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

Solution

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


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


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


 * 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