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Combinatorial Algorithms for Inverse Network Flow Problems
Ravindra K. Ahuja
Department of Industrial and Systems Engineering
University of Florida,
Gainesville, FL 32611, USA
ahuja@ufl.edu
James B. Orlin
Sloan School of Management
Massachusetts Institute of Technology
Cambridge, MA 02139, USA
jorlin@mit.edu
January, 1998
(Revised June 8, 2000)

Combinatorial Algorithms for Inverse Network Flow Problems
Ravindra K. Ahuja and James B. Orlin
ABSTRACT
An inverse optimization problem is defined as follows: Let S denote the set of feasible solutions
of an optimization problem P, let c be a specified cost vector, and x 0 ∈ S. We want to perturb the cost
vector c to d so that x 0 is an optimal solution of P with respect to the cost vector d, and ||d – c|| p is
minimum, where ||.|| p denotes some selected L p norm. In this paper, we consider inverse minimum cut
and minimum cost flow problems under the L 1 norm (where the objective is to minimize Σ j∈J w j |d j – c j |
for some index set J of variables), and under the L ∞ norm (where the objective is to minimize max{w j |d j –
c j |: j ∈ J}). We show that the unit weight (that is, w j = 1 for all j ∈ J) inverse minimum cut problem
under the L 1 norm reduces to solving a maximum flow problem, and under the L ∞ norm it requires
solving a polynomial sequence of minimum cut problems. The unit weight inverse minimum cost flow
problem under the L 1 norm reduces to solving a unit capacity minimum cost flow problem, and under the
L ∞ norm it reduces to solving a minimum mean cycle problem. We also consider the non-unit weight
versions of inverse minimum cut and minimum cost flow problems under the L ∞ norm.
1. INTRODUCTION
Let P denote the following optimization problem: min{cx : x ∈ S}, where S is the set of feasible
solutions. Let J denote the index set of the vector x. Let x 0 ∈ S denote a feasible solution of P that we
wish to make optimal for P by perturbing the cost vector c. We call a cost vector d inverse feasible for P
with respect to the solution x 0 if x 0 is an optimal solution of P when the cost vector c is replaced by the
cost vector d. The inverse problem under the L 1 norm is to find an inverse feasible cost vector d * of P for
which ||d * – c|| 1 = Σ j∈J | d* j - c j | is minimum among all inverse feasible cost vectors d. The inverse
problem under the L ∞ norm is to find an inverse feasible cost vector d * of P for which ||d * – c|| ∞ =
max{| d* j - c j | : j ∈ J} is minimum. We refer to d * an optimal cost vector for the inverse problem. In this
paper, we refer to the inverse problem under the L 1 norm as the inverse problem, and the inverse problem
under the L ∞ norm as the minimax inverse problem.
In this paper, we study inverse and minimax inverse minimum cut and minimum cost flow
problems. We consider a directed network G = (N, A) with N as the node set and A as the arc set. Let n =
|N| and m = |A|. Each arc (i, j) ∈ A has an associated cost c ij , an associated capacity u ij > 0, and an
associated weight w ij > 0. Let C = max{|c ij |: (i, j) ∈ A}, let U = max{u ij : (i, j) ∈ A}, and let W =
max{w ij : (i, j) ∈ A}. Table 1 gives a list of problems considered in this paper and the problem to which
1

the considered inverse problem reduces. In the table, whenever the running time involves C, U, or W, we
assume that arc costs, arc capacities, or arc weights are integer, respectively.
Inverse network flow problems under the L 1 norm have been studied in the past by several
researchers. Zang and Liu [1996] studied inverse minimum cost flow problems; Yang, Zhang and Ma
[1997], and Zhang and Cai [1998] have studied the inverse minimum cut problems. Ahuja and Orlin
[1998] develop a unified framework for inverse linear programming problems under both L 1 and L ∞
norms from which several inverse network flow problems are derived as special cases. In this paper, we
present combinatorial algorithms for solving inverse network flow problems as well as combinatorial
justifications for the algorithms. This contrasts with the linear programming based approaches suggested
in the literature. We believe that the combinatorial arguments provide additional insight and
understanding of inverse network flow problems.
