$, soSumming this inequality over $i$ gives $f(\pi )\ge {\sum }_i{\ell }_{i,\pi (i)}$. Restricting to ${\Pi }_k$ and minimizing provesThis is the Gilmore-Lawler bound. Its row minima may come from mutually incompatible assignments, so equality is not automatic. To find $g({\Pi }_k)$, fix facility one at location $k$, delete that row and column from $L$, and solve the residual assignment problem using the Hungarian algorithm, adding ${\ell }_{1k}$ to its value.

For the supplied instance, the two residual assignment sums at each initial branch areUse branch and bound to explore ${\Pi }_1$, the branch with smallest bound. Its minimizing linear assignment is $\pi =(1,2,3)$. Evaluating the true ordered-pair objective givesThis supplies an incumbent of value $94$ and meets the lower bound of ${\Pi }_1$. Alternatively, after branching on facility two, the other leaf $(1,3,2)$ has row-sum bound $31+20+48=99$ and is pruned. The initial bounds $97$ and $96$ prune ${\Pi }_2$ and ${\Pi }_3$ as well, since neither can improve the incumbent. ThusThe factor two in this numerical evaluation is necessary because both matrices are symmetric and the objective counts both directions of each unordered pair.

Represent a feasible allocation by a label for each item saying which bidder receives it. Every such label vector gives disjoint bundles covering all items, so the following heuristic optimization methods can search without violating allocation feasibility.

In local search, begin with any feasible partition and repeatedly make an improving move, such as transferring one item between bidders or exchanging two items. For a transfer $j:i\to r$, evaluate the changeStop when no allowed move improves the objective. Several starting partitions can avoid dependence on one local optimum, but no guarantee of a global optimum follows for general bids.

In simulated annealing, use the same feasible neighborhood but also accept a worsening move of loss $\Delta >0$ with probability $e^{-\Delta /T}$ at temperature $T$. Begin with a relatively high temperature, gradually cool, and retain the best allocation seen. Worsening moves can cross a barrier between local optima; a practical finite cooling schedule does not guarantee optimality.

In tabu search, retain a short memory of recent transfers or exchanges and temporarily prohibit their reversal. Select the best admissible move, possibly worsening the objective, to escape a local optimum. Permit an otherwise tabu move if it improves the best solution found, an aspiration rule. For the winner determination problem, bidder/item pairs or recent reverse transfers provide convenient tabu attributes. All three methods preserve the partition, but for arbitrary bids they are heuristics rather than algorithms with a proved constant approximation ratio.

The supplied hint interchanges its item and bidder indices in the first-value identity. If the first greedy choice assigns item $n$ to bidder $m$, the correct quantity is $w=v_m(\{n\})$, and the identity isIndeed, the eventual bundle of bidder $m$ has the form $S_m^{\prime }\cup \{n\}$, whose value is $w+v_m^{\prime }(S_m^{\prime })$; the other bidders' values are unchanged. Every later marginal contribution for bidder $m$ is exactly its marginal in the residual problem, and this is also true for every other bidder. Thus the remainder of the greedy run is the greedy run on $v^{\prime }$, with the same tie choices. The residual values are nonnegative and increasing; they remain submodular set functions as allowed in the question.

We prove the greedy half-approximation for submodular welfare by induction on the number of items. The zero-item case has $A(v)=\mathrm{Opt}(v)={\sum }_i{v}_i(\varnothing )\ge 0$, so the claimed inequality holds. This base case also handles the fact that the question does not explicitly normalize empty-bundle values to zero.

For the induction step, letbe the first greedy marginal. Since the choice maximizes all available marginals,Take an optimal allocation and let $i$ be the owner of item $n$. If $i\ne m$, remove that item from $i$'s bundle and give it to $m$. By submodularity, the loss to its old owner isMonotonicity makes the recipient's gain nonnegative. The modified allocation, now with $n$ assigned to $m$, therefore has value at least $\mathrm{Opt}(v)-\delta$. If $i=m$, no modification is necessary and this bound still holds. Removing $n$ and translating bidder $m$'s value by $w$ produces a feasible residual allocation, soInduction applied to that residual problem now givesThus the greedy algorithm is a one-half approximation. Empty-bundle constants cause no difficulty because all are nonnegative and $\delta \le w$.

