Simple Pooling Problem
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The NLP Procedure

Example 7.7 Simple Pooling Problem

The following optimization problem is discussed in Haverly (1978) and in Liebman et al. (1986, pp. 127–128). Two liquid chemicals, $X$ and $Y$, are produced by the pooling and blending of three input liquid chemicals, $A$, $B$, and $C$. You know the sulfur impurity amounts of the input chemicals, and you have to respect upper limits of the sulfur impurity amounts of the output chemicals. The sulfur concentrations and the prices of the input and output chemicals are:

  • Chemical $A$: Concentration = 3%, Price= $6

  • Chemical $B$: Concentration = 1%, Price= $16

  • Chemical $C$: Concentration = 2%, Price= $10

  • Chemical $X$: Concentration $\leq $ 2.5%, Price= $9

  • Chemical $Y$: Concentration $\leq $ 1.5%, Price= $15

The problem is complicated by the fact that the two input chemicals $A$ and $B$ are available only as a mixture (they are either shipped together or stored together). Because the amounts of $A$ and $B$ are unknown, the sulfur concentration of the mixture is also unknown.

You know customers will buy no more than 100 units of X and 200 units of Y. The problem is determining how to operate the pooling and blending of the chemicals to maximize the profit. The objective function for the profit is

\begin{eqnarray*} \emph{profit} & = & \emph{cost}(x) \times \emph{amount}(x) + \emph{cost}(y) \times \emph{amount}(y) \\ & - & \emph{cost}(a) \times \emph{amount}(a) - \emph{cost}(b) \times \emph{amount}(b) - \emph{cost}(c) \times \emph{amount}(c) \end{eqnarray*}

There are three groups of constraints:

  1. The first group of constraint functions is the mass balance restrictions illustrated by the graph. These are four linear equality constraints:

    • $ \mi {amount}(a) + \mi {amount}(b) = \mi {pool\_ to\_ x} + \mi {pool\_ to\_ y}$

    • $ \mi {pool\_ to\_ x} + \mi {c\_ to\_ x} = \mi {amount}(x)$

    • $ \mi {pool\_ to\_ y} + \mi {c\_ to\_ y} = \mi {amount}(y)$

    • $ \mi {amount}(c) = \mi {c\_ to\_ x} + \mi {c\_ to\_ y}$

  2. You introduce a new variable, $ pool\_ s$, that represents the sulfur concentration of the pool. Using $ pool\_ s$ and the sulfur concentration of $C$ (2%), you obtain two nonlinear inequality constraints for the sulfur concentrations of $X$ and $Y$, one linear equality constraint for the sulfur balance, and lower and upper boundary restrictions for $ pool\_ s$:

    • $\mi {pool\_ s} \times \mi {pool\_ to\_ x} + 2 \, \mi {c\_ to\_ x} \leq 2.5 \, \mi {amount}(x)$

    • $\mi {pool\_ s} \times \mi {pool\_ to\_ y} + 2 \, \mi {c\_ to\_ y} \leq 1.5 \, \mi {amount}(y)$

    • $ 3 \, \mi {amount}(a) + 1 \, \mi {amount}(b) = \mi {pool\_ s} \times (\mi {amount}(a) + \mi {amount}(b))$

    • $1 \leq \mi {pool\_ s} \leq 3$

  3. The last group assembles the remaining boundary constraints. First, you do not want to produce more than you can sell; and finally, all variables must be nonnegative:

    • $\mi {amount}(x) \leq 100, \quad \mi {amount}(y) \leq 200$

    • $\mi {amount}(a), \mi {amount}(b), \mi {amount}(c), \mi {amount}(x), \mi {amount}(y) \geq 0$

    • $\mi {pool\_ to\_ x}, \mi {pool\_ to\_ y}, \mi {c\_ to\_ x}, \mi {c\_ to\_ y} \geq 0$

There exist several local optima to this problem that can be found by specifying different starting points. Using the starting point with all variables equal to 1 (specified with a PARMS statement), PROC NLP finds a solution with $ \mi {profit}=400$: 

proc nlp all;
   parms amountx amounty amounta amountb amountc
         pooltox pooltoy ctox ctoy pools = 1;
               bounds 0 <= amountx amounty amounta amountb amountc,
               amountx <= 100,
               amounty <= 200,
          0 <= pooltox pooltoy ctox ctoy,
          1 <= pools <= 3;
   lincon amounta + amountb = pooltox + pooltoy,
          pooltox + ctox = amountx,
          pooltoy + ctoy = amounty,
          ctox + ctoy    = amountc;
   nlincon nlc1-nlc2 >= 0.,
           nlc3 = 0.;
   max f;
   costa = 6; costb = 16; costc = 10;
   costx = 9; costy = 15;
   f = costx * amountx + costy * amounty
       - costa * amounta - costb * amountb - costc * amountc;
   nlc1 = 2.5 * amountx - pools * pooltox - 2. * ctox;
   nlc2 = 1.5 * amounty - pools * pooltoy - 2. * ctoy;
   nlc3 = 3 * amounta + amountb - pools * (amounta + amountb);
run;

The specified starting point was not feasible with respect to the linear equality constraints; therefore, a starting point is generated that satisfies linear and boundary constraints. Output 7.7.1 gives the starting parameter estimates.

