Algorithm Design

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Chapter 11 Approximation Algorithms
analysis, therefore, we wig need a lower bound on the optimum--a quantity
with the guarantee that no matter how good the optimum is, it cannot be less
than this bound.
There are many possible lower bounds on the optimum. One idea for a
lower bound is based on considering the total processing time ~j tj. One of
the m machines must do at least a 1/m fraction of the total work, and so we
have the following.
(11.1) The optimal makespan is at least
1
T* >_ -- E t
. j.
1
There is a particular kind of case in which this lower bound is much too
weak to be useful. Suppose we have one job that is extremely long relative to
the sum of all processing times. In a sufficiently extreme version of this, the
optimal solution will place this job on a machine by itself, and it will be the
last one to finish. In such a case, our greedy algorithm would actually produce
the optimal solution; but the lower bound in (11.1) isn’t strong enough to
establish this.
This suggests the following additional lower bound on T*.
(11.2) The optimal rnakespan is at least T* > maxl
Now we are ready to evaluate the assignment obtained by our greedy
algorithm.
(iL3) AIgO~thra Greedy2Bal~he produCe~ an assign~ent b[ ]obs to ~achines
wtth makespan T_ 2T ,
Prooi. Here is the overall plan for the proof. In analyzing an approximation
algorithm, one compares the solution obtained to what one knows about the
optimum--in this case, our lower bounds (11.1) and (11.2). We consider a
machine Mi that attains the maximum load T in our assignment, and we ask:
What was the last job j to be placed on Mi ~. If tj is not too large relative t.o m6st
of the other jobs, then we are not too far above the lower bound (11.1). And,
if tj is a very large job, then we can use (11.2). Figure 11.2 shows the structure
of this argument.
Here is how we can make this precise. When we assigned job j to Mi, the
machine Mi had the smallest load of any machine; this is the key property
of our greedy algorithm. Its load just before this assignment was Ti - tp and
since this was the smallest load at that moment, it follows that every machine
11.1 Greedy Algorithms and Bounds on the Optimum: A Load Balancing Problem
I~
he contribution of
e last job alone is
most the optimum
I~
ust before adding
e last job, the load
n Mi was at most
e optimum.
Figure 11.2 Accounting for the load on machine Mi in two parts: the last job to be
added, and all the others.
had load at least Ti - t i. Thus, adding up the loads of all machines, we have
Y~.I~ TI~ >- m(Ti - ti), or equivalently,
Ti - t]