5In general, you only need to track enough sequence number bits to determine which
message in the current window to which the sequence number refers. Since there is
only 1 possibility, you need log(1) = 0 bits.
7Suppose the protocol has be in operation for some time. The sender is in state “Wait for
call from above” (top left hand quarter) and the receiver is in state “Wait for 0 from
below”. The senarios for corrupted data and corrupted arck are shown below.
12 The packet length is L, the transmission rate is R, and the round trip time is RT T .
Then on page 215, Usender =.00027. To find the number of packets nto reach 90%,
.9 = .00027nor 3333.33.
16 a) Here we have a window size of N= 3. Suppose the receiver has received packet k−1,
and has ACKed that and all preceeding packets. If all of these ACKs have been
received by the sender, then the sender’s window is [k, k +N−1]. Suppose next
that none of the ACKs have been received at the sender. In this second case the
sender’s window contains k−1. The sender’s window size is thus [k−N, k −1].
By these arguments, the sender’s window is of size 3 and begins somewhere in the
range [k−N, k].
b) If the receiver is waiting for packet k, then it has recieved (and ACKed) packet
k-1 and the N-1 packets before that. If none of these N ACKs have been yet
received by the sender, then ACK messages with values of [k−N, k −1] may
still be propagating back. Because the sender has sent packets [k−N, k −1], it
must be the case that the sender has already received an ACK that is less than
k−N−1. Thus the range of in-flight ACK values can range from k−N−1 to
k−1.
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