...
minRTT = propagation delay + extra delay (due to extra circular routes , congestion and router delays)
?delta(T) measured= ?delta(t + ?t0 ) + delta(t0)
(Pseudo-distance)
PD = ?delta(T) measured . ?alpha
(Actual distance)
D = ?delta(T) . ?alpha
PD = (delta(?T) + ?delta(T0)). ?alpha
PD = D + ?delta(T0) .c .................................(1)
D = actual distance from the landmark.
C = speed of light
? alpha = X(c) i.e. Speed of digital info in fiber optic cable
X = factor of c with which digital info travels in fiber optic cable.
?delta(T) = actual propagation delay along the greater circle router/paths.
?delta(T0) = the extra delay causing overestimation.
...
PD1= (sqrt(XL1 - XH )^2 + (YL1 - YH)^2) + ?delta(T0) . ? alpha .................... (A)
Similarly for other 2 landmarks:
PD2= (sqrt(XL2 - XH )^2 + (YL2 - YH)^2) + ?delta(t0) . ? alpha .................... (B)
PD3= (sqrt(XL3 - XH )^2 + (YL3 - YH)^2) + ?delta(t0) . ?alpha .....................(C)
We need to linearize (A), (B) & (C) to solve them
...
f(x0) . ( x- x0) f ' (x0) . (x - x0)
f( x ) = f (x0) + - ---------- + ---------------------
1! 2!
...
f ( x ) = f (x0) + f(x0)(x - x0)
put (x-x0) = delta(x) ?x
f ( x ) = f (x0) + f' (x0) delta ( x ) .......................(3)
...
d(EstDi) . delta ( x ) d(EstDi) . delta( y )
PDi = Est Di + - ---------- + - -----------------
dX dY
...
(Xest - Xli) (Yest - Yi)
ODi = Est Di + ---------. delta ( x ) + ---------------- - delta ( y) + c. delta(To)
dX dY
Now we need to solve ( x , y , ?delta(to))
Solution from (4) is put in eq(D) to get new estimations.
Hx,Hy becomes the new estimated position
Reference
http://www.ece.cmu.edu/research/publications/2003/CMU-ECE-2003-038.pdf