I am trying to create a linear optimization model. I have a set that looks like this:
si=[1,51,39,400909,1244]
sj=[31,47,5]
The numbers in this set represent codes. I am trying to loop through the set to add a constraint to my model, but I do not want to loop through the sets using their values, I want to loop through the sets based on their indices.
Here is the code I have now:
si=[1,51,39,400909,1244]
sj=[31,47,5]
c= [3 5 2;
4 3 5;
4 5 3;
5 4 3;
3 5 4]
b= [80;
75;
80;
120;
60]
# x_ij >= 0 ∀ i = 1,...,5, j = 1,...,3
#defVar(m, x[i in si,j in sj] >= 0)
#setObjective(m,Min,sum{c[i,j]*x[i,j],i in si, j in sj})
# ∀j = 1,...,3
for j in sj
#addConstraint(m, sum{x[i,j],i in si} <= 480)
end
for i in si
#addConstraint(m, sum{x[i,j],j in sj} >= b[i])
end
I keep getting an error because the numbers in the sets are too big. Does anyone know how to loop through the indices instead? Or does anyone have another way to do this?
I am also having trouble printing my solution. Here is my code:
for i in n
for j in p
println("x",i,",",j,"= ", getValue(x[i,j]))
end
end (incorporating Iain Dunning's answer from below)
However the output only reads
Objective value: 1165.0
x5,3= 0.0
Do you know how to fix the output so I can read the values of my variables?
The code you have posted doesn't work because you are trying to index c by, e.g. 400909,47. Try this:
n = length(si)
p = length(sj)
#variable(m, x[i=1:n,j=1:p] >= 0)
#objective(m,Min,sum{c[i,j]*x[i,j],i=1:n,j=1:p})
for j in 1:p
#constraint(m, sum{x[i,j],i=1:n} <= 480)
end
for i in 1:n
#constraint(m, sum{x[i,j],j=1:p} >= b[i])
end
Related
I'm trying to translate a C routine from an old sound synthesis program into R, but have indexing issues which I'm struggling to understand (I'm a beginner when it comes to using loops).
The routine creates an exponential lookup table - the vector exptab:
# Define parameters
sinetabsize <- 8192
prop <- 0.8
BP <- 10
BD <- -5
BA <- -1
# Create output vector
exptab <- vector("double", sinetabsize)
# Loop
while(abs(BD) > 0.00001){
BY = (exp(BP) -1) / (exp(BP*prop)-1)
if (BY > 2){
BS = -1
}
else{
BS = 1
}
if (BA != BS){
BD = BD * -0.5
BA = BS
BP = BP + BD
}
if (BP <= 0){
BP = 0.001
}
BQ = 1 / (exp(BP) - 1)
incr = 1 / sinetabsize
x = 0
stabsize = sinetabsize + 1
for (i in (1:(stabsize-1))){
x = x + incr
exptab [[sinetabsize-i]] = 1 - (BQ * (exp(BP * x) - 1))
}
}
Running the code gives the error:
Error in exptab[[sinetabsize - i]] <- 1 - (BQ * (exp(BP * x) - 1)) :
attempt to select less than one element in integerOneIndex
Which, I understand from looking at other posts, indicates an indexing problem. But, I'm finding it difficult to work out the exact issue.
I suspect the error may lie in my translation. The original C code for the last few lines is:
for (i=1; i < stabsize;i++){
x += incr;
exptab[sinetabsize-i] = 1.0 - (float) (BQ*(exp(BP*x) - 1.0));
}
I had thought the R code for (i in (1:(stabsize-1))) was equivalent to the C code for (i=1; i< stabsize;i++) (i.e. the initial value of i is i = 1, the test is whether i < stabsize, and the increment is +1). But now I'm not so sure.
Any suggestions as to where I'm going wrong would be greatly appreciated!
As you say, array indexing in R starts at 1. In C it starts at zero. I reckon that's your problem. Can sinetabsize-i ever get to zero?
It is not possible for Julia to solve it when I use a[i, j] = 1. how can i get julia to solve this problem?
using JuMP
using GLPK
u = [1 2 3 ; 1 2 3 ; 1 2 3]
m = Model(GLPK.Optimizer)
#variable(m, a[1:3,1:3], Bin)
#objective(m, Max, sum(u[i,j]*a[i,j] for i=1:3, j=1:3))
#constraint(m, [a[i,j]=1], sum(a[i:j][i:j]) == 1)
solution = optimize!(m)
opt_value = value.(a)
in the line
#constraint(m, [a[i,j]=1], sum(a[i:j][i:j]) == 1)
1) You're trying to set a variable, not to test equality, instead use
a[i,j] == 1
2) i and j are undefined. Without a minimal example to run, I would say according to the previous line of your code, I would say something like
sum(<what-to-sum-here> for i=1:3, j=1:3)
Or loop on the list of index you want to use if not the proper one.
I am working in an optimization problem (A*v = b) where I would like to rank a set of alternatives X = {x1,x2,x3,x4}. However, I have the following normalization constraint: |v[i] - v[j]| <= 1, which can be in the form -1 <= v[i] - v[j] <= 1.
