Viral

Theorem 5.10. For the assignment problem on a graph G and for a discrete uncertainty set U = {c1, . . . , cm}, we can polynomially reduce Problem (M2)

to Problem (M3_{) with at least m + k − 1 scenarios, for any fixed k < m.}

Proof. Instead of (M2) we consider the equivalent Problem (M2₀) (see Theo- rem 3.1). The proof is similar to the previous proofs. For a given instance of the min-max assignment problem, i.e. a bipartite graph G = (V, W, E) and an uncertainty set U = {c1, . . . , cm}, we define the graph Gask := (Vkas, Wkas, Ekas)

with Vas

k := V ∪ {v1, . . . , vk−1}, Wkas= W ∪ {w1, . . . , wk−1} and

E_kas:= E ∪ {fij = {vi, wj} : i, j = 1, . . . , k − 1} .

The latter construction is shown in Figure 5.3. Note that any feasible solution on Gas_k must induce a perfect matching in G and a perfect matching in the subgraph with nodes v1, . . . , vk−1 and w1, . . . , wk−1. The idea of the proof is

to define scenarios on Gas

k that force the min-max-min problem to choose k

solutions x(i) such that for each i = 1, . . . , k − 1 solution x(i) uses edge fii

v1 v2 v3 w1 w2 w3 G

Figure 5.3: The graph Gas

k for k = 4.

this end, we define scenarios ¯c1, . . . , ¯cm on Gask by extending every scenario

ci ∈ U by zero on the edges fii for all i = 1 . . . , k − 1 and by 2M on all other

edges. Here M can be chosen as

M := m X i=1 |E| X j=1 |(ci)j| + 1.

Then we add scenarios d1, . . . dk−1 such that in scenario di the edges fjj

have cost 3M for j 6= i and edge fii has cost −2M . Furthermore for any i

each edge fij for j = 1, . . . , k − 1 and j 6= i has cost 3M . All other edges

have cost 0. Now we choose a solution x(1)_{, . . . x}(k) _{of the min-max-min}

assignment problem on Gas

k , such that for each i = 1, . . . , k − 1 solution x(i)

uses edge fii and none of the edges fjj for j 6= i and solution x(k) uses all

edges fjj for each j = 1, . . . , k − 1. Note that there always exists a perfect

matching in Gas

k with the latter properties if a perfect matching in the original

graph G exists. For solutions with the latter properties the objective value is max c∈{¯c1,...,¯cm,d1,...dk−1} min i=1,...,kc > x(i) < M,

since on ¯ci the minimum is attained by x(k) and therefore, by the definition

of M , it is strictly lower than M . On the other hand on di the minimum is

attained by x(i) _{and is smaller than −M . Therefore we know that every opti-}

mal solution of the min-max-min problem on the graph Gas

k must contain for

each edge i = 1 . . . , k − 1 a solution with the property of x(i) since otherwise if this is not true every solution has to use one of the edges fij with j 6= i

which all have cost 3M in scenario di. Therefore in scenario di the minimum

mini=1,...,kc>x(i) is greater than M and therefore

max c∈{¯c1,...,¯cm,d1,...dk−1} min i=1,...,kc > x(i) ≥ M

which cannot be optimal by the observation above. By the same reasoning applied to the scenarios ¯ci, in every optimal solution there must be contained

a solution which uses all edges fii for i = 1, . . . , k − 1. So let x(1), . . . , x(k)

be an optimal solution with the properties like above. Then we have max c∈{d1,...,dk−1} min i=1,...,kc > x(i) < −M by the reasoning above. On the other hand

max c∈{¯c1,...,¯cm} min i=1,...,kc > x(i) = max c∈{c1,...,cm} c>x(k) > −M and therefore min

x(1)_,...,x(k)_{∈Xc∈{¯c1,...,¯}max_{cm,d1,...dk−1}}i=1,...,kmin c >

x(i) = min

x∈Xc∈{cmax1,...,cm} c>x. and the min-max optimal solution must be contained inx(1)_{, . . . , x}(k) _.

Corollary 5.11. For any fixed k ∈ N and fixed m > k, Problem (M3_{) is}

NP -hard for the assignment problem for uncertainty sets U with |U | = m. Proof. The Problem (M2_{) for the minimum assignment problem is NP -hard}

for m ≥ 2 (see Table 3.1). From Theorem 5.10 we derive that the min-max- min variant of the same problem is NP -hard if the number of scenarios is at least k + 1.

Note that by Proposition 5.3 for k ≥ m Problem (M3_{) can be solved in}

polynomial time for the assignment problem. Therefore the restriction to m > k is necessary.

Corollary 5.12. For any fixed k ∈ N, Problem (M3_{) with discrete U is}

strongly NP -hard for the assignment problem.

Proof. The min-max variant of the assignment problem is strongly NP -hard if the number of scenarios is not constant (see Table 3.1). The result fol- lows since all numbers in the construction in the proof of Theorem 5.10 are polynomial in |U |.