SOLVING COMBINATORIAL PROBLEMS as formulations of Linear Complementarity Problemsa Giacomo Patrizi
[email protected]
Dipartimento di Scienze Statistiche Sapienza Universita` di Roma, Italy a
Joint work with Laura Di Giacomo
Paper presented at C.O.R.E. (Center for Operations Research and Economerics), Universite` Catholique de Louvain, Belgium, 23 Feb. 2009
LCP algorithm and applications – p.1/43
Introduction
Aim:
• • •
Present algorithms to solve general LCPs, Solve different computational problem types, Theoretical Developments and convergence results.
The Linear Complementarity Problem is a prototype algorithm for the solution of many application problems.
• • • •
Combinatorial problems.
•
Applicable as quadratic subproblem in sequential optimization.
Bimatrix Games, Market Equilibrium, Engineering Problems, Optimal stopping, classification, Statistical Problems, LCP is a statement of the KKT conditions for a quadratic program with linear inequality constraints
LCP algorithm and applications – p.2/43
Outline •
Define the LCP and parametric solution technique
• • • •
•
a
Polynomial time solutions cannot be guaranteed because a parametric linear programming problem in one scalar variable must be solved. Computational Results for general problem classes: • Satisfiability Problems, • Knapsack Problems, • Emergency Service Problems • Quadratic problems with linear inequality constraints
Theoretical developments and convergence results for certain general combinatorial problems solved in polynomial time.
• • • •
Many LCP solution techniques are solvable limited to particular matrix classesa A general parametric solution technique is presented,
Knapsack Problems, Subset Problems
Research Directions Derivations Conclusions Cottle, R. W., J.-S. Pang, R. Stone,The Linear Complementarity Problem, Academic Press, Boston,
1992
LCP algorithm and applications – p.3/43
The Linear Complementarity Problem Given M ∈ Rn×n , x, q ∈ Rn . Consider LCP (1)
Mx + q
≥
0
(2)
x
≥
0
(3)
xT (M x + q)
=
0
Consider the following slack LCP formulated in a straightforward way from the LCP (1 3):
1 a
(4)
0 M
!
x0 x
!
(5)
(6)
“
x0 , x T
”
1 a
0 M
!
x0 x
!
+
+
0 q
!
≥
0
!
≥
0
=
0
x0 x !! 0 q
LCP algorithm and applications – p.4/43
The Linear Complementarity Problem (cont.)
Lemma 1:(Mangasarian 1978) Let a be any vector. If x solves the LCP (1 - 3) then (xo , xT ) solves the LCP (4 - 6) then xo = 0 and moreover x solves the LCP (1 - 3)a
•
LCP be not trivial and feasible, if: (7)
•
Define a positive vector a ∈ Rn to embed LCP in an augmented space: (8)
• •
a
0 < β = inf {eT x | M x + q ≥ 0, x ≥ 0} < ∞
ai = di − M in{0, Mij ,
qi | Mij ∈ Mi,. } β
∀i, j = 1, 2, ..., n
di is an arbitrary (small) parameter ∀i ∈ N Select a parameter α as an upper bound to the sum of the elements of the vector x, i.e. α > eT x. Mangasarian, O. L., Characterization of linear complementarity problems as linear programs, Mathe-
matical Programming Study 7, North-Holland, Amsterdam, pp.74-87
LCP algorithm and applications – p.5/43
Parametric LCP
Define an affine transformation of variables and then use this on the LCP u0 u
(9)
!
=
1 0
−eT
eT
!
Inn
!
!
x0 x
+
!
α 0
To get: 1 a
(10)
M + aeT
u0 u
!
+
(11)
(12)
“
u0 , u T
”
1 a
eT M + aeT
!
u0 u
!
+
−α q − aα
!
≥
0
!
≥
0
=
0
u0 u !!
−α q − aα
which is the form to be solved.
