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I have the following constraint I'm trying to model in Mixed Integer Programming with Python's PuLP module:

Given linear programming variables: x1,x2,y1,y2 where x1, x2, y1, y2 eventually solve to integer values

if (x1<=y2 and y1<=x2) then a=1 else b=0

I'm not sure how to handle the Logical AND in the IF condition. If the AND wasn't present I know I have to use the Big-M notation.

SashaZd
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  • I think the tag "linear-programming" was appropriate. Yes, this is a mixed integer *linear* programming problem but it is still linear. – ayhan Jan 10 '17 at 17:57

2 Answers2

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First of all, this is not Linear Programming but rather Mixed Integer Programming, since an AND constraint is not linear and neither is an implication. I also assumed that both a and b are binary variables. You can then reformulated your problem as follows:

x1    >  y2 + m*z1
y1    >  x2 + m*z2
a + 1 >= z1 + z2
a     <= z1
a     <= z2
a - b >= 0

Here, m needs to be some (negative) lower bound, i.e. m < x1-y2 and m < y1-x2. Both z1 and z2 are binary variables. To get around the < inequality you might want to add some small epsilon to the first two constraints:

x1    >= y2 + (m-eps)*z1 + eps
y1    >= x2 + (m-eps)*z2 + eps
SashaZd
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mattmilten
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  • Thanks! I changed the question to Mixed Integer Programming instead of the original (incorrect) Linear Programming. Also added a new tag for the same since there wasn't a mixed-integer-programming tag from before. – SashaZd Jan 10 '17 at 17:57
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    In this, `if a=1 then b=0`, and `if a=0 then b=1`. I want `b=0` only if we're in the `else` condition. And `a=1` only if we're in the `if-then` condition. In other cases, all values are allowed. Would `a-b>=0` as the last constraint work instead of `b=1-a`? – SashaZd Jan 10 '17 at 18:20
  • Yes, I suppose your formulation is correct. I misinterpreted your initial constraint. – mattmilten Jan 11 '17 at 10:35
  • Your formulation doesn't enforce the else condition. Only the if-condition. I put in a better answer below that uses a `(1-z)` extra condition on each of the first two conditions to make sure it works. I've still upvoted your answer because I used the z1, z2 concept in addition to the new constraints to allow for the AND condition. – SashaZd Jan 11 '17 at 22:22
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I found a formulation that works the IF-THEN-ELSE irrespective of the problem given. In the later part of the answer, I'm using the z1, z2 variables as described in @mattmilten's answer to handle the AND condition in the if statement

Assume the problem is the following specification: 

if α > 0 then β >= 0 else γ >= 0

then, 

α - z * U_α <= 0          # (1)
α - (1 - z)(L_α - 1) > 0  # (2)
β - (1 - z)L_β >= 0       # (3)
γ - z * L_γ >= 0          # (4)

where, 

L_α, L_β, L_γ # are constant lower bounds on α, β, γ (or values smaller than the lowest value they can take) 
U_α           # is a constant lower bounds on α 
z             # is a LP variable that can take values {0,1}

if z==1

Then equations (1) and (4) are redundant, and the then condition or (3) is enforced

if z==0

Then equations (2), and (3) are redundant, and the else condition or (4) is enforced

For this problem

We run this twice, the first time with α=α1 and the second with α=α2.

where, 

α1 = y2 - x1
z1 = decision variable for α1 with values {0,1} 
α2 = y1 - x2
z2 = decision variable for α2 with values {0,1} 

β # Currently unnecessary for my particular question. 
γ # Currently unnecessary for my particular question.

So our constraints become: 

α1 - z1 * U_α1 <= 0          # (1-1)
α1 - (1 - z1)(L_α1 - 1) > 0  # (1-2)
α2 - z2 * U_α2 <= 0          # (2-1)
α2 - (1 - z2)(L_α2 - 1) > 0  # (2-2)

If z1=1, then the first part of our if-condition is true. ie. x1<=y2

If z2=1, then the second part of our if-condition is true. ie. x2<=y1

Now, using @mattmilten's formulation for ensuring both conditions: 

a + 1 >= z1 + z2
a     <= z1
a     <= z2
a - b >= 0

This ensures that both z1 and z2 have to be >= 1 in order for a=1. If a=1 then b can be either b=1 or b=0 without violating the last condition.

If a=0 then we're in the else condition, and therefore b has to be 0.

SashaZd
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