ECEN 602 Solution Homework 4 F2015 - ECEN 602 Fall 2015:...

ECEN 602 Fall 2015: Solution Homework 4 Due: Oct 6
1
Solution Homework 4 ECEN 602
1. Peterson and Davie 5
th
Edition (P&D) 2.24 (Chapter 2, Problem 24) (16%)
Bandwidth × (roundtrip) delay is about 125KBps × 2.5s = 312 KB, or 312
packets. The window size should be this large.
If we are using a Go-Back-n protocol (i.e., in P&D notation RWS = 1), the
sequence number modulus must be greater than the window size (i.e., m >
n in my notes or in P&D notation MaxSeqNum ≥ SWS + 1). This means
m > 312 or 2 8 < 312 < 2 9 , so we need 9 bits.
For Selective Reject/Acknowledgement (i.e., RWS = SWS), the sequence
number modulus must be greater than twice the window size (i.e., m > 2n
in my notes or in P&D notation SWS < (MaxSeqNum + 1)/2 ). Thus m >
624 or 2 9 < 624 < 2 10 , so we need 10 bits.
2. P&D 2.28 (16%)

ECEN 602 Fall 2015: Solution Homework 4 Due: Oct 6
2
3. P&D 2.31 (16%)
The right diagram, for part (b), shows each of frames 4-6 timing out after a
2 × RTT timeout interval; a more realistic implementation ( e.g., TCP)
would probably revert to SWS = 1 after losing packets, to address both
congestion control and the lack of ACK clocking.
4. (a) (8%) Let T
min
be the minimum transmission time for data frames and T
d
be