I'm using Julia with JuMP and Gurobi to solve an optimisation problem.
Two of my constraints have "if/else" like properties. I need values to be less than the sum of to real numbers but if the real number is larger than one I need it to be equal to one.
I made a function which returns the sum if they are smaller than one and the sum if they are not. I did a similar solution to my other constraint.
#constraint(m, [l in Locations, i in Locations, r in Rotor_size, wd in WD],
wake_rot[l, wd, r] <= isone(wake_loc, wd, l, i, r))
with isone is
function isone(wake_loc, wd, l, i, r)
a = wake_loc[wd, l, i] + Adj_wake[r]
if a > 1
return 1
else
return a
end
end
My model is unbounded(or infeasible) and I'm wondering if this can be the the source to my problems.
Related
I'm trying to solve a linear relaxation of a problem I've already solved with a Python library in order to see if it behaves in the same way in Xpress Mosel.
One of the index sets I'm using is not the typical c=1..n but a set of sets, meaning I've taken the 1..n set and have created all the combinations of subsets possible (for example the set 1..3 creates the set of sets {{1},{2},{3},{1,2},{2,3},{1,2,3}}).
In one of my constraints, one of the indexes must run inside each one of those subsets.
The respective code in Python is as follows (using the Gurobi library):
cluster=[1,2,3,4,5,6]
cluster1=[]
for L in range(1,len(cluster)+1):
for subset in itertools.combinations(cluster, L):
clusters1.append(list(subset))
ConstraintA=LinExpr()
ConstraintB=LinExpr()
for i in range(len(nodes)):
for j in range(len(nodes)):
if i<j and A[i][j]==1:
for l in range(len(clusters1)):
ConstraintA+=z[i,j]
for h in clusters1[l]:
restricao2B+=(x[i][h]-x[j][h])
model.addConstr(ConstraintA,GRB.GREATER_EQUAL,ConstraintB)
ConstraintA=LinExpr()
ConstraintB=LinExpr()
(In case the code above is confusing, which I suspect it to be)The constraint I'm trying to write is:
z(i,j)>= sum_{h in C1}(x(i,h)-x(j,h)) forall C1 in C
in which the C1 is each of those subsets.
Is there a way to do this in Mosel?
You could use some Mosel code along these lines (however, independently of the language that you are using, please be aware that the calculated 'set of all subsets' very quickly grows in size with an increasing number of elements in the original set C, so this constraint formulation will not scale up well):
declarations
C: set of integer
CS: set of set of integer
z,x: array(I:range,J:range) of mpvar
end-declarations
C:=1..6
CS:=union(i in C) {{i}}
forall(j in 1..C.size-1)
forall(s in CS | s.size=j, i in C | i > max(k in s) k ) CS+={s+{i}}
forall(s in CS,i in I, j in J) z(i,j) >= sum(h in s) (x(i,h)-x(j,h))
Giving this some more thought, the following version working with lists in place of sets is more efficient (that is, faster):
uses "mmsystem"
declarations
C: set of integer
L: list of integer
CS: list of list of integer
z,x: array(I:range,J:range) of mpvar
end-declarations
C:=1..6
L:=list(C)
qsort(SYS_UP, L) ! Making sure L is ordered
CS:=union(i in L) [[i]]
forall(j in 1..L.size-1)
forall(s in CS | s.size=j, i in L | i > s.last ) CS+=[s+[i]]
forall(s in CS,i in I, j in J) z(i,j) >= sum(h in s) (x(i,h)-x(j,h))
I'm trying to solve a model using Julia-JuMP. The following is the outline of the model that I created. Here, z[i,j] is a binary variable and d[i,j] is the cost for which z[i,j]=1.
My constraint creates an infinite number of constraint and hence I need to use a separation algorithm to solve it.
First, I solve the model without any constraint, so the answer to all variables z[i,j] and d[i,j] are zero.
Then, I'm including the separation algorithm (which is given inside the if condition). Even though I'm including if z_value == 0, z_values are not passing to it.
Am I missing something in the format of this model?
m = Model(solver=GurobiSolver())
#variable(m, z[N,N], Bin)
#variable(m, d[N,N]>=0)
#objective(m, Min, sum{ d[i,j]*z[i,j], i in N, j in N} )
z_value = getvalue(z)
d_value = getvalue(d)
if z_value == 0
statement
elseif z_value == 1
statement
end
#constraint(m, sum{z[i,j], i in N, j in N}>=2)
solve(m)
println("Final solution: [ $(getvalue(z)), $(getvalue(d)) ]")
You're multiplying z by d which both are variables, hence your model is non-linear,
Are the costs d[i,j] constant or really a variable of the problem ?
