The following example has been adapted from the example "A Multicommodity Transshipment Problem with Fixed Charges" in Chapter 4: The LP Procedure in SAS/OR 14.1 User's Guide: Mathematical Programming Legacy Procedures.
This example illustrates the use of PROC OPTMODEL to generate a mixed integer linear program to solve a multicommodity network
flow model with fixed charges. Consider a network with nodes N, arcs A, and a set C of commodities to be shipped between the nodes. The commodities are defined in the data set COMMODITY_DATA
, as follows:
title 'Multicommodity Transshipment Problem with Fixed Charges'; data commodity_data; do c = 1 to 4; output; end; run;
Shipping cost is for each of the four commodities c across each of the arcs . In addition, there is a fixed charge for the use of each arc . The shipping costs and fixed charges are defined in the data set ARC_DATA
, as follows:
data arc_data; input from $ to $ c1 c2 c3 c4 fx; datalines; farm-a Chicago 20 15 17 22 100 farm-b Chicago 15 15 15 30 75 farm-c Chicago 30 30 10 10 100 farm-a StLouis 30 25 27 22 150 farm-c StLouis 10 9 11 10 75 Chicago NY 75 75 75 75 200 StLouis NY 80 80 80 80 200 ; run;
The supply (positive numbers) or demand (negative numbers) at each of the nodes for each commodity c is shown in the data set SUPPLY_DATA
, as follows:
data supply_data; input node $ sd1 sd2 sd3 sd4; datalines; farm-a 100 100 40 . farm-b 100 200 50 50 farm-c 40 100 75 100 NY -150 -200 -50 -75 ; run;
Let define the flow of commodity c across arc . Let if arc is used, and 0 otherwise. Since the total flow on an arc must be at most the total demand across all nodes , you can define the trivial upper bound as
This model can be represented using the following mixed integer linear program:
Constraint (balance_con) ensures conservation of flow for both supply and demand. Constraint (fixed_charge_con) models the fixed charge cost by forcing if for some commodity .
The PROC OPTMODEL statements follow:
proc optmodel; set COMMODITIES; read data commodity_data into COMMODITIES=[c]; set <str,str> ARCS; num unit_cost {ARCS, COMMODITIES}; num fixed_charge {ARCS}; read data arc_data into ARCS=[from to] {c in COMMODITIES} <unit_cost[from,to,c]=col('c'||c)> fixed_charge=fx; print unit_cost fixed_charge; set <str> NODES = union {<i,j> in ARCS} {i,j}; num supply {NODES, COMMODITIES} init 0; read data supply_data nomiss into [node] {c in COMMODITIES} <supply[node,c]=col('sd'||c)>; print supply; var AmountShipped {ARCS, c in COMMODITIES} >= 0 <= sum {i in NODES} max(supply[i,c],0); /* UseArc[i,j] = 1 if arc (i,j) is used, 0 otherwise */ var UseArc {ARCS} binary; /* TotalCost = variable costs + fixed charges */ min TotalCost = sum {<i,j> in ARCS, c in COMMODITIES} unit_cost[i,j,c] * AmountShipped[i,j,c] + sum {<i,j> in ARCS} fixed_charge[i,j] * UseArc[i,j]; con flow_balance {i in NODES, c in COMMODITIES}: sum {<(i),j> in ARCS} AmountShipped[i,j,c] - sum {<j,(i)> in ARCS} AmountShipped[j,i,c] <= supply[i,c]; /* if AmountShipped[i,j,c] > 0 then UseArc[i,j] = 1 */ con fixed_charge_def {<i,j> in ARCS, c in COMMODITIES}: AmountShipped[i,j,c] <= AmountShipped[i,j,c].ub * UseArc[i,j]; solve; print AmountShipped; create data solution from [from to commodity]={<i,j> in ARCS, c in COMMODITIES: AmountShipped[i,j,c].sol ne 0} amount=AmountShipped; quit;
Although the PROC LP example used M = 1.0e6
in the FIXED_CHARGE_DEF constraint that links the continuous variable to the binary variable, it is numerically preferable
to use a smaller, data-dependent value. Here, the upper bound on AmountShipped[i,j,c]
is used instead. This upper bound is calculated in the first VAR statement as the sum of all positive supplies for commodity
c. The logical condition AmountShipped[i,j,k].sol ne 0
in the CREATE DATA statement ensures that only the nonzero parts of the solution appear in the SOLUTION
data set.
