The OPTLP Procedure

Example 10.1 Oil Refinery Problem

Consider an oil refinery scenario. A step in refining crude oil into finished oil products involves a distillation process that splits crude into various streams. Suppose there are three types of crude available: Arabian light (a_l), Arabian heavy (a_h), and Brega (br). These crudes are distilled into light naphtha (na_l), intermediate naphtha (na_i), and heating oil (h_o). These in turn are blended into two types of jet fuel. Jet fuel j_1 is made up of 30% intermediate naphtha and 70% heating oil, and jet fuel j_2 is made up of 20% light naphtha and 80% heating oil. What amounts of the three crudes maximize the profit from producing jet fuel (j_1, j_2)? This problem can be formulated as the following linear program:

$\text{[math]}$

The constraints "blend1" and "blend2" ensure that j_1 and j_2 are made with the specified amounts of na_i and na_l, respectively. The constraint "blend3" is actually the reduced form of the following constraints:

$\text{[math]}$

where h_o1 and h_o2 are dummy variables.

You can use the following SAS code to create the input data set ex1:

data ex1;
input field1 $ field2 $ field3$ field4 field5 $ field6 ;
datalines;
NAME        .          EX1      .     .         .
ROWS        .          .        .     .         .
 N          profit     .        .     .         .
 E          napha_l    .        .     .         .
 E          napha_i    .        .     .         . 
 E          htg_oil    .        .     .         .
 L          blend1     .        .     .         .
 L          blend2     .        .     .         .
 L          blend3     .        .     .         .
COLUMNS     .          .        .     .         .
.           a_l        profit   -175  napha_l   .035
.           a_l        napha_i  .100  htg_oil   .390
.           a_h        profit   -165  napha_l   .030
.           a_h        napha_i  .075  htg_oil   .300
.           br         profit   -205  napha_l   .045
.           br         napha_i  .135  htg_oil   .430
.           na_l       napha_l  -1    blend2    -1
.           na_i       napha_i  -1    blend1    -1
.           h_o        htg_oil  -1    blend3    -1
.           j_1        profit   350   blend1    .3
.           j_1        blend3   .7    .         . 
.           j_2        profit   350   blend2    .2
.           j_2        blend3   .8    .         .
BOUNDS      .          .        .     .         .
UP          .          a_l      110   .         .
UP          .          a_h      165   .         .
UP          .          br       80    .         .
ENDATA      .          .        .     .         .
;

You can use the following call to PROC OPTLP to solve the LP problem:

proc optlp data=ex1 
   objsense  = max 
   solver    = primal 
   primalout = ex1pout 
   dualout   = ex1dout
   printfreq = 1;
run;
%put &_OROPTLP_;

Note that the OBJSENSE=MAX option is used to indicate that the objective function is to be maximized.

The primal and dual solutions are displayed in Output 10.1.1.

Output 10.1.1 Example 1: Primal and Dual Solution Output

The OPTLP Procedure

Primal Solution

Obs	Objective Function ID	Variable Name	Variable Type	Objective Coefficient	Upper Bound	Variable Value	Variable Status	Reduced Cost
1	profit	a_l	D	-175	110	110.000	U	10.2083
2	profit	a_h	D	-165	165	0.000	L	-22.8125
3	profit	br	D	-205	80	80.000	U	2.8125
4	profit	na_l	N	0	1.7977E308	7.450	B	0.0000
5	profit	na_i	N	0	1.7977E308	21.800	B	0.0000
6	profit	h_o	N	0	1.7977E308	77.300	B	0.0000
7	profit	j_1	N	350	1.7977E308	72.667	B	0.0000
8	profit	j_2	N	350	1.7977E308	33.042	B	0.0000

The OPTLP Procedure

Dual Solution

Obs	Objective Function ID	Constraint Name	Constraint Type	Constraint Lower Bound	Constraint Upper Bound	Dual Variable Value	Constraint Status	Constraint Activity
1	profit	napha_l	E	.	.	0.000	L	0.00000
2	profit	napha_i	E	.	.	-145.833	U	0.00000
3	profit	htg_oil	E	.	.	-437.500	U	0.00000
4	profit	blend1	L	.	.	145.833	L	-0.00000
5	profit	blend2	L	.	.	0.000	B	-0.84167
6	profit	blend3	L	.	.	437.500	L	0.00000

The progress of the solution is printed to the log as follows.

Output 10.1.2 Log: Solution Progress

The OPTLP Procedure

NOTE: The problem EX1 has 8 variables (0 free, 0 fixed).

NOTE: The problem has 6 constraints (3 LE, 3 EQ, 0 GE, 0 range).

NOTE: The problem has 19 constraint coefficients.

WARNING: The objective sense has been changed to maximization.

NOTE: The OPTLP presolver value AUTOMATIC is applied.

NOTE: The OPTLP presolver removed 3 variables and 3 constraints.

NOTE: The OPTLP presolver removed 6 constraint coefficients.

NOTE: The presolved problem has 5 variables, 3 constraints, and 13 constraint

coefficients.

NOTE: The PRIMAL SIMPLEX solver is called.

Objective Entering Leaving

Phase Iteration Value Variable Variable

2 1 1.5411014E-8 j_1 blend1 (S)

2 2 2.6969274E-8 j_2 blend2 (S)

2 3 5.2372044E-8 br blend3 (S)

2 4 1347.916667 blend2 (S) br

NOTE: Optimal.

NOTE: Objective = 1347.91667.

NOTE: The data set WORK.EX1POUT has 8 observations and 10 variables.

NOTE: The data set WORK.EX1DOUT has 6 observations and 10 variables.

Note that the %put statement immediately after the OPTLP procedure prints value of the macro variable _OROPTLP_ to the log as follows.

Output 10.1.3 Log: Value of the Macro Variable _OROPTLP_

STATUS=OK SOLUTION_STATUS=OPTIMAL OBJECTIVE=1347.9166667

PRIMAL_INFEASIBILITY=2.888315E-15 DUAL_INFEASIBILITY=0

BOUND_INFEASIBILITY=0 ITERATIONS=4 PRESOLVE_TIME=0.00 SOLUTION_TIME=0.00

The value briefly summarizes the status of the OPTLP procedure upon termination.