If so you need to use a non-linear solver
I want to model that departures from a node can only take place in a "every n hours" manner. I've started to model this using two variables - starttime[i,j,k] shows when vehicle k departured i with j as destination, x[i,j,k] is a binary variable having value 1 if vehicle k drove from i to j, and 0 otherwise. The model is:
maximize maxdrive: sum{i in V, j in V, k in K} traveltime[i,j]*x[i,j,k];
subject to TimeConstraint {k in K}:
sum{i in V, j in V} (traveltime[i,j]+servicetime[i])*x [i,j,k] <= 1440;
subject to StartTime{i in V,j in V, k in K}:
starttime[i,j,k] + traveltime[i,j] - 9000 * (1 - x[i,j,k]) <= starttime[j,i,k];
subject to yvar{i in V, j in V}:
sum{k in K} x[i,j,k] <= maxVisits[i,j];
subject to Constraint1{i in V, j in V, k in K, g in V, h in K}:
starttime[i,j,k] + TimeInterval[i]*x[i,j,k] <= starttime[i,g,h];
The constraint in question is "Constraint1" where i is the origin node, j the destination node, and k is the vehicle. The index g is used to show that the later departure can be to any destination node. TimeInterval corresponds to the interval intended, i.e. if TimeInterval at i is 2 hours, the starttime of the next vehicle to departure from i must not be less than 2 hours from previous departure. The origins corresponds to specific products (only available from said origin node) whereas I want the vehicles to not be bounded to a specific origin node - they should be able to jump between nodes to utilize backhauling etc. In other words, I want to conduct this constraint without having restraints on the vehicles themselves but rather the origin nodes.
The objective function to "maximize the traveltime" may seem strange, but the objective function is rather obsolete really. If the constraints are met, the solution is adequate. To maximize traveltime is merely an attempt to "force" the x variables to become 1.
The question is: how can I do this? With this formulation, all x[i,j,k] variables dissappears from the answer (without this constraint, some of the binary variables x becomes 1 and the other 0. The solution meets the maxVisits requirement. With the constraint all x variables becomes 0 and all starttimes becomes 0 as well. MINTO (The solver) doesn't state that the problem is infeasible either.
Also, how to separate the vehicles so the program recognizes that it is a comparison between all departures? I would rather to not include time dimensions, at it would give so much more variables.
EDIT: After trying a new model using a non-linear solver I've seen some strange results. Specifically, I'm using the limit 1440 (minutes) as an upper bound as to for how long a vehicle can operate each day. Using this model below the solution is 0 for every variable, but the starttime for all combinations of i,j,k is 720 (half of 1440). Does anyone have any clue in regards of what causing this solution? How did this constraint remove the link between starttime being higher than 0 requiring that x must be 1.
subject to StartTimeSelf{i in V, j in V, k in K, g in K, h in V}:
starttime[i,j,k]*x[i,j,k] + TimeInterval[i]*x[i,j,k] + y[i,k] <= starttime[i,h,g]*x[i,j,k];
I have a piece of code which is a significant bottleneck:
do s = 1,ns
msum = 0.d0
do k = 1,ns
msum = msum + tm(k,s)*f(:,:,k)
end do
m(:,:,s) = msum
end do
This is a simple matrix-vector product m=tm*f (where f is length k) for every x,y.
I thought about using a BLAS routine but i am not sure if any allows multiplying along a specific dimension (k). Do any of you have any good advice?
Unfortunately you do not mention the actual shape of f, i.e. the number of x and y. Since you mention this piece of code to be a bottleneck, you can and should replace msum and use the memory m(:,:,s) and spare the first step in you loop, e.g.
do s = 1,ns
m = tm(k,1)*f(:,:,k)
do k = 2, ns
m(:,:,s) = m(:,:,s) + tm(k,s)*f(:,:,k)
end do
end do
Secondly, a more general appraoch
There are ns summations of nK 2D matrices f(:,:,1:nK) by means of scalar factors that are stored in tm(:,1:ns). The goal is to store these sums in m(:,:,1:ns). Why not sum up element-wise wrt x and y to exploit contiguuos memory sections by means of the result? You already mentioned that you can redesign such that k is the first dimension in f, i.e. f(k,:,:).
Considering only the desired outcome, you ought to have ns 2D matrices m(:,:,1:ns) that are independent of each other (outer loop remains at it is). Lets drop this dimension for a moment. The problem then becomes:
m(:,:) = \sum_{k=1}^{ns} tm_k * f_k(:,:)
We should thus sum over k, e.g. have f(k,:,:) to determine m(:,:) as follows (note that I am adding the outer loop for s again):
nK = size(f, 1) ! the "k"s
nX = size(f, 2) ! the "x"s
nY = size(f, 3) ! the "y"s
m = 0.d0
do s = 1, ns
do ii = 1, nY
call DGEMV('N', nK, nY, &
1.d0, f(:,:,nY), 1, tm(:,s), 1, &
1.d0, m(:,nY,s), 1)
end do !ii
end do !s
See the documentation of DGEMV for more details on its usage.
Of course, the above advice of excluding the first step of the loop to spare the initialization by means of zeros may be applied at well.
Let's say we have a set : {1, 2, ..., n}.
How many subsets of order R exist S = {a_i1, a_i2, ...a_iR} that sum up to a certain number S?. What is the recursion for this problem?
Just define method to solve original problem. Parameters it receives are:
max number to use (n),
subset size (R),
subset sum (S),
and returns number of combinations.
To implement this method, first we have to check is it possible to make this request. It is not possible to fulfill task if:
subset size is larger than number of possible elements (R > n)
maximal possible sum is smaller than S. n + (n-1) + ... + (n-R+1) < S => R*((n-R) + (R+1)/2) < S
After that it is enough to try all possibilities for larger element that will go in subset. In python style it should be implemented like:
def combinations(n, R, S):
if R > n or R*((n-R) + (R+1)/2) < S:
return 0
c = 0
for i in xrange(R, n+1): # try i as maximal element in subset. It can go from R to n
# recursion n is i-1, since i is already used
# recursion R is R-1, since we put i in a set
# recursion S is S-i, since i is added to a set and we are looking for sum without it
c += combinations(i-1, R-1, S-i)
return c