Inverse problem considered:
Unit weight inverse minimum cut problem
under the L 1 norm
Unit weight inverse minimum cut problem
under the L ∞ norm
General weight inverse minimum cut problem
under the L ∞ norm
Unit weight inverse minimum cost flow problem
under the L 1 norm
Unit weight inverse minimum cost flow problem
under the L ∞ norm
General weight inverse minimum cost flow problem
under the L ∞ norm
Reduces to solving the following problem:
A maximum flow problem
O(log(nU)) minimum cut problems
O(log(nUW)) minimum cut problems
A unit capacity circulation problem
A minimum mean cycle problem
A minimum cost-to-weight ratio problem
(also known as the tramp steamer problem)
Table 1. Summary of problems considered in this paper and the results obtained.
2. THE INVERSE MINIMUM CUT PROBLEM
In this section, we study the inverse minimum cut problem. Consider a network G = (N, A) where
u ij ’s denote arc capacities, and s and t are two specified nodes, called the source and sink nodes,
respectively. In the network G, we define a cut as a set of arcs whose deletion disconnects the network
into two or more components, and such that no subset of arcs in it has this property. An s-t cut is a cut
that partitions the node set into exactly two parts of which one part, say S, contains the node s and another
part, S = N – S, contains node t. An alternate method to represent the s-t cut is by the node partition it
creates and we represent it as [S,S]. Let (S,S) denote the set of forward arcs in the cut, that is, (S,S) =
{(i, j) ∈ A : i ∈ S and j ∈ S }, and ( SS , ) denote the set of backward arcs in the cut, that is, ( SS , ) = {(i,
j) ∈ A : i ∈ S and j ∈ S}. We define the capacity of the s-t cut [S,S] as the sum of the capacities of the
forward arcs in the cut. We denote it by u[S,S] , that is, u[S,S] = ∑ u
{( ij , ) ∈( SS , )} ij . The minimum cut
problem is to determine an s-t cut of minimum capacity. The inverse minimum cut problem is to modify
2

the arc capacity vector u to d * 0 0
so that the cut [ S , S ] is a minimum cut with respect to the capacity
vector d * and ||d * - u|| 1 = Σ (i,j)∈A | d* ij - u ij | is minimum.
In this section, we will consider the network G with varying arc capacity vectors. We
subsequently refer to the network G with capacity vector z as G(z). We will use the following variant of
the max-flow min-cut theorem (see, for example, Ahuja, Magnanti and Orlin [1993]) in our analysis in
this section:
0 0
Property 1. An s-t cut [ S , S ] is a minimum cut in G(d) if and only if there exists a feasible flow x
from node s to node t in G(d) that “saturates” the cut
0 0
[ S , S ] , that is, xij = d ij for each arc (i, j) ∈
( S
0 , S
0 ) and xij = 0 for each arc (i, j) ∈ ( S , S )
0 0 .
Let the network G′ = (N, A′) be obtained from G = (N, A) by deleting the backward arcs in the
cut [ S
0 , S
0 ] . In other words, A ′ = A\( 0
S , 0
S ) . Let the capacity vector u′ for arcs in A ′ be defined as u' ij
= u ij .
0 0
0 0
Lemma 1. The cut [ S , S ] is a minimum cut in G(d) if and only if [ S , S ]
G′(d′), where
d ' ij = d ij for each arc (i, j) ∈ A′.
is a minimum cut in
Proof. Suppose that x saturates the cut [ S
0 , S
0 ] in G(d) and therefore [ S
0 , S
0 ] is a minimum cut in
G(d). Let x′ be a feasible flow in G′(d′) obtained by setting ' 0 0
x ij = x ij for each arc (i, j) ∈ A\ ( S , S ) .
Then the flow x′ saturates the cut [ S
0 , S
0 ] in G ′ (d ′) and hence [ 0
S , 0
S ] is a minimum cut in G ′ (d ′).
0 0
Now suppose that [ S , S ] is a minimum cut in G ′ (d ′) and the flow x′ saturates it. Let x be the flow in
the network G(d) obtained from x′ by setting x ij = x ' ij for each arc (i, j) ∈ A\
0 0
( S , S ) and xij = 0 for each
(i, j) ∈ ( S
0 , S
0 ) . It is easy to see that the flow x saturates the cut [ S
0 , S
0 ] in G(d) and therefore [ S
0 , S
0 ]
is a minimum cut in G(d).
