Interdicting the activities of a linear program — A parametric analysis

EUROPEAN JOURNAL OFOPERATIONAL RESEARCH

ELSEVIER

European Journal of Operational Research 86 (1995) 580-591

Theory and Methodology

Interdicting the activities of a linear program A parametric analysis Maw-Sheng Chern *, Kao-Ch~ng Lin Department of Industrial Engineering, National Tsing Hua University, Hsinchu 30043, Taiwan, ROC Received September 1992; revised November 1993

Abstract In this paper, we consider two interdiction problems for a linear program. These generalize the problems proposed by Fulkerson and Harding and by Golden for the shortest path problem. These problems also provide equilibrium analysis for a system in which some kind of resource can be used to change the original equilibrium. We show that these two problems can be solved simultaneously by performing parametric analysis of a linear program with bounded variables. We also consider their applications for the interdiction of flow networks. In particular, we propose an algorithm for solving the related parametric network flow problem. Keywords: Linear programming; Interdiction problems; Flow networks; Parametric analysis

1. Introduction

Let LPp be the linear program given as follows.

(LPp) .~(Cl,

C2 ....

,Cn) =

subject to

minimize x

• CjXj j=]

~

a i j x j >_bi,

i = 1..... m,

(1)

j=l xj>0,

j=l

. . . . . n,

w h e r e x = ( x j ) , cj is the unit cost o f activity j, a n d c o n s t r a i n t s (1) a r e the t e c h n i c a l r e q u i r e m e n t s o f the activities. In this p a p e r , we c o n s i d e r two p r o b l e m s in which t h e o p t i m a l v a l u e o f this l i n e a r p r o g r a m is i n c r e a s e d via i n c r e a s i n g t h e unit costs o f t h e activities. W i t h o u t loss o f generality, it is a s s u m e d that p r o b l e m L P e has a feasible solution.

* Corresponding author. 0377-2217/95/$09.50 © 1995 Elsevier Science B.V. All rights reserved SSDI 0377-2217(94)00057-J

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Let lj > 0 be the cost for increasing the unit cost of activity j by one unit. For a given budget/3, the first problem is to increase the unit cost of activity j by yj, j = 1. . . . . n, in such a way that Y'.7~lljyj is not greater than /3 and the minimum cost of the resulting linear program is maximized. This problem is mathematically stated as follows. (LPI1) 9(13) = maximize Y subject to

-2"(cl +Yl,

c2+Y2 .....

Cn + Y,,)

~ ljyj < fl, j=l

yj>_O,

j= 1,...,n,

where y = (yj). For a given specified value ~-, the second problem is to increase the unit cost of activity j by yj, j = 1. . . . . n, in such a way that the minimum cost of the resulting linear program is not less than r and E~= ll~yj is minimized. This problem is mathematically stated as follows. (LPI2) ~'(p) = minimize y subject to

• liYj

j=l

"Z~(Cl + Y l , C2 +Y2,"" 'Cn + Y~) >--~',

yj>0,

j = l . . . . ,n.

Problems LPI1 and LPI2 can be interpreted as interdiction problems of linear program LPv. Consider a system in which there are a defender and an interdictor. The defender wishes to perform his task at the minimum cost and his problem is formulated as LPv. However, the interdictor wishes to increase the difficulty of the defender's task. Specifically, the interdiction activity will cause the increase of the unit costs of the defender's activities. In problem LPI1, the interdictor maximizes the difficulty of the defender's task with the available budget. However, in problem LPI2, the interdictor increases the difficulty of the defender's task to a specified level with the minimum cost. Thus, these two problems can be used in the planning of military actions [7,8,11]. It is well-known [9] that the linear program can provide an equilibrium analysis for a system. Thus, problems LPI1 and LPI2 can also provide equilibrium analyses for a system in which some kind of resource can be used to change the original equilibrium. We note that if problem LPv is considered as a classical input-output analysis problem of a firm (cf. Section 9.2 of [9]), then problem LPI1 provides an equilibrium analysis of the firm when a capital can be used to change the original equilibrium. Moreover, it can be used in determining the interest rate of the capital [11]. For this case, we may allow a negative lower bound on the decision variable yj. There are some articles (e.g. Fulkerson and Harding [7], Golden [8], McMasters and Mastin [13]) concerning with the interdiction of networks. Fulkerson and Harding [7] consider the interdiction of the shortest path between two specified nodes subject to a budget constraint. They show that this problem can be solved by performing the parametric analysis of a special case of the minimum cost network flow problem with respect to the amount of flow from the source to the sink. Moreover, this parametric problem can be solved by using the minimum cost flow routine [6]. Golden [8] considers the problem of interdicting the shortest path to a specified level at the minimum cost. He shows that this problem can be reduced to a minimum cost network flow problem. He also considers the case that there are multiple destinations. McMasters and Mastin [13] consider an interdiction problem for the flow network in which

