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[네트위크관리] 라우터 프로그램 평가A좋아요

LAB_aRouter> enableRouter# configure terminalRouter(config)# hostname LAB_aLAB_a(config)# enable secret classLAB_a(config)# line console 0LAB_a(config-line)# loginLAB_a(config-line)# password ciscoLAB_a(config-line)# exitLAB_a(config)# interface ethernet 0LAB_a(config-if)# ip address 192.5.5.1 255.255.255.0LAB_a(config-if)# no shutdownLAB_a(config-if)# interface ethernet 1LAB_a(config-if)# ip address 205.7.5.1 255.255.255.0LAB_a(config-if)# no shutdownLAB_a(config-if)# interface serial 0LAB_a(config-if)# ip address 201.100.11.1 255.255.255.0LAB_a(config-if)# clock rate 56000LAB_a(config-if)# no shutdownLAB_a(config-if)# exitLAB_a(config)# router ripLAB_a(config-router)# network 192.5.5.0LAB_a(config-router)# network 205.7.5.0LAB_a(config-router)# network 201.100.11.0Router Program라우터이름은미리설정되어있다고가정한다LAB_a(config-router)#exitLAB_a(config)# ip host LAB_a 192.5.5.1 205.7.5.1 201.100.11.1LAB_a(config)# ip host LAB_b 219.17.100.1 199.6.13.1 201.100.11.2LAB_a(config)# ip host LAB_c 223.8.151..7.1 199.6.13.2LAB_a(config)# ip host LAB_d 210.93.105.1 204.204.7.2LAB_a(config)# ip host LAB_e 210.93.105.2LAB_a(config)# show runLAB_bRouter> enableRouter# configure terminalRouter(config)# hostname LAB_bLAB_b(config)# enable secret classLAB_b(config)# line console 0LAB_b(config-line)# loginLAB_b(config-line)# password ciscoLAB_b(config-line)# exitLAB_b(config)# interface ethernet 0LAB_b(config-if)# ip address 219.17.100.1 255.255.255.0LAB_b(config-if)# no shutdownLAB_b(config-if)# interface serial 0LAB_b(config-if)# ip address 199.6.13.1 255.255.255.0LAB_b(config-if)# clock rate 56000LAB_b(config-if)# no shutdownLAB_b(config-if)# interface serial 1LAB_b(config-if)# ip address 201.100.11.2 255.255.255.0LAB_b(config-if)# no shutdownLAB_b(config-if)# exitLAB_b(config)# router ripLAB_b(config-router)# network 219.17.100.0LAB_b(config-router)# network 199.6.13.0LAB_b(config-router)# network 201.100.11.0LAB_b(config-router)# exitLAB_b(config)# ip host LAB_a 192.5.5.1 205.7.5.1 201.100.11onfig)# ip host LAB_b 219.17.100.1 199.6.13.1 201.100.11.2LAB_b(config)# ip host LAB_c 223.8.151.1 204.204.7.1 199.6.13.2LAB_b(config)# ip host LAB_d 210.93.105.1 204.204.7.2LAB_b(config)# ip host LAB_e 210.93.105.2LAB_b(config)# show runLAB_cRouter> enableRouter# configure terminalRouter(config)# hostname LAB_cLAB_c(config)# enable secret classLAB_c(config)# line console 0LAB_c(config-line)# loginLAB_c(config-line)# password ciscoLAB_c(config-line)# exitLAB_c(config)# interface ethernet 0LAB_c(config-if)# ip address 223.8.151.1 255.255.255.0LAB_c(config-if)# no shutdownLAB_c(config-if)# interface serial 0LAB_c(config-if)# ip address 204.204.7.1 255.255.255.0LAB_c(config-if)# clock rate 56000LAB_c(config-if)# no shutdownLAB_c(config-if)# interface serial 1LAB_c(config-if)# ip address 199.6.13.2 255.255.255.0LAB_c(config-if)# no shutdownLAB_c(config-if)# exitLAB_c(config)# router ripLAB_c(config-router)# network 223.8.151.0LAB_c(config-router)# network 204.204.7.0LAB_c(config-router)# n9.6.13.0LAB_c(config-router)# exitLAB_c(config)# ip host LAB_a 192.5.5.1 205.7.5.1 201.100.11.1LAB_c(config)# ip host LAB_b 219.17.100.1 199.6.13.1 201.100.11.2LAB_c(config)# ip host LAB_c 223.8.151.1 204.204.7.1 199.6.13.2LAB_c(config)# ip host LAB_d 210.93.105.1 204.204.7.2LAB_c(config)# ip host LAB_e 210.93.105.2LAB_c(config)# show runLAB_dRouter> enableRouter# configure terminalRouter(config)# hostname LAB_dLAB_d(config)# enable secret classLAB_d(config)# line console 0LAB_d(config-line)# loginLAB_d(config-line)# password ciscoLAB_d(config-line)# exitLAB_d(config)# interface ethernet 0LAB_d(config-if)# ip address 210.93.105.1 255.255.255.0LAB_d(config-if)# no shutdownLAB_d(config-if)# interface serial 1LAB_d(config-if)# ip address 204.204.7.2 255.255.255.0LAB_d(config-if)# no shutdownLAB_d(config-if)# exitLAB_d(config)# router ripLAB_d(config-router)# network 210.93.105.0LAB_d(config-router)# network 204.204.7.0LAB_d(config-router)# exitLAB_d(config)# ip host LAB_a 192.5.5.1 205.7.00.11.1LAB_d(config)# ip host LAB_b 219.17.100.1 199.6.13.1 201.100.11.2LAB_d(config)# ip host LAB_c 223.8.151.1 204.204.7.1 199.6.13.2LAB_d(config)# ip host LAB_d 210.93.105.1 204.204.7.2LAB_d(config)# ip host LAB_e 210.93.105.2LAB_d(config)# show runLAB_eRouter> enableRouter# configure terminalRouter(config)# hostname LAB_eLAB_e(config)# enable secret classLAB_e(config)# line console 0LAB_e(config-line)# loginLAB_e(config-line)# password ciscoLAB_e(config-line)# exitLAB_e(config)# interface ethernet 0LAB_e(config-if)# ip address 210.93.105.2 255.255.255.0LAB_e(config-if)# no shutdownLAB_e(config-if)# exitLAB_e(config)# router ripLAB_e(config-router)# network 210.93.105.0LAB_e(config-router)# exitLAB_e(config)# ip host LAB_a 192.5.5.1 205.7.5.1 201.100.11.1LAB_e(config)# ip host LAB_b 219.17.100.1 199.6.13.1 201.100.11.2LAB_e(config)# ip host LAB_c 223.8.151.1 204.204.7.1 199.6.13.2LAB_e(config)# ip host LAB_d 210.93.105.1 204.204.7.2LAB_e(config)# ip host LAB_e 210.93.105.2LAB_e(conf run

