Jorge,
First of all, I really do think that questions such as these should be posted
directly to R-help.
More people will then see it with a greater chance of getting useful replies.
I am subscribed to R-help so will see posts.
I have cc'ed this reply to R-help so that you may get more answers than I can
give you.
I know absolutely nothing about transportation problems.
For comments see inline.
On 27-11-2012, at 00:24, Jorge I Velez wrote:
> Dear Dr. Hasselman,
>
> Please receive my apologies in advance for contacting you directly, but I
> have seen your replies in R-help and just thought that, perhaps, you could
> give me a hand. I am facing an optimization problem and unfortunately can not
> find a way to get around. I will try to explain myself as good as I can.
>
> Let us consider an optimization problem in which it is of interest to
> distribute resources from 2 to 3 points. The costs associated are as follows:
>
> Destination
> From 1 2 3 Total
> 1 1 3 4 300
> 2 3 2 3 200
> Total 150 250 100 500
>
> If X_i (i = 1, 2, ..., 6) is the cost of shipping X units from/to the ith
> combination (i.e., i = 1 means from shipping from point 1 to destination 1; i
> = 2 from point 1 to destination 2 and so on), the formulation in R would be
> as follows:
>
This description is confusing. So x4 means shipping from point to destination
4 (you only have 3 destinations)?
> require(lpSolve)
> f <- matrix(c(1, 3, 4, 3, 2, 3), ncol = 3, byrow = TRUE)
> row.rhs <- c(300, 200)
> col.rhs <- c(150, 250, 100)
> row.signs <- rep("==", length(row.rhs))
> col.signs <- rep("==", length(col.rhs))
> D <- lp.transport(f, "min", row.signs, row.rhs, col.signs, col.rhs)
> D$solution
>
> So far so good until this point. However, in addition to spend *all* the
> resources as above, in my application I would like to set up a couple of more
> constraints as follows.
>
> # row and column constraints
> x1 + x2 + x3 == 300 # row 1
> x4 + x5 + x6 == 200 # row 2
> x1 + x4 == 150 # col 1
> x2 + x5 == 250 # col 2
> x3 + x6 == 100 # col 3
> x1 + x2 + x3 + x4 + x5 + x6 == 500 # all available
>
But these constraints are already satisfied in the solution.
> Now, if w is a vector representing some constants (the length of w is the
> number of destinations from which we ship stuff), I would like to include the
> following three constraints, where r is fixed and known:
>
I don't understand "the number of destinations from which we ship stuff".
Shouldn't "from" be "to"?
> w1*x1 + w2*x4 == r # col 1
> w1*x2 + w2*x5 == r # col 2
> w3*x3 + w6*x6 == r # col 3
>
vector w is length 6 but you only have three destinations (from the initial
description)
I don't understand what you are doing here.
> Another constraint that I would like to impose is that, by column, the number
> of X_i's greater than zero should be at least two. Furthermore, the
> maximization function should not be the coefficients in the f matrix, but the
> standard deviation of g = c(l1, l2, l3) where
>
> l1 = w1*x1 + w2*x4
> l2 = w1*x2 + w2*x5
> l3 = w3*x3 + w6*x6
>
How does this relate to your previous formula?
> In my application I have up to 100 columns and 50 rows and I am able to
> include all the constraints but not the number of zeros neither the new
> function to optimize. Could you please give some advice me on how to do both?
> I have been reading a bit on quadratic programing as the function to
> optimize is the standard deviation, but I do not really know how to set up
> the constraints, let alone the complete problem.
>
>From this description I gather that you want to optimize (minimize?) the
>standard deviations, perhaps a weighted sum?
> Thank you very much in advance for any help you can provide me.
>
There is far too much unclear about your problem to give any sensible advice,
assuming that I have any.
Berend
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