Steve Jacobsen (jacobsen@ee.ucla.edu), Khosrow Moshirvaziri (moshir@ee.ucla.edu)
LINEAR REVERSE CONVEX MINIMIZATION PROBLEM (m=7; ntot=27, of which 13 are 
                                            involved in the reverse convex
                                            fuction):

** This problem is taken from "An Algorithm for Optimizing Network Flow
** Capacity Under Economies-of-Scale", JOTA, (1975), by Bansal and Jacobsen
** The optimal x, stated below, found by edge search is the same as 
** that reported in the above reference.

max        v                                    
   Ef     -vq =0
   f - x      < u
    g(x)      < B
          x,f >0                                            

where           

    E is the 7x13 node-arc incidence matrix of a directed graph,
    the m-vector q=(1,0,0,0,0,0,-1)', f is the 13-vector of flow
    variables, x is the 13-vector of arc capacity increments, u is the
    13-vector of initial arc capacities, g is the cost of the capacity
    increment vector x, v is the value of the flow from node0 to
    noden, and B is the value of the budget that's available for 
    capacity increases.  The problem is to allocate the
    budget B to the arcs of the graph so as to maximize the value of
    the flow from the source, node0, to the sink, noden.
    
    

E=
       0,1 0,2 0,4 1,3 1,5 2,1 2,3 2,5 3,4 3,n 4,5 4,n 5,n
node0    1   1   1   0   0   0   0   0   0   0   0   0   0
node1   -1   0   0   1   1  -1   0   0   0   0   0   0   0
node2    0  -1   0   0   0   1   1   1   0   0   0   0   0
node3    0   0   0  -1   0   0  -1   0   1   1   0   0   0
node4    0   0  -1   0   0   0   0   0  -1   0   1   1   0
node5    0   0   0   0  -1   0   0  -1   0   0  -1   0   1
noden    0   0   0   0   0   0   0   0   0  -1   0  -1  -1


  
B = 10

u =  1   
     2 
     3  
     3  
     5  
     0  
     5  
    20  
    15 
    10 
    10 
     6 
     4 
     
g(x) = x(0,1)/10+x(0,2)/4+sqrt(x(0,4))+(-1+sqrt(1+2*x(1,3)))+x(1,5)**0.75             
      +sqrt(x(2,1))+sqrt(x(2,3)/2)+sqrt(x(2,5))+x(3,4)**0.25+x(3,n)/3+x(4,5)/4
      +x(4,n)+sqrt(x(5,n)) 
     

optimal capacity expansion variables x =

 x(0,1) =  7.000000000000000e+00
 x(0,2) =  2.015603865200235e+01
 x(0,4) =  0.000000000000000e+00
 x(1,3) =  0.000000000000000e+00
 x(1,5) =  0.000000000000000e+00
 x(2,1) =  0.000000000000000e+00
 x(2,3) =  0.000000000000000e+00
 x(2,5) =  0.000000000000000e+00
 x(3,4) =  0.000000000000000e+00
 x(3,n) =  0.000000000000000e+00
 x(4,5) =  0.000000000000000e+00
 x(4,n) =  0.000000000000000e+00
 x(5,n) =  1.815603865200235e+01

 v* = -3.315603865200236e+01