공학/기술| 2001.11.26| 5페이지| 1,000원| 조회(777)

ccna, 네트위크관리, ROUTER

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디지탈공학 연습문제 풀이

9.3 (state Reduction) Given the state diagram in Figure Ex 9.3, determine which states should be combined to determine the reduced state diagram. You may use row matching or implication charts.[Ex9.3]SOL)Input sequencePresent stateNext stateOutputX=0X=1X=0X=1Reset0100011011S0S1S400S1S2S100S2S1S600S3S1S300S4S5S400S5S2S100S6S5S301S1S1-S2S4-S1S2S1-S1S4-S6S2-S1S1-S6S3S1-S1S4-S3S2-S1S1-S3S1-S1S6-S3S4S1-S5S4-S4S2-S5S1-S4S1-S5S6-S4S1-S5S3-S4S5S1-S2S4-S1S2-S2S1-S1S1-S2S6-S1S1-S2S3-S1S5-S2S4-S1S6S0S1S2S3S4S5S0-S4, S1-S5, S3-S4 는 서로 등가임.9.4(state Reduction) Given the state diagram in Figure Ex 9.4, draw the fully reduced state diagram. State succinctly what strings cause the recognizer to output a 1.SOL)Present stateNext stateOutput stateX=0X=1X=0X=1AAC00BDB01CDB00DBC00EEF00FGB00GBF00BCA-DC-BDA-BC-CD-BB-CEA-EC-FD-EB-FB-EC-FFA-GC-BD-GB-BB-GC-BE-GF-BGA-BC-FD-BB-FB-BC-FE-BF-FG-BB-FABCDEFA’=A/E C’=C/F D’=D/F9.7(State Assignment) Given the state diagram in Figure Ex 9.7 , select a good state assignment, justifying your answer in terms of the state assignment guidelines.SOL)High Priority : (C,E) , (B,E)Medium Priority : (C,D) , (A,D)Lowest Priority : 0/0 : (B,C,D,E)1/0 : (A,B,C)Q1 Q0Q2000111100AEB1DC9.14(State Partitioning) Figure Ex9.14 gives a state diagram with nine states. Show how to partition the state diagram into three communicating state machines, consisting of the state groups S0, S1, ; S3, S4, S5 ; and S6, S7, S6.SOL)Chapter9 Finite State Machine OptimizationPAGE PAGE 2