At each of $n$ stages evaluate the marginal for at most $m$ bidders and $n$ remaining items. This requires $O(m{n}^2)$ value queries, and polynomial additional work with bundle bit-vectors and cached current values. Hence it is a polynomial time approximation algorithm in the value-oracle model, assuming polynomial time bid evaluations. If every bundle bid is explicitly listed, this query bound is polynomial in that larger explicit input as well. The exponentially many possible bundles do not require enumerating allocations.

A Nash equilibrium is a profile in which each player's strategy is a best response to all the others: no unilateral change increases that player's payoff. In this routing game the payoff can be taken as minus delay. A pure strategy is a single route; a mixed strategy is a probability distribution over routes, with different players randomizing independently. A non-zero-sum game does not require the players' payoffs to add to a fixed total. Nash's theorem guarantees a mixed Nash equilibrium for every finite game with finitely many pure strategies. Pure equilibria need not exist in general; the congestion game here does have them, since a unilateral delay change equals the change in its finite congestion potential.

Put $x=n_1+n_3$ and $y=n_2+n_3$. The three current delays, in the order $ABD,ACD,ABCD$, areFor a player using the first route, switching to the second makes the new second-route load $y+1$, while switching to the third retains that player's use of $AB$ and adds one user to $CD$. The two comparisons are thereforeThey reduce to $x\le y+1$ and $y\ge 74$. Reversing the two outer routes gives $y\le x+1$ and $x\ge 74$ for each second-route user. A third-route user who switches to the first retains their own use of $AB$, and one who switches to the second retains their own use of $CD$. The corresponding comparisons give $y\le 75$ and $x\le 75$. Thus the necessary and sufficient conditions, together with nonnegative integer counts summing to $n$, areThey are sufficient because each used route has been compared with both alternatives, and necessary because each comparison is an available unilateral deviation. The unit changes in loads matter in a finite-player congestion game.

For $n=100$, the counts $(25,25,50)$ give $x=y=75$ and every player's delay is $15/4$. They satisfy all the conditions. To find every pure Nash equilibrium, first note that $n_3=0$ is impossible: if either outer route is empty, its other route users fail the lower-load condition; if both are used, $x,y\ge 74$ contradict $x+y=100$. Thus $n_3>0$, so $x,y\le 75$. Since $x=100-n_2$ and $y=100-n_1$, this forces $n_1,n_2\ge 25$. Both are positive, so the remaining conditions force $x,y\ge 74$ and hence $n_1,n_2\le 26$. Their differences are at most one, so every such pair works. The complete equilibrium count vectors areEach vector also represents many pure profiles obtained by choosing which labeled players use each route. Therefore the displayed equilibrium is not unique, even as an aggregate count vector.

A coordinated allocation with fifty players on each outer route and nobody on $ABCD$ gives every player delay $7/2$, strictly better than $15/4$. It is not a Nash equilibrium: a first-route user can switch to $ABCD$ and obtain delayThis exhibits the conflict between coordinated welfare and unilateral incentives. In fact $7/2$ is below every user's delay in all four pure equilibrium count vectors above.

For a symmetric Nash equilibrium when $n=100$, let each player independently choose the two outer routes with probability $q$ each and the middle route with probability $1-2q$. Conditional on choosing the first route, that player certainly uses $AB$, while each of the other $99$ players uses it with probability $1-q$. The expected delays of the first and third routes are consequentlyThe second has the same expected delay as the first by symmetry. Equalizing these givesAll three probabilities are positive and all three pure routes have expected delay $15/4$, so every route is a best response to the other players' common mixture. This proves a symmetric mixed Nash equilibrium. Using probabilities $1/4,1/4,1/2$ would omit the player's own-load correction and would not make all three expected delays equal.

For completeness, the same calculation works for every positive number of players. If $n\le 75$, every player choosing $ABCD$ is a symmetric pure Nash equilibrium. For $75<n<149$, a symmetric mixed equilibrium hasFor $n\ge 149$, each player choosing either outer route with probability one half is a symmetric mixed equilibrium. In that last regime the unused middle route has expected cost at least that of the outer routes, since their difference is $-3/4+(n+1)/200\ge 0$. These formulas agree at the regime boundaries.

