Allocating percentages

[Fred Smith]

I have percentages in B2:B30 which add up to 100%, and an amount to allocate
in A2 (for example, 8). However, the results must be integers, and must add
up to A2. The simple formula is =round($A$2*B2,0) copied down, but the
rounding doesn't always produce the right total. How do I ensure that the
total always equals A2?

Thanks,
Fred

 

[Joe User]

I have percentages in B2:B30 which add up to 100%,
and an amount to allocate in A2 (for example, 8).
However, the results must be integers, and must add up to A2. The simple
formula is =round($A$2*B2,0)
copied down, but the rounding doesn't always produce
the right total. How do I ensure that the total always
equals A2?

I believe I already answered that in Lilyput's thread. I'll paraphrase....

This is a common quantization problem, i.e. the result of rounding "long"
decimal fractions to fewer decimal places (or integers). There are no
perfect solutions.

Consider the following simple example. You have 3 dollar bills, and you
want to award them to 4 people in the proportion to their contributions,
which is 25% each. It can't be done! At least, not fairly. Someone must
get zero.

One simple (but flawed) approach is to put the following formulas into C2
and C3, say, and copy C3 down through C30:

C2: =ROUND($A$2*B2,0)

C3: =MIN($A$2 - SUM($C$2:C2), ROUND($A$2*B3,0))

That approach is flawed because it is more unfair to the people represented
by the later cells.

We might ameliorate the unfairness by randomizing B2:B30 (and associated
columns), using the formula above, then reordering C2:C30 according to the
original order. I would use a UDF for that.

 

[Joe User]

PS....

One simple (but flawed) approach is [....]

I seem to recall posting (or at least preparing for posting or email) a
solution to another similar problem that ameliorated the unfairness by
ensuring that everyone gets at least one (if they deserve it), then everyone
gets at least two (if they deserve it), etc per the OP's request.

Some people might consider that to be more fair, even if it means that a 50%
contributor might get the same as everyone else due to quantization.

I don't remember if my approach worked. But if that's something that
Lilyput would like to consider, I can search for it.

(It's a difficult search for me. So I don't want to start doing it unless
there is some interest.)


----- original message -----

 

[Joe User]

Errata....

We might ameliorate the unfairness by randomizing
B2:B30 (and associated columns), using the formula
above, then reordering C2:C30 according to the original order. I would
use a UDF for that.

I suspect most people (those represented by B2:B30) would not consider that
to be less unfair.

It would probably be more fair to effectively sort B2:B30 in descending
order, then apply the distribution. Again, you could reorder C2:C30
according to the original order, if need be; and in that case, I would use a
UDF.


----- original message -----

 

[Joe User]

Errata (again)....

One simple (but flawed) approach is to put the following
formulas into C2 and C3, say, and copy C3 down through C30:
C2: =ROUND($A$2*B2,0)
C3: =MIN($A$2 - SUM($C$2:C2), ROUND($A$2*B3,0))

I think C30 should have the formula:

C30: =$A$2 - SUM($C$2:C29)

That may be necessary to remedy the case where SUM(C2:C29)+ROUND(A2*B3,0) is
less than A2.

At least, I __think__ that is a possibility. In any case, the
non-generalization cannot hurt, even it proves to be surperfluous.


----- original message -----

 

[Ron Rosenfeld]

I have percentages in B2:B30 which add up to 100%, and an amount to allocate
in A2 (for example, 8). However, the results must be integers, and must add
up to A2. The simple formula is =round($A$2*B2,0) copied down, but the
rounding doesn't always produce the right total. How do I ensure that the
total always equals A2?

Thanks,
Fred

I have run into a similar situation recently, trying to completely distribute a
pot of money according to a predetermined percentage distribution.

I would use the formula you have in B2:B29

However, for B30, I would use the formula =a2-sum(b2:b29)

I believe, although I don't know how to prove it mathematically, that this
should result in no one being off by more than one unit, from what they might
receive if the units were smaller.

I would also use what is euphemistically termed bankers rounding, where the
rounding is done towards the nearest even number for a value that is half
between. That produces a less biased result than the round half up method,
which always rounds up when at the halfway point.

The VBA Round function uses "banker's rounding". The Excel worksheet function
uses algebraic rounding.

For my purposes, the round to even method gives a result that is suitable. But,
depending on the distribution of your numbers, other methods may be appropriate
for you, and the round half up may be OK, even with the bias it introduces.



--ron

 

[Joe User]

I would also use what is euphemistically termed bankers rounding

It is not likely to make much of any different. VBA Round(x,0) and Excel
ROUND(x,0) differ only when MOD(x,1) is exactly 0.5 (and INT(X) is even).

(Techically, when MOD(x,1) is exactly 0.5 within 15 significant digits for
Excel ROUND, notwithstanding defects.)

I would use the formula you have in B2:B29
However, for B30, I would use the formula =a2-sum(b2:b29)

Consider A2=15, B2:B11=10%, and B12:B30=0%. Put =ROUND($A$2*B2,0) into C2,
copy down through C29, and put =$A$2-SUM(B2:B29) into C30.

C2:C11=2, which sums to 20(!). C12:C29=0, but C30=-5(!). True, the sum is
15; but some of the line item values are incorrect.

(And that is true whether you use ROUND or VBA Round.)