Output 7.7.1: Starting Estimates

PROC NLP: Nonlinear Maximization

Optimization Start
Parameter Estimates
N Parameter Estimate Gradient
Objective
Function 		Gradient
Lagrange
Function 		Lower
Bound
Constraint 		Upper
Bound
Constraint
1 amountx 1.363636 9.000000 -0.843698 0 100.000000
2 amounty 1.363636 15.000000 -0.111882 0 200.000000
3 amounta 0.818182 -6.000000 -0.430733 0 .
4 amountb 0.818182 -16.000000 -0.542615 0 .
5 amountc 1.090909 -10.000000 0.017768 0 .
6 pooltox 0.818182 0 -0.669628 0 .
7 pooltoy 0.818182 0 -0.303720 0 .
8 ctox 0.545455 0 -0.174070 0 .
9 ctoy 0.545455 0 0.191838 0 .
10 pools 2.000000 0 0.068372 1.000000 3.000000


Value of Objective Function = 3.8181818182


Value of Lagrange Function = -2.866739915

PROC NLP: Nonlinear Maximization

Optimization Results
Parameter Estimates
N Parameter Estimate Gradient
Objective
Function 		Gradient
Lagrange
Function 		Active
Bound
Constraint 		 
1 amountx -1.40474E-11 9.000000 0 Lower BC  
2 amounty 200.000000 15.000000 -8.88178E-16 Upper BC  
3 amounta 5.561161E-16 -6.000000 0 Lower BC  
4 amountb 100.000000 -16.000000 1.065814E-14    
5 amountc 100.000000 -10.000000 -1.77636E-15    
6 pooltox 7.024225E-12 0 0 Lower BC  
7 pooltoy 100.000000 0 1.776357E-15    
8 ctox -2.10716E-11 0 1.776357E-15 Lower BC LinDep
9 ctoy 100.000000 0 5.329071E-15    
10 pools 1.000000 0 4.973799E-14 Lower BC LinDep


The starting point satisfies the four equality constraints, as shown in Output 7.7.2. The nonlinear constraints are given in Output 7.7.3.

Output 7.7.2: Linear Constraints

PROC NLP: Nonlinear Maximization

Linear Constraints
1 2.2204E-16 : ACT 0 == + 1.0000 * amounta + 1.0000 * amountb - 1.0000 * pooltox - 1.0000 * pooltoy
2 0 : ACT 0 == - 1.0000 * amountx + 1.0000 * pooltox + 1.0000 * ctox        
3 -1.11E-16 : ACT 0 == - 1.0000 * amounty + 1.0000 * pooltoy + 1.0000 * ctoy        
4 -1.11E-16 : ACT 0 == - 1.0000 * amountc + 1.0000 * ctox + 1.0000 * ctoy        


Output 7.7.3: Nonlinear Constraints

PROC NLP: Nonlinear Maximization

Values of Nonlinear Constraints
Constraint Value Residual Lagrange
Multiplier 		 
[ 5 ] nlc3 0 0 4.9441 Active NLEC  
[ 6 ] nlc1_G 0.6818 0.6818 .    
[ 7 ] nlc2_G -0.6818 -0.6818 -9.8046 Violat. NLIC  

PROC NLP: Nonlinear Maximization

Values of Nonlinear Constraints
Constraint Value Residual Lagrange
Multiplier 		 
[ 5 ] nlc3 1.11E-15 1.11E-15 6.0000 Active NLEC  
[ 6 ] nlc1_G 4.31E-16 4.31E-16 . Active NLIC LinDep
[ 7 ] nlc2_G 0 0 -6.0000 Active NLIC  


Output 7.7.4 shows the settings of some important PROC NLP options.

Output 7.7.4: Options

PROC NLP: Nonlinear Maximization

Minimum Iterations 0
Maximum Iterations 200
Maximum Function Calls 500
Iterations Reducing Constraint Violation 20
ABSGCONV Gradient Criterion 0.00001
GCONV Gradient Criterion 1E-8
ABSFCONV Function Criterion 0
FCONV Function Criterion 2.220446E-16
FCONV2 Function Criterion 1E-6
FSIZE Parameter 0
ABSXCONV Parameter Change Criterion 0
XCONV Parameter Change Criterion 0
XSIZE Parameter 0
ABSCONV Function Criterion 1.340781E154
Line Search Method 2
Starting Alpha for Line Search 1
Line Search Precision LSPRECISION 0.4
DAMPSTEP Parameter for Line Search .
FD Derivatives: Accurate Digits in Obj.F 15.653559775
FD Derivatives: Accurate Digits in NLCon 15.653559775
Singularity Tolerance (SINGULAR) 1E-8
Constraint Precision (LCEPS) 1E-8
Linearly Dependent Constraints (LCSING) 1E-8
Releasing Active Constraints (LCDEACT) .


