In my attached R function, I was wondering how to solve for mdes (suppose it is unknown) which is currently one of the input values IF everything else is known?
Is it also possible to solve for mdes and power (both currently input values) IF everything else is known?
foo <- function(A = 200, As = 15, B = 100,Bs = 10,iccmax = 0.15,mdes = .25,SD = 1.2,power = 80)
{
tail <- 2
alpha <- 5
inv_d <- function(mdes) {
c(mean_dif = 1, Vmax = 2/mdes^2)
}
SDr <- 1/SD
pars <- inv_d(mdes)
mean_dif <- pars[[1]]
Vmax <- pars[[2]]
zbeta <- qnorm((power/100))
zalpha <- qnorm(1-(alpha/(100*tail)))
maxvarmean_difhat <- (mean_dif / (zbeta + zalpha))**2
ntreat <- sqrt((A/As)*((1-iccmax)/iccmax))
ncont <- sqrt((B/Bs)*((1-iccmax)/iccmax))
costpertreatcluster <- A + (As*ntreat)
costperconcluster <- B + (Bs*ncont)
gtreat <- (sqrt(A*iccmax) + sqrt(As*(1-iccmax)))**2
gcon <- (sqrt(B*iccmax) + sqrt(Bs*(1-iccmax)))**2
pratio <- sqrt(gtreat/gcon)
budgetratio <- 99999
budgetratio <- ifelse( ((pratio <= SD) & (pratio >= SDr)), pratio**2, ifelse((pratio > SD), pratio*SD, pratio*SDr))
fraction <- budgetratio/(1 + budgetratio)
mmvnumer <- 99999
mmvnumer <- ifelse( ((pratio <= SD) & (pratio >= SDr)),
gcon*Vmax*(1+(pratio**2)),
ifelse((pratio > SD),
gcon*Vmax*(((pratio*SD)+1)**2/((SD**2)+1)),
gcon*Vmax*(((pratio*SDr)+1)**2/((SDr**2) + 1))) )
budget <- mmvnumer/maxvarmean_difhat
treatbudget <- fraction*budget
conbudget <- (1-fraction)*budget
ktreat <- treatbudget/costpertreatcluster
kcont <- conbudget/costperconcluster
ktreatrup <- ceiling(ktreat)
kcontrup <- ceiling(kcont)
ktreatplus <- ifelse(pmin(ktreatrup,kcontrup) < 8, ktreatrup + 3, ktreatrup + 2)
kcontplus <- ifelse(pmin(ktreatrup,kcontrup) < 8, kcontrup + 3, kcontrup + 2)
budgetplus <- (ktreatplus*costpertreatcluster) + (kcontplus*costperconcluster)
return(c(ncont = ncont, kcont = kcontplus,
ntreat = ntreat, ktreat = ktreatplus, budget = budgetplus))
}
#--------------------------------------------------------------------------------
# EXAMPLE OF USE:
foo()
ncont kcont ntreat ktreat budget
7.527727 73.000000 8.692270 62.000000 33279.051347
Define a function of one variable as
p0 = foo()
fn1 = function(x) sum((foo(mdes=x) - p0)^2)
and find a minimum that should be 0, and which corresponds to your mdes = 0.25 input!
optimize(fn1, c(0.0, 1.0))
## $minimum
## [1] 0.2497695
## $objective
## [1] 0
For two variables, this is more difficult, as the function has many local minima and is ill-defined outside certain regions. Applying optim() you will need well-chosen starting points.