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the interdictor reduces the maximum flow from the source to the sink via reducing the arc capacities. For the planar network, they propose an algorithm by using the relationship of cuts and paths in topological duality. In Section 2, we show that for the general case, the parametric analysis of problem LPI1 with respect to/3 and that of problem LPI2 with respect to ~- can be solved simultaneously by performing parametric analysis of a linear program. In Section 3, we consider the case that problem LPe is formulated for the minimum cost network flow problem [1,6,10]. We show that for this case, the related parametric linear program is reduced to a parametric minimum cost network flow problem. In Section 4, we propose an algorithm to solve this parametric minimum cost network flow problem. The proposed algorithm can also be modified to solve a large scale parametric linear program with special structure in the constraint matrix.

2. Interdicting the activities of a linear program - A parametric analysis

The dual problem of LPp is given as follows. (LPD) rn

. ~ ( c l , c 2 . . . . . On) = maximize

E biwi i-I rn

w

subject to

Eaijwi<_Cj,

j = l . . . . . n,

i=l

w i > O,

i = 1. . . . . m,

where w = (w i) are the dual variables associated with constraints (1). Note that problem LPo may be considered as a formulation of problem LPe from the interdictor's view point [2]. Thus, problem LPI1 can be formulated from the interdictor's view point as follows. (LPI1 D) rn

maximize (w,y)

subject to

E biwi i=1

~-'~ai/wi-Yj
j=

1. . . . .

n,

(2)

i=l n

E ljyj <-/3,

j=l

wi>O, ),1.>0,

i= l,...,m, j = l . . . . . n,

In the following, we will show that the parametric analysis of this problem with respect to /3 can be solved by performing the parametric analysis of the following linear program with respect to A. (LPR e) n

minimize f

~ cjfj j=l n

subject to

Y'. a i J j > Abi, j=l

i = 1 . . . . . m,

(3)

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fj
j = l . . . . . n,

fj>_O,

j=l .... ,n,

583

(4)

where f = ( f ) . The dual problem of L P R p is given as follows. (LPR o) m

maximize (u,v)

~_, Abiu i - ~, ljvy i=1

j=l

m

subject to

~_,aiyui-vy
j = l . . . . ,n,

i=l

ui>O,

i=1 ..... m,

vj>O,

j = l . . . . . n,

where u = (u i) and v = ( v ) are the dual variables associated with constraints (3) and (4), respectively. Note that for a given value of A, the feasible region of problem L P R e is bounded. Therefore, the objective value of problem L P R p cannot be infinite for any given value of A. After performing the parametric analysis of problem LPR e with respect to A, we find a partition S : = {S 1 = (0, A1]. . . . . S k = (Ak_ 1, Ak],..., Sp = (Ap_ 1, oo)} of the interval (0, oo) [4]. For each interval S k ~ S "~, either there is a basis of problem LPR e which is optimal for A ~ S k or the problem has no feasible solution with A ~ Sk. The latter case can occur only when k =p. Moreover, for k --- 1 , . . . , p - 1, we find an optimal solution f(k) of problem L P R p with A = Ak. Let (u (k), v (k)) be the optimal solution of problem LPR o with A = Ak such that f(k) and (u (k), v (k)) satisfy the K a r u s h - K u h n - T u c k e r conditions [4]. We then have the following theorems. I ,,(k) . Theorem 1. (u (k), v (k)) is an optimal solution of problem LPI1 o with /3 = ~'n ,..y= 1.yvy

Proof. The dual problem of LPI1 o, i.e. problem LPI1 formulated from the defender's view point, is given

as follows. (LPIIp) n

minimize (x,p)

subject to

E jxj +13p

j=1

~ aijxj>_bi,

i= l,...,m,

j=l

--Xj + ljp >_ O, p>0,

xy>0,

j = 1 . . . . . n,

(5)

j = l . . . . . n,

where p is the dual variable associated with constraint (2). It is clear that (f(k)/Ak, 1 / h k) and (u (k), v (k)) are feasible solutions of problems LPI1 e and LPI1 o with / 3 - w,-,j=l,j~j t ,,(k), respectively. Moreover, they satisfy the K a r u s h - K u h n - T u c k e r conditions. Thus, they are optimal solutions of problems LPI1 e and ! ,,(k) LPI1 o with/3 = ~'n ,.,j= ~,y~: , respectively. Hence, the theorem holds. [] Theorem 2. v (k) is an optimal solution of problem LPI2 with z