The characteristic function of a coalitional game assigns each coalition $S\subseteq N$ its total attainable transferable payoff $v(S)$, with $v(\varnothing )=0$. In a transferable utility game the members may divide that total among themselves. This game-theoretic characteristic function is not the characteristic function of a probability distribution.

The Shapley value assigns to player $i$ the average marginal contribution that player makes over all uniformly random orderings of the players. If $P_i^{\pi }$ is the predecessor set in ordering $\pi$, thenFor any ordering, summing all contributions telescopes to $v(N)$, so the Shapley value is efficient. It treats interchangeable players equally, pays a null player zero, and is additive in the characteristic function. These properties give its standard fairness axioms and uniquely characterize the Shapley value. The random-order interpretation explains why every position and predecessor coalition receives a principled weight rather than privileging a particular bargaining order.

For a convex cooperative game, couple a uniform ordering of $N$ with its restriction to $T$. The restricted ordering is uniform on $T$. For $i\in T$, the restricted predecessor set is $P_i^{\pi }\cap T\subseteq P_i^{\pi }$, so increasing marginal contributions giveAveraging proves Shapley population monotonicity in a convex game:The comparison is meaningful for players who participate in the restricted game; no restricted value is defined for a player outside $T$. Summing over $i\in T$ and applying efficiency in that restricted game givesTogether with efficiency for $N$, these are exactly the core of a cooperative game inequalities. Hence the Shapley value belongs to the core of a convex game. This proves the requested stability, not just individual rationality.

Now label the entrepreneur by $0$ and the $n\ge 1$ workers by $1,\dots ,n$. A coalition without the entrepreneur has value zero; one containing the entrepreneur and $k$ workers has value $p(k)$. Workers are interchangeable, so their Shapley values agree. Fix one worker. Their contribution is zero if the entrepreneur appears after them. If the entrepreneur and exactly $k$ of the other $n-1$ workers appear before them, it is $p(k+1)-p(k)$. For a fixed choice of those workers, the $k+1$ predecessors can be ordered in $(k+1)!$ ways and the remaining $n-k-1$ successors in $(n-k-1)!$ ways. Therefore the probability of this event, summing over the $\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$ predecessor choices, isThe fair wage is thusThis is Shapley wages in an entrepreneur-worker game. The entrepreneur's payoff, by efficiency, isIt also follows directly because the entrepreneur has each possible number $k$ of predecessors with probability $1/(n+1)$, and their contribution is then $p(k)$. When $p(k)=\alpha k$, the differences are all $\alpha$ and ${\sum }_{k=0}^{n-1}(k+1)=n(n+1)/2$, so

Finally suppose $p$ is convex and nondecreasing on nonnegative integers, so $d_k=p(k+1)-p(k)$ is nonnegative and nondecreasing. We verify that the entire coalitional game is convex. The entrepreneur's contribution to a set of $k$ workers is $p(k)$, which increases with that set. A worker's contribution is zero when the entrepreneur is absent and is $d_k$ when the entrepreneur and $k$ other workers are present. Enlarging a predecessor set either leaves both contributions zero, introduces the entrepreneur and a nonnegative contribution, or increases $k$ and hence $d_k$. Every player's marginal contributions therefore increase, giving a convex cooperative game. The result already proved yields the complete Shapley allocation lies in the core, including the entrepreneur's payoff.

One can also verify the core inequalities directly. Nonentrepreneur coalitions receive $kw\ge 0$, their value. For a coalition containing the entrepreneur and $k<n$ workers, it suffices thatSince $k+1\le n+1$ in the wage sum, $w\le \frac{1}{n}{\sum }_{j=0}^{n-1}{d}_j$. Nondecreasing $d_j$ makes this full average no greater than the average of the final $n-k$ differences, which is the displayed right side. Thus $x_0+kw=p(n)-(n-k)w\ge p(k)$; the grand coalition has equality by efficiency. This confirms the core conclusion without assuming $p(0)=0$.