r - Optimization with Constraints

Question

I am working with the output from a model in which there are parameter estimates that may not follow a-priori expectations. I would like to write a function that forces these utility estimates back in line with those expectations. To do this, the function should minimize the sum of the squared deviance between the starting values and the new estimates. Since we have a-priori expections, the optimization should be subject to the following constraints:

B0 < B1
B1 < B2
...
Bj < Bj+1 

For example, the raw parameter estimates below are flipflopped for B2 and B3. The columns Delta and Delta^2 show the deviance between the original parameter estimate and the new coefficient. I am trying to minimize the column Delta^2. I've coded this up in Excel and shown how Excel's Solver would optimize this problem providing the set of constraints:

Beta    BetaRaw    Delta    Delta^2    BetaNew
B0       1.2       0        0          1.2
B1       1.3       0        0          1.3
B2       1.6       -0.2     0.04       1.4
B3       1.4       0        0          1.4
B4       2.2       0        0          2.2

After reading through ?optim and ?constrOptim, I'm not able to grok how to set this up in R. I'm sure I'm just being a bit dense, but could use some pointers in the right direction!

3/24/2012 - Added bounty since I'm not smart enough to translate the first answer.

Here's some R code that should be on the right path. Assuming that the betas start with:

betas <- c(1.2,1.3,1.6,1.4,2.2)

I want to minimize the following function such that b0 <= b1 <= b2 <= b3 <= b4

f <- function(x) {
  x1 <- x[1]
  x2 <- x[2]
  x3 <- x[3]
  x4 <- x[4]
  x5 <- x[5]

 loss <- (x1 - betas[1]) ^ 2 + 
         (x2 - betas[2]) ^ 2 + 
         (x3 - betas[3]) ^ 2 + 
         (x4 - betas[4]) ^ 2 +
         (x5 - betas[5]) ^ 2    

  return(loss)
}

To show that the function works, the loss should be zero if we pass the original betas in:

> f(betas)
[1] 0

And relatively large with some random inputs:

> set.seed(42)
> f(rnorm(5))
[1] 8.849329

And minimized at the values I was able to calculate in Excel:

> f(c(1.2,1.3,1.4,1.4,2.2))
[1] 0.04

Answer 1

1. Since the objective is quadratic and the constraints linear, you can use solve.QP.

It finds the b that minimizes

(1/2) * t(b) %*% Dmat %*% b - t(dvec) %*% b 

under the constraints

t(Amat) %*% b >= bvec. 

Here, we want b that minimizes

sum( (b-betas)^2 ) = sum(b^2) - 2 * sum(b*betas) + sum(beta^2)
                   = t(b) %*% t(b) - 2 * t(b) %*% betas + sum(beta^2).

Since the last term, sum(beta^2), is constant, we can drop it, and we can set

Dmat = diag(n)
dvec = betas.

The constraints are

b[1] <= b[2]
b[2] <= b[3]
...
b[n-1] <= b[n]

i.e.,

-b[1] + b[2]                       >= 0
      - b[2] + b[3]                >= 0
               ...
                   - b[n-1] + b[n] >= 0

so that t(Amat) is

[ -1  1                ]
[    -1  1             ]
[       -1  1          ]
[             ...      ]
[                -1  1 ]

and bvec is zero.

This leads to the following code.

# Sample data
betas <- c(1.2, 1.3, 1.6, 1.4, 2.2)

# Optimization
n <- length(betas)
Dmat <- diag(n)
dvec <- betas
Amat <- matrix(0,nr=n,nc=n-1)
Amat[cbind(1:(n-1), 1:(n-1))] <- -1
Amat[cbind(2:n,     1:(n-1))] <-  1
t(Amat)  # Check that it looks as it should
bvec <- rep(0,n-1)
library(quadprog)
r <- solve.QP(Dmat, dvec, Amat, bvec)

# Check the result, graphically
plot(betas)
points(r$solution, pch=16)

2. You can use constrOptim in the same way (the objective function can be arbitrary, but the constraints have to be linear).

3. More generally, you can use optim if you reparametrize the problem into a non-constrained optimization problem, for instance

b[1] = exp(x[1])
b[2] = b[1] + exp(x[2])
...
b[n] = b[n-1] + exp(x[n-1]).

There are a few examples here or there.