The NLP Procedure 
The following example is taken from the user’s guide of GINO (Liebman et al. 1986). A simple network of five roads (arcs) can be illustrated by the path diagram:
The five roads connect four intersections illustrated by numbered nodes. Each minute vehicles enter and leave the network. Arc refers to the road from intersection to intersection , and the parameter refers to the flow from to . The law that traffic flowing into each intersection must also flow out is described by the linear equality constraint
In general, roads also have an upper capacity, which is the number of vehicles which can be handled per minute. The upper limits can be enforced by boundary constraints
Finding the maximum flow through a network is equivalent to solving a simple linear optimization problem, and for large problems, PROC LP or PROC NETFLOW can be used. The objective function is
and the constraints are
The three linear equality constraints are linearly dependent. One of them is deleted automatically by the PROC NLP subroutines. Even though the default technique is used for this small example, any optimization subroutine can be used.
proc nlp all initial=.5; max y; parms x12 x13 x32 x24 x34; bounds x12 <= 10, x13 <= 30, x32 <= 10, x24 <= 30, x34 <= 10; /* what flows into an intersection must flow out */ lincon x13 = x32 + x34, x12 + x32 = x24, x24 + x34 = x12 + x13; y = x24 + x34 + 0*x12 + 0*x13 + 0*x32; run;
The iteration history is given in Output 6.9.1, and the optimal solution is given in Output 6.9.2.
Iteration  Restarts  Function Calls 
Active Constraints 
Objective Function 
Objective Function Change 
Max Abs Gradient Element 
Ridge  Ratio Between Actual and Predicted Change 


1  *  0  2  4  20.25000  19.2500  0.5774  0.0313  0.860  
2  *  0  3  5  30.00000  9.7500  0  0.0313  1.683 
Optimization Results  

Parameter Estimates  
N  Parameter  Estimate  Gradient Objective Function 
Active Bound Constraint 
1  x12  10.000000  0  Upper BC 
2  x13  20.000000  0  
3  x32  10.000000  0  Upper BC 
4  x24  20.000000  1.000000  
5  x34  10.000000  1.000000  Upper BC 
Finding a traffic pattern that minimizes the total delay to move vehicles per minute from node 1 to node 4 introduces nonlinearities that, in turn, demand 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), assuming vehicles per minute have to be moved through the network. The objective function is now
and the constraints are
Again, the default algorithm is used:
proc nlp all initial=.5; min y; parms x12 x13 x32 x24 x34; bounds x12 x13 x32 x24 x34 >= 0; lincon x13 = x32 + x34, /* flow in = flow out */ x12 + x32 = x24, x24 + x34 = 5; /* = f = desired flow */ t12 = 5 + .1 * x12 / (1  x12 / 10); t13 = x13 / (1  x13 / 30); t32 = 1 + x32 / (1  x32 / 10); t24 = x24 / (1  x24 / 30); t34 = 5 + .1 * x34 / (1  x34 / 10); y = t12*x12 + t13*x13 + t32*x32 + t24*x24 + t34*x34; run;
The iteration history is given in Output 6.9.3, and the optimal solution is given in Output 6.9.4.
Iteration  Restarts  Function Calls 
Active Constraints 
Objective Function 
Objective Function Change 
Max Abs Gradient Element 
Ridge  Ratio Between Actual and Predicted Change 


1  0  2  4  40.30303  0.3433  4.44E16  0  0.508 
Optimization Results  

Iterations  1  Function Calls  3 
Hessian Calls  2  Active Constraints  4 
Objective Function  40.303030303  Max Abs Gradient Element  4.440892E16 
Ridge  0  Actual Over Pred Change  0.5083585587 
Optimization Results  

Parameter Estimates  
N  Parameter  Estimate  Gradient Objective Function 
Active Bound Constraint 
1  x12  2.500000  5.777778  
2  x13  2.500000  5.702479  
3  x32  2.77556E17  1.000000  Lower BC 
4  x24  2.500000  5.702479  
5  x34  2.500000  5.777778 
The active constraints and corresponding Lagrange multiplier estimates (costs) are given in Output 6.9.5 and Output 6.9.6, respectively.
Linear Constraints Evaluated at Solution  

1  ACT  0  =  0  +  1.0000  *  x13    1.0000  *  x32    1.0000  *  x34 
2  ACT  4.4409E16  =  0  +  1.0000  *  x12  +  1.0000  *  x32    1.0000  *  x24 
3  ACT  0  =  5.0000  +  1.0000  *  x24  +  1.0000  *  x34 
First Order Lagrange Multipliers  

Active Constraint  Lagrange Multiplier 

Lower BC  x32  0.924702 
Linear EC  [1]  5.702479 
Linear EC  [2]  5.777778 
Linear EC  [3]  11.480257 
Projected Gradient  

Free Dimension 
Projected Gradient 
1  4.440892E16 
The projected Hessian matrix (shown in Output 6.9.8) is positive definite, satisfying the secondorder optimality criterion.
Projected Hessian Matrix  

X1  
X1  1.535309013 
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