The iteration history, given in Output 7.7.5, does not show any problems.

Output 7.7.5: Iteration History

PROC NLP: Nonlinear Maximization


Dual Quasi-Newton Optimization


Modified VMCWD Algorithm of Powell (1978, 1982)


Dual Broyden - Fletcher - Goldfarb - Shanno Update (DBFGS)


Lagrange Multiplier Update of Powell(1982)

Iteration   Restarts Function
Calls 		Objective
Function 		Maximum
Constraint
Violation 		Predicted
Function
Reduction 		Step
Size 		Maximum
Gradient
Element
of the
Lagrange
Function
1   0 19 -1.42400 0.00962 6.9131 1.000 0.783
2 ' 0 20 2.77026 0.0166 5.3770 1.000 2.629
3   0 21 7.08706 0.1409 7.1965 1.000 9.452
4 ' 0 22 11.41264 0.0583 15.5769 1.000 23.390
5 ' 0 23 24.84607 1.78E-15 496.1 1.000 147.6
6   0 24 378.22829 147.4 3316.8 1.000 840.4
7 ' 0 25 307.56787 50.9338 607.9 1.000 27.143
8 ' 0 26 347.24475 1.8328 21.9882 1.000 28.482
9 ' 0 27 349.49273 0.00915 7.1826 1.000 28.289
10 ' 0 28 356.58291 0.1083 50.2554 1.000 27.479
11 ' 0 29 388.70709 2.4280 24.7997 1.000 21.114
12 ' 0 30 389.30094 0.0157 10.0473 1.000 18.647
13 ' 0 31 399.19199 0.7996 11.1866 1.000 0.416
14 ' 0 32 400.00000 0.0128 0.1534 1.000 0.00087
15 ' 0 33 400.00000 7.38E-11 2.43E-10 1.000 365E-12

Optimization Results
Iterations 15 Function Calls 34
Gradient Calls 18 Active Constraints 10
Objective Function 400 Maximum Constraint Violation 7.381118E-11
Maximum Projected Gradient 0 Value Lagrange Function -400
Maximum Gradient of the Lagran Func 4.973799E-14 Slope of Search Direction -2.43334E-10

FCONV2 convergence criterion satisfied.


The optimal solution in Output 7.7.6 shows that to obtain the maximum profit of $400, you need only to produce the maximum 200 units of blending $Y$ and no units of blending $X$.

Output 7.7.6: Optimization Solution

PROC NLP: Nonlinear Maximization

Optimization Results
Parameter Estimates
N Parameter Estimate Gradient
Objective
Function 		Gradient
Lagrange
Function 		Active
Bound
Constraint 		 
1 amountx -1.40474E-11 9.000000 0 Lower BC  
2 amounty 200.000000 15.000000 -8.88178E-16 Upper BC  
3 amounta 5.561161E-16 -6.000000 0 Lower BC  
4 amountb 100.000000 -16.000000 1.065814E-14    
5 amountc 100.000000 -10.000000 -1.77636E-15    
6 pooltox 7.024225E-12 0 0 Lower BC  
7 pooltoy 100.000000 0 1.776357E-15    
8 ctox -2.10716E-11 0 1.776357E-15 Lower BC LinDep
9 ctoy 100.000000 0 5.329071E-15    
10 pools 1.000000 0 4.973799E-14 Lower BC LinDep


Value of Objective Function = 400


Value of Lagrange Function = 400


The constraints are satisfied at the solution, as shown in Output 7.7.7

Output 7.7.7: Linear and Nonlinear Constraints at the Solution

PROC NLP: Nonlinear Maximization

Linear Constraints Evaluated at Solution
1 ACT 0 = 0 + 1.0000 * amounta + 1.0000 * amountb - 1.0000 * pooltox - 1.0000 * pooltoy
2 ACT 3.8975E-17 = 0 - 1.0000 * amountx + 1.0000 * pooltox + 1.0000 * ctox        
3 ACT 0 = 0 - 1.0000 * amounty + 1.0000 * pooltoy + 1.0000 * ctoy        
4 ACT 0 = 0 - 1.0000 * amountc + 1.0000 * ctox + 1.0000 * ctoy        

