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srtt k+1 = (1-a) srtt k + artt drtt k+1 = (1-b) drtt k + b|srtt k - rtt k |

Suppose that the last SampleRTT in a TCP connection is equal to I sec. The current value of Timeoutlnterval for the connection will necessarily be >= 1 sec. True. srtt k+1 = (1-a) srtt k + artt drtt k+1 = (1-b) drtt k + b|srtt k - rtt k | Rto = srtt k + 4*drtt k+1

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srtt k+1 = (1-a) srtt k + artt drtt k+1 = (1-b) drtt k + b|srtt k - rtt k |

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  1. Suppose that the last SampleRTT in a TCP connection is equal to I sec. The current value of Timeoutlnterval for the connection will necessarily be >= 1 sec. True • srttk+1 = (1-a)srttk + artt • drttk+1 = (1-b)drttk + b|srttk - rttk| • Rto = srttk + 4*drttk+1 • Or perhaps Rto= srttk+1 + 4*drttk+1 • Plug eq for drtt into eq for rto • Rto = srtt + 4((1-b)drttk + b|srttk - rttk|) • To make rto as small as possible, suppose that drttk=0 • Rto = srtt + 4(b|srttk - rttk|) • B=1/4, so • Rto = srtt + (|srttk - rttk|) • Since we are seeking a small rto, srtt<=rtt • Rto = srtt + (rttk-srttk) = rttk • when Rto = srttk+1 + 4*drttk+1 • Rto = (1-a)srttk + artt + (rttk-srttk) = srttk + a(rtt-srtt) + (rttk-srttk) • = a(rtt-srtt) + rttk= a(rtt-srtt) + 1>1

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