Computer Networks 3th (컴퓨터통신 3판 솔루션 ) Larry L.Peterson Bruce S.Davie 다운로드
Computer Networks 3th (컴퓨터통신 3판 솔루션 ) Larry L.Peterson Bruce S.Davie 다운로드
Computer Networks 3th (컴퓨터통신 3판 솔루션 ) Larry L.Peterson Bruce S.Davie
Computer Networks 3판 솔루션 입니다. Larry. L. Peterson 저 2004. 8. 1 Ch1 ~ Ch9
SOLUTIONS MANUAL
Chapter 1
1
Solutions for Chapter 1
3. Success here depends largely on the ability of ones search tool to separate out the chaff. I thought a naive search for Ethernet would be hardest, but I now think it’s MPEG. Mbone www.mbone.com ATM www.atmforum.com MPEG try searching for “mpeg format”, or (1999) drogo.cselt.stet.it/mpeg IPv6 playground.sun.com/ipng, www.ipv6.com Ethernet good luck. 5. We will count the transfer as completed when the last data bit arrives at its destination. An alternative interpretation would be to count until the last ACK arrives back at the sender, in which case the time would be half an RTT (50 ms) longer. (a) 2 initial RTT’s (200ms) + 1000KB/1.5Mbps (transmit) + RTT/2 (propagation) ? 0.25 + 8Mbit/1.5Mbps = 0.25 + 5.33 sec = 5.58 sec. If we pay more careful attention to when a mega is 106 versus 220 , we get 8,192,000 bits/1,500,000 bits/sec = 5.46 sec, for a total delay of 5.71 sec. (b) To the above we add the time for 999 RTTs (
자료출처 : http://www.ALLReport.co.kr/search/Detail.asp?pk=10974932&sid=sanghyun7776&key=
[문서정보]
문서분량 : 106 Page
파일종류 : PDF 파일
자료제목 : Computer Networks 3th (컴퓨터통신 3판 솔루션 ) Larry L.Peterson Bruce S.Davie
파일이름 : Computer Networks 3th 데이터통신 솔루션.pdf
키워드 : Computer,Networks,3th,컴퓨터통신3판,솔루션,Larry,L,Peterson,Bruce,S
자료No(pk) : 10974932
Computer Networks 3th (컴퓨터통신 3판 솔루션 ) Larry L.Peterson Bruce S.Davie
Computer Networks 3판 솔루션 입니다. Larry. L. Peterson 저 2004. 8. 1 Ch1 ~ Ch9
SOLUTIONS MANUAL
Chapter 1
1
Solutions for Chapter 1
3. Success here depends largely on the ability of ones search tool to separate out the chaff. I thought a naive search for Ethernet would be hardest, but I now think it’s MPEG. Mbone www.mbone.com ATM www.atmforum.com MPEG try searching for “mpeg format”, or (1999) drogo.cselt.stet.it/mpeg IPv6 playground.sun.com/ipng, www.ipv6.com Ethernet good luck. 5. We will count the transfer as completed when the last data bit arrives at its destination. An alternative interpretation would be to count until the last ACK arrives back at the sender, in which case the time would be half an RTT (50 ms) longer. (a) 2 initial RTT’s (200ms) + 1000KB/1.5Mbps (transmit) + RTT/2 (propagation) ? 0.25 + 8Mbit/1.5Mbps = 0.25 + 5.33 sec = 5.58 sec. If we pay more careful attention to when a mega is 106 versus 220 , we get 8,192,000 bits/1,500,000 bits/sec = 5.46 sec, for a total delay of 5.71 sec. (b) To the above we add the time for 999 RTTs (
자료출처 : http://www.ALLReport.co.kr/search/Detail.asp?pk=10974932&sid=sanghyun7776&key=
[문서정보]
문서분량 : 106 Page
파일종류 : PDF 파일
자료제목 : Computer Networks 3th (컴퓨터통신 3판 솔루션 ) Larry L.Peterson Bruce S.Davie
파일이름 : Computer Networks 3th 데이터통신 솔루션.pdf
키워드 : Computer,Networks,3th,컴퓨터통신3판,솔루션,Larry,L,Peterson,Bruce,S
자료No(pk) : 10974932
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