## i can't solve this formulas....

hello, i have problem here.

> restart;
> with(linalg);
> NULL;
> fungsi1 := sum(d1[h]+b1[h], h = 1 .. 7);
> fungsi2 := sum(sum(d2[h, t]+b2[h, t], t = 1 .. 23), h = 1 .. 7);
> fungsi3 := sum(sum(d3[h, t]+b3[h, t], t = 1 .. 23), h = 1 .. 7);
> fungsi4 := sum(d4[k]+b4[k], k = 1 .. 3);
> fungsi := fungsi1+fungsi2+fungsi3+fungsi4;
> NULL;
> NULL;
> k1 := seq(sum(X[h, t], t = 1 .. 23) >= 9, h = 1 .. 6);
> k2 := seq(sum(Y[h, t], t = 1 .. 23) >= 2, h = 1 .. 6);
> k3 := seq(sum(Z[h, t], t = 1 .. 23) >= 2, h = 1 .. 6);
> NULL;
> k4 := seq(seq(X[h, t]+Y[h, t]+Z[h, t] <= 1, h = 1 .. 6), t = 1 .. 23);
> NULL;
> k5 := seq(seq(Z[h, t]+Z[h+1, t] <= 1, h = 1 .. 6), t = 1 .. 23);
> NULL;
> k6 := seq(sum(X[h, t]+Y[h, t]+Z[h, t], t = 1 .. 23) >= 5, h = 1 .. 7);
> k7 := seq(sum(X[h, t]+Y[h, t]+Z[h, t], t = 1 .. 23) <= 6, h = 1 .. 7);
> NULL;
> k8 := seq(sum(X[h, t], t = 1 .. 23)+b1[h]-d1[h] <= 15, h = 1 .. 6);
> k9 := seq(sum(Y[h, t], t = 1 .. 23)+b1[h]-d1[h] <= 4, h = 1 .. 6);
> k10 := seq(sum(Z[h, t], t = 1 .. 23)+b1[h]-d1[h] <= 4, h = 1 .. 6);
> NULL;
> k11 := seq(seq(Y[h, t]+Y[h+1, t]+b2[h, t]-d2[h, t] <= 1, t = 1 .. 23), h = 1 .. 6);
> NULL;
> k12 := seq(seq(Z[h, t]+Z[h+1, t]+b3[h, t]-d3[h, t] <= 1, t = 1 .. 23), h = 1 .. 6);
> NULL;
> k13 := sum(X[7, t], t = 1 .. 23)+b4[1]-d4[1] = 2;
> k14 := sum(Y[7, t], t = 1 .. 23)+b4[1]-d4[1] = 2;
> k15 := sum(Z[7, t], t = 1 .. 23)+b4[1]-d4[1] = 2;
> with(Optimization);
[ImportMPS, Interactive, LPSolve, LSSolve, Maximize, Minimize,

NLPSolve, QPSolve]
> CodeTools:-Usage(LPSolve(fungsi, {k1, k10, k11, k12, k13, k14, k15, k2, k3, k4, k5, k6, k7, k8, k9}, assume = {integer, nonnegative}));
Error, (in Optimization:-LPSolve) no feasible point found for LP subproblem

why it can be? please i need help.

## How to extract specific values from solution given...

If binary constraints are imposed on an optimization problem and LPSolve presents a solution, is it possible to extract the variables that have zero or one assigned to them? This would be most useful if there are many variables, for example...

If a solution is returned that looks like ...

[x[001]=0, x[101]=1, x[201]=0, x[301]=1, ....], how can I filter those solutions that equal zero?

## How to extract specific decision variable solution...

I was curious to know if one can extract a specific solution from a LPSolve routine.

As an example, consider the following output to a constrained linear problem. The objective value is 8 and the decision variable values (binary) are given.

Sol := [8, [w[1, 1] = 1., x[0, 0, 1] = 0, x[0, 1, 1] = 1, x[0, 2, 1] = 0, x[1, 0, 1] = 0, x[1, 1, 1] = 0, x[1, 2, 1] = 1, x[2, 0, 1] = 0, x[2, 1, 1] = 0, x[2, 2, 1] = 0, y[0, 0] = 0., y[0, 1] = 0., y[1, 1] = 2.]]

I am interested to know if we can isolate any variable value from this solution. I know that Sol[1] will return 8, and Sol[2] will return the remaining terms. But what if I wanted, say, x[1,2,1] alone?

## How to obtain >1 solution in LPSolve...

Greetings

How can I get LPSolve to output the unique subsets of {3,1,1,2,2,1} which sum to 5 (no recycling of set values)

partition.mw

(the code is a Yury/Love hybrid).

## How do I solve the example from Maple Help LPSolve...

When I try the example from Maple Help for LPSolve (I use Windows)

with(Optimization);
LPSolve(-4*x-5*y, {0 <= x, 0 <= y, x+2*y <= 6, 5*x+4*y <= 20});

I do not get the same solution like in the example: [-19., [x = 2.66666666666667, y = 1.66666666666667]]

Warning, problem appears to be unbounded
[0., [x = HFloat(0.0), y = HFloat(0.0)]]

My Professor uses the same version, but with Linux and do not have such problems. Why my installation does not solve the standart Help example?

Thank you

## Optimization which uses solutions of system...

how to do optimization for two equations in terms of two variables

LPSolve({eq1}, {eq2}, assume = {nonnegative});

eq1 is a rational function and eq2 is a very large rational function

after run , it return error, objective function must be specified as a linear polynomial or vector

da := [LengthSplit(Flatten([[1,m],[2,m2],[seq([i+1,close3[i][1]], i=2..4)]]),2)];
f := PolynomialInterpolation(da, z):
solution := solve(f=-z, z, explicit);
zz := [x1,x2,x3,x4];
sigma := symMonomial(zz); #sigma := [x1+x2+x3+x4, x1*x2+x1*x3+x1*x4+x2*x3+x2*x4+x3*x4, x1*x2*x3+x1*x2*x4+x1*x3*x4+x2*x3*x4, x1*x2*x3*x4]
sys1 := subs([x1=solution[1],x2=solution[2],x3=solution[3],x4=solution[4]], sigma[1]):
sys2 := subs([x1=solution[1],x2=solution[2],x3=solution[3],x4=solution[4]], sigma[2]):
sys3 := subs([x1=solution[1],x2=solution[2],x3=solution[3],x4=solution[4]], sigma[3]):
sys4 := subs([x1=solution[1],x2=solution[2],x3=solution[3],x4=solution[4]], sigma[4]):
da := [seq([i,close3[i][1]], i=1..5)];

with(Optimization):
LPSolve(sys1, {sys2}, assume = {nonnegative});

## Set the variable of equation to be a Boolean compo...

Hi all,

I have some "boolean variable" constraint equation like this:

a1*x1+a2*x2+...+an*xn>=b1*y1+b2*y2+...+bn*yn

where a1,a2,...,an and b1, b2, ..., bn are 1 or -1

These equations will be used in LPSolve or the other command to find a group of parameters which can fit them.

Now I used for-loop to deal with this kind of question, for example:

But there are more than 10 boolean variables in my case and It's very inefficient. On the other hand, using for-loop to determine the equation we solve in the command will lead to great confusion.

I think there should be some ways able to solve this kind of "boolean variables" question in Maple, such as, through assume command to define the type of "boolean variable".

But I have no idea how to do it.