Example 14.2 Network Flow and Delay

The following example is taken from the user’s guide of the GINO program (Liebman et al.; 1986). A simple network of five roads (arcs) can be illustrated by a path diagram.

The five roads connect four intersections illustrated by numbered nodes. Each minute, vehicles enter and leave the network. The parameter refers to the flow from node to node . The requirement that traffic that flows into each intersection must also flow out is described by the linear equality constraint

     

In general, roads also have an upper limit on the number of vehicles that can be handled per minute. These limits, denoted , can be enforced by boundary constraints:

     

The goal in this problem is to maximize the flow, which is equivalent to maximizing the objective function , where is

     

The boundary constraints are

     
     

and the flow constraints are

     
     
     

The three linear equality constraints are linearly dependent. One of them is deleted automatically by the optimization subroutine. The following notation is used in this example:

     

Even though the NLPCG subroutine is used, any other optimization subroutine would also solve this small problem. The following code finds the maximum flow:

   proc iml;
      title 'Maximum Flow Through a Network';
      start MAXFLOW(x);
         f = x[4] + x[5];
         return(f);
      finish MAXFLOW;

      con = {  0.  0.  0.  0.  0.   .   . ,
              10. 30. 10. 30. 10.   .   . ,
               0.  1. -1.  0. -1.  0.  0. ,
               1.  0.  1. -1.  0.  0.  0. ,
               1.  1.  0. -1. -1.  0.  0. };
      x = j(1,5, 1.);
      optn = {1 3};
      call nlpcg(xres,rc,"MAXFLOW",x,optn,con);

The optimal solution is shown in the following output.

Optimization Results
Parameter Estimates
N Parameter Estimate Gradient
Objective
Function
Active
Bound
Constraint
1 X1 10.000000 0 Upper BC
2 X2 10.000000 0  
3 X3 10.000000 1.000000 Upper BC
4 X4 20.000000 1.000000  
5 X5 0 0 Lower BC

Finding the maximum flow through a network is equivalent to solving a simple linear optimization problem, and for large problems, the LP procedure or the NETFLOW procedure of the SAS/OR product can be used. On the other hand, finding a traffic pattern that minimizes the total delay to move vehicles per minute from node 1 to node 4 includes nonlinearities that need nonlinear optimization techniques. As traffic volume increases, speed decreases. Let be the travel time on arc and assume that the following formulas describe the travel time as decreasing functions of the amount of traffic:

     
     
     
     
     

These formulas use the road capacities (upper bounds), and you can assume that vehicles per minute have to be moved through the network. The objective is now to minimize

     

The constraints are

     
     
     
     
     

In the following code, the NLPNRR subroutine is used to solve the minimization problem:

   proc iml;
      title 'Minimize Total Delay in Network';
      start MINDEL(x);
         t12 = 5. + .1 * x[1] / (1. - x[1] / 10.);
         t13 = x[2] / (1. - x[2] / 30.);
         t32 = 1. + x[3] / (1. - x[3] / 10.);
         t24 = x[4] / (1. - x[4] / 30.);
         t34 = 5. + .1 * x[5] / (1. - x[5] / 10.);
         f = t12*x[1] + t13*x[2] + t32*x[3] + t24*x[4] + t34*x[5];
         return(f);
      finish MINDEL;

      con = {  0.  0.  0.  0.  0.   .   . ,
              10. 30. 10. 30. 10.   .   . ,
               0.  1. -1.  0. -1.  0.  0. ,
               1.  0.  1. -1.  0.  0.  0. ,
               0.  0.  0.  1.  1.  0.  5. };

      x = j(1,5, 1.);
      optn = {0 3};
      call nlpnrr(xres,rc,"MINDEL",x,optn,con);

The optimal solution is shown in the following output.

Optimization Results
Parameter Estimates
N Parameter Estimate Gradient
Objective
Function
Active
Bound
Constraint
1 X1 2.500000 5.777778  
2 X2 2.500000 5.702479  
3 X3 5.09575E-18 1.000000 Lower BC
4 X4 2.500000 5.702480  
5 X5 2.500000 5.777778  

The active constraints and corresponding Lagrange multiplier estimates (costs) are shown in the following output.

Linear Constraints Evaluated at Solution
1 ACT 4.4409E-16 = 0 + 1.0000 * X2 - 1.0000 * X3 - 1.0000 * X5
2 ACT 4.4409E-16 = 0 + 1.0000 * X1 + 1.0000 * X3 - 1.0000 * X4
3 ACT 0 = -5.0000 + 1.0000 * X4 + 1.0000 * X5        

First Order Lagrange Multipliers
Active Constraint Lagrange
Multiplier
Lower BC X3 0.924702
Linear EC [1] 5.702479
Linear EC [2] 5.777778
Linear EC [3] 11.480257