공학/기술| 2000.11.14| 6페이지| 1,000원| 조회(1,254)

무어, 디지탈, 회로설계, 머신, 밀리

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간단한 조합논리회로

디지털 실험 보고서제목간단한 조합논리 회로학번I20960960학년2학년실험조1조날짜2000년9월21일이름오재필1목적간단한 조합논리회로의 기초를 아는것과 조합논리를 이용하여 회로를 구성할 수 있는지를 아는것이다.이것을 통해 실제 TTL gate써서 구현해 낼 수 있는능력과 조합논리의 기본을 아는 것이 목적이다.2내용①이론입력변수ABC출력변수F최소항mi최대항Mi0001A’B’C’ ; m0A+B+C; M00010A’B’C ; m1A+B+C’; M10100A’BC’ ; m2A+B’+C; M20111A’BC ; m3A+B’+C’; M31001AB’C’ ; m4A’+B+C ; M41010AB’C ; m5A’+B+C’ ; M51100ABC’ ; m6A’+B’+C ; M61110ABC ; m7A’+B’+C’ ; M7SOP형:F(A,B,C)=∑(m0, m3, m4 )=A’B’C’+A’BC+AB’C’POS형:F(A,B,C)=∏(M1 M2 M5 M6 M7 )=(A+B+C’)ᆞ(A+B’+C)ᆞ(A’+B+C’)ᆞ(A’+B’+C’)②내용입력 변수가 A,B,C 이고 출력변수가 F인 three-input majority gate를 설계하라. Majority gate의 출력은 입력변수 중 다수(majority) 가 1이 될 때 출력이 1이 된다···곱의 합(SOP) 형태와 합의 곱(POS) 형태로 majority gate를 나타내어라.SolABCF*************1*************11111K-MAPABC*************10111SOP식은 AB+BC+ACGate 로나타내면간단한 조합논리회로 설계PAGE PAGE VII

공학/기술| 2000.11.03| 8페이지| 1,000원| 조회(957)