Values of Nonlinear Constraints
Constraint Value Residual Lagrange
Multiplier 		 
[ 5 ] nlc3 1.11E-15 1.11E-15 6.0000 Active NLEC  
[ 6 ] nlc1_G 4.31E-16 4.31E-16 . Active NLIC LinDep
[ 7 ] nlc2_G 0 0 -6.0000 Active NLIC  

Linearly Dependent Active Boundary
Constraints
Parameter N Kind
ctox 8 Lower BC
pools 10 Lower BC

Linearly Dependent
Gradients of Active
Nonlinear Constraints
Parameter N
nlc3 6


The same problem can be specified in many different ways. For example, the following specification uses an INEST= data set containing the values of the starting point and of the constants COST, COSTB, COSTC, COSTX, COSTY, CA, CB, CC, and CD: 

data init1(type=est);
   input _type_ $ amountx amounty amounta amountb
         amountc pooltox pooltoy ctox ctoy pools
         _rhs_ costa costb costc costx costy
         ca cb cc cd;
   datalines;
parms  1 1 1 1 1 1 1 1 1 1
       . 6 16 10 9 15  2.5 1.5 2. 3.
;

proc nlp inest=init1 all;
   parms amountx amounty amounta amountb amountc
         pooltox pooltoy ctox ctoy pools;
   bounds 0 <= amountx amounty amounta amountb amountc,
               amountx <= 100,
               amounty <= 200,
          0 <= pooltox pooltoy ctox ctoy,
          1 <= pools <= 3;
   lincon amounta + amountb = pooltox + pooltoy,
          pooltox + ctox = amountx,
          pooltoy + ctoy = amounty,
          ctox + ctoy    = amountc;
   nlincon nlc1-nlc2 >= 0.,
           nlc3 = 0.;
   max f;
   f = costx * amountx + costy * amounty
       - costa * amounta - costb * amountb - costc * amountc;
   nlc1 = ca * amountx - pools * pooltox - cc * ctox;
   nlc2 = cb * amounty - pools * pooltoy - cc * ctoy;
   nlc3 = cd * amounta + amountb - pools * (amounta + amountb);
run;

The third specification uses an INEST= data set containing the boundary and linear constraints in addition to the values of the starting point and of the constants. This specification also writes the model specification into an OUTMOD= data set: 

data init2(type=est);
   input _type_ $ amountx amounty amounta amountb amountc
                 pooltox pooltoy ctox ctoy pools
                 _rhs_ costa costb costc costx costy;
   datalines;
parms      1   1  1  1  1    1   1   1  1  1
           .   6 16 10  9   15 2.5 1.5  2  3
lowerbd    0   0  0  0  0    0   0   0  0  1
           .   .  .  .  .    .   .   .  .  .
upperbd  100 200  .  .  .    .   .   .  .  3
           .   .  .  .  .    .   .   .  .  .
eq         .   .  1  1  .   -1  -1   .  .  .
           0   .  .  .  .    .   .   .  .  .
eq         1   .  .  .  .   -1   .  -1  .  .
           0   .  .  .  .    .   .   .  .  .
eq         .   1  .  .  .    .  -1   . -1  .
           0   .  .  .  .    .   .   .  .  .
eq         .   .  .  .  1    .   .  -1 -1  .
           0   .  .  .  .    .   .   .  .  .
;

proc nlp inest=init2 outmod=model all;
   parms amountx amounty amounta amountb amountc
         pooltox pooltoy ctox ctoy pools;
   nlincon nlc1-nlc2 >= 0.,
           nlc3 = 0.;
   max f;
   f = costx * amountx + costy * amounty
       - costa * amounta - costb * amountb - costc * amountc;
   nlc1 = 2.5 * amountx - pools * pooltox - 2. * ctox;
   nlc2 = 1.5 * amounty - pools * pooltoy - 2. * ctoy;
   nlc3 = 3 * amounta + amountb - pools * (amounta + amountb);
run;

The fourth specification not only reads the INEST=INIT2 data set, it also uses the model specification from the MODEL data set that was generated in the last specification. The PROC NLP call now contains only the defining variable statements: 

proc nlp inest=init2 model=model all;
   parms amountx amounty amounta amountb amountc
         pooltox pooltoy ctox ctoy pools;
   nlincon nlc1-nlc2 >= 0.,
           nlc3 = 0.;
   max f;
run;

All four specifications start with the same starting point of all variables equal to 1 and generate the same results. However, there exist several local optima to this problem, as is pointed out in Liebman et al. (1986, p. 130). 

proc nlp inest=init2 model=model all;
   parms amountx amounty amounta amountb amountc
         pooltox pooltoy ctox ctoy = 0,
         pools = 2;
   nlincon nlc1-nlc2 >= 0.,
           nlc3 = 0.;
   max f;
run;

This starting point with all variables equal to 0 is accepted as a local solution with $ \mi {profit} = 0$, which minimizes rather than maximizes the profit.

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