An Enumerative Approach for Multilevel
Quadratic Integer Programming problems
Abstract: An algorithm is proposed in this paper for solving multilevel quadratic integer programming problems with linear constraints. A convergent algorithm based on Stackelberg strategy is employed to solve (MQIP) problem which does not increase the complexities of the problem under consideration for a given choice of the variables under the control of the upper level decision maker. The problem is solved considering the relationship between feasible solutions to the problem and bases of the coefficient submatrix associated to variables of lower levels. An enumerative algorithm is proposed which finds a global solution to the problem.
Keywords: Quasi concave functions, Non-convex optimization, Multilevel programming, Enumerative method.
1. INTRODUCTION
Multilevel Integer Quadratic Programming (MIQP) problems involve sequential or multistage decision making. In this paper, we consider the Multilevel Quadratic Programming Problem (MQPP) as
(MQPP) 1
1 11 12
max ( ) ( ) ( )
X Z X Z X Z X
11 1 11 2 13 3 1 12 1 12 2 12 3 1
(c X d X e X )(c X d X e X )
where X2 solves
2
2 21 22
max ( ) ( ) ( )
X Z X Z X Z X
21 1 21 2 23 3 2 22 1 22 2 21 3 2
(c X d X e X )(c X d X e X )
for a given X1
where X3 solves
3
3 31 32
max ( ) ( ) ( )
X Z X Z X Z X
31 1 31 2 31 3 3 32 1 32 3 32 3 3
(c X d X e X )(c X d X e X )
for given X1, X2 subject to X (X1, X2, X3)S
where 1 2 3
1 , 2 and 3
n
n n
X R X R X R are the variables controlled by the first level, second level and third level respectively.
An (MLPP) is mathematically formulated as
(MLPP)
1
1 1
max ( ,..., K)
X Z X X
2
2 1
max ( ,..., K)
X Z X X
1
max ( ,..., )
K K K
X Z X X
subject to X (X X1, 2,...,XK)S
where Zj F Xj( 1,...,XK); j1, 2,..,K be non-linear with respect to X X1, 2,...,XK where
1 1 2
1 ( 11, 12,..., 1n),...., K ( K, K ,..., Knk)
X x x x X x x x ; S is a convex feasible set.
1. The system has interacting decision making units within a hierarchical structure.
2. Each subordinate level performs its policies after knowing completely the decisions of superior levels. 3. Each unit maximizes net benefits independently of other units, but may be influenced by actions and
reactions of those units.
4. The MLPP problem considered in this paper has K decision makers located at K different hierarchical levels, each independently controls a set of decision variables and each DM has q q( 2) objective functions at each level. In this paper we focus on optimizing a quasiconvex function.
A number of algorithms have been proposed to minimize concave functions based upto the fact that concave functions have an extreme point optimal solution. The most common among them are branch and bound methods [7, 15, 17], cutting plane method [2, 8, 9, 19].
A bibliography of references on bilevel and multilevel programming problems in both linear and non-linear cases can be found in [18] K. Mathur and M.C. Puri [11] developed an algorithm for ranking the extreme points. In 1999 [4], an enumerative algorithm was proposed by Calvete and Gale that finds a global optimal solution to the bilevel programming problem. Considering the relationship between points of the inducible region and bases of the coefficient submatrix associated to variables of the second level, an algorithm is proposed here that finds a global optimum solution to the MLPP in a finite number of steps.
2. MATHEMATICAL FORMULATION OF PROBLEM
In this paper, a three level programming problem in which the government's objectives are atleast in partial conflict with the two sectors viz. industry and consumers, the policy makers face an optimization 'problem subject to the optimization problem for industries as well as the consumers. The problem is mathematically stated as :
1
1 11 12
max ( ) ( ) ( )
X Z X Z X Z X
11 1 11 2 11 3 1 12 2 12 3 12 3 1
(c X d X e X )(c X d X e X )
where X2 solves
2
2 21 22
max ( ) ( ) ( )
X Z X Z X Z X
21 1 21 2 21 3 2 22 1 22 2 22 3 2
(c X d X c X )(c X d X e X )
for a given X1
where X2 solves
3 3 31 32
max ( ) ( ) ( )
X Z X Z X Z X
31 1 31 2 31 3 3 32 1 32 2 32 3 3
(c X d X e X )(c X d X e X )
Subject to A X1 1A X2 2A X3 3 b
1, 2, 3 0
X X X and integers.
