Expression 11: "t" Subscript, "f" , Baseline equals StartFraction, 112 Over 500 times 10 cubed , EndFractiontf=112500·103
equals=
0.0 0 0 2 2 40.000224
11
t_p: time to transmit a packet
12
Expression 13: "t" Subscript, "p" , Baseline left parenthesis, "x" , right parenthesis equals "t" Subscript, "f" , Baseline times "l" left parenthesis, "x" , right parenthesis left brace, 0 less than or equal to "x" less than or equal to 63 , right bracetpx=tf·lx0≤x≤63
13
r: round trip time
14
Expression 15: "r" equals 0.1r=0.1
0.10.1
0.10.1
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f: time to transmit the update
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Expression 17: "f" left parenthesis, "x" , right parenthesis equals left parenthesis, "t" Subscript, "p" , Baseline left parenthesis, "x" , right parenthesis plus "p" times "t" Subscript, "p" , Baseline left parenthesis, "x" , right parenthesis plus "r" , right parenthesis "n" left parenthesis, "x" , right parenthesisfx=tpx+p·tpx+rnx
17
we are trying to minimize the time of the update f(x) -> the optimal value of x is where the derivative f'(x) is 0
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Expression 19: "f" prime left parenthesis, "x" , right parenthesis equals 0f′x=0
19
Label: left parenthesis, 13.8 7 2 , 73.9 8 3 , right parenthesis13.872,73.983
Label:
20
x must be an integer -> optimal x is 14
21
most time is convenient to work with bytes -> optimal x multiple of 8 is 16 (2 bytes)
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left parenthesis, 13.8 7 2 , 73.9 8 3 , right parenthesis13.872,73.983