= W m l~'i-i h ,,(k) • -.i=

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584

Proof. Similar to that for problem LPI1, problem LPI2 can be formulated from the interdictor's view point as follows. (LPI2 o) r/

minimize (w,y) subject to

ljyj j=l

~-'.aijwi--yj
j = 1.... ,n,

(6)

i=1

~ biw i ~ 'r, i=1

w~>O,

i = 1 . . . . . m,

yj_>O,

j = 1 . . . . . n.

The dual problem of this linear program, i.e. problem LPI2 formulated from the defender's view point, is given as follows. (LPI2 e) maximize (x,T)

subject

":3' -

~ cjx~ j= 1

to

~ aijx j >_"Ybi, j=l

T>_O,O
i = 1. . . . , m, j=l

.... ,n,

where 3' is the dual variable associated with constraint (6). It is clear that (f(k), hk ) and (u eke, v ¢~)) are "~ lOiUi optimal solutions of problems LPI2p and LPI2 D with ~- = Ei= - (k), respectively. Hence, the theorem holds. [] T h e o r e m 1 and T h e o r e m 2 show that the parametric analysis of problem LPI1 with respect to/3 and that of problem LPI2 with respect to ~" can be solved simultaneously by performing the parametric analysis of problem L P R e with respect to h. For a fixed value of /3 (or z), we may find the optimal solution of problem LPI1 (or LPI2) by using property that 9-(/3) (or ~ ( r ) ) is a piecewise linear function. For example, the optimal solution of problem LPI1 with/3 = aET= lljuJ k) + (1 - a)Z7_ lljVJ k+ 1) is a v (k) + (1 - a ) v (k+l), where 0 < a < 1.

3. Interdicting flow networks The minimum cost network flow problem [1,4,6,10] is one of the most widely studied network problems. Let G ( N ; E ) be a directed network with the node set N and the arc set E. In this problem, each node i ~ N is associated with a number b i which is the available supply of some commodity (if b i > 0) or the required demand for the commodity (if b i < 0). Nodes with b i > 0 are called sources, and nodes with b i < 0 are called sinks. If bi = 0, then none of the commodity is available at node i and none is required. In this case, node i is called an intermediate node. Without loss of generality, it is assumed that b = (b i) ~ 0 and E, ~ Nbi = 0. For each (i, j) ~ E, let Xij be the amount of flow on the arc, Cij be the

M.-S. Chern,K.-C.Lin / EuropeanJournalof OperationalResearch86 (1995)580-591

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unit shipping cost along the arc, and u o. be the capacity of the arc. The problem is to transship the available supplies through the network to satisfy the specified demands at the minimum cost. This problem can be formulated as the linear program given as follows. (FPe) minimize x

E CijXij (i,j)EE

subject to

xi/(i,j)~E

~

x/i=bi,

i~N,

(j,i)~E

O~_~Xij<_~Uij,

(i, j) ~ E .

This problem and its special cases (e.g. the shortest path problem, the maximum flow problem and the transportation problem) have been applied in the design of military logistics systems [6,10]. However, in a conflict situation, the opponent wishes to interdict the network to increase the user's logistic cost. From the results in Section 2, for this case, problems LPI1 and LPI2 can be formulated as follows. (PFI1 e) minimize (x,p) subject to

~, CijXij "}- ~ p (i,j)~E

~

xii-

(i,j)~E

~

x/i=bi,

i~N,

(j,i)~E

p>O, xi/<_Plij, O
(i,j)~E.

(FPI2 e) maximize "f'Y -- Z CijXij (x,~,) (i,j)~E subject to

xij(i,j)~E

~

x/i= yb i, i ~ N,

(j,i)~E

2/>0, xi: < yuij, O
subject to

E CijXij (i,j)EE Y'.

(i,j)~E

xij -

~

xii = ,~bi,

i s N,

(j,i)~E

Xij <_~t~Uij , 0 <<_Xij<_~lij ,

( i,

j)

~ E.

