Nested if statements better formulation

New to Julia
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12

I was just sketching this. The following looks complete to me:

using Optim

# This is, btw not efficient given that you only use consecutive periods.
# Instead of the `^period[i]` it would be more efficient to loop as in
# pv = 0.0
# R = 1 + irr
# cumR = R
# for j = 1 : length(cf)
#   cumR /= R
#   pv += cf[j] / cumR
# end
function npv(irr,cf,period)
    sum(cf[i]/(1+irr)^period[i] for i in 1:length(cf))
end

function irr(cf)
	# Get non-zero indices
	idx = findall(x -> x != 0.0, cf);
	# Make sure solution may exist
	if length(idx) < 2
		return NaN
	elseif !(any(x -> x > 0.0, cf)  &&  any(x -> x < 0.0, cf))
		return NaN
	end

	lb= [0.0,1.0,-1.0,-1.0e8]  
	up=[1.0,1.0e8,0.0,0.0] 
	done = false;
	j = 1;
	result = 0.0;
	f(x) = npv(x, cf[idx], 0 : length(idx) - 1)
	while !done
	  result = optimize(x -> f(x)^2, lb[j], up[j], Brent());
	  if Optim.minimum(result) <= 1e-9
		done = true
	  elseif j == length(lb)
	    # No more bounds to try. Solution does not exist.
		done = true
		result = NaN
	  else
	    # Next set of bounds to try
		j += 1;
	  end
	end
	return result
end

irr([-100,10,110])

Instead of checking whether the first entry is non-zero etc, I am simply checking whether there are any positive and negative entries.