approximate algorithm for combinatorial optimization with two parameters
David Blokh,
Dept. of Industrial Engineering and LU'~U~~""'-'"""'L'V"""V,
Ben-Gurion University of the Negev, Beer-Sheva 84105, Israel.
Gregory Gutin*
Dept. of Mathematics and Statistics,
BruneI University, Uxbridge, Middlesex UB8 3PH, U.K.
and Dept. of Mathematics and Computer IJ~Jl'-'Uv"',
Odense University, DK-5230, Odense, Denmark.
Abstract
We call a minimum cost restricted time combinatorial optimization (MCRT) problem any problem that has a finite set P, finite family S of subsets of P, non-negative threshold h, and two non-negative real-valued functions y : (say, cost) and x :
p...-?
R+ (say, time). One seeks a solution F* E S with y(F*) min{y(F): F E S,x(F):S
h}, where x(G)L:gEGx(g), y(G) = L:gEGy(g)
and G E S. We also assume that for the corresponding minimum cost problem there is an efficient exact or approximate algorithm. We describe a very simple approximate algo- rithm for any MCRT problem. Though our algorithm is not polynomial in general, we provide some evidence that the algorithm may be fairly fast in many cases.1 Introduction
It is well known that there is a large variety of different real-world optimization problems (d. [2, 8, 14]). Thus, along with development of very effective algorithms solving some important optimization problems, it seems to be a right strategy to design algorithms for quite general optimization problems. In this paper, we propose
* Corresponding author.
a very UL>,~V.LLUL"LLH for a wide of different combinatorial V P U H l U L I < k U J . V H
Problems from this are often appear in both of combinatorial (d. [1,3, 4,5,6,9, 11, 13, our is not polynomial in (which seems to be
HLL/--,",-"HOJL'" to expect in such we some evidence that the
algorithm may be fast in many
We call a minimum cost restricted time combinatorial optimization
any n r { , h l p T r l that has a finite set finite of subsets of
threshold h, and two non-negative real-valued functions y time). One seeks a solution F* E x(F') h}, where x(G) x(g), y(G)
that for the C01Te,;po,ndmg is an efficient or approximate alj1(Oflthm"
assume that the last A well known ex<unPle
This n r { " , h l p r n
Others de vel-
o , L I " , \ J U l U H 1 U " may bound for the ratio of a found solution that ae]:>erlas on both instance of a MCRT problem and solution. It is difficult to a prac-
tical algorithm with performance ratio for any
MCRT an a ratio may better evaluate
ratio of the found solution than a "uniform" guaranteed one.
2
We first an informal description of our We have assumed that there is effective algorithm
A
for the minimum cost problem.A
we can find two solution F andH
such thaty(F) =
min{y(T):T S}
andx( Ii) min{ x(T): T E If
:s;
h, then F is an optimal solution. Ifx( H)
>
h, then there is no solution for our problem. we may assumethat
x(H) :s;
h< x(F).
The main idea of the algorithm is to applyA
to a "linear combination" of F and H in order to obtain a feasible solution which is better thanH.
RepresentF
andH
as points with coordinates(x(F),y(F))
and(x(H),y(H))
in a rectangular Cartesian coordinate system. We shall use the equation of the straight3
If x( F) h, then If x(H) h, then and c x(P)y(H) - x(H)y(F).
A find G so that w( 0)
=
.;:r.I"1T:H,n when
xC
G):s;
hand H is afor the
hand F := G,
of our <hLr.VL""U.J..ll
with coordinates and let
R
be another:s;
y(F):s;
y(R):s;
y(H) and ax(R)+
< c} where a, band c are the algorithm. Let G
I-
be the solution f01.md in 3 the algorithm F and H above and let the line that is parallel to the line F H and contains intersect the lines H Rand FR in the
M and respectively Then
x( H)
:s;
<
P:roof: Since M and N are and
:s;
x(R) inequalitiesx(H) :s; x(M),
x(M):s;
x(G):s;
x(N).x(G) y(G)
(1) (2) belonging to the segments [H R] and [F R], re- x(F), the x-coordinates of M and N satisfy the
:s;
x(F). G lies on the segment Therefore, (1) holds. Analogously, we can prove (2). D.y
F
x Fig. 1
Theorem 3.2 After every performance of Step 5 the value of x(F) +y(H) decreases by positive number.
Proof:
Suppose that Step 3 of the algorithm is performed T
+
1 times, T Let Fi , Hi and Gi be the current values of F, Hand G in the i-th performance of Step 3 (i 1,2, ... , T). Ri , Mi and Ni are defined recursively as follows. Rl denotes the point with coordinates X(Rl) X(Hl) and y(Rl) y(FI)' For i=
1,2, ... , T, Mi and Ni denote the points where the line that is parallel to the line FJ1i and contains Giintersects the lines Ht~ and FiRi. For i
=
2,3, ... , T, ~=
if X(Gi-l ) > hand Rt Nz-1 if x( Gi -I ) h.We first prove by induction on i that
x ( Hi) ::; X ( Mi) ::; X ( G i) ::; X ( Ni ) ~ x ( Fi ) ,
y(Fi) y(Ni ) ~ y(Gi ) ::; y(Mi) ~ y(Hi) for every i = 1,2, ... , T.