Clearly, this parametric problem can be solved by using the conventional method [4] for parametric linear programs. However, for large scale networks, the memory space required for solving this problem by using the conventional method is tremendous. Moreover, it may be over the available memory space provided by an ordinary computer. For example, for a network with 1500 nodes and 5730 arcs, the

586

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memory space required is more than 49 MB. In the last decade, many (theoretically or practically) efficient algorithms for the minimum cost network flow problem have been proposed (cf. Ahuja et al. [1]). Some codes based on these algorithms can solve an instance with thousands of nodes and arcs in a few seconds. In the next section, we will propose an algorithm for solving problem F P R e. The proposed algorithm uses a minimum cost network algorithm as a subroutine and requires almost the same memory as that for the original minimum cost flow problem. This also generalizes the e-flow approach proposed by Ruhe [14] for the maximum flow problem with parametric capacities.

4. An algorithm for the parametric minimum cost network flow problem Note that problem F P R e with A = 0 has the unique feasible solution x ~°) = 0. Let /~max be the largest value of A such that problem F P R e has a feasible solution. Given the optimal solution x ¢k) of problem F P R e with A = Ak, we define an accompanying minimum cost network flow problem as follows.

(AV7)) minimize x

subject to

where 0

Lij =

-M

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587

Theorem 4. Let ~(k) be an optimal solution o f problem AF(e k). Then, - M < y,~k) < M f o r all (i, j ) ~ E. Proof. We prove this theorem by contradiction. Suppose that there is an arc, say (p, q), such that 2(k) = M. Since M is sufficiently large, this implies that there is a cycle C in G ( N ; E ) such that the unit Pq shipping cost along C in the same direction as (p, q) is negative. Moreover, for (i, j) ~ C, if (i, j) has the same direction as (p, q) then $~k) > Uii; otherwise, 2}k) < 0. We note that if > uij, then by (8), x i( k ) < AkUij and xij( k ) < l~. We also note that if 2~k) < 0, then by (7), x~k) > 0. Thus, given the solution x '(k) of problem FPRp with A = Ak, a positive flow may be augmented along C in the same direction as (p, q) to reduce the total cost. This contradicts to the assumption that x (k) is an optimal solution of problem FPRp with A = )t k . Therefore, 2~k) < M for all (i, j) ~ E. Similarly, we can show that ..pq~(k)~ - M for all (i, j) ~ E. Hence, the theorem holds. [] Note that the dual problem of

AF (k) is given as

follows.

(A 5 E bi'tT"i-

maximize (¢r,v,y)

i~N

subject to

~'~ - Trj - vij + ytj = c~j,

Let (~-',

E Uijuij -b E LijYij (i,j)~E (i,j)¢E

yij > 0,

(i,j)~E,

vii > 0,

(i,j)~E.

i ~ N,

v', y') be the optimal solution of problem AF(Dk) such that

v;j(U/j-2~))= 0 ,

(i,

j) ~ E ,

y;j(Lij-ff}~))=O,

(i,j)~E.

(9) (10)

Then, we have the following results. Lemma 1. I f y~j > O, then Zij = . ~ k ) = x~k) = O. Proof. If y~j > 0, then by (10), L~i = 2 ~ ). By (7) and Theorem 4, this can occur only when Lij = 0 --x~k). Hence, the lemma holds. [] Lemma 2. I f x ~ >> O, then Y~i = O. Proof. This lemma follows directly from Lemma 1. !

__

v

[]

!

Lemma a. I f x} k) = lij, then vii - 7ri - 7ri - cij ~_~O. Proof. Note that ~r~ - ~'~ - cij = v~i - y~j. If x~k) = lij > 0, then by Lemma 2, y!,j = 0. Therefore, zr~ - zr~ cij = v~j > 0 and the lemma holds. [] Let AA = min{AA1, AX2, A~3} ,

(11)

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588

where (k)

-(k) _

AAI=

min [(hkUi.--Xij (i,j)~E I,\ 1

)//(xij

AA 2 =

min [(l i. _ (i,j)~E ~.\ J

AA 3 =

min ( - x ! k ~ / ~ ) l ~ + ~ < 0 } . (i,j)~E~ --~1 +-'~1 ,--tj

Uij)lX}~)>Uij},

- ( k ) I-~f ) ) xit(k) )lxij

O\

,[

It is clear that for A ~ [A k, Ak + AA], x' = x <~) + (A - Ak)$(k) is a feasible solution of p r o b l e m F P R p . Note that the dual p r o b l e m of F P R p is given as follows. ( F P R D) maximize (~r,v,y)

E AbiTriE Auijviji~N (i,j)~E

subject to

77"i -- TFj -- Vij -- Yij <- Cij,

E litYit (i,j)~E

i ~ N,

Yit>_O, (i,j)~E, vii>O,

(i,j)~E.

