讲义通信网作业

：L/(L+58)= IPpacketsize=1500bytes=1460+headover(TCP/IP)Numberofpackets=|-1.5Mbytes/1460-|=1028Theamountofoverhead=1028*40=Theefficiency= /(41120+ )=97.33%Theoverhead=41120/(41120+ )=2.67% datarate=Efficiency=50%;→IPpacketload(SDU)=IPheader+TCPheader=40bytes;Totaldatarate=datarate/50%=2*8000bytes/second=16000bytes/second;Delay=40bytes/8000bytes/second=5ms NZR01极性反了，0→nochange1change,bandwithistwice.Ingeneral,weonlyconsidertheextremesituation; → 110000001100 101100001010 101100001011 010011110100 10000mode(FFFF)=(1+FFFF)mod(FFFF)=1FFFF+FF00=1FEFF1FEFFmod(FFFF)=FF00FEFF+(1FEFF>>16) FEFF+1=FF001B1B0mod(FFFF)=B1B1 ;pcode(x)=q(x)g(x);g2(x)=x3+x2+1;g(1)=(1+1)g2(x)=0;isx7+1dividedbyg(x)?Note:Itisbetterforbinarydivisionthanforpolynomialdivision.Thelengthofinfoisn’t110101or 0x7e0xfe0x240x7d0x7d0x62 note：theflagisn’t112211212R*N/2+R/2*N/2=10M*80%-→N<=106R*N=10M*80%-→N=80Ttotal=N(k+h)/R+(L-1)(k+h)/R=(N+L-N~=TtotalN(k+h)/R+(L-1)(k+h)/R(m/k+L-1)(k+h)/RTtotalk求导取极值，得-m/k2(k+h)/R+(m/k+L-1)/R=0→k=(mh/(L-(Next,NodeNodeNodeNode(-1,(-1,(-1,(-1,(-1,1(-1,(-1,2 3 4 NodeNodeNodeNodeNodeBefore1234Start4 4、 Next12422.40262<Area=4pi*r*r=4pi*6400*6400km2=1.6384pi*1018cm2<263,soitneedsAhostinanorganizationhasanipaddress150.32.64.34andasubnetmask255.255.240.0.whatistheaddressofthesubnet？whatistherangeofIPaddressesthatahostcanhaveonthis 150.32.64.34 addressofthe therangeofIPaddresses:→150.32.64.1~~Thenet-idofaClassCaddressis24bits(21bits+3bitsfortheclassc).Thesubnetshallbe3bitsandthehostidshallbe5bitsNetworkHostMask: 使用A、 C、172.16.4.0/22D、：sameordifferentnetworkIPpacket IP 20Maxdatalength:200-20=Thedatalengthofeachfragmentshallbeamultipleof8bytesSothemaxnumberofdat