Now, try what I suggested initially. Put
=MIN($A$2-SUM($C$2:C2),ROUND($A$2*B3,0)) into C3 and copy down through C30.

C2:C8=2 and C9=1, which sums to 15. C10:C30=0.

However, as I noted subsequently, I believe there are cases when
=A2-SUM(B2:B29) in C30 also needed. "The exercise is left to the student"
.


----- original message -----

 

[Joe User]

Errata (yet again)....

Sorry for the incessant responses.

One simple (but flawed) approach is to put the following formulas into C2
and C3, say, and copy C3 down through C30: C2: =ROUND($A$2*B2,0)
C3: =MIN($A$2 - SUM($C$2:C2), ROUND($A$2*B3,0))

I think it would be better if C3 is:

=MIN($A$2,ROUND($A$2*SUM($B$2:B2),0)) - SUM($C$2:C2)

especially when B2:B30 is sorted in descending order.

And then I believe it is unnecessary to make a special case of C30 (my
previous suggestion of $A$2-SUM($C$2:C29)).

However, that is still flawed in some respects. Consider Lilyput's example
(below).

(By the way, did you notice that her percentages do not add up to 100%? I
assume there is a typo. So I changed her 2nd-to-last entry from 2.00% to
2.40%.)

When B2:B30 is not sorted, the non-zero entries would be:

27.90% 2
1.50% 1
1.40% 1
1.60% 1
10.50% 1
9.50% 1
2.90% 1

But it might seem counter-intuitive that people with 1.4%, 1.5% and 1.6%
each get 1, but people with 3.0%, 3.6%, 3.7% and 5.0% get 0.

Sorting B2:B30 in descending order has slightly better results, to wit:

27.90% 2
10.50% 1
9.50% 1
3.70% 1
2.80% 1
2.00% 1
1.40% 1

But there are still inexplicable anomalies; I mean anomalies for which
people will not understand the explanation.


-----
Lilyput's data:

A2:
8

B2:B30:
1.50%
2.30%
1.80%
1.00%
5.40%
1.40%
2.10%
1.40%
3.60%
0.90%
1.70%
2.00%
1.60%
1.90%
10.50%
3.70%
1.10%
1.40%
2.10%
9.50%
2.80%
0.80%
2.90%
0.60%
2.40%
2.70%


----- original message -----

 

[Ron Rosenfeld]

It is not likely to make much of any different. VBA Round(x,0) and Excel
ROUND(x,0) differ only when MOD(x,1) is exactly 0.5 (and INT(X) is even).

(Techically, when MOD(x,1) is exactly 0.5 within 15 significant digits for
Excel ROUND, notwithstanding defects.)



Consider A2=15, B2:B11=10%, and B12:B30=0%. Put =ROUND($A$2*B2,0) into C2,
copy down through C29, and put =$A$2-SUM(B2:B29) into C30.

C2:C11=2, which sums to 20(!). C12:C29=0, but C30=-5(!). True, the sum is
15; but some of the line item values are incorrect.

(And that is true whether you use ROUND or VBA Round.)

Now, try what I suggested initially. Put
=MIN($A$2-SUM($C$2:C2),ROUND($A$2*B3,0)) into C3 and copy down through C30.

C2:C8=2 and C9=1, which sums to 15. C10:C30=0.

However, as I noted subsequently, I believe there are cases when
=A2-SUM(B2:B29) in C30 also needed. "The exercise is left to the student"
.

I think the kind of rounding that needs to be used will depend on the data and
the expected results. As you wrote, there is no single perfect method.

I was not clear last night, but I think the goal of the rounding method used
should attempt to result in no distribution being off by more than one unit.

For dividing pots of money, it may usually be the case that round half to even,
or round half up, is good enough. For other types of data and divisions, it
may be better to use stochastic rounding, where halves are randomly rounded up
or down; or even dithering.
--ron

 

[Joe User]

For dividing pots of money, it may usually be the case
that round half to even, or round half up, is good enough.

You fail to acknowledge and seem to fail to notice that I demonstrated by
example that it is not.

Moreover, when rounding dollars-and-cents to dollars, the rounding algorithm
("banker's" or normal) makes a difference only when the amount ends in
exactly 50 cents. Yes, that does depend on the data. But in general, that
is unusual.


----- original message -----

 

[Ron Rosenfeld]

You fail to acknowledge and seem to fail to notice that I demonstrated by
example that it is not.

I quoted your comment about there being no perfect method. But if it makes you
feel better, I explicitly acknowledge than one can construct sets of data and
rules for division for which the simpler rounding algorithms are inadequate,
and the data set and rules you posted earlier are an example of that
inadequacy.

Your example, although demonstrating this point is, in my opinion, unrealistic.

I think it more likely that those who are dividing a pot of $15 with a rule of
10% to 10 people, and 0% to 10 people, rounded to the nearest $1, would come up
with a different rule.

The situation I am dealing with has to do with dividing a much larger pot
amongst many fewer people, and the maximum deviation from perfect has been a
mere penny, using simple rounding algorithms.
--ron

 

[Ron Rosenfeld]

The situation I am dealing with has to do with dividing a much larger pot
amongst many fewer people, and the maximum deviation from perfect has been a
mere penny, using simple rounding algorithms.

And, of course, the deviation could be greater depending on the various
parameters that make up the rules; in which case, a more involved algorithm
would be required; than merely rounding the first n computations and taking the
difference for the last.
--ron