1 2
1 , 2
n n
X R X R and 3
3 n
X R are the variables controlled by the leader and the first and second follower respectively.
1
11, 12, 21, 22, 31, 32 n
c c c c c c R ; 2
12, 22, 12, 22, 31, 32 n
d d d d d d R ;
3
12, 22, 12, 22, 31, 32 n
e e e e e e R ;
1 2 3 1 2 2R
; 3
1 2
1 , 2 , 3 ,
m n m n m n m
A R A R A R b R .
where S = (X X1, 2,X3) : A X1 1A X2 2A X3 3b X X, 1, 2,X30} is non-empty and compact, A3 has full row
rank and mn2.
The projection of S onto Rn1 is denoted by
1
1{ 1 : ( 1, 2, 3) } n
The projection of S onto Rn1n2 is denoted by
1 2
2 {( 1, 2) : ( 1, 2, 3) } n n
S X X R X X X S
For each X1S1 the feasible region of first follower’s problem is denoted by
2 3
1 2 3 2 2 3 3 1 1 2 3
( ) {( , ) n n : , , 0}
S X X X R A X A X b A X X X
For each X X1, 2Si the feasible region of the second followers' problems is denoted by
3
1 2 3 3 3 1 1 2 2 3
( , ) { n : , 0}
S X X X R A X b A X A X X
Let V and V1 2 be the sets of indices of first level and second level controlled variables respectively.
Also, the optimal solution set to the first and the second followers’ problem is assumed to be singleton.
The inducible region of the relaxed leader’s problem is given by
2 3 1 2 3
{( , , ) : 0; ,
IR X X X X X X solve
2 3
21 1 21 2 21 3 2 22 1 22 2 21 3 1
,
max ( )( )
X X
c X d X e X c X d X e X s.t. A X1 1A X2 2 A X3 3 b X; 2 0, X3 0
The inducible region of the first follower's problem is
1 2 3 2 3 3
1 ( , , ); 0, 0;
IR X X X X X X solves
3
31 1 31 2 31 3 3 32 1 32 2 32 3 3
max ( )( )
X c X d X e X c X d X e X
s.t. A X1 1A X2 2A X3 3b X, 30
The following lemma is proved:Lemma: The optimal solution to the three level programming problem (TPP) occurs at an extreme point of S, provided S is regular.
Proof: Since Zi1( )X and Zi2( )X are positive X S and i = 1, 2, 3 therefore Zj(X) is both quasiconcave and
quasiconvex on S for i = 1, 2, 3.
Hence, the objective function at each level of (TPP) is quasiconcave in nature so that optimal solution occurs at an extreme point.
Note: (TPP) is obtained from (TIPP) by relaxing the condition that X1, X2, X3 are integers. 3. ALGORITHMIC DEVELOPMENT
In order to obtain optimal integer solution to (TIPP), Gomory’s cut is applied.
For each X1S1, a point of (IR1) is obtained by solving.
QPP(X1): 2
21 2 21 3 1 22 2 22 3 1
max( )( )
X d X e X d X e X
Subject to (X2,X3)S X( 1)
Where 1 c X21 12 and 2 c X22 12. Also, a point of (IR2) is obtained by
1 2
QPP(X X, ) : 3
31 3 3 32 3 3
max ( )( )
Subject to X3S X X( 1, 2);
where 3 c X31 1d X31 23 and 3 c X32 1d X32 23
An extreme point solution X3 of. S X X( 1, 2) can be found that solves QPP X X( 1, 2) and the solution so
obtained (X X1, 2, X3) belongs to (IR2).
Further, a basis B of A3 can be associated to each point of (IR1) and (IR2) because a basis B of A3 is associated to X3. Hence only this basis needs to be considered.
In order to determine the points of the inducible region (IR1) and (IR2), there must exist X1S1 and
1 2 2
(X , X )S such that B is also a feasible basis of QPP X( 1, X2).
