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

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ECEN 602 Fall 2015: Solution Homework 4Due: Oct 61Solution Homework 4 ECEN 6021. Peterson and Davie 5thEdition (P&D) 2.24 (Chapter 2, Problem 24) (16%)Bandwidth × (roundtrip) delay is about 125KBps × 2.5s = 312 KB, or 312packets. The window size should be this large.If we are using a Go-Back-n protocol (i.e., in P&D notation RWS = 1), thesequence number modulus must be greater than the window size (i.e., m >n in my notes or in P&D notation MaxSeqNumSWS + 1). This meansm > 312 or 28< 312 < 29, so we need 9 bits.For Selective Reject/Acknowledgement (i.e., RWS = SWS), the sequencenumber modulus must be greater than twice the window size (i.e., m > 2nin my notes or in P&D notation SWS < (MaxSeqNum + 1)/2 ). Thus m >624 or 29< 624 < 210, so we need 10 bits.2. P&D 2.28 (16%)
ECEN 602 Fall 2015: Solution Homework 4Due: Oct 623. P&D 2.31 (16%)The right diagram, for part (b), shows each of frames 4-6 timing out after a2 × RTT timeout interval; a more realistic implementation (e.g.,TCP)would probably revert to SWS = 1 after losing packets, to address bothcongestion control and the lack of ACK clocking.4. (a) (8%) Let Tminbe the minimum transmission time for data frames andTdbe

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