Theorem 1. If the capacity vector d′ is an optimal solution for the inverse problem for G′, then the
capacity vector d * is an optimal solution for the inverse problem for G, where d * ij =
0 0
0 0
∈ A| ( S , S ) and d * ij = u ij for each arc (i, j) ∈ ( S , S ).
♦
d ' ij for each arc (i, j)
Proof. Suppose that the arc capacity vector d′ is optimal for the inverse problem for G′. Let d * ij = d' ij
for each arc (i, j) ∈ A\
0 0
( S , S ) and d * ij = u ij for each arc (i, j) ∈
0 0
( S , S ) . Observe that ||d* – u||1 = ||d′ –
u′|| 1 . Hence the optimal objective function value of the inverse problem for G is at most the optimal
objective function value for the inverse problem for G′. Now suppose that d* is optimal for the inverse
3

problem for G. Let d ' ij = d * ij for each arc (i, j) ∈ A\
0 0
( S , S ) . Then by Lemma 1, d ′ is inverse feasible
for G′. Moreover, ||d′ – u′|| 1 ≤ ||d* – u|| 1 and so the optimal objective function value for the inverse
problem for G is at least the optimal objective value for the inverse problem for G′.
♦
Theorem 1 allows us to solve the inverse problem for G by solving the inverse problem for G′.
Before explaining how to solve the inverse problem for G′, we introduce a new term and establish one
more lemma. Let Excess(S, d′) be the capacity of the cut [S, S] in G′ with respect to capacity vector d′
minus the maximum s-t flow in G′ with respect to d′. Thus d′ is inverse feasible for G′ if and only if
Excess(S 0 , d′) = 0.
Lemma 2. The optimal objective function value to the inverse minimum cut problem for G′ is at least
Excess(S 0 , u′).
Proof. Let
[S', S '] be a minimum cut in G′(u′). Then, Excess(S 0 , u′) is the capacity of the cut [S, S]
minus the capacity of the cut [S,S]. Alternatively,
'
'
∑ - ∑ ' ' 0 0 = Excess(S 0 , u′) (1)
0 0 ' ' '
(i,j) ∈(S , S )\(S , S ) u ij
(i,j) (S , S )\(S , S ) u '
∈
ij
Let d′ be an optimal solution to the inverse problem for G′. Then, the cut
cut in G′(d′). Consequently,
0 0
(S , S ) is a minimum
∑ - 0 0 ' ' '
∑ ij ≥ 0. (2)
' ' 0 0
'
(i,j) ∈(S , S )\(S , S ) d ij
(i,j) ∈(S , S )\(S , S ) d
Adding (1) and (2) yields
0 0 ' ' ' '
∑ (i,j) (S , S )\(S , S ) (u ij − d ij ) + ' ' 0 0
'
∑
'
(i,j) (S , S )\(S , S ) (d ij − u ij ) ≥ Excess(S 0 , u′), (3)
∈
∈
or, alternatively,
∑ || d −u || 1 + ∑ ' ' 0 0 || d'
− u ' || 1 ≥ Excess(S0 , u′). (4)
0 0 ' ' ' '
(i,j) ∈(S , S )\(S , S ) ij ij
(i,j) ∈(S , S )\(S , S )
It follows from (4) that ||d′ – u′|| 1 ≥ Excess(S 0 , u′), thereby establishing the lemma.
ij
ij
Theorem 2. The optimal value to the inverse minimum cut problem for G′ is Excess(S 0 , u′).
Proof. Let x′ be a maximum s-t flow in G′(u′). Let d ' ij = ' 0 0
x ij for each arc (i, j) ∈ [ S , S ] , and d ' ij = u ij
for each arc (i, j) ∈ A\[ S
0 , S
0 ] . Note that the flow x′ saturates the cut [ S
0 , S
0 ] in G(d ′), and hence d′ is
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inverse feasible. Also note that ||d′ – u′|| = Excess(S 0 , u′), and so by Lemma 2, the capacity vector d′ is
optimal for the inverse problem.
♦
We have explained in Theorem 1 how can we obtain an optimal solution d′ for the inverse
problem for G′ into an optimal solution d * for G. We have thus shown that the inverse minimum cut
problem reduces to solving a maximum flow problem. Currently, the fastest strongly polynomial bound to
solve the maximum flow problem is O(nm log(n 2 /m)) and is due to Goldberg and Tarjan [1986]. The
best weakly polynomial bound to solve the maximum flow problem is O(min{n 2/3 , m 1/2 }m log(n 2 /m) log
U) and is due to Goldberg and Rao [1997].