Let X3B represent the variables of X3 associated with the basis B.
i.e. 3 1( 1 1 2 2)
B
X B b A X A X
10, 20
X X ,
1
1 1 2 2
( ) 0
B b A X A X .
Let V1, V2 and V3 be the set of indices controlled by the first level, second level and third level problems respectively.
Before a new iteration begins, check the optimality condition (OC):
(OC): 31( 4 32) 32( 3 31) ( 3 31)( 4 32) 0 3
j j j j j j j j
B B B B
Z z e Z z e z e z e j V
Where 31 j
e and 32 j
e are the jth components of vectors e31 and e32 respectively and 31 and 32
B B
e e are m-row
vectors of e31 and e32 associated to the basic variables of B;
3 j B
Z and Z4jB are the jth components of Z3B and Z4Brespectively;
The point of IR corresponding to the basis B is obtained by solving the indefinite quadratic programming problem.
QPP(B)
1
11 1 11 2 11 3 1 12 1 12 2 12 3 1
max( )( )
X
c X d X e X c X d X e X
Subject to
1 1 2 2 3 3B
A X A X A X b
1, 2, 3B 0
X X X and integers.
Note that while B is analyzed, variables of the third level not associated to B are equal to zero. Assume that QPP(B) is feasible and (OC) is verified.
Let X1 X1* be the solution so obtained. The points of (IR2) corresponding to the basis B can be obtained by solving the indefinite quadratic programming problem:
QPP1(B) : max (d X21 21e X21 3B 2)(d X32 2e X32 3B 2)
subject to A X2 2A X3 3B b A X1 1* 2, 3B 0
X X and integers;
2 c X21 1 2, 2 c X22 1 2
Again the variables of the third level not associated to B are equal to zero while B is being analyzed.
The optimal solution so obtained X* (X1*, X2*, X3*) is the best point of (IR2).
We now search for a new basis to improve the values of Z1 and Z2 obtained so far.
The following lemma is proved.
Lemma : Any basis from A3 capable of providing a point of (IR2) better than X* must include atleast one vector whose index belongs to the set
3 1 2
{ : }
C j V T jC and jC , where C1 {j V3 T L: 1j 0}
and C2 {j V3 T L: 2j 0}
1 12 11 12 11
11( 12) 12( 11) ( 12)( 11)
j j j j
j j j j j
L Z z d Z z c z d z c
is the jth reduced cost coefficient in the optimal integer solution of QPP(B) and
2 22 21 21 22
21( 22) 22( 21) ( 21)( 22)
j j j j
j j j j j
L Z z d Z z c z c z d
is the jth reduced cost coefficient in the optimal integer solution of QPP1(B).
Proof: Let Z X1( *) denote the value of the leader's (first level) objective function at X
*
Corresponding to X*, the matrix [A1 A3] is decomposed into [B N] where B is an (mm) basis matrix associated to the basic variables of X*.
Let XB* be a basic feasible solution and ˆXB be the new basic feasible solution obtained by entering aj into the basis and departing br.
*
ˆ
i i r
ij
B B B
rj
Y
X X X
Y
and
*
ˆ r 0
r B B rj X X Y
i.e. XˆBi XˆBr Yij and XˆBr where
r j B r X Y .
Given Z11* c X1TB B* 1 and Z12* D X1TB B* 2 the new value of the objective function is
1( )ˆ 11( )ˆ 12( )ˆ
Z X Z X Z X
1 1 1 1
1 1
ˆ
ˆ ˆ
ˆ
i i i i
m m
B B B B
i i
c X d X
* 1 1 1 ˆ ( )i i r
m
B B ij B
i i r
c X Y c
* 1 1 1 ˆ ( )i i r
m
B B ij B
i i r
d X Y d
* 1 1 1 ( ) i i mB B ij ij
i i r
c X Y c
* 1 1 1 ( ) i i mB B ij ij
i i r
d X Y d
since ˆ1
r
B ij
c c and ˆ1
r
B ij
d d )
= * 1 1
1 1
i i i
m m
IB IB B ij ij
i i
c X c Y c
* 1 1 1 1i i i
m m
IB IB B ij ij
i i
d X d Y d
* 11 * 12
11 12
(Z zj cij)(Z zj dij)
* * * 12 * 11 11 12
11 12 ( 11( j ij) 12( j j) ( j ij)( j ij)
Z Z Z z d Z z c z c z d
* * 1
11 12 j
Z Z L
* *
11 12 ( 0 and ij 0)
Z Z L
* 1( )
Z X
Hence, Z X1( )ˆ Z X1( *)
So, for improvement in the solution of the first level, we should take into consideration those a sj' for which
1
0
j
L .
