Calculate Networking Speed for 1500KB Packet

[Cyborg31]

Homework Statement

I would like to send a packet of size 1500 kilobytes from point A to
point B. There are a total of 3 links that it has to cross, of length 1000,
2000, 3000m respectively. The bandwidth on those links is 100, 1000,
and 500 kilobytes/second. Assume that there is no processing time,
but they have to wait about 1 second at each intermediate router due
to congestion. Assuming that a bit travels at about half the speed of
light in the medium, and that a router must have the entire packet
before it can process it, how long must I wait to get my packet?

Homework Equations

The Attempt at a Solution

I'm just trying to see if I'm understanding this correctly.

If Router A to Router B has a 1000m physical link with a bandwidth of 100 KB per second, then it takes 1 second to push 100 KB of data on to the physical line (of 1000m length). Once it is pushed on the physical line, and I have calculated that one bit takes 0.0000067 seconds to travel 1000m, that 100 KB of data takes 0.0000067 seconds to travel to Router B (which is 1000m away).

There is 1 second in congestion delay so it'd take 2 seconds (one second to push it on the line and one second in queue) + 0.0000067 seconds of physical travel tie for every 100 KB of data from Router A to Router B. So total time from A to B is 15 * (2 + 0.0000067) seconds.

 

[Valkarie]

To get from Router B to Router C, it has a 2000m physical link with a bandwidth of 1000 KB per second. So it takes 1/1000th of a second to push 1 KB of data on the line (of 2000m length). Once it is pushed on the physical line, and I have calculated that one bit takes 0.0000067 seconds to travel 2000m, that 1000 KB of data takes 0.0000067 seconds to travel to Router C (which is 2000m away).There is 1 second in congestion delay so it'd take 2 seconds (one second to push it on the line and one second in queue) + 0.0000067 seconds of physical travel tie for every 1000 KB of data from Router B to Router C. So total time from B to C is 15 * (2 + 0.0000067) seconds.Finally to get from Router C to Router D, it has a 3000m physical link with a bandwidth of 500 KB per second. So it takes 1/500th of a second to push 1 KB of data on the line (of 3000m length). Once it is pushed on the physical line, and I have calculated that one bit takes 0.0000067 seconds to travel 3000m, that 500 KB of data takes 0.0000067 seconds to travel to Router D (which is 3000m away).There is 1 second in congestion delay so it'd take 2 seconds (one second to push it on the line and one second in queue) + 0.0000067 seconds of physical travel tie for every 500 KB of data from Router C to Router D. So total time from C to D is 15 * (2 + 0.0000067) seconds.The total time from A to D is 45 * (2 + 0.0000067) seconds.

 

[Mene]

To calculate the networking speed, we need to divide the size of the packet (1500 KB) by the total time it takes to travel through the network, including the congestion delays. So the networking speed would be 1500 KB / (15 * (2 + 0.0000067)) seconds, which is approximately 66.666 KB per second.