stop and wait計算frame大小

有一點對點的連線,採用「停與等」(stop and wait)的方式傳送資料。其位元速率(bit rate)為6 kbps,以及單向的傳遞延遲時間(propagation delay)為30 msec。請問欲達到系統效率在50%以上,訊框(frame)的長度至少要多少?

1 Answer

  • Anonymous
    1 decade ago
    Favorite Answer

    1.定義 a = frame propogation delay/frame transmission time= T(p)/T(f)

    = 30 x 10(-3)/[L/6x10(3)]=180/L

    2.Utilization = T(f)/[2T(p) T(f)] = 1/[2a 1]

    U=1/[1 (360/L)]>= 0.5 故L>=360

    an efficiency of at least 50% requires a frame size of at least 360 bits.

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