If X* is a feasible solution of QPP(B), then L j 0 j V1 and j T.
Now, let Z X2( *) denote the value of the second level objective function at X
*.
Also [A2 A3] can be decomposed into [B N] where B is an (mm) basis matrix associated to the basic variables of X*. Let XB* be the b.f.s obtained by phase 1 of the simplex method.
Let ˆXB be the new b.f.s. obtained by entering aj into the basis and departing br.
Therefore, the new value of the second level objective function is
2( )ˆ 21( )ˆ 22( )ˆ
Z X Z X Z X
It can be proved in the same manner (as above) that
* 2( )ˆ 2( )
Z X Z X .
So, to improve, the second level objective function, we must consider those aj's for which L2j 0.
If X* solves QPP1(B) then L2j 0 j V2 and j T.
If C,Z1and Z2 cannot be improved and hence the current best integer point is optimum to (TIPP).
If we have previously built sets C C1, 2,...,Ci, the new bases B should include atleast one index from each sets
1 2
, ,..., i
C C C . Let G1 { }Ci .
Suppose that QPP(B) is infeasible. Then this basis is no longer we are interested in.
Let H denote the set of indices associated with the bases B. Then the new basis should not include all vectors with indices in the set H.
Now, if we have previously built sets H H1, 2,...,Hi, the new basis should not include all vectors with indices in each of these sets.
Let G2 U H{ i}.
NOTE: In order to select indices which form the new basis. It is suggested to solve for wj.
(P1): 1, 1, , 1
0, otherwise
j j j
j
j C C G
w d
1, ,
1,
0, otherwise
j
j j j j
j j
j H H G
w d
j j
w m
, wj {0, 1}, j V 3.Remark: If the basis so formed has rank k < m, then BN [ ,B Nˆ ] where ˆB is a matrix of independent vectors of B and N being a matrix of (m k) vectors of A3 is a basis from A3.
The optimal solution of the problem
1
1 11 12
max ( ) ( ) ( )
X Z X Z X Z X
11 1 11 2 11 3 1
(c X d X e X )
(c X12 1d X12 2e X12 31)
where X2 solves
2 2 1 2 3 max ( , , )
X Z X X X
21 1 21 2 21 3 2
(c X d X e X )
(c X22 2d X22 2e X22 32) for a given X1,
subject to (X X1, 2,X3)S,
becomes a lower bound on the optimum of (TIPP).
So, if an integer point of (IR2) provides this optimum, then that point is the optimum solution to (TIPP).
In this paper, the leader's (first level) problem is considered as
(QPP) 1
1 11 12
max ( ) ( ) ( )
X Z X Z X Z X
subject to (X X1, 2,X3)S
and the first followers' problem is considered as
(QPP1) 2
2 21 22
max ( ) ( ) ( )
X Z X Z X Z X , for a given X1 subject to X = (X X1, 2,X3)S .
DESCRIPTION OF THE ALGORITHM Step 0 : Solve the problem
1
1 1 2 3
max ( , , )
X Z X X X
subject to (X X1, 2,X3)S.
0.1 If it does not have a feasible solution, stop. (TIPP) is not feasible.
0.2 If the optimal solution is not an integer solution, then apply Gomory's cutting plane method to find an integer solution.
Let X*(X1*,X2*,X3*) be an optimal integer solution.
Step 1: Put X X1* in (QPP1). Let
1 1 2 3
(X ,X ) be an optimal solution to (QPP1).
1.1 If X2* X21, go to step 3. Otherwise go to step 2.
Step 2: If X2* X12 find the second best solution of Leader's problem (QPP) and move back to step 1. Step 3: Put X X1*,X2 X2* in (QPP2).
Let X3* be its optimal integer solution.
If X3*Xˆ3, then (X1*,X2*,X3*) is the optimal integer solution to (TIPP).
