The OPTLP Procedure
Example 13.1 Oil Refinery Problem
(View the complete code for this example.)
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:
subject to
and
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:
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
algorithm = primal
primalout = ex1pout
dualout = ex1dout
logfreq = 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 13.1.1.
Output 13.1.1: Example 1: Primal and Dual Solution Output
| Primal Solution |
| Obs | Objective Function ID | RHS ID | Variable Name | Variable Type | Objective Coefficient | Lower Bound | Upper Bound | Variable Value | Variable Status | Reduced Cost |
|---|---|---|---|---|---|---|---|---|---|---|
| 1 | profit | a_l | D | -175 | 0 | 110 | 110.000 | U | 10.2083 | |
| 2 | profit | a_h | D | -165 | 0 | 165 | 0.000 | L | -22.8125 | |
| 3 | profit | br | D | -205 | 0 | 80 | 80.000 | U | 2.8125 | |
| 4 | profit | na_l | N | 0 | 0 | 1.7977E308 | 7.450 | B | 0.0000 | |
| 5 | profit | na_i | N | 0 | 0 | 1.7977E308 | 21.800 | B | 0.0000 | |
| 6 | profit | h_o | N | 0 | 0 | 1.7977E308 | 77.300 | B | 0.0000 | |
| 7 | profit | j_1 | N | 350 | 0 | 1.7977E308 | 72.667 | B | 0.0000 | |
| 8 | profit | j_2 | N | 350 | 0 | 1.7977E308 | 33.042 | B | 0.0000 |
| Dual Solution |
| Obs | Objective Function ID | RHS ID | Constraint Name | Constraint Type | Constraint RHS | Constraint Lower Bound | Constraint Upper Bound | Dual Variable Value | Constraint Status | Constraint Activity |
|---|---|---|---|---|---|---|---|---|---|---|
| 1 | profit | napha_l | E | 0 | . | . | 0.000 | U | 0.00000 | |
| 2 | profit | napha_i | E | 0 | . | . | -145.833 | U | 0.00000 | |
| 3 | profit | htg_oil | E | 0 | . | . | -437.500 | U | 0.00000 | |
| 4 | profit | blend1 | L | 0 | . | . | 145.833 | L | 0.00000 | |
| 5 | profit | blend2 | L | 0 | . | . | 0.000 | B | -0.84167 | |
| 6 | profit | blend3 | L | 0 | . | . | 437.500 | L | -0.00000 |
The progress of the solution is printed to the log as follows.
Output 13.1.2: Log: Solution Progress
| 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. |
| NOTE: The MPS read time is 0.01 seconds. |
| WARNING: The objective sense has been changed to maximization. |
| NOTE: The LP presolver value AUTOMATIC is applied. |
| NOTE: The LP presolver time is 0.00 seconds. |
| NOTE: The LP presolver removed 3 variables and 3 constraints. |
| NOTE: The LP presolver removed 6 constraint coefficients. |
| NOTE: The presolved problem has 5 variables, 3 constraints, and 13 constraint |
| coefficients. |
| NOTE: The LP solver is called. |
| NOTE: The Primal Simplex algorithm is used. |
| Objective Entering Leaving |
| Phase Iteration Value Time Variable Variable |
| P 2 1 0.000000E+00 0 j_1 blend1 (S) |
| P 2 2 2.022784E-03 0 j_2 blend2 (S) |
| P 2 3 3.902347E-03 0 br blend3 (S) |
| P 2 4 4.025073E-03 0 a_l a_l |
| P 2 5 1.202248E+03 0 blend2 (S) br |
| P 2 6 1.347921E+03 0 |
| D 2 7 1.347917E+03 0 |
| NOTE: Optimal. |
| NOTE: Objective = 1347.9166667. |
| NOTE: The Primal Simplex solve time is 0.00 seconds. |
| 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 13.1.3: Log: Value of the Macro Variable _OROPTLP_
| STATUS=OK ALGORITHM=PS SOLUTION_STATUS=OPTIMAL OBJECTIVE=1347.9166667 |
| PRIMAL_INFEASIBILITY=3.552714E-15 DUAL_INFEASIBILITY=0 BOUND_INFEASIBILITY=0 |
| ITERATIONS=7 PRESOLVE_TIME=0.00 SOLUTION_TIME=0.00 |
The value briefly summarizes the status of the OPTLP procedure upon termination.
Alternatively, you can use an MPS-format file instead of a data set. Using the MPS-format file is typically faster than using the data set for large instances. You can use the following file ex1.mps:
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 mpsfile='ex1.mps'
objsense = max
algorithm = primal
primalout = ex1pout
dualout = ex1dout
printlevel = 0
logfreq = 1;
run;
%put &_OROPTLP_;
The output is the same as when you use the data set for input.