LCP algorithm and applications – p.6/43
Parametric LCP solution method •
Set: A = (M + aeT )
(13)
•
Take sT as the positive eigenvector of matrix A, the cost coefficients of the objective function is: cT = s T A
(14)
•
The LP to solve is: c T u + cT z + c 0 u 0
=
M inZ
Au − (I − A)z + au0 + q − aα
≥
0
(17)
eT u + eT z + u0 − α
=
0
(18)
u0 , u, z
≥
0
(15) (16)
•
s.t.
The LP to be solved with parametric variation of c0 a
• •
a
s>0
General solution method to LCPs. LP with parametric variation of a scalar variable is not provable to be solvable in polynomial time.
Patrizi, G., The Equivalence of an LCP to a parametric Linear Program with a Scalar Parameter,
E.J.O.R., 51(1991), p.367-386
LCP algorithm and applications – p.7/43
Parametric LCP solution method (cont.2)
Theorem 1 A nontrivial and feasible LCP has a solution x∗ such that the sum of the solution elements is equal to or less than a given parameter, i.e. α ≥ eT x∗ if and only if the parametric linear programming problem with a scalar parameter c0 , given by: cT u + c T z + c 0 u 0
=
M inZ
Au − (I − A)z + au0 + q − aα
≥
0
(21)
eT u + eT z + u0 − α
=
0
(22)
u0 , u, z
≥
0
(19) (20)
s.t.
has a dual nondegenerate solution for some c0 with all the elements of the vector z non basic in the primal solution. A is defined as in equation (13), assuming that T T ˆ Aj,J A−1 J,I ≥ 0 for some j ∈ J. Also c is defined as in equation (14) with s the left positive eigenvector corresponding to the largest eigenvalue of the positive matrix A. From the primal optimal solution (u∗ , 0, u∗0 ) to (15) - (18) which satisfies the conditions, a solution to the original LCP is obtained immediatly by placing x∗ = u∗ .
LCP algorithm and applications – p.8/43
Parametric LCP solution method ( cont. 3) •
Suppose there is an optimal solution for (15) - (18) for a given c0 the solution is dual nondegenerate and the auxilliary variables z are all nonbasic.
•
The primal-dual relationship in the form Min-Min may be stated for the optimal solution as follows z≥0
(I − AT )y − ey0 + c ≥ 0
u≥0
−AT y − ey0 + c ≥ 0
u0 ≥
−aT y − y0 + c0 ≥ 0
w = Au − (I − A)z + au0 + (q − aα) ≥ 0 eT u + eT z + u0 − α = 0 M inZ = cT u + cT z + c0 u0
•
y≥0 y0
f ree
M inW = (q − aα)T y − αy
The results follow immediatly. For,
• • •
if uk > 0 then −AT k y − y 0 + ck = 0 as zk = 0 so yk > 0 hence wk = 0 and thus uk wk = 0
LCP algorithm and applications – p.9/43
Combinatorial Problems
• •
Discrete structures, render Mathematical Analysis difficult to apply.
•
Definition 1 A linear combinatorial optimization problem is a combinatorial optimization problem, limited to variables which can take on only binary values (0,1) defined over a polyhedral constraint set and selected through a linear objective function.
•
Let c ∈ Rn b ∈ Rm be vectors and C a matrix of order m×n then the linear combinatorial optimization problem may be defined as:
Embedding combinatorial problems as continuous variables allows to determine their solution.
