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