Algorithm Design

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Chapter 7 Network Flow
tree one computes for G may not in fact be the "real" minimum spanning
tree.
Given error parameters ~ > 0 and/~ > 0, and a specific edge e’ = (u, v),
you would like to be able to make a claim of the following form.
(,) Even if the cost of each edge were to be changed by at most s (either
increased or decreased), and the costs of k of the edges other than e’ were
further changed to arbitrarily different values, the edge e’ would still not belong
to any minimum spanning tree of G.
Such a property provides a type of guarantee that e’ is not likely to belong
to the minimum spanning tree, even assuming significant measurement
error.
Give a polynomial-time algorithm that takes G, e’, e, and k, and decides
whether or not property (*) holds for e’.
41. Suppose you’re managing a collection of processors and must schedule
a sequence of jobs over time.
The jobs have the following characteristics. Each job ] has an arrival
time aj when it is first available for processing, a length g1 which indicates
how much processing time it needs, and a deadline d i by which it must
be finished. (We’ll assume 0 < ~i < d1 - ai.) Each job can be run on any
of the processors, but only on one at a time; it can also be preempted
and resumed from where it left off (possibly after a delay) on another
processor.
Moreover, the collection of processors is not entirely static either:
You have an overall pool of k possible processors; but for each processor
i, there is an interval of time % t~] during which it is available; it is
unavailable at all other times.
Given all this data about job requirements and processor availability,
you’d like to decide whether the jobs can all be completed or not. Give a
polynomial-time algorithm that either produces a schedule completing all
jobs by their deadlines or reports (correctly) that no such schedule exists.
You may assume that all the parameters associated with the problem are
integers.
Example. Suppose we have two jobs J1 and J2. J1 arrives at time 0, is due
at time 4, and has length 3. J2 arrives at time 1, is due at time 3, and has
length 2. We also have two processors P1 and P2. P~ is available between
times 0 and 4; Pz is ava~able between Omes 2 and 3. In this case, there is
a schedule that gets both jobs done.
¯ At time 0, we start job J~ on processor P~.
42.
43.
¯ At time 1, we preempt J1 to start J2 on P~.
¯ At time 2, we resume J~ on P~. (J2 continues processing on P1.)
Exercises
¯ At time 3, J2 completes by its deadline. P2 ceases to be available, so
we move J1 back to P1 to finish its remaining one unit of processing
there.
¯ At time 4, J1 completes its processing on P1.
Notice that there is no solution that does not involve preemption and
moving of jobs.
Give a polynomial-time algorithm for the following minimization ana-
!ogue of the Maximum-Flow Problem. You are given a directed graph
G = (V, E), with a source s s V and sink t ~ V, and numbers (capacities)
*(v, w) for each edge (v, w) ~ E. We define a flow f, and the value of a flow,
as usual, requiring that all nodes except s and t satisfy flow conservation.
However, the given numbers are lower bounds on edge flow--that
is, they require that f(v, w) > ~.(v, w) for every edge (v, w) ~ E, and there is
no upper bound on flow values on edges.
(a) Give a polynomial-time algorithm that finds a feasible flow of minimum
possible value.
(b) Prove an analogue of the Max-Flow Min-Cut Theorem for this problem
(i.e., does min-flow = max-cut?).
You are trying to solve a circulation problem, but it is not feasible. The
problem has demands, but no capacity limits on the edges. More formally,
there is a graph G = (V, E), and demands dv for each node v (satisfying
~v~v dv = 0), and the problem is to decide ff there is a flow f such that
f(e) > 0 and f~n(v) - f°Ut(v) = du for all nodes v V. Note that this problem
can be solved via the circulation algorithm from Section 7.7 by setting
ce = +oo for all edges e ~ E. (Alternately, it is enough to set ce to be an
extremely large number for each edge--say, larger than the total of all
positive demands dv in the graph.)
You want to fix up the graph to make the problem feasible, so it
would be very useful to know why the problem is not feasible as it stands
now. On a closer look, you see that there is a subset’U of nodes such that
there is no edge into U, and yet ~v~u du > 0. You quickly realize that the
existence of such a set immediately implies that the flow cannot exist:
The set U has a positive total demand, and so needs incoming flow, and
yet U has no edges into it. In trying to evaluate how far the problem is
from being solvable, you wonder how big t~e demand of a set with no
incoming edges can be.
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