(23)
M ax[cT x : Cx + b
≥
0]
(24)
x
∈
{0, 1}n
LCP algorithm and applications – p.10/43
Combinatorial Problems (cont.2)
Any linear optimization combinatorial problem can be represented by an LCP as:
0
−I B @ C cT
(25)
0 0 0
(26)
(27)
“
xT
ΓT
µ
”T
00
−I BB @@ C cT
0 0 0
10
1
0
1
e x 0 C C B CB 0 A@ Γ A + @ b A −k µ 0 1 0 x C B @ Γ A µ 11 1 0 10 e x 0 CC C B CB 0 A @ Γ A + @ b AA −k µ 0
≥
0
≥
0
=
0
Top row: (28)
xT (x − e) = 0
LCP algorithm and applications – p.11/43
Numerical Results ( Combinatorial Problems (cont. 3))
Theorem 2 a The linear combinatorial optimization problem has an optimal solution with a value of the objective function equal to k∗ , if the linear complementarity problem has a solution with the given parameter k = k∗ and the variables Γ and µ equal to zero. Conversely, let the linear complementarity problem have a solution with Γ and µ equal to ˆ This will be a solution to the linear combinatorial zero and a given value of k = k. ˆ which solves the linear complementarity optimization problem, if there is no k > k problem. Solve the LP by dichotomous search on k, for each recursion:
• •
a
for general combinatorial problems the parametric LP may be used, for some classes of combinatorial problems one LP will solve the problem if the conditions on the LP are satisfied, (see Solving LCP and Combinatorial Problems in polynomial time.) Di Giacomo, L., G. Patrizi,E. Argento, Linear Complementarity as a General Solution Method to Com-
binatorial Problems, Journal of Computing, 19(2007), 1, 73 - 79
LCP algorithm and applications – p.12/43
Numerical Results ( Satisfiability Problems in Normal Conjunctive Form ) •
A satisfiability problem in normal conjunctive form is represented as the intersection of a set of M clauses, where each clause is the union of a set of literals, which may receive the value of true (1) or false (0).
•
Representing these literals in a clause as boolean variables, each literal can be indicated in the affermative form by xi , i= 1, 2, ..., n or in the negated form as (1 − xi ).
•
Thus without loss of generality, any clause k ∈ M may be represented with the index set of affermative litterals indicated by Ik and the index set of negated litterals, indicated by Jk and for it to be true, it must receive values of the boolean variable such that the following inequality holds:
(29)
X
i∈Ik
xi +
X
(1 − xj ) ≥ 1
j∈Jk
LCP algorithm and applications – p.13/43
Numerical Results ( Satisfiability Problems in Normal Conjunctive Form (cont.2 )) All clauses must be true for the expression to be true, so write:
(30)
Cx + b
≥
0
(31)
xi
∈
{0, 1}
(32)
−1
≤
bk ≤ Jk − 1, k = 1, ..., M.
The LCP formulation has no objective function in these problems:. M =
(33)
−I C
0 0
!
q=
e b
!
Results are given:a . a
Mayer, J., I. Mitterreiter, F. J. Radermacher, Running Time experiments on some Algorithms for Solving
Propositional Satisfiability Problems, Annals of Operations Research,55(1995), 139 - 178
LCP algorithm and applications – p.14/43
Numerical Results ( Satisfiability Problems in Normal Conjunctive Form (cont.3) )
Table 1: Name
Selected Results of the solution of Satisfiability Problems: time in secs. Litterals
Clauses
GLPK
XPRESS
FORTMP
LCP
ulm027r0
25
64
5.00
2.00
3.00
0.53
ulm054r1
46
128
6.00
4.00
3.00
2.72
ulm081r1
92
256
5.00
3.00
2.00
4.12
ulm189r1
161
448
7.00
4.00
4.00
145.49
ulmbc040
40
340
6.00
3.00
4.00
2.96
ulmbp160*
160
485
%
4.00
63.00
5762.73
ulmbs180
180
540
%
5.00
18.00
9558.82
% solution not feasible, * instance is logical contradiction
LCP algorithm and applications – p.15/43
Numerical Results ( Satisfiability Problems in Normal Conjunctive Form 6 )
Table 2: Name
Selected Results of the solution of Satisfiability Problems: time in secs. Litterals
Clauses
GLPK
XPRESS
FORTMP
LCP
real2b12
15
111
6.00
4.00
2.00
0.21
real2l12
15
119
6.00
4.00
2.00
0.17
real2q11
16
100
5.00
4.00
3.00
0.24
real2x11
14
77
6.00
4.00
3.00
0.15
hole8*
72
297
150.00
4.00
27.00
518.06
hole9*
90
415
%
6.00
193.00
5770.87
hole10*
110
561
%
3.00
301.00
19505.87
% solution not feasible, * instance is logical contradiction
LCP algorithm and applications – p.16/43
Numerical Results (Knapsack Problems )
•
consider a knapsack problem: M ax{cT x s.t.
