Consider a distribution problem (from Jensen and Bard 2003) with three supply plants ( – ) and five demand points ( – ). Further information about the problem is as follows:

Additionally, there is a 5% “shipping loss” on each of the arcs between supply and demand nodes.

You can model this scenario as a generalized network. Since there are both fixed and varying supply and demand supdem values, you can transform this to a case where you need to address missing supply and demand simultaneously. As seen from Output 7.14.1, we have added two artificial nodes, Y and Z, with missing S supply value and missing D demand value, respectively. The extra production capability is depicted by arcs from node Y to the corresponding supply nodes, and the extra revenue generation capability of the demand points (and scrap revenue for ) is depicted by arcs to node Z.

The following SAS data set has the complete information about the arc costs, multipliers, and node supdem values:

data dnodes;
   input _node_ $ _sd_ ;
   missing S D;
datalines;
S1   700
S2     0
S3   200
D1  -200
D2  -300
D3  -200
D4  -150
D5   100
Y      S
Z      D
;

data darcs;
   input _from_ $ _to_ $ _cost_ _capac_ _mult_;
datalines;
S1  D1    3   200  0.95
S1  D2    3   200  0.95
S1  D3    6   200  0.95
S1  D4    7   200  0.95
S2  D1    7   200  0.95
S2  D2    2   200  0.95
S2  D4    5   200  0.95
S3  D2    6   200  0.95
S3  D4    4   200  0.95
S3  D5    7   200  0.95
D4  D3    4   200  0.95
Y   S2   10   300  .
Y   S3   14   100  .
S1  Z   -5    700  .
D2  Z   -20   100  .
D3  Z   -20   100  .
D5  Z   -25   250  .
;

You can solve this problem by using the following call to PROC NETFLOW:

title1 'The NETFLOW Procedure';
proc netflow 
   nodedata = dnodes 
   arcdata  = darcs 
   conout   = dsol;
run;

The optimal solution is displayed in Output 7.14.2.