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]
<span>The equivalent of the TTL(Time to Live) field in an IPv4 header is known as the Hop Limit in an IPv6 header.
</span>The IPv6 header is a streamlined version of the IPv4
header. The field Hop Limit has the size of 8 bits and indicates the maximum number of links
over which the IPv6 packet can travel before being discarded.