(34)
•
dT x ≤ b |x ∈ {0, 1}}
Problems of various sizes were generated by sampling the value of an object and its encumbrance from different random probability distributions:
• •
uniform distribution geometric distribution has density function fn = p(1 − p)n−1 . Mean is µ=
• • •
1 , p
Variance is µ2 =
1−(1−p)2 p3
for the chosen p.
For low values p distribution is similar to a uniform one, for higher values of p it peaks and values are more bunched, The time to solve the problems in the LCP algorithm does not depend on their structure but purely on their size and compare this with the times of the other routines.
LCP algorithm and applications – p.17/43
Numerical Results (Knapsack Problems 2 )
Table 3:
Mean of 5 Knapsack Problems, various routines and distributions ( secs.)
Distribution
size
GLPK
XPRESS
FORTMP
LCP
Uniform
100
6.00
5.00
3.40
23.53
200
6.25
5.25
4.75
64.64
500
9.50
3.50
3.50
390.17
Geometric p=0.4
100
222.30
6.20
7.20
22.24
Geometric p=0.5
100
7320.00
5.00
12.33
24.18
Geometric p=0.6
100
%
744.20
11.20
22.82
Geometric p=0.8
100
%
379.20
%
25.41
200
%
240.00
%
85.09
100
%
3607.00
%
24.80
200
%
945.00
%
72.06
Geometric p=0.9
% not solved
LCP algorithm and applications – p.18/43
Numerical Results (The Emergency Service Problem)
•
•
Emergency service problem:
•
Distribute optimally over a territory a limited number of services to ensure a given service level.
•
Locations often of two types of points: service points and facility points.
Index sets of demand nodes( I ), service nodes ( J), facility nodes ( J’ ) and time period considered ( T ).
• • •
yikt = 1 if demand node i ∈ I is not covered by k services at time t; zht = 1 if a facility node is covered at h ∈ J ′ at time t; xjt ∈ N : number of services assigned to node j ∈ J ∪ J ′ at time t. • Here integer expressed as an expansion in binary form.
•
In the table the column headings are: ’dd’ number of demand nodes, ’Ser’ number of service nodes, ’Pds’ number of periods,’Veh’ number of vehicles available,’Cplex’ runtime, ’LCP’ this routine.
•
The routines behave properly if there are relative abundant number of emergency vehicles for a given demand and behave very poorly if the vehicles must be be used intensily for the emergencies, while the LCP algorithm is stable as before.
LCP algorithm and applications – p.19/43
Numerical Results ( The Emergency Service Problem 2 )
bit T X N X X
Q(pt , qt , (k − 1))(1 − qt )qtk−1 ait yikt
=
M inZ
≥
bit
xjt
≤
pt
xjt − ct zht X zht
≤
0
≤
lt
(40)
xjt
∈
integer
(41)
yikt
∈
{0, 1}
(42)
zht
∈
{0, 1}
(35)
i=1 t=1 k=1
(36)
X
j∈Mit
(37)
xjt +
bit X
yikt
k∈
X
j∈J
(38) (39)
h∈J ′
LCP algorithm and applications – p.20/43
Numerical Results ( The Emergency Service Problem 4 )
Table 4:
Solution Emergency vehicle problem
Name
dd.
Ser
Pds
Veh
Ceplx∗
LCP
Veh
Ceplx∗
LCP
Mirchandani
10
10
6
2.67+
7.77
0.77
..
..
..
Torregas
30
30
3
7.67+
5.55
62.96
..
..
..
Florence1
30
30
6
40
8.88
585.40
8
30.8
318.91
Florence2
80
80
2
40
11.00
397.95
8
21567
319.91
Florence3
95
95
2
40
13.20
742.39
8
Rome1
100
100
2
40
16.5
242.77
8
Rome2
120
120
1
40
8.8
457.49
8
..
