Linear optimization of difference subject to constraints - r

Question

I have a vector of values (x1,x2,x3,x4,x5,x6,x7) and i want to create a vector which minimizes a new unknown vector (y1,y2,y3,y4,y5,y6,y7) such that I can minimize ||x-y||^2. I also want to create this new vector subject to the constraints that x1+x2+x3+x4+x5=x6 and x1+x2+x3+x4=x7. I tried to use constrOptim but I do not think I have the right inputs. Any help would be greatly appreciated!
Would it be best to come up with a set of values and then use a nls model to predict them? How would I do that?
Thank you!!

Answer 1

We assume that what the question actually intended was that y is known and we want to get x with the indicated constraints.
Note that nls does not work for zero residual problems and since no data was provided in the question we don't know whether that is the case here or not so we first present two solutions that can handle that and then finally we show an nls for the non-zero residual case. We use y shown below in (1) as our test input for (1) and (2) and it does have zero residuals. For (3), the nls solution, we use a different y which does not lead to zero residuals.
Here are some alternative solutions:
1) lm We define x5_to_x7 which maps the first 5 components of x to the entire 7-element vector. Because x5_to_x7 is a linear operator it corresponds to a matrix X which we form and then use in lm:
# test data
y <- c(1:5, sum(1:5), sum(1:4))
x5_to_x7 <- function(x5) c(x5, sum(x5), sum(x5[1:4]))
X <- apply(diag(5), 1, x5_to_x7)
fm <- lm(y ~ X + 0)
giving:
coef(fm)
## X1 X2 X3 X4 X5
## 1 2 3 4 5
all.equal(x5_to_x7(coef(fm)), y)
## [1] TRUE
2) optim Alternatively we can use optim by defining a residual sum of squares function and solve it using optim where y and x5_to_x7 are as above:
rss <- function(x) sum((y - x5_to_x7(x))^2)
result <- optim(numeric(5), rss, method = "BFGS")
giving:
> result
$par
[1] 1 2 3 4 5
$value
[1] 5.685557e-20
$counts
function gradient
18 12
$convergence
[1] 0
$message
NULL
> all.equal(x5_to_x7(result$par), y)
[1] TRUE
3) nls If y were such that the residuals are not zero then it would be possible to use nls as suggested in the question.
y <- 1:7
fm1 <- lm(y ~ X + 0)
fm2 <- nls(y ~ x5_to_x7(x), start = list(x = numeric(5)))
all.equal(coef(fm1), coef(fm2), check.attributes = FALSE)
## [1] TRUE