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answer 3.1 and 3.2

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Exercise 3. We are given a directed graph G = (V, E) with a source s and a sink t, where every edge e has a (finite, non-negative) lower capacity le and a (finite or infinite) upper capacity ce, and where no directed edges enters or leave t. Furthermore, a number k ≥ 0 is given. All these numbers are integers. We want to find a flow f in G that respects the capacity constraints and has exactly the value val(f) k (or show that no such flow exists in G). We define ko = max{Σe=(sv) le, Σe=(v,t) le}. Obviously, k ≥ ko is a necessary condition for such a flow to exist. - Note that, unlike the ordinary Maximum Flow problem, the flow value k is prescribed here, and certain edges must (at least) carry some prescribed positive amounts of flow. Despite these differences, we would like to solve this problem variant via a reduction to Maximum Flow. For the following, recall the reductions LZ-red and CF-red from Lecture Notes 5. 3.1. Let G(k) be the graph obtained from G by introducing the demands d(s) := −k, d(t) := +k, and d(v) = 0 for all other nodes v. Explain in a few lines why the following equivalence is true: G has a flow f with val(f) = k if and only if G(k) has a circulation. 3.2. Let H(k) be the graph obtained from G(k) by LZ-red, where k ≥ ko. Explain in a few lines why the following statements about H(k) are true: The d(v) for v‡s, t do not depend on k. We still have d(s) ≤ 0 and d(t) ≥ 0. 3.3. Let J(k) be the graph obtained from H(k) by CF-red, where k ≥ ko. We denote the new source and sink by s' and t', respectively. (Note that s and t are retained as usual nodes.) Show that G has a flow f with val (f) k if and only if the maximum s’ – t' flow in J(k) saturates all the edges at s' and t'. Advice: Consider the whole chain of reductions and use the already known facts about LZ-red and CF-red from Lecture Notes 5. You should not need to write much. From 3.3 it follows that computing this flow in J(k) and checking saturation solves the problem introduced above.
Transcribed Image Text:Exercise 3. We are given a directed graph G = (V, E) with a source s and a sink t, where every edge e has a (finite, non-negative) lower capacity le and a (finite or infinite) upper capacity ce, and where no directed edges enters or leave t. Furthermore, a number k ≥ 0 is given. All these numbers are integers. We want to find a flow f in G that respects the capacity constraints and has exactly the value val(f) k (or show that no such flow exists in G). We define ko = max{Σe=(sv) le, Σe=(v,t) le}. Obviously, k ≥ ko is a necessary condition for such a flow to exist. - Note that, unlike the ordinary Maximum Flow problem, the flow value k is prescribed here, and certain edges must (at least) carry some prescribed positive amounts of flow. Despite these differences, we would like to solve this problem variant via a reduction to Maximum Flow. For the following, recall the reductions LZ-red and CF-red from Lecture Notes 5. 3.1. Let G(k) be the graph obtained from G by introducing the demands d(s) := −k, d(t) := +k, and d(v) = 0 for all other nodes v. Explain in a few lines why the following equivalence is true: G has a flow f with val(f) = k if and only if G(k) has a circulation. 3.2. Let H(k) be the graph obtained from G(k) by LZ-red, where k ≥ ko. Explain in a few lines why the following statements about H(k) are true: The d(v) for v‡s, t do not depend on k. We still have d(s) ≤ 0 and d(t) ≥ 0. 3.3. Let J(k) be the graph obtained from H(k) by CF-red, where k ≥ ko. We denote the new source and sink by s' and t', respectively. (Note that s and t are retained as usual nodes.) Show that G has a flow f with val (f) k if and only if the maximum s’ – t' flow in J(k) saturates all the edges at s' and t'. Advice: Consider the whole chain of reductions and use the already known facts about LZ-red and CF-red from Lecture Notes 5. You should not need to write much. From 3.3 it follows that computing this flow in J(k) and checking saturation solves the problem introduced above.
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