디지털, 실험, 조합논리회로

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확률과 렌덤 연습문제 풀이 1장 평가A좋아요

2.Repert I when the marble is drawn without replacing the first marble.S={(R,G)(R,B)(G,R)( G,B)(B,R)(B,G)},[단 B=blue, G=green, R=red]P ={ 1} over {6 }14.The probability of winning on a single toss of the dice is p. A starts, and if the sum is seven or eleven, then he wins. If the sum is two, three, or twelve, then he loses. If the sum is anything else, then he continues throwing until he either throws that number again (in which case he wins) or he throws a seven(in which case he loses). Calculate the probability that the plater wins.E = 주사위를 던져서나올수의 확률F = 주사위의 합이 7, 11 나올확률7 = {(1,6),(2,5),(3,4),(4,3),(5,2),(6,1)} 11 = {(5,6),(6,5)}F` = 주사위의 합이 2, 3, 20 나올 확률2 = (1,1) 3 = {(1,2),(2,1)} 20 = 0A가 이길 확률 P(E F) ={ 8} over {36 }{ 2} over {9 }B가 이길 확률 P(E F) ={ 3} over { 36}{ 1} over { 12}18.Assume that each child that is born is equally likely to be a boy or a girl. If a family has two children what is the probability that both are girls given that (a) the eldest is girl, (b) at least one is a girl?E=둘다 여자일 확률 S={(G,G)(B,B)(G,B)(B,G)}{ 1} over {4 }F=첫째 아이가 여자일 확률={ 1} over {2 }P=(E|F)={ P(EF)} over {P(F) }= { { 1} over {4 } } over { { 1} over {2 } } = { 1} over {2 }E=둘다 여자일 확률 S={(G,G)(B,B)(G,B)(B,G)}{ 1} over {4 }F=여자아이가 한명만 태어날확률={ 2} over {4 } = { 1} over {2 }D(E|F)={ P(EF)} over {P(E) }= { { 1} over {4 } } over { {3 } over {4 } }{ 1} over {3 }36.consider two boxes, one containing one black and one white marble, the other , two black and one white marble. A box is selected at random and a marble is drawn at random from the selected box. What is the probability that the marble is black?B=Box1이 검은색일 확률={ 1} over {2 }H=box2가 검은색일 확률={ 2} over {3 }P(H|B)={ P(HB)} over { P(B)} = { P(B|H)P(H)} over {P(B|H)P(H)+P(B|Hc) P(Hc)}= { {1} over {2 } { 1} over {2 } } over { { 1} over {2 } { 1} over { 2} + { 2} over {3 } { 1} over {2 } }={ { 1} over { 4} } over { { 7} over {12 } }={ 3} over {7 }38.Urn 1 contains two white balls and black ball, while urn 2 contains one white ball and five black balls. One ball is drawn at random from urn 1and placed in urn 2. A ball is then drawn from urn 2. It happens to be white. What is the probability that transferred ball was white?확률 및 렌덤 과정F=하얀색 공이 첫 번째 항아리로부터 던져질 확률={ 2} over { 3}B=검은색 공이 첫 번째 항아리로부터 던져질 확률={ 1} over { 3}W= 옮긴 뒤 흰공을 뽑을 확률P(F|W)={ D(W|F)P(F)} over {P(W|F)P(F) +P(W|B)P(B)}= { {2 } over {7 } { 2} over {3 } } over { ({ 2} over { 7} {2 } over {3 }) +( {1 } over { 7} {1 } over { 3} ) }= { 4} over {5 }40.(a)Gambler has in his pocket a fair coin and a two-headed coin. He selects one of the coins at random, and when he flips it, it shows heads. What is the probability that it is he fair coin? (b) Suppose that he flip the same coin a second time and again it shows heads. Now what is the probability that it is the fair coin? (c)suppose that he flops the same coin a third time and it is shows tails. Now what is the probability that it is the fair coin?F=Fair coin이 선택될 경우T=two-headed cion이 선택될 경우H=앞면이 나올 확률(a)P(F|H)={ P(H|F)P(F)} over { P(H|F)P(F)+P(H|T)P(T)} ={ { 1} over {2 } {1 } over {2 } } over { {1 } over {2 } {1 } over {2 }+1 {1 } over {2 } }={ { 1} over {4 } } over { {3 } over {4 } } = { 1} over {3 }(b) P(F|HH)={ { 1} over {4 } { 1} over { 2} } over { { 1} over {4 } { 1} over {2 } +1 {1 } over {2 } }={ { 1} over { 8} } over { { 5} over { 8} }(c)뒷면이 나올 확률 수 있는 경우는 fair coin 이므로 P=142.there are three coins in a box. One is a two-headed coin, another is a fair coin, and the third is a biased coin which comes up heads 75 percent of the time. When one of the three coin is selected at random and fliped. it shows heads. What is the probability that it was the two-headed coin?T=two-headed 동전F=fair동전H=앞면이 75% 나오게한 동전C=two-headed 동전이나올 확률{ 1 { 1} over { 2} } over { 1 { 1} over { 3}+ { 1} over { 2} { 1} over { 3} + { 3} over {4 } { 1} over {3 } } = { 4} over { 9}44.Urn I has five white and seven black balls. Urn 2 has three white and twelveblack balls. We flip a fair coin. If the outcome is heads, then a ball from urn 1 is selected, Suppose that a white ball is selected. What is the probability that the coin landed tail?H=동전의 앞면이 나올 경우T=동전의 뒷면이 나올 경우W=흰공이 나올 경우P(T|W)={ P(W|T)P(T)} over {P(W|T)P(T)+P(W|H)P(H) }={ {3 } over {15 } { 1} over {2 } } over { { 3} over { 15} {1 } over {2 }+ {5 } over {15 } {1 } over {2 } }

자연과학| 2000.11.02| 3페이지| 1,000원| 조회(684)

대구대학교, 확률과, 렌덤

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