My code is as follows:
import cvxpy as cp
n = len(X) #set of alternatives
v = cp.Variable(n)
objective = cp.Minimize(cp.sum_squares(A*v - b))
constraints = [0 <= v]
#Normalization condition -1 <= v[i] - v[j] <= 1
for i in range(n):
for j in range(n):
constraints = [-1 <= v[i]-v[j], 1 >= v[i]-v[j]]
prob = cp.Problem(objective, constraints)
# The optimal objective value is returned by `prob.solve()`.
result = prob.solve()
# The optimal value for v is stored in `v.value`.
va2 = v.value
Which outputs:
[-0.15 0.45 -0.35 0.05]
Result, which is not close to what should be and even have negative values. I think, my code for the normalization contraint most probably is wrong.
You are not appending your constraints, instead you are overwriting them each time. Instead of this line
constraints = [-1 <= v[i]-v[j], 1 >= v[i]-v[j]]
You should have
constraints += [-1 <= v[i]-v[j], 1 >= v[i]-v[j]]
For cleanliness you may want to change this
for i in range(n):
for j in range(n):
To only consider each pair once:
for i in range(n):
for j in range(i+1, n):
I want to solve a mathematical problem in a fastest possible way.
I have a set of natural numbers between 1 to n, for example {1,2,3,4,n=5} and I want to calculate a formula like this:
s = 1*2*3*4+1*2*3*5+1*2*4*5+1*3*4*5+2*3*4*5
as you can see, each element in the sum is a multiplications of n-1 numbers in the set. For example in (1*2*3*4), 5 is excluded and in (1*2*3*5), 4 is excluded. I know some of the multiplications are repeated, for example (1*2) is repeated in 3 of the multiplications. How can I solve this problem with least number of multiplications.
Sorry for bad English.
Thanks.
Here is a way that does not "cheat" by replacing multiplication with repeated addition or by using division. The idea is to replace your expression with
1*2*3*4 + 5*(1*2*3 + 4*(1*2 + 3*(1 + 2)))
This used 9 multiplications for the numbers 1 through 5. In general I think the multiplication count would be one less than the (n-1)th triangular number, n * (n - 1) / 2 - 1. Here is Python code that stores intermediate factorial values to reduce the number of multiplications to just 6, or in general 2 * n - 4, and the addition count to the same (but half of them are just adding 1):
def f(n):
fact = 1
term = 2
sum = 3
for j in range(2, n):
fact *= j
term = (j + 1) * sum
sum = fact + term
return sum
The only way to find which algorithm is the fastest is to code all of them in one language, and run each using a timer.
The following would be the most straightforward answer.
def f(n):
result = 0
nList = [i+1 for i in range(n)]
for i in range(len(nList)):
result += reduce(lambda x, y: x*y,(nList[:i]+nList[i+1:]))
return result
Walkthrough - use the reduce function to multiply all list's of length n-1 and add to the variable result.
If you just want to minimise the number of multiplications, you can replace all the multiplications by additions, like this:
// Compute 1*2*…*n
mult_all(n):
if n = 1
return 1
res = 0
// by adding 1*2*…*(n-1) an entirety of n times
for i = 1 to n do
res += mult_all(n-1)
return res
// Compute sum of 1*2*…*(i-1)*(i+1)*…*n
sum_of_mult_all_but_one(n):
if n = 1
return 0
// by computing 1*2*…*(n-1) + (sum 1*2*…*(i-1)*(i+1)*…*(n-1))*n
res = mult_all(n-1)
for i = 1 to n do
res += sum_of_mult_all_but_one(n-1)
return res
Here is an answer that would work with javascript. It is not the fastest way because it is not optimized, but it should work if you want to just find the answer.
function combo(n){
var mult = 1;
var sum = 0;
for (var i = 1; i <= n; i++){
mult = 1;
for (var j = 1; j<= n; j++){
if(j != i){
mult = mult*j;
}
}
sum += mult;
}
return (sum);
}
alert(combo(n));
I am trying to add constraints to a linear optimization problem in Julia using JuMP. I am using the sum{} function however, I am having trouble with some of the constraints. Does anyone know how to write "for all" in JuMP (the upside down A)? Here is the code I have so far:
using JuMP
m = Model()
c= [3 5 2 ; 4 3 5 ; 4 5 3 ; 5 4 3 ; 3 5 4]
#variable(m, x[i=1:5,j=1:3] >= 0)
#objective(m,Min,sum{c[i,j]*x[i,j],i=1:5,j=1:3})
for i=1:5
#constraint(m, sum{x[i,j],i,j=1:3} <= 480)
end
What I am trying to get is this:
I am trying to use the for loop as a substitute of "for all i from 1 to 5" however I keep getting errors. Is there another way to do this?
In mathematical notation, you sum across i, and do so for each j.
In Julia/JuMP, you can think of "∀" as being a for loop ("for all"),
and a "Σ" as being a sum{ }:
using JuMP
m = Model()
c= [3 5 2;
4 3 5;
4 5 3;
5 4 3;
3 5 4]
# x_ij >= 0 ∀ i = 1,...,5, j = 1,...,3
#variable(m, x[i=1:5,j=1:3] >= 0)
#objective(m,Min,sum{c[i,j]*x[i,j],i=1:5,j=1:3})
# ∀j = 1,...,3
for j in 1:3
#constraint(m, sum{x[i,j],i=1:5} <= 480)
end