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

## Solution

a) dprop = ${\displaystyle {\frac {M}{S}}}$

b) dtrans = ${\displaystyle {\frac {L}{R}}}$

c) dprop + dtrans = ${\displaystyle {\frac {M}{S}}}$ + ${\displaystyle {\frac {L}{R}}}$

d) At t=dtrans, the bit would have moved along the link a distance of:
d = S*dtrans
The link distance M is given by:
M = S*dprop
Since dprop > dtrans, the bit will still be on the link going towards the other end of the link.

e) At t = dtrans, the bit would have moved along the link a distance of:
d = S*dtrans
The link distance M is given by:
M = S*dprop
Since dprop < dtrans then M < d, hence the bit would have already reached the other end of the link