Let v * = (vii) and y * = ( y i j ) be defined as follows. 0,

if x ~ ) = li/,

Vij

Vijt ,

otherwise,

y.. =

"ITi'

and

tJ

-

Tfj'

-

" Cij = Uit

0,

if

x}~

~ =

lij,

(12)

otherwise.

It is clear that (rr', v *, y * ) is a feasible solution of p r o b l e m F P R D. T h e following t h e o r e m shows that x' and (~", v *, y * ) are optimal solutions of p r o b l e m s F P R e and F P R D for A ~ [A k, A~ + AA], respectively. 5. For A ~ [ A k, Ak + AA], X ' = x t k ) + ( A F P R p and F P R D , respectively.

Theorem

Ak).~
Proof. It is sufficient to show that x' and (~-', v *, y *) satisfy the K a r u s h - K u h n - T u c k e r (i) T o show that vi~(Auij -x'it) = 0 for (i, j) ~ E. Case 1: If "'(~) = lit, then vi~ = 0 and vi*j(Auit -x'it) = O. Case 2: If x ) = A~uit < lit, then vi~ = v~t and U,t = uit. T h e r e f o r e , , , , , -(k) ) = v;t(

-

conditions.

= o.

Case 3: If x}~ ~ < AkU,j and x ~ > < lit, then ~ t = M. By (9) and T h e o r e m 4, this implies that vij = vit = 0 and vi~ (Auit - x'it) = O. (ii) T o show that Yi~(lit - x'it) = 0 for (i, j) ~ E. Case 1: If x}k) lij , then y.Z,j _- t/~/. If v,t" = 0, then Yit*(lit -- x;t) = 0. Otherwise, by (8) and (9), x[~ > - = U~t = 0 =

and yi~ ( lij -- Xtij ) = yi; ( lij -- ( X}f ' -t- (1~ -- i~k ) X~k ' ) ) = yi; ( lij - lij - ( A - Ak)0 ) = 0 .

M.-S. Chern, K.-C. Lin / European Journal of Operational Research 86 (1995) 580-591 *

*

589

t

Case 2: If ,%.-{k)4= lit , then Y;t = 0 and Yiy(lit --Xij) = O. (iii) TO show that x'ij(cij - ~'~ + ~'~ + vi~ + Yi~) = 0 for (i, j) e E. Case 1: If x}k) = liy, then Ui~ = 0 and yi7 -- ~'~ - ~r~ - %. Therefore, X;j( Cij -- 7I'~ -1- "l'i'~ "Jr- Ui~ -F yi~ ) = X'ij( Cij -- "IT; -I- "rl'~ d- "IT~ -- 77r~ -- Cij ) = x;jO = O.

Case 2: If 0
~'i'; Jr

71"~ -Jr- U;j -- y~j) = XtijO = O.

Case 3: If x } k ) = o , then vi~ =v:j and Yi~ = 0 . (i) If y~j > 0, then by Lemma 1, xit, =xij(k) + (A -Ak)ff}k) = 0. Therefore, X~y(Cij -- ~ + ~ + Ui; +Yi~) : O. (ii) If y~j = 0, then

X;j( Cij -- 77";q- "B'~q- Ui~ -I- Yi~ ) = X;j( Cij -- "17"~+ ~ + U;j -- Y~t) = x;tO = O. Hence, the theorem holds.

[]

Theorem 6. A A > 0.

Proof. It is sufficient to show that (i) If ~(') > nit, then x ~ ) < Akuit; (ii) If :~:(/k)> 0, then x}k) < lit; and (iii) If ~ k ) < 0 , then x~k) > 0. If 2}~') > uij, then by (8), U/j = M. Thus, x}k) < Akuij and (i) follows. If ~}k) > 0, then by (8), Uii = uij or M. Thus, x}~) < lit and (ii) follows. If if}k) < 0, then by (7), Lit = - M . Thus, x}k)> 0 and (iii) follows. [] Corollary 1. Problem AF(e k) has no feasible solution if and only if Ak = Area~.