Our algorithm developed in this section is the same as obtained by Zhang and Cai [1998] and
Ahuja and Orlin [1998]; however, our proof is based on arguments involving combinatorial properties of
the max flow and min cut. We point out that the weighted version of the inverse minimum cut problem
cannot be transformed to a maximum flow problem; in fact, it is a minimum cost flow problem (Zhang
and Cai [1998] and Ahuja and Orlin [1998]).
3. THE MINIMAX INVERSE MINIMUM CUT PROBLEM
In this section, we study the unit weight minimax inverse minimum cut problem. In this problem,
0 0
the objective is to modify the capacity vector u to d* so that the s-t cut [ S , S ] becomes a minimum cut
in the network G and max{| d* ij – u ij |: (i, j) ∈ A} is minimum. We will show that if all arc capacities are
integer, then we can solve the minimax inverse minimum cut problem as a sequence of O(log(nU))
minimum cut problems. We will use the same notation in this section as used in Section 2.
0 0
To solve the inverse problem, we first delete the backward arcs in the cut [ S , S ] and denote the
resulting network by G ′ . Theorem 1 allows us to solve the inverse problem for G by solving it for G ′ . We
* *
next obtain a minimum cut [ S , S ] in G′ by solving a maximum flow problem. If
0 0
u′[ S , S ] =
* * 0 0
u′[ S , S ], then [ S , S ] is also a minimum cut in G ′ and d * = c; otherwise, arc capacities must be
modified.
Suppose that the optimal objective function value of the minimax inverse minimum cut problem
is δ′, that is, ||d′ - u|| ∞ = δ′. Observe that there exists an optimal solution of the inverse problem where the
modified capacity d ij
' of each arc (i, j) ∈
0 0
(S , S ) satisfies d ij
' = max{0, u ij - δ′} and the modified
capacity d ' ij of each arc (i, j) ∉
0 0
(S , S ) satisfies d ' ij = u ij + δ ′. To see this, notice that (i) if d ' ij > max{0,
0 0
u ij - δ′} for some arc (i, j) ∈ (S , S ) , then we can decrease d ij
' to (u ij - δ′) and
0 0
[ S , S ] remains a
minimum cut in G ′ , and (ii) if d ' ij < u ij + δ′ for some arc (i, j) ∉
0 0
(S , S ) , then we can increase d ' ij to (u ij
0 0
+ δ′) and (S , S ) remains a minimum cut in G ′ ; in either case the objective function value of the inverse
problem, ||d′ - u|| ∞ , does not increase. Let G ′ (δ) denote the network obtained from G ′ by defining the arc
capacities in the following manner:
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dij( δ)
=
R
S|
T|
max{ 0, uij
−δ} 0 0
for each arc ( i, j) ∈(S , S ),
uij
+ δ
0 0
for each arc (, i j) ∉(S , S ).
Each value of δ for which
0 0
[ S , S ] is a minimum cut in G′(δ) gives an inverse feasible capacity
vector d(δ). Let F(δ) be the capacity of the cut
0 0
[ S , S ] in G′(δ) minus the capacity of the minimum cut.
The minimax inverse problem is to find the minimum value of δ, say δ′, for which F(δ) = 0. It is easy to
see that F(δ) > 0 for all δ < δ′, and F(δ) = 0 for all δ ≥ δ′. Accordingly, one may use binary search in the
interval [0, U] to determine the value of δ′. We now claim that δ′ is a rational number whose denominator
is at most 2m. If the claim is true, we may terminate the binary search when the search interval is of
length at most 1/(4m 2 ) since such a search interval can contain at most one rational number with
denominator at most 2m. This implies that the number of iterations of the binary search is O(log m 2 U).
We now prove the claim. At the minimum value λ′ where F(δ′) = 0, there are two cuts
0 0
[ S , S ]
* *
and [ S , S ] that are both minimum cuts. As a function of δ, the capacity of these two cuts are a 1 + s 1 δ
and a 2 + s 2 δ, where s 1 ≠ s 2 , and s 1 ≥ - m, s 2 ≤ m, and all coefficients are integral. Thus a 1 + s 1 δ′ = a 2 +
s 2 δ′. Solving for δ′, we obtain a rational number with denominator at most 2m.