(3) (4)
The case i = 1 follows from Lemma 3.1 and the definition of
R
I . Let i 2: 2. Then (3, and (4) follow from the definition ofR
i ) induction hypothesis, and Lemma 3.1.To prove the theorem, it is sufficient to show that for every i
=
1,2, "0' Tif x(Gi )::::; h, then y(Gi )
<
y(Gj ), where j = max{k:-1:::;
k<
i,x(Gk )::; h}, (5)and
if h, where m
o k i,x(Ch)
h}, (6)j and m are defined in (5) and (6).
(4), y(G
i ), S-point if there C
An is minimal if there is no S-point (x",
y")
such that x"x"+-
y"x' +
y'. Let m(S) denote the number of minimal S-points.every solution obtained in 3, y( G)) a minimal from Theorem that after 5 we obtain distinct elements G of
1)
Step 3 of the algorithm is performed at most m( S)+
1 times.2) The algorithm
+
time} where is the timef:mmVI.f?T.:1.r1] of
If the functions x and y then we can obtain a bound the number of iterations of 3 to be still performed
terminated fitVhere
~"u"',uu~u"'"~""U of the ~""'.~L". VH""H
every ) each of the functions
be the current values of points P and H in an iter- Then the number of iterations of 3 to be still
- y(Fo),
V f J " H J . 1 U J solution of and G be a solution found
algorithm w.r.t G the performance ratio
fl.l.lIlJr".I.lI.""II. A find solutions. If our algorithm has
a Vf?1rtorTrUl/rl.f:f' Step
4
and G is the found solution} then r( 0) Sr(
G)y(
G) /(y
(G)+ ---~---c---:-""-:----.
Proof: Let the line G F intersect the line x h in the point B and let A have coordinates (h,
y(
G)). Then the triangle G BA contains all optimal solutions. Indeed, no feasible solution can be to the right of the line AB, no optimal solution can be above the line AG, and no solution can belong to a line parallel toGF
and "below"GF
such solutionG'
would haveax(G') + by(G')
< c).Hence, for an optimal solution
G*, y(B)
~y(G*).
Now it is easy to check thatr( G) y( G)/y(B) = r( G).
D.4 Computational
We have tested the behavior of our algorithm for the minimum restricted time
as~ngnrrlerlt problem: given two n X n-matrices C and T with non-
negative entries and a non-negative real h, find a permutation 7r E Sn so that
Ci1r(t) min{Lf=l Cicr(i) . a E Sn, h}, where Sn is the set of
all permutation on {I, 2, ... , n}. We generated 60 instances of the mini- mum cost constrained time assignment problem with n 6,8 and 10. The entries of matrices varied from 0 to 99. The value of h was calculated by the formula h = max{30n, nJ-Lj2}, where J-L is random integer from the interval [0,99]. For only five of the instances, Step 3 was not performed at all: two instances with n 6 did not have any solution, an instance with n = 8 and two instances with n 10 only needed 1 to find an optimal solution. Some results obtained for the rest 55 are reflected in Table 1: we show the number of instances with specified real ratio of the found solution (i.e.
r(
G)) and specified quality of the bound r(G) (i.e. q(G)=
r(G)jr(G)). Note that (for the 55 instances) the number of i, of Step 3 varied from 2 to 5 i = 5 only for two instances).Table 1: Number of instances with specified rand q.
n 1 1 < r ~ 1.1 T 1.1 q 1.1 1.1 < q 1.2 q > 1.2
6 14 3 1 11 7 0
8 8 8 3 9 7 3
10 7 7 4 16 2 0
We have also tested the behavior of our algorithm and Gvozdev's algorithm [3J for the two parameters minimum spanning tree (in complete graph) problem: given two symmetric n x n-matrices C [Cij] and T
=
[tij] with non-negative entries and a non-negative realh,
find a spanning treeF = (V, E)
in the complete graphKn
on vertices{1,2, ...
,n} such that Cij = min{LijEE1Cij:E'
E E,LijEE,tij ~h},
where E is the family of all edge sets of spanning trees in Kn.The matrices C and T were generated analogously to the corresponding matrices in the assignment problem above. They have orders n=20,30,40,50 and 60. The value of h was calculated by the formula h = max{25n, O.4nfL}, where J-L is a random integer from the interval [0,99].
Table 2 reflects, in a sense, the quality of solutions found by our algorithm. Not knowing optimal solutions in most of cases, we measure the quality by the bound if (see Proposition 3.5). One can see that the quality of the solutions steadily increases while the order n grows.
Table Number of instances with
Table 3 shows some data which can be compare our algorithm with In Table 3, ml is the number of instances when the costs of solutions found by our algorithm and Gvozdev's coincide, m2
(m3,
is the number when solutions obtained by our v"LF,~.LLULCLLLtime Gvozdev's ULF,V.LLUU.H.L.
siderably more time than our COL!',VL."H.UH
~), in Gvozdev's algorithm (before time COJD.S11IDLln;g.
Table 3: Comparison of our
about half of requires con- C01TI]::mtatllon of the parameters .:\ and
iterations the procedure) is quite
and Gvozdev's algorithm.
~3
!---+~~---
20 17 1 2 7.2 9 13.4 30 10 6 4 8.7 10 16.2 40 6 7 7 9.6 11 18.5 50 6 6 8 10.3 11 21.2 60 3 12 5 10.9 12 22.0 70 3 14 3 11.0 12 22.3
~--~--- ---~
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(Received 23/11/95; revised 16/5/96)