If X3* Xˆ3, stop. (X X X1*, 3*, ˆ3) is the current best integer point of (IR2). Set G1 ,G2 .
Step 4: Solve QPP(B)
4.2 If QPP(B) is feasible, then this optimal solution be compared with the current best point of (IR2) and updated.
Let (X1*,X3*,X3*) be the optimal integer solution.
Construct C1 {j V3 T L: 1j 0}. Step 5 : Solve QPP1 (B) for given X1 = X1.
5.1 Let its optimal integer solution be (X2**, X3**).
5.2 If X2**X20, find the next best solution of QPP(B). Go to step 4.
5.3 If X2**X20, construct C2 {j V3 T L: 1j 0}. Step 6 : Compute C {j V3 T: jC and j1 C2}.
6.1 If C, stop. The optimal solution to (TIPP) is the current best integer point of (IR2). Otherwise G1 G1{ }C
Step 7 : Solve (P1).
7.1 If (P1) is infeasible, stop. The current best integer point of (IR2) is optimal to (TIPP).
Step 8 : If (P1) is feasible, construct B by solving (OC). If no solution exists for (OC), compute H. Set
2 2 { }
G G H . Go to Step 8. Otherwise go to step 4. Numerical Example
Consider the three level integer quadratic programming problem
(TIPP)
1
1 2 3 1 2 3 4
max( 2 4)( 2 1)
X x x x x x x x
where x2 solves
2
1 2 4 2 3
max(2 1)( 2 1)
X x x x x x for a given x1.
where x3 and x4 solve
3 4
1 2 3 4 1 3 4
,
max ( 2 5 3 20)(2 3 2 70)
X X x x x x x x x
subject to
1 2 3 5
3x 7x x x 10
1 2 6
14x 4x x 6
1 2 3 4 7 5
x x x x x
1 2 4 8
2x x 2x x 8
1, 2,..., 8 0
1 j
c 1 1 2 0 0 0 0 0
1 j
d 1 1 1 2 0 0 0 0
CB DB VB XB x1 x2 x3 x4 x5 x6 x7 x8
0 0 x5 1 5 11/2 0 0 1 0 1 1/2
0 0 x6 6 14 4 0 0 0 1 0 0
2 1 x3 9 2 3/2 1 0 0 0 1 ½
0 2 x4 4 1 ½ 0 1 0 0 0 ½
Z11=22 11 1
j j
z c 3 2 0 0 0 0 2 1
Z12=18 12 1
j j
z d 5 7/2 0 0 0 0 1 3/2
Z11=22 11 1
j j
z c 3 2 0 0 0 0 2 1
Z12=18 12 1
j j
z d 5 7/2 0 0 0 0 1 3/2
1 j
L 164 113 0 0 0 0 58 51
5 7 20 (3/2)
We note that L1j 0 j.
Therefore, the optimal integer solution is given by
* * * * * * * * * * * *
1 2 3 1 2 3 4 5 6 7 8
( , , ) ( , , , , , , , )
X X X X x x x x x x x x
= (0, 0, 9, 4, 1, 6, 0, 0)
Put x1* 0 and solve the first follower's problem
QPP1(x1*) 2
2 8 4 2 3
max ( 2 1)( 2 1)
X Z x x x x
subject to
2 3 5
7x x x = 10
2
4x + x6 = 6
2 3 4
x x x
x7 = 5
2 2 4
x x x8= 8
x1, x3, ..., x8 0 and integers.
The optimal solution is (X11,X21, X31)( ,x11 x x x x12, 31, 14, 15, x x61, 71, x81)
* * * * * * * *
1 2 3 4 5 6 7 8
(0, 0, 9, 4,1, 6, 0, 0) ( ,x x x, , x , x , x , x , x )
For a given X1 0, X2 0 solve the second follower's problem QPP2(X1*, X2*) defined as
QPP2
3 4
3 4 3 4
,
max(5 3 20)( 3 2 70)
X X
x x x x x3 + x5 = 10
x6 = 6 x3 x4 +x7 = 5 2x4 +x8 = 8 x3, x4, x5, x6, x7, x8 = 0 and integers.