To summarize, we obtain a bound of log(4m 2 U) = O(log(nU)) on the number of iterations
performed by the binary search algorithm. At each search point, we need to solve a minimum cut problem
on a network with n nodes, m arcs, and where arc capacities can be non-integer (but can be made integer
by multiplying arc capacities by a number no larger than 4m 2 ). This gives an overall bound of O(T(n, m,
m 2 U) log(nU)) on the running time of the algorithm, where T(n, m, α) is the time needed to solve a
minimum cut problem on a network with n nodes, m arcs, and maximum (integer) arc capacity α.
Currently, T(n, m, α) = O(nm log(n 2 /m)), due to Goldberg and Tarjan [1986], is the best strongly
polynomial bound, and T(n, m, α) = O(min{n 2/3 , m 1/2 }m log(n 2 /m) log α), due to Goldberg and Rao
[1997], is the best available weakly polynomial bound.
The running time of our algorithm is not strongly polynomial. However, we can use the
techniques described in Radzik [1993] to show that we can solve the minimax inverse minimum cut
problem by solving O(n) minimum cut problems. This gives a strongly polynomial algorithm to solve the
minimax inverse minimum cut problem.
Weighted Version
We next study the weighted version of the minimax inverse minimum cut problem. In this
problem, the objective function of the inverse problem is to minimize max{w ij | d ' ij – u ij |: (i, j) ∈ A} where
0 0
w ij > 0 for each arc (i, j) ∈ A. As in the unit weight case, we delete the backward arcs in [S ,S ] to
obtain the network G′. Let δ′ denote the optimal objective function value of the inverse problem. Using
arguments similar to the unit weight case, it can be shown that there exists an optimal solution of the
6

inverse problem where d ' ij = max{0, u ij - δ′/w ij } for every arc (i, j) ∈
every arc (i, j) ∉
manner:
dij( δ)
=
R
S|
T|
0 0
(S ,S ) and d ' ij = u ij + δ′/w ij for
0 0
(S ,S ) . We define the network G′(δ) with arc capacities defined in the following
max{ 0, uij
−δ
/ wij} for each arc ( i, j) ∈ 0 0
(S , S ),
uij
+ δ / wij
for each arc ( i, j) ∉ 0 0
(S , S ).
We next use the binary search for δ in the interval [0, UW] to determine the minimum value δ′ of
0 0
δ, for which [S , S ] is a minimum cut in G ′ and terminate the binary search when the length of the
search interval is less than 1/4m 2 W 2 . At termination, d′ = d(δ′). The running time of this method is
O(T(n, m, m 2 W 2 U) log(nUW)), where T(n, m, α) is the time needed to solve a minimum cut problem with
n nodes, m arcs, and maximum arc capacity α. The running time of our algorithm is again not strongly
polynomial. However, we can use the techniques described in Radzik [1993] to show that we can solve
the minimax inverse minimum cut problem by solving O(m) minimum cut problems.
4. THE INVERSE MINIMUM COST FLOW PROBLEM
In this section, we study the inverse minimum cost flow problem. In the minimum cost flow
problem on a network G = (N, A), each arc (i, j) ∈ A has an associated cost c ij and an associated capacity
u ij , and each node i has an associated supply/demand b(i). We assume for notational convenience that for
any node pair i and j, both (i, j) and (j, i) do not belong to A. If b(i) ≥ 0, then node i is a supply node;
otherwise it is a demand node. The problem concerns determining the least cost shipment that meets the
demands at demand nodes of the network by the available supplies at the supply nodes by sending a flow
that honors arc capacities. In the inverse minimum cost flow problem, we are given a feasible flow x 0
which we wish to make an optimal flow by modifying the cost vector c to d * in a manner such that
||c – d * || 1 is minimum.
For the simplicity of presentation, we shall assume that x 0 = 0. There is no loss of generality in
this assumption because we can always satisfy it through a transformation of variables. If x 0 ≠ 0, then we
replace the flow x by y + x 0 . It is easy to see that in the transformed problem, we want to make y 0 = 0 as
an optimal solution of the minimum cost flow problem by perturbing the arc costs.
