In Chapter 3 we showed that strong equivalence is a desirable property when we use the continuous relaxation and penalty function techniques to solve a nonlinear 0-1 program- ming problem. The QAP relaxation with the quadratic penalty function possesses such a property. As we also mentioned in Section 3.7, we will solve a single relaxation problem of form (3.16) for a sufficiently large penalty parameter as described in Theorem 3.5.
By the strong equivalence, we are guaranteed to obtain an integer solution to the QAP by solving (3.16). Nevertheless, two fundamental issues need to be addressed:
1. If we start from an arbitrary point and solve (3.16), chances are that we will only get an “average” solution to the QAP. Thus we need to choose a starting point for (3.16) so we can on the average obtain a “good” solution to the QAP.
2. Since there is no guarantee that we will obtain the optimal solution to the QAP by solving (3.16), if possible, we would like to use a current solution to find a better one.
In the discussions that follow in this chapter, we will denote the objective function of the QAP by q(x), i.e.,
q(x) = 1 2x
4.1.1
The Algorithmic Framework
In this section we will outline the framework of our proposed algorithm for the QAP. The framework as depicted in Figure 4.1 addresses the two issues brought up in the above.
In step (1) of the framework, we solve the following quadratic programming relax- ation: minimize x q(x) (QPR) subject to Lx=b x≥0
to obtain a starting pointx0. In step (2), we start from x0 and solve the QAP relaxation
with the quadratic penalty function (3.16) for the penalty parameter µ to obtain an initial solution to the QAP. That is, we solve
minimize x q(x) +µx T(e−x) (RXP) subject to Lx=b x≥0.
Since there is no guarantee that the resulting solution will be optimal to the QAP or even a “good” one, we would like to employ a method to find a better one. Let us denote theincumbent solution, or the current best solution, by ˆx. It is obvious that any solution to the QAP with a smaller objective value, if it exists, must satisfy
(4.1) q(x)≤q(ˆx)−
for some small positive . This suggests that we add (4.1) to (RXP) and solve the resulting more restricted problem. We refer to (4.1) as a quadratic cut. More specifically, we set the incumbent solution ˆxin step (3), and in step (4) starting from x0 we solve the
relaxation with the penalty and quadratic cut: minimize x q(x) +µx T(e−x) (RPC) subject to x∈Ω where Ω is defined as (4.2) Ω ={x∈Rn :Lx=b, q(x)≤q(ˆx)−, x≥0}.
Begin
(1) Solve (QPR) forx0.
(2) Starting fromx0, solve (RXP).
(3) Set or update incumbent
solution ˆx. Setk= 0.
(4) Starting fromx0, solve (RPC)
or (DPC).
Found an integer solution?
(5) Incrementkby 1. Generate a
random starting pointxk.
(6) Starting fromxk, solve (TPC).
Found an integer solution? k= Ξ? End No Yes No Yes Yes No
Alternatively, we can replace the objective function of (RPC) with a slightly different one. That is, in step (4) we start from x0 and solve the minimal distance problem with the penalty and quadratic cut:
minimize
x P(x;ρ),kx−x
0k2+ρxT(e−x) (DPC)
subject to x∈Ω
where ρ is chosen to be greater than one so that ∇2
xP(x;ρ) is negative definite. From the form of P(x;ρ), ∇2
xP(x;ρ) is either negative definite or positive definite (except for
ρ= 1). If∇2
xP(x;ρ) were positive definite, (DPC) would have a unique optimum, which is unlikely to be an integer solution. We will explain whykx−x0k2 is used in the objective
function in Section 4.4.
Since the optimal solution to either (RPC) or (DPC) satisfies (4.1), if it is integer it will be a better solution than ˆx. However, even though the objective functions of (RPC) and (DPC) are concave, the presence of the quadratic cut allows the possibility that an optimal solution to (RPC) or (DPC) may not be at an integer point. That is, strong equivalence does not hold for (RPC) or (DPC). If we obtain an integer solution, we go back to step (3), update ˆx and solve (RPC) or (DPC) again. Otherwise, knowing that we are unable to obtain a better solution to the QAP by solving (RPC) or (DPC) from the given starting point, we try different points in Ω to see whether any of them would lead to an improved solution. To this end, we randomly generate a point xk in Ω in step (5) and in step (6) we solve thequartic penalty problem with the quadratic cut:
minimize
x kx◦(e−x)k
2
(TPC)
subject to x∈Ω
starting fromxk. We will discuss why we use the quartic penalty function as the objective function in Section 4.4. If we find an integer solution to (TPC), we go back to step (3), update ˆx, and carry on from there. Otherwise, if we fail to find an integer solution to (TPC), we go back to step (5), generate a different random starting point, and re-solve (TPC) with the new starting point. We define the number of stalled cuts as the number of successive times we fail to find an integer solution to (TPC). We terminate when the number of stalled cuts reaches a prescribed number Ξ.
In the following discussions, we will collectively refer to (QPR), (RXP), (RPC), (DPC) and (TPC) as the relaxation problems, and (RPC), (DPC) and (TPC) as the
relaxation problems with the quadratic cut.
There are several techniques or components that are indispensable to our proposed algorithm for the QAP. We will describe the convex transformation of the objective