The optimal solution to QPP2(X1*, X2*) is given by
2 2 2 2 2 2 2 2 2 2 2
1 2 3 1 2 3 4 5 6 7 8
(X , X , X )(x , x , x , x , x , x , x , x ) = (0, 0, 5, 0, 5, 6, 0,8)
Set G1 and G2 .
* *
1 2
Now, the basis B1 is comprised of the vectors with induces 3, 5, 6 and 8. Next, we solve the problem QPP(B1) given by
QPP(B1)
1
1 3 1 3
max ( 2 4)( 1)
X x x x x
s.t. 3x1x3
x5 = 10
1 6
14x x = 6
1 3
x x = 5
1 8
2x x = 8
1, 3, 5, 6, 8 0
x x x x x and integers
The optimal integer solution is
0 0 0 0 0 0
1 3 5 6 8
( , , , , ) (0, 5, 5, 6, 8)
X x x x x x with Z1 = 84
It should be noted that variables of the third level not associated to B1, which are not considered while solving problem QPP(B1) are included in the construction of the set C1.
The reduced cost Lj of x4 is negative in the optimal solution of problem QPP(B1) and therefore, we
get
3
1 { 3 : j 0} {4}
C j V T L For x1x10 0, we solve QPP1(B1)
2 2 3
max( 2 x 1)(x 2x 1)
QPP(B1) subject to 7x2 x3 x5 = 10
2
4x
x6 = 6
2 3
x x = 5
2
x x8 = 8 2, 3, 5, 6, 8 0
x x x x x and integers. The optimal integer solution to obtained is
** ** ** ** ** **
2 ( 2 , 3 , 5 , 6 , 8 ) (0, 5, 5, 6, 3)
X x x x x x with Z2 = 11.
Here, x*0 x2**0. Again, the variables of the third level, not associated to B1, are not considered while
solving the problem QPP1(B1) and are included for constructing the set C2
2
2 { 3 : j 0} {4}
C j V T L Hence, C {j V T3\ : jC1 j C2} {4} .
1 1 { } {4}
G G C .
Now, solve the problem (P1) to obtain a new base for obtaining a better point of IR2,
(P1) : w4 1
3 4 5 6 7 8 4
w w w w w w
1 {0,1}
w .
Choose w3w4 w5 w6 1.
Therefore, the new bases B2 is associated to variables x x x3, 4, 5 and x6 and having rank 4.
32 31 31 32
31( 32) 32( 31) 0( 31)( 32) 0
j j j j
j j j j
Z Z d Z z c z c z d
3 {3, 4, 5, 6, 7, 8}
j V
Here begins the new iteration for the base B2.
Solve QPP(B2) defined by
QPP(B2) max(x12x34)( x1 x3 2x41)
subject to
1 3 5
3x x x
= 10
1
14x
x6 = 6
1 3 4
x x x = 5
1
2x x4 = 8
1, 3, 4, 5, 6 0
x x x x x and integers. The optimal integer solution is
4 4 4 4 4 4
1 3 4 5 6
( , , , , ) (0, 9, 4, 1, 6)
X x x x x x with Z1 = 612 > 84. Update the current best integer point of inducible region as (0, 0, 9, 4, 1, 6, 0, 0).
So, C1 {j V T L3\ : 2j 0} { } For x14 0, solve QPP1(B2);
QPP1(B2)
2
2 4 2 3
max( 2 1)( 2 1)
X x x x x
subject to 7x2 x3 x5 = 10
2
4x x6 = 6
2 5 4
x x x = 5
2
x 2x4 = 8
2, 3, 4, 5, 6 0
x x x x x and integers.
The optimal integer solution is X2**(x**2 , x3**, x4**, x5**, x6**)(0, 9, 4, 1, 6) with Z2**95Z2*
Here, x2**x02 0
3
2 { 3\ : j 0}
C j V T L .
So, C {j V3\T: jC1 j C2}.
Hence, the current best integer point is (0, 0, 9, 4, 1, 6, 0, 0) with max Z1 = 396; max Z2 = 95; max Z3 = 1855.
So, it happens to be the optimal integer solution for (TIPP).
Acknowledgement: I am grateful to the referees for their valuable support in the improvement of the paper. Reference
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