%
350.33
Rome3
220
220
1
40
33.0
1825.16
8
..
%
1891.96
Rome4
150
150
1
40
16.5
507.94
8
..
%
488.14
Rome5
200
200
1
40
27.50
2229.12
8
..
%
1443.40
∗
..
%
116.6
678.26 761.91
Machine equivalent times in secs.( Dongarra) ratio cplex/Lcp= 110 + Average of emergency vehicles available over the periods. % not solved in 3000 secs of actual machine time. LCP algorithm and applications – p.21/43
Numerical Results ( continuous Problems)
• •
Behaviour of algorithm with general LCPs
• • • •
Class (1) should be easily solved,
•
The run times are stable to solve the LCPs whatever their structure. Only their size is relevent to the run times, which indicates the correctness and reliability of this algorithm.
a
Problems generated randomly with coefficient matrices, feasible set of inequalities, for unit vector and variables non negative : 1. Positive definite or positive semi definite matrices, 2. Negative definite or negative semi definite matrices, 3. Indefinite matrices Class (2) should be solved only by enumeration methods, Class (3) may be solved sporadically, The results that follow imply that the classes Q, Qo , P, Po do not discriminate the solvability in these classes of LCP, if the described algorithm is applieda
Cottle, R., W., et al. chp. 3
LCP algorithm and applications – p.22/43
Numerical Results ( continuous Problems) (cont. 2)
Table 5: Size Type
Results for the solution of 140 random instances of LCPs
30
50
100
250
No
av.sec.
No
av.sec.
No
av.sec.
No
av.sec.
Total No
Positive
6
0.06
3
0.28
6
3.47
5
109.37
20
Negative
12
0.08
3
0.38
6
7.00
7 (4)
121.51
32
Indefinite
28
0.07
26(3)
0.33
16
5.18
15
111.99
88
total
46
..
35
..
28
..
31
..
140
LCP algorithm and applications – p.23/43
LCP and Combinatorial Problems in Polynomial time Consider the following theorem: Theorem 3 a The LCP (48)-(51) has a solution if and only if the LP:
(43)
M in
(r T + sT M )x
s.t.
M x + q ≥ 0, x≥0
is solvable for some nonnegative vectors h, r, s and Z-matrices Z 1 , Z 2 , which must satisfy the following conditions : (44)
(a)
M Z 1 = Z 2 + qhT
(45)
(b)
r T Z 1 + sT Z 2 ≥ 0
(46)
(c)
r T Z 1 + sT Z 2 + h T > 0
(47)
(d)
r+s>0
Furthermore, each solution of the LP solves the LCP. a
Mangasarian, O., L., Simplified Characterizations of Linear complementarity Problems Solvable as
Linear Programs, Maths. O.R., 4(1979), 3, pp. 268 - 273
LCP algorithm and applications – p.24/43
Knapsack Problem: Theoretical Development •
Consider the knapsack problem:
(48)
Maximise
V
=
n X
pˆj x ˆj
n X
w ˆj x ˆj
j=1
(49)
s.t.
b
≥
j=1
(50)
b
≥
w ˆj x ˆj
j = 1, 2, · · · , n
(51)
x ˆj
∈
{0, 1}
j = 1, 2, · · · , n
LCP algorithm and applications – p.25/43
Knapsack Problem: Theoretical Development (cont.1)
•
Express KP as a system of inequalities: (52)
ˆ pˆT x ˆ−k
≥
0
(53)
−w ˆT x ˆ+b
≥
0
(54)
0
≤
x ˆ ≤ e {Binary}
• •
with x ˆ has x ˆj = {0, 1}, pˆ = {ˆ pj }, w ˆ = {w ˆj },
•
Let in (53) for α > 0, pj = αˆ pj , , wj =
•
No change in feasibility or optimality will occur
j = 1, 2, · · · , n.