Our approach relies on the concept of the residual network G(x 0 ) defined with respect to a given
flow x 0 . To construct it, we consider each arc (i, j) ∈ A one by one, and add arcs to G(x 0 ) in the following
0
manner: (i) if x ij < uij , then we add the arc (i, j) of cost c ij to A(x 0 0
); (ii) if x ij > 0, then we add the arc (j,
0
i) with cost –c ij ; and (iii) 0 < x ij < uij , then we add the arcs (i, j) and (j, i) with costs c ij and –c ij
respectively. We also use the following well known result from the network flow theory.
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Property 2. The flow x 0 is an optimal flow of the minimum cost flow problem if and only if the residual
network G(x 0 ) contains no negative cycles (that is, negative cost directed cycles).
It is easy to see that for x 0 = 0, G(x 0 ) = G. We shall henceforth refer to G(x 0 ) by G. Property 2
implies that x 0 is an optimal flow in G if and only if G contains no negative cost cycle. Let π be any
vector of size n and c
π
ij
denote the arc reduced costs defined as cij
π = c ij - π i + π j . The following property
follows from this definition.
Property 3. For any directed cycle W in G, Σ (i,j)∈W cij
π = Σ (i,j)∈W
c ij . Consequently, if c π ij ≥ 0 for each
arc (i, j) ∈ A, then G does not contain any negative cycle.
Theorem 3. Let v* be the minimum cost of a collection of arc disjoint cycles in G. Then –v* is the
optimal objective function value for the inverse minimum cut problem.
Proof. Let W = {W 1 , W 2 , … , W K } denote any collection of arc-disjoint negative cycles in the network
G. Let c(W K ) denote the cost of the cycle W k , that is, c(W K ) = Σ (i,j)∈Wk c ij , and c(W) denote the cost of
K
the collection, that is, c(W) = ∑ k=1 c(W k ) . Observe that in an optimal cost vector d * , the sum of the
costs of arcs in any cycle W k must increase by at least -c(W K ) units in order to eliminate this negative
cycle. Hence
Σ (i,j)∈W k | d * ij - c ij | ≥ - c(W K ). (5)
Since all the cycles W 1 , W 2 , … , W K are arc-disjoint, it follows from (5) that
||d * - c|| 1 = Σ (i,j)∈A || d * ij - c ij || 1 ≥
K * k= 1 (i,j) ∈W || d ij − c ij || 1
k
∑ ∑ ≥ - k
K
∑ k=1 c(W ) = -c(W), (6)
establishing that - c(W) is a lower bound on || d * - c || 1 . Thus, – v* ≤ || d * - c || 1 . We will next establish
that this lower bound is achievable, which will prove the theorem.
* * *
1 2 K
Suppose that W* = { W,W,...,W } is a minimum cost collection of arc-disjoint cycles in G
with total cost equal to v*. This collection defines a flow in a unit capacity network (that is, the network
where each arc has capacity equal to 1) obtained by setting x ij = 1 for each arc (i, j) ∈ W* and x ij = 0 for
each arc (i, j) ∉ W*. By the duality theory for the minimum cost flow problem, there exists a vector π of
optimal dual variables, and arc reduced costs
cij
π = c ij - π i + π j so that
cij
π ≤ 0 for all (i, j) ∈ W*, (7a)
cij
π ≥ 0 for all (i, j) ∉ W*. (7b)
8

Let
*
d ij = c ij -c π ij for (i, j) ∈ W*, and let
*
d ij = c ij otherwise. Then, (i) for any arc (i, j) ∈ W*, d * ij -
π(i) + π(j) = (c ij - cij
π ) - π(i) + π(j) = cij
π - cij
π = 0, and (ii) for all any arc (i, j) ∉ W*, d * ij - π(i) + π(j) = c ij -
π(i) + π(j) = c π ij ≥ 0 (from (7b). We thus observe that with respect to the arc cost vector d * , the reduced
cost of each arc (i, j) ∈ A is nonnegative, and from Property 3, the flow x 0 = 0 is an optimal solution,
establishing that d * is inverse feasible. Moreover,
||d * - c|| 1 = -Σ (i,j)∈W * cij
π =
K
k= 1 (i,j) ∈W* cπ ij k
K
− ∑ ∑ = − ∑k= 1 ∑ (i,j) ∈W* c ij = -c(W) = - v * , (8)
k
where the first equality follows from the manner we define d*, the second equality follows from the fact
that W = { W,W,...,W }, the third equality follows from Property 3, and the fourth equality follows
* * *
1 2 K
from the definition of W * , and the fourth equality follows from the fact that c(W * ) = v*. The fact |d * - c| =
-c(W * ) in view of (6) implies that d * is an optimal cost vector for the inverse minimum cost flow problem.♦
Recall that we assumed at the beginning of the section that for any pair of nodes i and j in G, both
(i, j) and (j, i) are not present in A. We also assumed that x 0 = 0 and to satisfy this assumption, we needed
to perform a transformation of variables. This transformation would create some pairs of nodes i and j,
both the arcs (i j) and (j, i). We need to ensure that in this case d ij = -d ji , since the pair (i, j) and (j, i) refers
to same arc in G. Now notice that in our algorithm, we change arc costs of those arcs only that have cij
π <
0. Also notice that if c π ij < 0, then c π
ji > 0. Consequently, we change only one arc cost in the pair, and
count it just once in our objective. But when c ij changes then c ji must also change too by an equal and
opposite amount. If an arc cost c ij changes, its modified reduced cost becomes zero, and so the modified
reduced cost of the arc (j, i), and the solution continues to satisfy the optimality condition. This point also
applies to the inverse minimum cost flow problem under the L ∞ norm considered in the next section.
