Answer: bandwidth = 0.10 MB/s
Explanation:
Given
Total Time = Compression Time + Transmission Time
Transmission Time = RTT + (1 / Bandwidth) xTransferSize
Transmission Time = RTT + (0.50 MB / Bandwidth)
Transfer Size = 0.50 MB
Total Time = Compression Time + RTT + (0.50 MB /Bandwidth)
Total Time = 1 s + RTT + (0.50 MB / Bandwidth)
Compression Time = 1 sec
Situation B:
Total Time = Compression Time + Transmission Time
Transmission Time = RTT + (1 / Bandwidth) xTransferSize
Transmission Time = RTT + (0.40 MB / Bandwidth)
Transfer Size = 0.40 MB
Total Time = Compression Time + RTT + (0.40 MB /Bandwidth)
Total Time = 2 s + RTT + (0.40 MB / Bandwidth)
Compression Time = 2 sec
Setting the total times equal:
1 s + RTT + (0.50 MB / Bandwidth) = 2 s + RTT + (0.40 MB /Bandwidth)
As the equation is simplified, the RTT term drops out(which will be discussed later):
1 s + (0.50 MB / Bandwidth) = 2 s + (0.40 MB /Bandwidth)
Like terms are collected:
(0.50 MB / Bandwidth) - (0.40 MB / Bandwidth) = 2 s - 1s
0.10 MB / Bandwidth = 1 s
Algebra is applied:
0.10 MB / 1 s = Bandwidth
Simplify:
0.10 MB/s = Bandwidth
The bandwidth, at which the two total times are equivalent, is 0.10 MB/s, or 800 kbps.
(2)
. Assume the RTT for the network connection is 200 ms.
For situtation 1:
Total Time = Compression Time + RTT + (1/Bandwidth) xTransferSize
Total Time = 1 sec + 0.200 sec + (1 / 0.10 MB/s) x 0.50 MB
Total Time = 1.2 sec + 5 sec
Total Time = 6.2 sec
For situation 2:
Total Time = Compression Time + RTT + (1/Bandwidth) xTransferSize
Total Time = 2 sec + 0.200 sec + (1 / 0.10 MB/s) x 0.40 MB
Total Time = 2.2 sec + 4 sec
Total Time = 6.2 sec
Thus, latency is not a factor.
