Example 3.8: A Simple Integer Program

The LP Procedure

Example 3.8: A Simple Integer Program

Recall the linear programming problem presented in the "Introduction to Mathematical Programming" chapter. In that problem, a firm produces two products, chocolates and gumdrops, that are processed by four processes: cooking, color/flavor, condiments, and packaging. The objective is to determine the product mix that maximizes the profit to the firm while not exceeding manufacturing capacities. The problem is extended to demonstrate a use of integer-constrained variables.

Suppose that you must manufacture only one of the two products. In addition, there is a setup cost of 100 if you make the chocolates and 75 if you make the gumdrops. To identify which product will maximize profit, you define two zero-one integer variables, ICHOCO and IGUMDR, and you also define two new constraints,CHOCOLATE and GUM. The constraint labeled CHOCOLATE forces ICHOCO to equal one when chocolates are manufactured. Similarly, the constraint labeled GUM forces IGUMDR to equal one when gumdrops are manufactured. Also, you should include a constraint labeled ONLY_ONE that requires the sum of ICHOCO and IGUMDR to equal one. (Note that this could be accomplished more simply by including ICHOCO and IGUMDR in a SOSEQ set.) Since ICHOCO and IGUMDR are integer variables, this constraint eliminates the possibility of both products being manufactured. Notice the coefficients -10000, which are used to force ICHOCO or IGUMDR to 1 whenever CHOCO and GUMDR are nonzero. This technique, which is often used in integer programming, can cause severe numerical problems. If this driving coefficient is too large, then arithmetic overflows and underflow may result. If the driving coefficient is too small, then the integer variable may not be driven to one as desired by the modeler.

The objective coefficients of the integer variables ICHOCO and IGUMDR are the negatives of the setup costs for the two products. The following is the data set that describes this problem and the call to PROC LP to solve it:

   data;
      input _row_ $10. choco gumdr ichoco igumdr _type_ $ _rhs_;
      datalines;
   object        .25    .75   -100    -75 max        .
   cooking        15     40      0      0 le     27000
   color           0  56.25      0      0 le     27000
   package     18.75      0      0      0 le     27000
   condiments     12     50      0      0 le     27000
   chocolate       1      0 -10000      0 le         0
   gum             0      1      0 -10000 le         0
   only_one        0      0      1      1 eq         1
   binary          .      .      1      2 binary     .
   ;

   proc lp;
   run;

The solution shows that gumdrops are produced. See Output 3.8.1.

Output 3.8.1: Summaries and an Integer Programming Iteration Log

The LP Procedure

Problem Summary
Objective Function	Max object
Rhs Variable	_rhs_
Type Variable	_type_
Problem Density (%)	25.71

Variables	Number

Non-negative	2
Binary	2
Slack	6

Total	10

Constraints	Number

LE	6
EQ	1
Objective	1

Total	8

Output 3.8.1: (continued)

The LP Procedure

Integer Iteration Log
Iter	Problem	Condition	Objective	Branched	Value	Sinfeas	Active	Proximity
1	0	ACTIVE	397.5	ichoco	0.1	0.2	2	.
2	-1	SUBOPTIMAL	260	.	.	.	1	70
3	1	SUBOPTIMAL	285	.	.	.	0	.

Output 3.8.1: (continued)

The LP Procedure

Solution Summary
Integer Optimal Solution
Objective Value	285

Phase 1 Iterations	1
Phase 2 Iterations	5
Phase 3 Iterations	5
Integer Iterations	3
Integer Solutions	2
Initial Basic Feasible Variables	8
Time Used (seconds)	0
Number of Inversions	5

Epsilon	1E-8
Infinity	1.797693E308
Maximum Phase 1 Iterations	100
Maximum Phase 2 Iterations	100
Maximum Phase 3 Iterations	99999999
Maximum Integer Iterations	100
Time Limit (seconds)	120

Output 3.8.1: (continued)

The LP Procedure

Variable Summary
Col	Variable Name	Status	Type	Price	Activity	Reduced Cost
1	choco	DEGEN	NON-NEG	0.25	0	0
2	gumdr	BASIC	NON-NEG	0.75	480	0
3	ichoco		BINARY	-100	0	2475
4	igumdr	BASIC	BINARY	-75	1	0
5	cooking	BASIC	SLACK	0	7800	0
6	color		SLACK	0	0	-0.013333
7	package	BASIC	SLACK	0	27000	0
8	condiments	BASIC	SLACK	0	3000	0
9	chocolate		SLACK	0	0	-0.25
10	gum	BASIC	SLACK	0	9520	0

Output 3.8.1: (continued)

The LP Procedure

Constraint Summary
Row	Constraint Name	Type	S/S Col	Rhs	Activity	Dual Activity
1	object	OBJECTVE	.	0	285	.
2	cooking	LE	5	27000	19200	0
3	color	LE	6	27000	27000	0.0133333
4	package	LE	7	27000	0	0
5	condiments	LE	8	27000	24000	0
6	chocolate	LE	9	0	0	0.25
7	gum	LE	10	0	-9520	0
8	only_one	EQ	.	1	1	-75

The branch and bound tree can be reconstructed from the information contained in the integer iteration log. The column labeled Iter numbers the integer iterations. The column labeled Problem identifies the Iter number of the parent problem from which the current problem is defined. For example, Iter=2 has Problem=-1. This means that problem 2 is a direct descendent of problem 1. Furthermore, because problem 1 branched on ICHOCO, you know that problem 2 is identical to problem 1 with an additional constraint on variable ICHOCO. The minus sign in the Problem=-1 in Iter=2 tells you that the new constraint on variable ICHOCO is a lower bound. Moreover, because Value=.1 in Iter=1, you know that ICHOCO=.1 in Iter=1 so that the added constraint in Iter=2 is ICHOCO $\geq \lceil .1\rceil$ . In this way, the information in the log can be used to reconstruct the branch and bound tree. In fact, when you save an ACTIVEOUT= data set, it contains information in this format that is used to reconstruct the tree when you restart a problem using the ACTIVEIN= data set. See Example 3.10.

Note that if you defined a SOSEQ special ordered set containing the variables CHOCO and GUMDR, the integer variable ICHOCO and IGUMDR and the three associated constraints would not have been needed.

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