Teletraffic engineering/How does Internet telephony traffic differ/Solutions to Module 8

Solution

a) d_prop = ${\frac {M}{S}}$

b) d_trans = ${\frac {L}{R}}$

c) d_prop + d_trans = ${\frac {M}{S}}$ + ${\frac {L}{R}}$

d) At t=d_trans, the bit would have moved along the link a distance of:

d = S*d_trans

The link distance M is given by:

M = S*d_prop

Since d_prop > d_trans, the bit will still be on the link going towards the other end of the link.

e) At t = d_trans, the bit would have moved along the link a distance of:

d = S*d_trans

The link distance M is given by:

M = S*d_prop

Since d_prop < d_trans then M < d, hence the bit would have already reached the other end of the link