| The INTPOINT Procedure |
Example 2.5: Nonarc Variables in the Side Constraints
You can verify that the FACT2 MAR GIZMO constraint has a left-hand-side activity of 3,470, which is not equal to the _RHS_ of this constraint. Not all of the 3,750 chips that can be supplied to factory 2 for March production are used. It is suggested that all the possible chips be obtained in March and those not used be saved for April production. Because chips must be kept in an air-controlled environment, it costs one dollar to store each chip purchased in March until April. The maximum number of chips that can be stored in this environment at each factory is 150. In addition, a search of the parts inventory at factory 1 turned up 15 chips available for their March production.
Nonarc variables are used in the side constraints that handle the limitations of supply of Gizmo chips. A nonarc variable called f1 unused chips has as a value the number of chips that are not used at factory 1 in March. Another nonarc variable, f2 unused chips, has as a value the number of chips that are not used at factory 2 in March. f1 chips from mar has as a value the number of chips left over from March used for production at factory 1 in April. Similarly, f2 chips from mar has as a value the number of chips left over from March used for April production at factory 2 in April. The last two nonarc variables have objective function coefficients of 1 and upper bounds of 150. The Gizmo side constraints are
3*prod f1 19 mar + 4*prod f1 25 mar + f1 unused chips = 2615
3*prod f2 19 apl + 4*prod f2 25 apl + f2 unused chips = 3750
3*prod f1 19 apl + 4*prod f1 25 apl - f1 chips from mar = 2600
3*prod f2 19 apl + 4*prod f2 25 apl - f2 chips from mar = 3750
f1 unused chips + f2 unused chips -
f1 chips from mar - f2 chips from mar >= 0
The last side constraint states that the number of chips not used in March is not less than the number of chips left over from March and used in April. Here, this constraint is called CHIP LEFTOVER.
The following SAS code creates a new data set containing constraint data. It seems that most of the constraints are now equalities, so you specify DEFCONTYPE=EQ in the PROC INTPOINT statement from now on and provide constraint type data for constraints that are not "equal to" type, using the default TYPEOBS value _TYPE_ as the _COLUMN_ variable value to indicate observations that contain constraint type data. Also, from now on, the default RHSOBS value is used.
title2 'Nonarc Variables in the Side Constraints';
data con6;
input _column_ &$17. _row_ &$15. _coef_ ;
datalines;
prod f1 19 mar FACT1 MAR GIZMO 3
prod f1 25 mar FACT1 MAR GIZMO 4
f1 unused chips FACT1 MAR GIZMO 1
_RHS_ FACT1 MAR GIZMO 2615
prod f2 19 mar FACT2 MAR GIZMO 3
prod f2 25 mar FACT2 MAR GIZMO 4
f2 unused chips FACT2 MAR GIZMO 1
_RHS_ FACT2 MAR GIZMO 3750
prod f1 19 apl FACT1 APL GIZMO 3
prod f1 25 apl FACT1 APL GIZMO 4
f1 chips from mar FACT1 APL GIZMO -1
_RHS_ FACT1 APL GIZMO 2600
prod f2 19 apl FACT2 APL GIZMO 3
prod f2 25 apl FACT2 APL GIZMO 4
f2 chips from mar FACT2 APL GIZMO -1
_RHS_ FACT2 APL GIZMO 3750
f1 unused chips CHIP LEFTOVER 1
f2 unused chips CHIP LEFTOVER 1
f1 chips from mar CHIP LEFTOVER -1
f2 chips from mar CHIP LEFTOVER -1
_TYPE_ CHIP LEFTOVER 1
back f1 19 apl TOTAL BACKORDER 1
back f1 25 apl TOTAL BACKORDER 1
back f2 19 apl TOTAL BACKORDER 1
back f2 25 apl TOTAL BACKORDER 1
back f1 19 may TOTAL BACKORDER 1
back f1 25 may TOTAL BACKORDER 1
back f2 19 may TOTAL BACKORDER 1
back f2 25 may TOTAL BACKORDER 1
_TYPE_ TOTAL BACKORDER -1
_RHS_ TOTAL BACKORDER 50
;
The nonarc variables f1 chips from mar and f2 chips from mar have objective function coefficients of 1 and upper bounds of 150. There are various ways in which this information can be furnished to PROC INTPOINT. If there were a TYPE list variable in the CONDATA= data set, observations could be in the form
_COLUMN_ _TYPE_ _ROW_ _COEF_
f1 chips from mar objfn . 1
f1 chips from mar upperbd . 150
f2 chips from mar objfn . 1
f2 chips from mar upperbd . 150
It is desirable to assign ID list variable values to all the nonarc variables:
data arc6;
set arc5;
drop oldcost oldfc oldflow _flow_ _fcost_ ;
run;
data arc6_b;
input _name_ &$17. _cost_ _capac_ factory key_id $ ;
datalines;
f1 unused chips . . 1 chips
f2 unused chips . . 2 chips
f1 chips from mar 1 150 1 chips
f2 chips from mar 1 150 2 chips
;
proc append force
base=arc6 data=arc6_b;
run;
proc intpoint
bytes=1000000
printlevel2=2
nodedata=node0 arcdata=arc6
condata=con6 defcontype=eq sparsecondata
conout=arc7;
run;
The following messages appear on the SAS log:
NOTE: Number of nodes= 20 .
