Shell Programming and Scripting

Thank you very much for your great help.
assume we have a graph (time, y) axis:

y(i+1) t(i+1)
/
y /
*
/ t
/
y(i), t(i)


y is the new point on the line at time t.
rotuer1.dat
10 2
20 4
30 4
40 5
50 4
60 2
sender1.dat
15 2
30 5
45 4
60 3
Assume that the t is between t(i) =10 and t(i+1)=20, which is 15 from sender1.dat. t could have also the same value e.g., 30 and 60.
y = y(i) + {y(i+1) - y(i)} {t-t(i)}/ (t(i+1)-t(i))
y = 2 + {4-2} (15-10)/{20-10}

y = 3;

estimated value is at time t is:
e = 3 - 2 = 1
in this case, I did try what you told me, but it didn't work. I will try to fix now. I will let you know with good news soon.


again thank you very much...

So is "t" derived or is it read from a file or is it user-defined?

I tried t = (yi+ylast)/2 which seemed to produce reasonable results.

thank you very much for your reply...

yes, it is derived from the files. for example

As you said t = {t(i+1) - t(i)} / 2 and the new y = y(i) + {(y(i+1) - y(i))(t-t(i))}/{t(i+1) - t(i)}, if file "sender.dat" t has (15,25,35,45,55)

y
|-------------5-------------5
|---------4-------4----4 -----------4----4
|---------------------------------------------
|-2---2-------------------------2------------2
|------------------------------------------------> t
-10--15--20--25--30--35--40--45--50--55--60

For example: y(i) = 2, y(i+1)=4, t(i)=10, and y(i+1)=20;
the new interpolate value are given?
y(1) = 2 + {(4-2)(15-10)}/(20-10) = 3
y(2) = 4 + {(4-4)(25-20)/(30-20) = 4



how about when t != {t(i+1) - t(i)} / 2?
y
|
|=====4===4======4============4
|=3==========3=======3====3
|===============================2====2
|------------------------------------------------> t
-10---13---20---26--30---39--40-----50-52---60

Assume that we have the following time t (column 1 in each file) for the both nodes:
router.dat sender.dat
10 13
20 26
30 39
40 52
50
60


For example: y(i) = 2, y(i+1)=4, t(i)=10, and y(i+1)=20;
the new interpolate value are given:
y(1) = 3 + {(4-3)(13-10)}/(20-10) = 3.3
y(2) = 4 + {(4-4)(26-20)/(30-20) = 4
y(3) = 4 + {(3-4)(39-30)/(40-30) = 3.1


I did try this one, but it gave me wrong answer. I will try to change:
{ yi=$2; ti=$1; t=(yi + ylast)/2; y=ylast + ( yi - ylast ) * ( t - tlast ) / ( ti - tlast ); print t,ti,y; ylast=yi; tlast=ti; }

There's a big difference between

Code:

t=(t(i+1) - t(i)) / 2

and

Code:

t=(t(i+1) + t(i)) / 2

In the first, t represents the mean difference between the two ti values, whereas in the second, t represents the mean of the two ti values.

Anyway, you're en route to figuring out this problem. Good luck!

I am so sorry for that I sent the answer without check it. You right it is t=(t(i+1) + t(i)) / 2.
I would appreciate it if you could check this again.

y = y(i) + {(y(i+1) - y(i)) (t - t(i))} / {t(i+1) - t(i)}
I used the sort function, which produced the following results

90.0 0.00
120.0 22.95
180.0 21.11
240.0 21.94
270.0 40.00
300.0 19.95
360.0 30.00
420.0 21.11
450.0 45.45
480.0 22.51
540.0 30.00
600.0 58.11
630.0 45.45
660.0 46.48
720.0 40.00
720.0 45.58
780.0 49.12

where t is from the first file (router1.dat), as shown in the above output.
In this case, there are only one point between y(i+1) and Y(i). For example 90 120 180 140 270


awk '{ yi=$2; ti=$1; y=ylast + ( yi - ylast ) * ( t - tlast ) / ( ti - tlast ); print ti,y; ylast=yi; tlast=ti; }'
is it possible to use an array: t = ti[i+1] or any other options?

Again, thank you very much for your help.

Ah so you need multiple interprolations between t(i) and t(i+1)? I really don't know how to do that. OR rather, I can imagine several possibilities, depending on what you need.

Anyway, the results you posted look wrong. I suspect that "t" is still not being calculated correctly. An additional problem is that you have an oscillating input, not a non-decreasing one. Thus, you cannot rely on tlast, because your input might be at a local minimum. Interporlation as I understand it is only good for non-increasing or non-decreasing curves.

thank you very much for your reply. I did try to make it easy. so that it wouldn't make any problems.
router data
60.0 21.62
120.0 22.95
180.0 21.11
240.0 21.94
300.0 19.95
360.0 19.96
420.0 21.11
480.0 22.51
540.0 59.96
sender data
90.0 0.00
180.0 9.09
270.0 40.00
360.0 30.00
450.0 45.45
540.0 30.00

I had to change the simulation timer for the router and sender side. According to the results, there is only one interpolate point between y(i) and y(i+t). In this case, t is interpolate time, which is from router.dat file. could I pass the t value without sorting the two files? any suggestion is helpful. again thank you for your help.