The problem summary, solution summary, and the output from the two PRINT statements are shown in Output 8.2.1.
Output 8.2.1: Multicommodity Transshipment Problem with Fixed Charges Solution Summary
Multicommodity Transshipment Problem with Fixed Charges |
[1] | [2] | [3] | unit_cost |
---|---|---|---|
Chicago | NY | 1 | 75 |
Chicago | NY | 2 | 75 |
Chicago | NY | 3 | 75 |
Chicago | NY | 4 | 75 |
StLouis | NY | 1 | 80 |
StLouis | NY | 2 | 80 |
StLouis | NY | 3 | 80 |
StLouis | NY | 4 | 80 |
farm-a | Chicago | 1 | 20 |
farm-a | Chicago | 2 | 15 |
farm-a | Chicago | 3 | 17 |
farm-a | Chicago | 4 | 22 |
farm-a | StLouis | 1 | 30 |
farm-a | StLouis | 2 | 25 |
farm-a | StLouis | 3 | 27 |
farm-a | StLouis | 4 | 22 |
farm-b | Chicago | 1 | 15 |
farm-b | Chicago | 2 | 15 |
farm-b | Chicago | 3 | 15 |
farm-b | Chicago | 4 | 30 |
farm-c | Chicago | 1 | 30 |
farm-c | Chicago | 2 | 30 |
farm-c | Chicago | 3 | 10 |
farm-c | Chicago | 4 | 10 |
farm-c | StLouis | 1 | 10 |
farm-c | StLouis | 2 | 9 |
farm-c | StLouis | 3 | 11 |
farm-c | StLouis | 4 | 10 |
Problem Summary | |
---|---|
Objective Sense | Minimization |
Objective Function | TotalCost |
Objective Type | Linear |
Number of Variables | 35 |
Bounded Above | 0 |
Bounded Below | 0 |
Bounded Below and Above | 35 |
Free | 0 |
Fixed | 0 |
Binary | 7 |
Integer | 0 |
Number of Constraints | 52 |
Linear LE (<=) | 52 |
Linear EQ (=) | 0 |
Linear GE (>=) | 0 |
Linear Range | 0 |
Constraint Coefficients | 112 |
Solution Summary | |
---|---|
Solver | MILP |
Algorithm | Branch and Cut |
Objective Function | TotalCost |
Solution Status | Optimal |
Objective Value | 42825 |
Relative Gap | 0 |
Absolute Gap | 0 |
Primal Infeasibility | 0 |
Bound Infeasibility | 0 |
Integer Infeasibility | 0 |
Best Bound | 42825 |
Nodes | 1 |
Iterations | 45 |
Presolve Time | 0.00 |
Solution Time | 0.01 |
[1] | [2] | [3] | AmountShipped |
---|---|---|---|
Chicago | NY | 1 | 110 |
Chicago | NY | 2 | 100 |
Chicago | NY | 3 | 50 |
Chicago | NY | 4 | 75 |
StLouis | NY | 1 | 40 |
StLouis | NY | 2 | 100 |
StLouis | NY | 3 | 0 |
StLouis | NY | 4 | 0 |
farm-a | Chicago | 1 | 10 |
farm-a | Chicago | 2 | 0 |
farm-a | Chicago | 3 | 0 |
farm-a | Chicago | 4 | 0 |
farm-a | StLouis | 1 | 0 |
farm-a | StLouis | 2 | 0 |
farm-a | StLouis | 3 | 0 |
farm-a | StLouis | 4 | 0 |
farm-b | Chicago | 1 | 100 |
farm-b | Chicago | 2 | 100 |
farm-b | Chicago | 3 | 0 |
farm-b | Chicago | 4 | 0 |
farm-c | Chicago | 1 | 0 |
farm-c | Chicago | 2 | 0 |
farm-c | Chicago | 3 | 50 |
farm-c | Chicago | 4 | 75 |
farm-c | StLouis | 1 | 40 |
farm-c | StLouis | 2 | 100 |
farm-c | StLouis | 3 | 0 |
farm-c | StLouis | 4 | 0 |