NOTE: Number of supply nodes= 4 .
NOTE: Number of demand nodes= 4 .
NOTE: Total supply= 4350 , total demand= 4150 .
NOTE: Number of arcs= 64 .
NOTE: Number of nonarc variables= 4 .
NOTE: Number of <= side constraints= 1 .
NOTE: Number of == side constraints= 4 .
NOTE: Number of >= side constraints= 1 .
NOTE: Number of side constraint coefficients= 24 .
NOTE: The following messages relate to the equivalent
Linear Programming problem solved by the Interior
Point algorithm.
NOTE: Number of <= constraints= 1 .
NOTE: Number of == constraints= 25 .
NOTE: Number of >= constraints= 1 .
NOTE: Number of constraint coefficients= 160 .
NOTE: Number of variables= 72 .
NOTE: After preprocessing, number of <= constraints= 1.
NOTE: After preprocessing, number of == constraints= 24.
NOTE: After preprocessing, number of >= constraints= 1.
NOTE: The preprocessor eliminated 1 constraints from the
problem.
NOTE: The preprocessor eliminated 9 constraint coefficients
from the problem.
NOTE: 2 columns, 0 rows and 2 coefficients were added to
the problem to handle unrestricted variables,
variables that are split, and constraint slack or
surplus variables.
NOTE: There are 78 nonzero elements in A * A transpose.
NOTE: Of the 26 rows and columns, 15 are sparse.
NOTE: There are 87 nonzero superdiagonal elements in the
sparse rows of the factored A * A transpose. This
includes fill-in.
NOTE: There are 104 operations of the form
u[i,j]=u[i,j]-u[q,j]*u[q,i]/u[q,q] to factorize the
sparse rows of A * A transpose.
Iter Complem_aff Complem-ity Duality_gap Tot_infeasb Tot_infeasc Tot_infeasd
0 -1.000000 182185121 0.837584 55030 37757 47382
1 56730833 28510760 0.910015 5004.066395 3433.344938 6917.139928
2 9604878 2589633 0.665677 0 1.309672E-10 226.047223
3 328091 309636 0.191136 0 0 26.943297
4 135892 89815 0.063031 0 0 6.389904
5 62039 31604 0.022877 0 0 0
6 16881 7606.584128 0.005568 0 0 0
7 3753.426021 1918.980183 0.001408 0 0 0
8 709.549939 330.027670 0.000242 0 0 0
9 155.181759 36.697802 0.000026949 0 0 0
10 3.865957 0.421161 0.000000309 0 0 0
11 0.001557 0.000021177 1.557982E-11 0 0 0
NOTE: The Primal-Dual Predictor-Corrector Interior Point algorithm
performed 11 iterations.
NOTE: Objective = -1295542.742.
NOTE: The data set WORK.ARC7 has 68 observations and 14
variables.
NOTE: There were 68 observations read from the data set
WORK.ARC6.
NOTE: There were 8 observations read from the data set
WORK.NODE0.
NOTE: There were 31 observations read from the data set
WORK.CON6.
The optimal solution data set, CONOUT=ARC7, is given in Output 2.5.1.
proc print data=arc7;
var _tail_ _head_ _name_ _cost_ _capac_ _lo_
_flow_ _fcost_;
sum _fcost_;
run;
The optimal value of the nonarc variable f2 unused chips is 280. This means that although there are 3,750 chips that can be used at factory 2 in March, only 3,470 are used. As the optimal value of f1 unused chips is zero, all chips available for production in March at factory 1 are used. The nonarc variable f2 chips from mar also has zero optimal value. This means that the April production at factory 2 does not need any chips that could have been held in inventory since March. However, the nonarc variable f1 chips from mar has value of 20. Thus, 3,490 chips should be ordered for factory 2 in March. Twenty of these chips should be held in inventory until April, then sent to factory 1.
Output 2.5.1: CONOUT=ARC7
Copyright © 2008 by SAS Institute Inc., Cary, NC, USA. All rights reserved.