Question

We are sending a 30 Mbit file from source host A to destination host B. All links in the path between source and destination have a transmission rate of 10 Mbps. Assume that the propagation speed is 2 *10^8 meters/sec, and the distance between source and destination is 10,000 km. The end-to-end delay (transmission delay plus propagation delay) is:

Answer 1

`t=1.5* 10^(-4)\ s`

Step-by-step explanation:

Given:

• file size to be transmitted,
  `D=30\ Mb`

• transmission rate of data,
  `\dot D=10\ Mb.s^(-1)`

• propagation speed,
  `v=2* 10^8\ m.s^(-1)`

• distance of data transfer,
  `s=10000\ km=10^4\ m`

Now the delay in data transfer from source to destination for each 10 Mb:

`t'=(s)/(v)`

`t'=(10^4)/(2* 10^8)`

`t'=5* 10^(-5)\ s`

Now this time is taken for each 10 Mb of data transfer and we have 30 Mb to transfer:

So,

`t=3* t'`

`t=3* 5* 10^(-5)`

`t=1.5* 10^(-4)\ s`