To summarize, we have reduced the inverse minimum cost flow problem into a minimum cost
flow problem in a unit capacity network. This result was first established by Zhang and Liu [1996] and
was also obtained by Ahuja and Orlin [1998]. We obtain a different proof of the same result. The
minimum cost flow problem in a unit capacity network is in general easier to solve than the general
minimum cost flow problem. Using the successive shortest path algorithm, this minimum cost circulation
problem can be solved in O(m(m + n log n)) time (see, for example, Ahuja, Magnanti, and Orlin [1993]).
Using the cost scaling algorithm, this minimum cost circulation problem can be solved in O(min{n 5/3 ,
m 3/2 }log(nC)) time, using the algorithm due to Gabow and Tarjan [1989].
5. THE INVERSE MINIMUM COST FLOW PROBLEM UNDER THE L ∞ NORM
In this section, we study the minimax inverse minimum cost flow problem. In this problem, our
objective is to modify the cost vector c to d * so that the given solution x 0 becomes a minimum cost flow
in G and max{|
*
d ij – cij |: (i, j) ∈ A} is minimum. We will subsequently refer to the objective function of
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this inverse problem by ||d * - c|| ∞ . We will show that the minimax inverse minimum cost flow problem
reduces to solving a minimum mean cycle problem in the residual network G(x 0 ) = (N, A(x 0 )).
As in Section 4, we will assume that x 0 = 0, and if not the necessary transformation has been done
to satisfy this assumption. Under this assumption, G(x 0 ) = G, and our objective is to change arc costs in G
so that G does not contain any negative cost cycles. For any directed cycle W, we denote by c(W) the cost
of this cycle; that is, c(W) =
. Clearly, if G does not contain any negative cost cycle with c
∑
(i,j) ∈W c ij
as the arc cost vector, then d * = c; otherwise, arc costs must be modified to eliminate such negative
cycles. We will henceforth assume that G contains a negative cost cycle. Let W * be a minimum mean
cycle in G, that is, a directed cycle in G for which the mean cost given by c(W)/|W| is minimum among
all directed cycles W in the network. Let µ * = c(W * )/|W * |. By our assumption, µ * < 0. The following
property is well known for the minimum mean cycle problem (Karp [1978] and Karp and Orlin [1981]):
Property 4. Let W* be a minimum mean cycle in G, and µ* denote its mean cost. Then there exists a
π
vector π of node potentials so that c ij = µ * π
for each arc (i, j) ∈ W, and c ij ≥ µ * for each arc (i, j) ∉ W.
Theorem 4. Let µ* denote the mean cost of a minimum mean cycle in G. Then, the optimal objective
function value for the inverse minimum cost flow problem under the L ∞ norm is max(0, -µ*),
Proof. We solve the minimum mean cycle problem in G and choose π as in Property 4. If µ* ≥ 0, then
x0 = 0 is an optimum flow in G and the theorem is true. Suppose instead that µ* < 0, and W* is the
minimum mean cycle in G. Let λ* be the optimum solution to the minimax inverse problem. We first
claim that λ* ≥ -µ*. For, if we reduce the cost of each arc by an amount greater than -µ*, then W would
become a negative cycle and x0 = 0 would not be an optimal flow in G. Hence, λ* ≥ -µ*.
