Hello Xiaomi,
> The value range is the problem I don't know. So far, the benchmark gives
> the result that value ranges from 0 to 2000. However, the value may
> increase in larger benchmark.
to avoid rounding problems you should reduce your "M" to the smallest
possible number.
E.g. if you had a problem like
var Xmax{j in J}, >= 0, <= ub[j];
var B{J}, binary;
s.t. c1 {i in I} :
sum{j in J} a[i,j] * Xmax[j] >= 1;
s.t. c2 {j in J} :
Xmax[j] <= M * B[j];
you could write instead
var Xmax{j in J}, >= 0, <= ub[j];
var B{J}, binary;
s.t. c1 {i in I} :
sum{j in J} a[i,j] * Xmax[j] >= 1;
s.t. c2 {j in J} :
Xmax[j] <= B[j]
* max{i in I : a[i,j] > 0} ( if a[i,j] < 1 / ub[j] then ub[j] else 1 /
a[i,j]);
>
> By the way, is it the only way to formulate the total number of non-zero
> value among all Xmax***_*** by following constrants:
>
> 0<= Xmax***_*** <= M B,
> where M is a large number( I used 999,999 in my case), and B is a binary
> variable.
>
> By minimizing sum of B, I can get the minimum total number of non-zero
> value. Is there another way to do so?
>
Reformulation will depend on the rest of the problem,
e.g. if you had a problem like
param a{I,J}, >=0;
var Xmax{j in J}, >= 0, <= ub[j];
var B{J}, binary;
minimize obj:
sum{j in J} B[j];
s.t. c1 {i in I} :
sum{j in J} a[i,j] * Xmax[j] >= 1;
s.t. c2 {j in J} :
Xmax[j] <= M * B[j];
you could write instead
param a{I,J}, >=0;
var B{J}, binary;
minimize obj:
sum{j in J} B[j];
s.t. c1 {i in I} :
sum{j in J} B[j] * ub[j] * a[i,j] >= 1;
Best regards
Xypron
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