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Chapter 3
Introduction to Mixed Integer
Linear Problems
Many business problems need integer solutions (e.g. we need to decide how many em- ployees to assign to each shift, how many vehicles to purchase... ), so it is useful to introduce a new type of problem called (Mixed) Integer Linear Programming ((M)ILP): it is a linear programming problem where certain decision variables must assume only integer values. Hereafter we will often use ILP also to denote mixed integer linear problems.
For example, let’s resume the Blue Ridge Hot Tubs problem from the previous chapter:
max 350x 1 + 300x (^2) x 1 + x 2 200 9 x 1 + 6 x 2 1 ,566 (1,520) 12 x 1 + 16 x 2 2 ,880 (2,650) x 1 , x 2 0 x 1 , x 2 integer
whose optimal solution was already integer (x ⇤ 1 = 122, x ⇤ 2 = 78). However, by changing the two right hand sides (RHS) as in parentheses, we would get x ⇤ 1 = 116. 9444 , x ⇤ 2 =
- 9167 : so, integrality constraints have to be added to the model (from LP to ILP).
How can we address ILP?
- Solving its Linear Relaxation, i.e. eliminating the integrality constraints, and then rounding the obtained solution, e.g. x˜ 1 = 116, x˜ 2 = 77 (with profit = 63,700). However, this is not necessarily an optimum solution to the ILP. Notice that the optimum LP value, that is 64,306, is an upper bound to the optimum ILP value (for a maximization problem).
- Solving directly the ILP model (e.g. via the Excel solver).
26 3.1. Other examples
3.1 Other examples
3.1.1 A fixed-charge problem
Remington Manufacturing is planning its next production cycle. They produce 3 prod- ucts (say Product 1, Product 2 and Product 3), each of which must undergo machining, grinding and assembly operations to be completed. Table 3.1 summarizes the hours required by each unit of product, and the total hours available for each operation.
However, manufacturing units of Product 1 requires a setup operation on the production line that costs 1,000; similarly for Product 2 (800) and Product 3 (900) (fixed-charge costs). In order to solve the optimization problem we must determine the most profitable mix of products to produce.
Define the following decision variables:
- x (^) i : amount of Product i to be produced, i = 1, 2 , 3 ;
- y (^) i =
1 if x (^) i > 0 0 otherwise
, i = 1, 2 , 3.
The y (^) i are auxiliary binary variables, which will be used to express logical condi- tions.
Using these variables, an ILP model for Remington Manufacturing is:
max 48x 1 + 55x 2 + 50x 3 1000 y 1 800 y 2 900 y (^3) 2 x 1 + 3x 2 + 6x 3 600 6 x 1 + 3x 2 + 4x 3 300 5 x 1 + 6x 2 + 2x 3 400 x 1 50 y 1 linking x 2 67 y 2 constraints x 3 75 y (^3) x (^) i 0 , i = 1, 2 , 3 y (^) i 2 { 0 , 1 }, i = 1, 2 , 3
Prod 1 Prod 2 Prod 3 total hours machining 2 3 6 600 grinding 6 3 4 300 assembly 5 6 2 400 unitary profit 48 55 50
Table 3.1: Production requirements and availability for Remington Manufacturing
28 3.1. Other examples
- y (^3) k 2 { 0 , 1 }, k = 1, 2 , 3 for the additional constraint of Company 3.
The objective function, to be minimized, is the sum of the amounts of cement supplied by each company for each project, multiplied by their associated unitary costs, to be minimized:
min 120x 11 + 115x 12 + 130x 13 + 125x 14 +
- 100x 21 + 150x 22 + 110x 23 + 105x 24 +
- 140x 31 + 95x 32 + 145x 33 + 165x 34.
Each company may supply a total maximum amount of cement:
x 11 + x 12 + x 13 + x 14 525 x 21 + x 22 + x 23 + x 24 450 x 31 + x 32 + x 33 + x 34 550.
Each project must be supplied with a specific total amount of cement from any of the companies:
x 11 + x 21 + x 31 = 450 x 12 + x 22 + x 32 = 275 x 13 + x 23 + x 33 = 300 x 14 + x 24 + x 34 = 350.
The additional constraint placed by Company 1 on its bid can be expressed as follows: “if x (^1) j > 0 , then the associated boolean variable y (^1) j must be 1”:
x 11 525 y (^11) x 12 252 y (^12) x 13 525 y (^13) x 14 525 y 14 ;
and also “if y (^1) j = 1 then x (^1) j must be greater than or equal to 150”:
x 11 150 y (^11) x 12 150 y (^12) x 13 150 y (^13) x 14 150 y 14.
The additional constraint placed by Company 2 on its bid (“more than 200 tons to no more than one project”) can be expressed as follows:
Chapter 3. Introduction to Mixed Integer Linear Problems 29
x 21 200 + 250y (^21) x 22 200 + 250y (^22) x 23 200 + 250y (^23) x 24 200 + 250y (^24) y 21 + y 22 + y 23 + y 24 1.
The total amount of cement ordered from Company 3 can be only 200 (if y 31 = 1), 400 (if y 32 = 1), 550 (if y 33 = 1) or 0 (if y 31 = y 32 = y 33 = 0):
x 31 + x 32 + x 33 + x 34 = 200y 31 + 400y 32 + 550y (^33) y 31 + y 32 + y 33 1.
The overall ILP model is presented below. In order to implement the B&G Construction model as a spreadsheet model, we refer the interested reader to C. Ragsdale (2004): Section 6.16.
References C. Ragsdale (2004): Chapter 6
