Answer:
<em>I </em><em>think </em><em>the </em><em>answer </em><em>is </em><em>a </em><em>bullet.</em>
<em>Hope </em><em>this </em><em>helps</em>
Answer:
Latency of an object O is shown below.
Explanation:
W segment and stalled state transmits nothing and waits for acknowledgement. The latency is 2 R TT + the time required for server that are using to transmit the object + the amount of time when server is in stalled state. Let K be the number of window that is K= O/WS .The serveries stalled state where K-1 is period of ime with period lasting RTT-(W-1)S/R
latency = 2RTT +O/R +(K-1)[S/R +RTT - WS/R]
After combining the two case
latency = 2 RTT + O/R + (K-1)[S/R +RTT - WS/R]
where [x] means maximum of (x.0). This is the complete ananlysis of the static windows.
server time for transmit the object is (K-1)[S/R +RTT - WS/R]
Answer D isn't a successful budgeting strategy.