I know I can do:
p0 = foo() ; fn1 = function(x) sum((foo(power=x) - p0)^2)
optimize(fn1, c(0, 100))[[1]] ### >[1] 79.8817 almost 80 as `power` in input of `foo()`
to solve for power (suppose it was unknown) which is currently one of the input values in my below function (foo).
Question: But now suppose I know one of the OUTPUTS (budget) of foo, can I now solve back for power (which is one of the inputs) via optimization?
foo <- function(A = 200, As = 15, B = 100,Bs = 10, power = 80, iccmax = 0.15,mdes = .25,SD = 1.2)
{
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
I thought you could do this yourself, easily.
b0 = foo()[5] # budget: 33279.051347
fn2 = function(x) {
foo(power = x)["budget"] - b0
}
uniroot(fn2, c(70, 90))
## $root
## [1] 79.99041
## $f.root
## budget
## 0
This may not work for all input variables, depending on how much influence they have on the different outputs.