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…networking-chapter-4-lans-part2.md 2023-10-02 16:47:50

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src/content/blog/data-communications-and-networking-chapter-4-lans-part2.md
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Gyunseo Lee authored and Gyunseo Lee committed Oct 2, 2023
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Expand Up @@ -102,10 +102,31 @@ star topology에서는 station A가 station B에게 frame 하나를 전송하면
Standard Ethernet은 1-persistent CSMA/CD를 Media Access Method로 사용한다.
Ehternet에서 slot time은 bits로 정의된다.
slot time은 한 station이 **512 bits**를 보내는 데에 요구되는 시간이다. (Chapter 3의 CSMA/CD에서 왜 **512bits**를 보내는지 그 이유를 다뤘었다. $Minimum\;T_{fr}=2 \times T_p$이기 때문이다. 나중에 해당 블로그 글 링크로 걸기)
그렇다면, 10Mbps의 Standard Ethernet에서는 51.2μs가 걸린다. (거속시로 계속해 보면 금방 나온다.)
그렇다면, 10Mbps의 Standard Ethernet에서는 Time Interval이 51.2μs가 걸린다. (중학교 때 배운 거속시로 계산해 보면 금방 나온다.)
하기 그림을 보며 이해해 보자.
![](/src/assets/image/data-communications-and-networking-chapter-4-lans-part2-1696230966046.jpeg)
그렇다면 collision은 언제 일어날까?
최악의 경우를 생각해 보자.
station A가 있고, frame을 보내려는 destination station B가 있다고 하자.
두 station은 shared media에서 양 끝단에 존재한다고 하자.
두 station은 shared media에서 가장 양 끝단에 존재한다고 하자.
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12)"><text x="0" y="0" font-family="LocalFont, Segoe UI Emoji" font-size="20px" fill="#1e1e1e" text-anchor="start" style="white-space: pre;" direction="ltr" dominant-baseline="text-before-edge">1bit가 B가 거의 도달하기 전에...</text></g><g stroke-linecap="round"><g transform="translate(730.2994892642156 226.0384530141394) rotate(0 -254.65303044647447 -1.961345811072917)"><path d="M0.42 0.54 C-84.51 -0.34, -423.79 -3.69, -508.89 -4.46 M-0.82 -0.23 C-85.95 -1.06, -425.23 -3.01, -509.72 -3.49" stroke="#1e1e1e" stroke-width="1" fill="none"></path></g><g transform="translate(730.2994892642156 226.0384530141394) rotate(0 -254.65303044647447 -1.961345811072917)"><path d="M-480.58 -13.22 C-493.26 -11.64, -502.64 -6.53, -509.06 -2.64 M-480.72 -13.9 C-487.99 -11.9, -492.46 -9.05, -510.11 -2.5" stroke="#1e1e1e" stroke-width="1" fill="none"></path></g><g transform="translate(730.2994892642156 226.0384530141394) rotate(0 -254.65303044647447 -1.961345811072917)"><path d="M-480.7 7.3 C-493.47 1.58, -502.81 -0.61, -509.06 -2.64 M-480.84 6.63 C-488 4.35, -492.44 2.93, -510.11 -2.5" stroke="#1e1e1e" stroke-width="1" fill="none"></path></g></g><mask></mask><g transform="translate(297.2187173766614 238.7228883931066) rotate(0 130 12)"><text x="0" y="0" font-family="LocalFont, Segoe UI Emoji" font-size="20px" fill="#1e1e1e" text-anchor="start" style="white-space: pre;" direction="ltr" dominant-baseline="text-before-edge">B도 1bit를 보내기 시작하면</text></g><g transform="translate(211.05310216366303 315.903763069968) rotate(0 205 36)"><text x="0" y="0" font-family="LocalFont, Segoe UI Emoji" font-size="20px" fill="#1e1e1e" text-anchor="start" style="white-space: pre;" direction="ltr" dominant-baseline="text-before-edge">A에서 보낸 bit가 B에 도착직전에 </text><text x="0" y="24" font-family="LocalFont, Segoe UI Emoji" font-size="20px" fill="#1e1e1e" text-anchor="start" style="white-space: pre;" direction="ltr" dominant-baseline="text-before-edge">B에서 보낸 bit와 섞여서 A에 다시 도착하고</text><text x="0" y="48" 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위 그림과 같은 상황이 발생하게 된다.
그러니깐, collision은 slot time의 first half 동안에만 일어나게 된다.
이유는 위의 최악의 상황을 가정한 위의 그림을 보며 생각해 보자. (1bit가 B에 도달하기 직전에 B가 carrier sense를 하여, 자기도 512bits 중 1bit를 보내기 시작하고, 그래서 결국 $T_p$ 에 수렴한 시간 즉, half of slot time인 25.6μs에 collision이 발생하게 되는 것이다.)

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