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 MaxSeqNum≥SWS + 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%)