the 10 cars. Each of these cars has a propagation delay of 45 minutes (travel 75 km)
before arriving at the second tollbooth. Thus, all the cars are lined up before the
second tollbooth after 47 minutes. The whole process repeats itself for traveling
between the second and third tollbooths. It also takes 2 minutes for the third tollbooth
to service the 10 cars. Thus the total delay is 96 minutes.
b) Delay between tollbooths is 8*12 seconds plus 45 minutes, i.e., 46 minutes and 36
seconds. The total delay is twice this amount plus 8*12 seconds, i.e., 94 minutes and
48 seconds.
Problem 6
a) s m d prop / = seconds.
b) R L d trans / = seconds.
c) ) / / ( R L s m d
end to end
+ =
− −
seconds.
d) The bit is just leaving Host A.
e) The first bit is in the link and has not reached Host B.
f) The first bit has reached Host B.
g) Want
( ) 536 10 5 . 2
10 56
120
8
3
=
= = s
R
L
m km.