We now claim that there exist a vector d* with ||d* – c|| ∞ = -µ* such that x* is optimal with respect
to d*. Select d * ij = c ij - cij
π if cij
π < 0, and d * ij = c ij otherwise. Observe that we decrease costs of only those
arcs that have negative reduced costs and by an amount equal to c π ij . Also observe from Property 4 that
≤ -µ * for each arc (i, j) ∉ W*. It follows from these
observations that ||d* – c|| ∞ = -µ*. Finally, using the same arguments as in Theorem 3, it can be shown that
x0 is optimal with respect to d*. This establishes that d* is an optimal solution of the minimax inverse
minimum cost flow problem.
♦
- c π
ij
= -µ * for each arc (i, j) ∈ W* and - c π
ij
To summarize, we have shown above that the minimax inverse minimum cost flow problem
reduces to solving a minimum mean cycle problem. Currently, the best available strongly polynomial
time algorithm to solve the minimum mean cycle problem is an O(nm) algorithm due to Karp [1978], and
the best available weakly polynomial time algorithm is an O( n m log(nC)) algorithm due to Orlin and
Ahuja [1992].
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Weighted Version
We next study the weighted version of the minimax inverse minimum cost flow problem. In this
problem, the objective function of the inverse problem is to minimize max{w ij | *
d ij – c ij |: (i, j) ∈ A},
where w ij > 0 for each (i, j) ∈ A. As earlier, we will assume that x0 = 0 and the graph G contains a
negative cost directed cycle.
Consider any negative cycle W in G. Let c(W) denote the cost of this cycle. Then, in an optimal
cost vector d*, costs of the arcs in the cycle W must increase by at least –c(W) units in order to eliminate
this negative cycle. Suppose that we increase the cost of an arc (i, j) ∈ W by α ij units. Then, Σ (i,j)∈W α ij
≥ -c(W). The impact of this change on the objective function value will be max{w ij α ij : (i, j) ∈ W}. This
impact will be minimum when each w ij α ij is the same for every arc (i, j) ∈ W, say, w ij α ij = T.
Substituting this result in Σ (i,j)∈W α ij ≥ -c(W) yields Σ (i,j)∈W T/w ij
≥ -c(W), which can be restated as T ≥
-c(W)/τ(W), where τ(W) = Σ (i,j)∈W τ ij with τ ij = 1/w ij . We thus observe that each negative cycle W
provides a lower bound of -c(W)/τ(W) on the optimal solution value of the inverse problem. Let W*
denote a directed cycle in G with the smallest value of c(W)/τ(W); we call such a cycle the minimum costto-weight
ratio cycle. Let µ* = c(W*)/τ(W*). Then, -µ* gives the greatest lower bound on the optimal
objective function value of the inverse problem. While solving the minimum cost-to-weight ratio cycle in
G(x0) with c ij 's as arc costs and τ ij 's as arc weights, we obtain both µ* and a vector π so that -c π ij /τ ij ≤ -µ*
for every arc (i, j) ∈ A. We define the arc cost vector d* as d * ij = c ij - c π ij if c π ij < 0, and d * ij = c ij
otherwise. Using similar arguments as in the unit weight case, it can be shown that d* is an optimal cost
vector for the weighted minimax inverse minimum cost flow problem in G.
Consequently, the weighted minimax inverse minimum cost flow problem can be solved by
solving a minimum cost-to-weight ratio cycle problem. The minimum cost-to-weight ratio problem can
be solved in O(nm log(nCW)) time using Lawler's algorithm, or in O(n4 log n) time using Meggido's
[1979] algorithm, where C = max{|c ij | : (i, j) ∈ A} and W = max{|w ij | : (i, j) ∈ A}. It can also be solved
in O(
n m log2(CW)) time using Goldberg's [1995] shortest path algorithm.
ACKNOWLEDGEMENTS
We thank the referees and the Associate Editor for their detailed and insightful comments that
helped to improve the presentation and to remove some subtle bugs. We also acknowledge the help of
Don Wagner who raised some perceptive and fundamental questions that led to the pursuit of the research
reported in this paper. The first author gratefully acknowledges the support from the National Science
Foundation Grant DMI-9900087 and the second author acknowledges the support from the Office of
Naval Research under contract ONR N00014-98-1-0317 as well as the National Science Foundation
Grant DMI-9820998.
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