The following example shows how to use PROC OPTMODEL to solve the example “Shortest Path Problem” in Chapter 6: The NETFLOW Procedure in SAS/OR 12.3 User's Guide: Mathematical Programming Legacy Procedures. The input data set is the same as in that example.
title 'Shortest Path Problem'; title2 'How to get Hawaiian Pineapples to a London Restaurant'; data aircost1; input ffrom&$13. tto&$15. _cost_; datalines; Honolulu Chicago 105 Honolulu San Francisco 75 Honolulu Los Angeles 68 Chicago Boston 45 Chicago New York 56 San Francisco Boston 71 San Francisco New York 48 San Francisco Atlanta 63 Los Angeles New York 44 Los Angeles Atlanta 57 Boston Heathrow London 88 New York Heathrow London 65 Atlanta Heathrow London 76 ;
The following PROC OPTMODEL statements read the data sets, build the linear programming model, solve the model, and output
the optimal solution to a SAS data set called SPATH
:
proc optmodel; str sourcenode = 'Honolulu'; str sinknode = 'Heathrow London'; set <str> NODES; num _supdem_ {i in NODES} = (if i = sourcenode then 1 else if i = sinknode then -1 else 0); set <str,str> ARCS; num _lo_ {ARCS} init 0; num _capac_ {ARCS} init .; num _cost_ {ARCS}; read data aircost1 into ARCS=[ffrom tto] _cost_; NODES = (union {<i,j> in ARCS} {i,j}); var Flow {<i,j> in ARCS} >= _lo_[i,j]; min obj = sum {<i,j> in ARCS} _cost_[i,j] * Flow[i,j]; con balance {i in NODES}: sum {<(i),j> in ARCS} Flow[i,j] - sum {<j,(i)> in ARCS} Flow[j,i] = _supdem_[i]; solve; num _supply_ {<i,j> in ARCS} = (if _supdem_[i] ne 0 then _supdem_[i] else .); num _demand_ {<i,j> in ARCS} = (if _supdem_[j] ne 0 then -_supdem_[j] else .); num _fcost_ {<i,j> in ARCS} = _cost_[i,j] * Flow[i,j].sol; create data spath from [ffrom tto] _cost_ _capac_ _lo_ _supply_ _demand_ _flow_=Flow _fcost_ _rcost_=(if Flow[ffrom,tto].rc ne 0 then Flow[ffrom,tto].rc else .) _status_=Flow.status; quit;
The statements use both single-dimensional (NODES) and multiple-dimensional (ARCS) index sets. The ARCS index set is populated
from the ffrom
and tto
data set variables in the READ DATA statement. To solve a shortest path problem, you solve a minimum cost network flow problem
that has a supply of one unit at the source node, a demand of one unit at the sink node, and zero supply or demand at all
other nodes, as specified in the declaration of the _SUPDEM_ numeric parameter. The SPATH
output data set contains most of the same information as in the PROC NETFLOW example, including reduced cost and basis status.
The _ANUMB_ and _TNUMB_ values do not apply here.
The PROC PRINT statements are similar to the PROC NETFLOW example:
proc print data=spath; sum _fcost_; run;
The output is displayed in Output 6.9.1.
Output 6.9.1: Output Data Set
Obs | ffrom | tto | _cost_ | _capac_ | _lo_ | _supply_ | _demand_ | _flow_ | _fcost_ | _rcost_ | _status_ |
---|---|---|---|---|---|---|---|---|---|---|---|
1 | Honolulu | Chicago | 105 | . | 0 | 1 | . | 0 | 0 | . | B |
2 | Honolulu | San Francisco | 75 | . | 0 | 1 | . | 0 | 0 | . | B |
3 | Honolulu | Los Angeles | 68 | . | 0 | 1 | . | 1 | 68 | . | B |
4 | Chicago | Boston | 45 | . | 0 | . | . | 0 | 0 | 61 | L |
5 | Chicago | New York | 56 | . | 0 | . | . | 0 | 0 | 49 | L |
6 | San Francisco | Boston | 71 | . | 0 | . | . | 0 | 0 | 57 | L |
7 | San Francisco | New York | 48 | . | 0 | . | . | 0 | 0 | 11 | L |
8 | San Francisco | Atlanta | 63 | . | 0 | . | . | 0 | 0 | 37 | L |
9 | Los Angeles | New York | 44 | . | 0 | . | . | 1 | 44 | . | B |
10 | Los Angeles | Atlanta | 57 | . | 0 | . | . | 0 | 0 | 24 | L |
11 | Boston | Heathrow London | 88 | . | 0 | . | 1 | 0 | 0 | . | B |
12 | New York | Heathrow London | 65 | . | 0 | . | 1 | 1 | 65 | . | B |
13 | Atlanta | Heathrow London | 76 | . | 0 | . | 1 | 0 | 0 | . | B |
177 |
The log is displayed in Output 6.9.2.
Output 6.9.2: OPTMODEL Log
NOTE: There were 13 observations read from the data set WORK.AIRCOST1. |
NOTE: Problem generation will use 4 threads. |
NOTE: The problem has 13 variables (0 free, 0 fixed). |
NOTE: The problem has 8 linear constraints (0 LE, 8 EQ, 0 GE, 0 range). |
NOTE: The problem has 26 linear constraint coefficients. |
NOTE: The problem has 0 nonlinear constraints (0 LE, 0 EQ, 0 GE, 0 range). |
NOTE: The OPTMODEL presolver is disabled for linear problems. |
NOTE: The problem is a pure network instance. The ALGORITHM=NETWORK option is |
recommended for solving problems with this structure. |
NOTE: The LP presolver value AUTOMATIC is applied. |
NOTE: The LP presolver removed all variables and constraints. |
NOTE: Optimal. |
NOTE: Objective = 177. |
NOTE: The data set WORK.SPATH has 13 observations and 11 variables. |