It looks like the author made a mathematical error somewhere.
If you expand the sum-of-squares deviation
$$
S = ((b+m)-1)^2+ ((b+2m)-5)^2 + ((b+4m)-9)^2
$$
you get
$$
\begin{split}
S = & b^2+2 b m+ m^2 + 1 - 2 b - 2 m \\
+ & b^2+4 b m+ 4 m^2 + 25 - 10 b -20 m \\
+ & b^2+8 b m+16 m^2 + 81 - 18 b -72 m
\end{split}
$$
which reduces to
$$
3 b^2 + 14 b m + 21 m^2 + 107 - 30 b - 94 m
$$
which is the same as the author's expression, except the constant term, which doesn't matter anyway).
Now we need to try to minimize this by setting the derivatives of $S$ with respect to $b$ and $m$ to zero and solving the system.
$$
dS/db = 6 b + 14 m -30 \to 3 b +7 m-15 = 0
$$
$$
dS/dm = 14 b +42 m -94 \to 7 b + 21 m -47 = 0
$$
Solve
$$
\begin{split}
b & = (15-7m)/3 \\
0 & = 7 (15-7m)/3 + 21 m-47 \\
47 - 35 & = (-49/3 + 21) m \\
m & = (47-35)/(21-49/3) = 18/7
\end{split}
$$
R says this is indeed 2.571429 ...
Based on this link this seems to be from a Coursera course ... ? Maybe there was a mis-transcription of the data somewhere?
The other, independent way to do this calculation is to know that the estimated regression slope is equal to the sum of cross products ($\sum (y-\bar y) (x-\bar x)$) divided by the sum of squares ($\sum (x-\bar x)^2$).
g <- c(1,2,4)
g0 <- g - mean(g)
s <- c(1,5,9)
s0 <- s- mean(s)
sum(g0*s0)/(sum(g0^2))
## [1] 2.571429
If think if the shoe sizes were $\{1,11/3,9\}$ instead of $\{1,5,9\}$ then the slope would come out to 8/3 ...
One thing you could try is to change the optimizer. See Ben Bolker's comment at this github issue. The nlopt implementation of bobyqa is usually much faster than the default (at least whenever I try it).
library(nloptr)
defaultControl <- list(algorithm="NLOPT_LN_BOBYQA",xtol_rel=1e-6,maxeval=1e5)
nloptwrap2 <- function(fn,par,lower,upper,control=list(),...) {
for (n in names(defaultControl))
if (is.null(control[[n]])) control[[n]] <- defaultControl[[n]]
res <- nloptr(x0=par,eval_f=fn,lb=lower,ub=upper,opts=control,...)
with(res,list(par=solution,
fval=objective,
feval=iterations,
conv=if (status>0) 0 else status,
message=message))
}
system.time(test.model <- glmer(test.formula, data=test.case,
family='binomial', verbose=TRUE))
system.time(test.model2 <- update(test.model,
control=glmerControl(optimizer="nloptwrap2"))
Also, see this answer for more options and this thread from R-sig-mixed-models (which looks more relevant to your issue).
Edit:
I gave you some out-of-date info related to nloptr. In lme4 1.1-7 and up, nloptr is automatically imported (see ?nloptwrap). All you have to do is add
control = [g]lmerControl(optimizer = "nloptwrap") # +g if fitting with glmer
to your call.
Best Answer
NA as a coefficient in a regression indicates that the variable in question is linearly related to the other variables. In your case, this means that $Q3 = a \times Q1 + b \times Q2 + c$ for some $a, b, c$. If this is the case, then there's no unique solution to the regression without dropping one of the variables. Adding $Q4$ is only going to make matters worse.