KP as augmented LCP with two artificial variables γ1 ≥ 0, γ2 ≥ 0, and let x = Diag(p)ˆ x. w ˆj pj
k′ the LCP has no solution, then the solution of the LCP with the parameter k = k′ is an optimal solution to the KP and ˆ∗ = k′ , the solution to the knapsack problem. k α
Corollary 1 a . The optimal solution to the KP (48) - (51) may be determined by solving at most ⌈log2 (pT e)⌉ linear complementarity problems by a dichotomous search algorithm. a
Nemhauser, G., and L. Wolsey, Integer and Combinatorial Optimization, Wiley, New York, 1988
LCP algorithm and applications – p.28/43
Knapsack Problem: Theoretical Development (cont.4)
Consider the LCP (55)-(57) and rewite the coefficient matrix in a modified form. The solutions to the problem will not be altered, if the conditions γ1 = γ2 = 0 given by theorem 4 hold.
0 B @
(58)
−I eT −wT
0 −1 −1
(59)
(60)
“
xT , γ1 , γ2
”T
00
−I BB @@ eT −wT
0 −1 −1
10
1
0
1
p x 0 C C B CB 0 A @ γ1 A + @ −k A b γ2 1 1 0 x C B @ γ1 A γ2 11 1 0 10 p x 0 CC C B CB 0 A @ γ1 A + @ −k AA b γ2 1
≥
0
≥
0
=
0
LCP algorithm and applications – p.29/43
Knapsack Problem: Theoretical Development (cont.5)
For M , Z1 , Z2 the relation is respectively:
(61)
0
−I B @ eT −wT
0 −1 −1
Let σ = Diag(ρ)e with ρ =
10
−I 0 CB −eT 0 A@ −(w + e)T 1 1 2pj
0 −1 −1
1
0
I 0 C B 0 A=@ 0 0 1
0 1 0
1
0 C 0 A 1
j = 1, 2, · · · , n
(62)
(s1 )T
=
(s2 eT − s3 wjT + σ T ) > 0,
(63)
s2
=
s3 =
∀j = 1, 2, · · · , n
1 5maxj p
LCP algorithm and applications – p.30/43
Knapsack Problem: Theoretical Development (cont.6)
The LP can be specified as: (64)
M inW =
subject to:
(65)
(66)
0 B @
−I eT −wT
“ ” T −s1 + s2 e − s3 w x − (s2 + s3 )γ1 + s3 γ2
0 −1 −1
10
1
0
1
p x 0 C C B CB 0 A @ γ1 A + @ −k A b γ2 1 “ ” x , γ1 , γ 2
≥
0
≥
0
LCP algorithm and applications – p.31/43
Knapsack Problem: Theoretical Development (cont. 7)
Theorem 5 The following are equivalent: (i) The KP (48)-(51) has an optimal solution x ˆ∗ , V ∗ = k ∗ ,
•
Let the KP have a solution, then the LCP (55)-(57) x ˆ∗ has a complementarity solution: • The LCP has a complementarity solution with x = Diag(p)ˆx given x ˆ∗ = 0, 1 as xT (p − x) = 0, • the inequalities of the LCP are satisified. (ii) The LCP (55)-(57) has a solution with γ1 = 0, γ2 = 0 • By dichotomous research search algorithm the optimal knapsack function value will be determined in less than or equal to ⌈log2 (pT e)⌉. If the selected k is greater than eT x the LCP will not have a solution, while if k ≤ eT x then there is a complementarity solution to this LCP and by dichotomous research search the value k can be increased until k = eT x. • Again the LCP has a complementarity solution with x = Diag(p)ˆx given x ˆ∗ = 0, 1 as xT (p − x) = 0 • γ1 = 0, as eT x = k by dichotomous research algorithm and γ2 = 0 as b ≥ wT x so any positive value of γ2 would render a non complementarity solution.
LCP algorithm and applications – p.32/43
Knapsack Problem: Theoretical Development (cont. 8)
(iii) The LCP (58)-(60) has a solution with γ1 = 0, γ2 = 0
•
The objective function value of the LP (64)-(66) given by the solution of the P xj k∗ LCP (55)-(57) will be : W 1 = n − = − . j=1 2p 2 j
•
If k ≥ eT x the value of the objective function for this solution of the LP is W 2 > W 1 . Further if γ1 = 0 and γ2 ≥ 0 again the value of the objective function is W 3 ≥ W 2 > W 1
(iv) The LP (64)-(66) has an optimal solution x ≥ 0.
