Algorithm 5 Answer to Question 3 (pseudo-code) a «- 0.35 S < 0.0G
£7^0.20 ,
k* - (f )^ 3/* - (**)"
I, *- 0.&-Jc*
t <— 0 {Initialize the time counter}
d <— 100 {Any arhitraiy value greater than 0.005}
while d > 0.005 do
t = t + 1
it «- oyt
kt+\ •   (1   6)kt | it
f  * I end while
return t