Originally posted by: y82li

If y[t]=1 and yt+1=0, then xt+1=1, xt+2=1, xt+3=1;
or xt+1=1, xt+2=1, xt+3=1, xt+4=1, xt+5=1, xt+6=1;
or xt+1=1, xt+2=1, xt+3=1, xt+4=1, xt+5=1, xt+6=1, xt+7=1, xt+8=1, xt+9=1.

if y[t]=0, then xt+1=0.

x,y are binary variables

Thank you!
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Originally posted by: Eumpfenbach

working this through as a type but hopefully it will be clear.

introduce a new variable y(new). Constrain it as such:

y(new1) + 1 >= y(t) + (1 - y(t+1))
y(new1) <= y(t)
y(new1) <= (1 - y(t+1))

y(new1) is equal to 1 if y(t) = 1 and y(t+1) = 0.

Now create 3 more binary variables y(new2), y(new3), y(new4).

Constrain y(new1) = y(new2) + y(new3) + y(new4)

now

xt+1 + xt+2 + xt+3 = 3 * y(new2)
xt+1 + xt+2 + xt+3 + xt+4 + xt+5 + xt+6 = 6 * y(new3)
xt+1 + xt+2 + xt+3 + xt+4 + xt+5 + xt+6 + xt+7 + xt+8 +, xt+9 = 9 * y(new4)

Might not be the most efficient way to do this, but I think it should work.
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