•
Consider the optimal solution of the LP (64)-(66) indicated with the value of the objective function as W 1 This has the maximum value of the knapsack problem given by V ∗ = eT x and the knapsack contains the following x elements x ˆ∗j = pj = {0, 1} binary. j
¤
LCP algorithm and applications – p.33/43
Knapsack Problem: Theoretical Development (cont. 8)
Corollary 2 A given KP is solvable by a linear program, given by the LP (64)-(66) for some value of k determined by the dichotomous search procedurea . ¤ a
Nemhauser, G., and L. Wolsey, Integer and Combinatorial Optimization, Wiley, New York, 1988
LCP algorithm and applications – p.34/43
Subset sum Problem (SSP)
Definition 2 The SSP is given by the equation:
b=
(67)
n X
a ˆj x ˆj ,
xj ∈ 0, 1
j=1
where a ˆj > 0 and b
0,
0 1 0
1
0 C 0 A 1
∀j = 1, 2, · · · , n
(s1 )j 2 s3 = 1
LCP algorithm and applications – p.37/43
Subset sum Problem (SSP) (cont.4)
For some suitable value of ρ, s2 and s3 then the following LP based on the LCP can be solved. (76)
M inW =
„
eT T T T r 1 − s 1 + s2 e − s 3 2
«
x − (s2 + s3 )γ1 + s3 γ2
subject to:
(77)
(78)
0 B @
−I T
e 3 T − e2
0 −1 −1
0
10
1
0
1
a x C B CB C 0 A @ γ1 A + @ − 3b A b γ2 1 2 “ ” x , γ1 , γ 2
≥
0
≥
0
LCP algorithm and applications – p.38/43
Subset sum Problem (SSP) (cont.5) Theorem 6 The following are equivalent: (i) Let the SSP (67) have a solution x ˆ∗ = {0, 1},
•
Let the SSP have a solution, then the LCP (68)-(70) x ˆ∗ has a complementarity solution with γ1 = 0, γ2 = 0
•
The SSP has a complementarity solution with x = Diag(a)ˆ x∗ given x ˆ∗ = {0, 1} as xT (a − x) = 0,
•
the inequalities of the LCP are satisified so the LCP (??)-(70) has a solution and γ1 , γ2 = 0,
(ii) The LP (76)-(78) has an optimal solution x ≥ 0 such that γ1 , γ2 = 0,
•
The objective function value of the LP (76)-(78) given by the solution of the LCP (68)-(70) will be : W 1 = 78 eT x
•
If γ1 > 0 then γ2 ≥ 0 and the value of the objective function is W 2 > W 1 as γ1 =
eT x−b 3
≥ 0 which will require γ2 =
5(eT x−b) 6
≥0
(iii) The LP (64)-(66) has an optimal solution x ≥ 0.
•
Consider the optimal solution of the LP (76)-(78) indicated with the value of the objective function as W 1 This has the maximum value of the SSP problem x and the following elements x ˆ∗j = aj = {0, 1} binary will be in the required set. j
¤
LCP algorithm and applications – p.39/43
Computational Complexity Analysis The transformations indicated above are polynomially related to the size of the KP or SSP
• • •
Let C be the KP or SSP original matrices all of order 2 × n, P P ˆj the affine vector is of order 2 and assume w.l.g. k ≤ j pˆj and b ≤ j w the size of the objective function is at most:
(79)
0 1 X @ size(c) = 2(n + 2) + 1 + ⌈log2 (|pj | + 1)⌉A j
•
So the size of the LCP is: X X ⌈log2 (|ˆ pj |+1)⌉) (⌈log2 (|α|)+2( size(c, M, q) ≤ 6(n+2)+size(C, d)+2(n+1) j
j
(80)
•
After some manipulation there results:
(81)
size(c, M, q)