Proof. By Theorem 3, it is sufficient to show that if h k = Amax, then problem AFe(k) has no feasible solution. We prove it by contradiction. Suppose that problem AFp(k) has an feasible solution for h k -- Areax. Then, by Theorem 4, problem AFe(k) has a finite optimal solution ,~(k). Since b #= 0, ,~(k) ~ 0. Then, by Theorem 5 and Theorem 6, there exists h > hm~, such that problem FPRp has a feasible solution. This contradicts to the definition of hm~x. Hence, the corollary holds. [] In summary, we propose an algorithm for the parametric analysis of problem FPR e as follows. Algorithm PFP.

StepO. Set k = 0 , A0 = 0 a n d x <°)=0. Step 1. Find the optimal primal solution .~(k) and the optimal dual solution (~r', v', y') of problem A F (k). If the problem has no feasible solution, then Areax = )tk, and go to Step 5. Otherwise, go to Step 2. Step 2. Find AA by (11). Step 3. Set Ak+ 1 =A k + AA, x (k+l) =x(k) + AA.~(k), y ( k + l ) = y * , and k = k + 1, where y* is defined in (12). Step 4. Output Ak+j, X(k+l) and y~k+l). Go to Step 1. Step 5. Stop. From the proofs of the above theorems, we note that this algorithm can be easily modified to solve the large scale parametric linear program with special structure in its constraint matrix. Specifically, we only have to modify the initial value of A, problem AFp(k), and Eq. (11).

590

M.-S.

Chern, K.-C. Lin /

European Journal of Operational Research 86 (1995) 580-591

Cij; ~zij; ~ij 3;7;1

5

5;4;8!

~

0

~,~ :

-2

ei"%'~

12;i;3

20;o0;2

3

10;0o;6

0

(i

12;2;4

-6

10;6;5

Fig. 1. An example of the minimum cost network flow problem.

Example. Consider the network in Fig. 1, where the number associated with each node is the available supply or required demand of a commodity at the node. After performing the parametric analysis of problem FPRp with respect to A ~ [0, oo), we find that A1 = 1, A2 = 32 and A 3 = ~tmax = 1. By Theorem 1 and Theorem 2, the parametric analysis of FPI1p with respect to/3 and that of FPI2 e with respect to r, are given in Tables 1 and 2.

5. Conclusions The maximum flow problem [1,4,6] is a well-known special case of the minimum cost network flow problem. McMasters and Mastin [13] propose a meaningful interdiction problem for this case. We note that for a general network, similar to that for problem LPI1, McMasters and Mastin's problem can be formulated as a bi-linear program [11]. In addition, we may consider the binary interdiction problems which are mathematically stated as follows.

Table 1 The parametric analysis of FPIlp with respect to/3

/3

Y1"2

Y1"3

Y~3

Y:~4

Y3"4

Y;5

Y4"6

Y5"4

Y5"6

[0, 6]

0

/3

0

0

0

0

0

0

0

(6, 12] (12, oo)

0 0

6 /3 - 6

0 0

½/3 - 3 3

0 0

0 0

0 0

0 0

0 0

Table 2 The parametric analysis of FPI2 e with respect to

1" [191,221] (221,230] (230, oo)

Y1"2 0 0 0

Y1"3 ~(~- - 191) 6 ~-- 224

r Y;.*3

Y2"4

Y3"4

Y3"5

Y4"6

Y5"4

Y5"6

0

0

0

0

0

0

0

0 0

½(r - 221) 3

0 0

0 0

0 0

0 0

0 0

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591

(LPBI1) maximize y

. ~ ( c 1 + e l y 1 , c 2 + e 2 y 2 . . . . . cn +e~y~) n

subject to

Y'~/jyy
yj=0orl,

j = l . . . . ,n.

(LPBI2)

~ ljy i

minimize Y

subject to

j=l

. ~ ( c 1 + elY 1, c 2 + e2Y2,..., c n + enyn) >_ ~', y j = 0 or 1,

j = l . . . . . n.

Similar to that for problem LPI1, problems LPBI1 and LPBI2 can be formulated as 0-1 mixed integer linear programs. It has been shown that these two problems are NP-hard [11]. Ball et al. [3], Corley and Sha [5] and Malik et al. [12] have considered two special cases of these two problems. These problems are called the most vital arcs problem and the most important arcs problem, respectively.

Acknowledgement This work was supported in part by the National Science Council of the Republic of China under grant NSC-82-0415-E007-03.

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