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I have a variable t, and I would like to have the following constraints:

a <= t <= b 

and 

(c <= t <= d) or (e <= t <= f)

Here is the code I used in Julia:

using JuMP, Cbc, StatsBase
import Random

model = Model(with_optimizer(Cbc.Optimizer));

@variable(model, T, Int);

@constraint(model, 5 <= T <= 1000);
@constraint(model, (100 <= T <= 200) | (1100 <= T <= 1300) );

# which fails with error:
# ERROR: LoadError: In `@constraint(model, (100 <= T[i] <= 200) | (1100 <= T[i] <= 1300))`: Unrecognized sense |

Is there a way to linearize these constraints? Or, alternatively, is there a non-linear solver that can deal with these constraints?

Timothée HENRY
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1 Answers1

4

The special case:

5 ≤ T ≤ 1000
and
(100 ≤ T ≤ 200) or (1100 ≤ T ≤ 1300)  

is easy. The result is just:

100 ≤ T ≤ 200

In general:

a ≤ t ≤ b 
and 
(c ≤ t ≤ d) or (e ≤ t ≤ f)

(where a,b,c,d,e,f are constants) can be linearized using binary variables:

a ≤ t ≤ b 
c + (a-c)δ ≤ t ≤ d + (b-d)δ 
e + (a-e)(1-δ) ≤ t ≤ f + (b-f)(1-δ)
δ ∈ {0,1}

Now the bonus question: how to do

a ≤ t ≤ b 
and 
(c ≤ t ≤ d) or (e ≤ t ≤ f) or (g ≤ t ≤ h)

Again, the only variable here is t. All other quantities are constants. Below is a direct extension of what we did before:

a ≤ t ≤ b 
c + (a-c)δ1 ≤ t ≤ d + (b-d)δ1 
e + (a-e)δ2 ≤ t ≤ f + (b-f)δ2
g + (a-g)δ3 ≤ t ≤ h + (b-h)δ3
δ1+δ2+δ3 = 2
δ1,δ2,δ3 ∈ {0,1}
Erwin Kalvelagen
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