
New Member
Maximum absolute error when calculating a difference using two rounded down values.
I have a maths question from an old text book where I disagree with the answer given.
You have 2 pins. One is measured as 12mm and the other is 14mm.
The values measured are rounded down to the nearest mm.
That means the true length of each pin "l1" and "l2" by my calculation would be:
12 <= l1 < 13
14 <= l2 < 15
That means the actual error for each pin is:
1 < E1 <= 0
1 < E2 <= 0
That means that the maximum absloute error for each pin is just under 1mm.
My problem is what is the maximum absolute error when calculating the difference between the two pins?
Using the given values: 14mm  12mm = 2mm.
The largest l1 can be is just under 13mm and the lowest l2 can be is 14mm giving a difference of 14mm  13mm = 1mm.
The lowest l1 can be is 12mm and the largest l2 can be is 15mm giving a difference of 15mm  12mm = 3mm.
Therefore I calculate the true difference as:
1 < (l2l1) < 3
That means the actual error when calculating the difference is:
1 < E3 < 1
And so the maximum absolute error in calculating the difference is just under 1mm.
I believe the book says that maximum absolute error in calculating the difference should be just under 2mm (combining the two errors of each measurement) but I can't find a way to combine the errors in this case.
Anyway I needed to check if I was right or if there was something I'm missing before moving on. Thank you in advance for any replies.
Last edited by trampolinerist; 10292017 at 01:05 PM.

Senior Member
[tex]\text {Let the true length of first pin be } x.[/tex]
[tex]\text {Let the true length of second pin be } y.[/tex]
[tex]12 \le x < 13 \text { and } 14 \le y < 15.[/tex]
The estimated difference in length is [tex]14  12 = 2[/tex].
The true difference is [tex]y  x.[/tex]
[tex]\ 13 < \ x \le \ 12 \implies 1 < y  x < 3.[/tex]
The error is [tex]e = y  x  2 \implies \ 1 < e < 1.[/tex]
So yes [tex] e < 1.[/tex]
But the potential range of error has a span of [tex]1  (\ 1) = 2.[/tex]
I suspect that there is a difference in vocabulary.

New Member
Thank you. It was helpful to see somebody else work through the problem.
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