I have run across a math problem that is vexing me.  I think the
answer is probably very simple, but I've worked on the problem for
nearly an hour, and it's driving me crazy.  I need this answer for a
client meeting next week, but would really appreciate an answer as
soon as possible so that I can get moving with the rest of the
presentation.

Here is the problem:

I have eight different buldings with rooftop air conditioning units. 
All of the units together produce 100 tons of cooling.  Each of the
units are a different age and produce different amounts of cooling.  I
need the average age of the units by tons of cooling.

For example:  If Unit A produces 50 tons of cooling and is 10 years
old, and Unit B produces 50 tons of cooling and is 20 years old, then
the average age of the units (by tons of cooling) is 15 years.

Here are the units by tons of cooling and age:

Unit     Tons     Age (in years)
A        65.5     10
B         5.0     12
C         3.0      3
D         6.0     20
E         4.0      5
F         4.0     20
G         5.0      5
H         7.5      3

Five stars and a tip for the one who gets me the average age of the
units asap.  Please show your work.

-ddelphi

Hi ddelphi:

Thanks for the interesting question. What you are in need of here is
called a "weighted average", which can be computed by summing up the
weight times the age for each unit and then dividing through by the
total number of tons. In mathematical terms:

Average age by tons of cooling = Sum(weight_i * age_i)/Sum(weight_i),
for each i from A to H.

So, in your example, the calculation would be:

  (65.5*10+5*12+3*3+6*20+4*5+4*20+5*5+7.5*3)/(65.5+5+3+6+4+4+5+7.5)
= (655+60+9+120+20+80+25+22.5)/100
= (
= 9.96 years

---

Clarification of Answer by websearcher-ga on 13 Dec 2002 13:24 PST

PLEASE IGNORE FIRST ANSWER - IT IS WRONG. I'M NOT SURE HOW IT GOT
POSTED - GLITCH

Hi ddelphi:

Thanks for the interesting question. What you are in need of here is
called a "weighted average", which can be computed by summing up the
weight times the age for each unit and then dividing through by the
total weight. (In this example, the "weighting factor" in the
calculation is actually *weight*.)

In mathematical terms:

Average age by tons of cooling = Sum(weight_i * age_i)/Sum(weight_i),
for each i from A to H.

So, in your example, the calculation would be:

weighted average = (65.5*10 + 5*12 + 3*3 + 6*20 + 4*5 + 4*20 + 5*5 +
7.5*3) / (65.5 + 5 + 3 + 6 + 4 + 4 + 5 + 7.5)
= (655 + 60 + 9 + 120 + 20 + 80 + 25 + 22.5) / 100
= 991.5 / 100
= 9.915 

So, the average age of the units by tons of cooling is 9.915 years.

Some webpages where you can find further information include: 

Computing Weighted Averages
http://www.ccs.neu.edu/course/com3351/weighted-averages.html

Examples of Weighted Averages
http://mathforum.org/library/drmath/view/52803.html

I hope that this information has been of help in your search.       
      
If you need any clarification of the information I have provided,
please ask using the Clarification feature and provide me with
additional details as to what you are looking for. As well, please
allow me to provide you with clarification(s) *before* you rate this
answer.
           
Thank you.            
           
websearcher-ga           
           
         
Search Strategy (on Google):    
 
"weighted averages"
://www.google.com/search?hl=en&lr=&ie=ISO-8859-1&safe=off&q=%22weighted+averages%22

---

Request for Answer Clarification by ddelphi-ga on 20 Dec 2002 12:50 PST

This really isn't a request for clarification, but I didn't know how
to get hold of you without posting a new question.

I just wanted to let you know that my client was quite surprised to
learn the average age of his equipment and asked how I arrived at my
conclusion.  I told him "I got it on Google Answers," and he said
"Then it must be accurate!"

You and your fellow researcher have developed quite a good reputation,
and that reputation is well deserved.

Thanks again for the help.  I will be requesting you specifically when
I have questions in the future.

P.S. My client will sign the contract January 7th, and is the biggest
contract my company has ever had - in part thanks to Google Answers
and websearcher!

---

Clarification of Answer by websearcher-ga on 20 Dec 2002 13:00 PST

Hi ddelphi:

Thanks for getting in touch! You were absolutely correct to use the
Clarification feature for this. We Researchers are alerted to Requests
for Clarification but not to Comments. In fact, I missed the whole
back and forth that took place in the comments - I only found out
about it yesterday!

I'm glad that everyone was able to agree that my original answer was
the one that was wanted. Part of being a good Google Researcher is
being able to intuit, to a certain extent, what the asker wants, based
on their use of language and how they frame the question. I don't
*always* nail it the first time, but I'm certainly glad I did with
you. :-)

I'm gratified that you are getting your contract and glad I could play
some small part in that success.

Thanks again for all your kind words. 

websearcher-ga

Thanks webresearcher!  I knew that the answer was probably very
simple, but I'm a product of the public schools.  I wish the tip could
be bigger, but I'm putting this on my company credit card, and they
gripe enough about what I spend.

Thanks again...this will help me with my proposal.

Hi ddelphi:

Thanks for the nice comments, rating and the tip. Glad to have been of service. :-)

websearcher-ga

You should not have rated this answer so quickly -- It is WRONG.

A quick check of the answer will show why it is not correct.  The
"average" of the individual averages cannot exceed the range of the
averages.  They range from 6.55 to 0.2.  Therefore, the answer of 9.91
is obviously wrong.

The correct answer is the average of the average.  [SUM (average)]/n
         tons    years   average
A	65.5	10	6.55
B	5	12	0.416666667
C	3	3	1
D	6	20	0.3
E	4	5	0.8
F	4	20	0.2
G	5	5	1
H	7.5	3	2.5


Then add the averages and divide by the number of averages

6.55 + 0.416666667 + 1 + 0.3 + 0.8 + 0.2 + 1 + 2.5 = 12.77
12.77/8 = 1.595833333 tons/year

Looking back at the individual averages can see that this average of
the average os within the range and is close to amount that
represented most of the individual averages.

I hope this helps and you see this answer BEFORE you make your
presentation.

Remember, if you are only willing to pay $2.00, you will get an answer
of equal value.

Ok...now I'm confused.

I understand websearcher's answer, and it seems to make sense (in
fact, his answer was very close to my intuitive guess).  But socal's
math is confusing to me, and doesn't really seem to give an answer to
my question of what is the average age of the units by tons.  What is
the average age of the units then?  What does "1.595833333 tons/year"
mean in realtion to my question?

As for the price...well, I really didn't think this was a complicated
question, and would take someone with better math skills than me only
a few minutes to calculate.  I think $2.00 and a $1.50 tip was quite
reasonable for something that I thought would only take a few minutes
to calculate.

So...anyone else care to chime in?  What is the answer to my question
("what is the average age of the units by ton?").  Man, I wish I would
have studied harder in college instead of drinking a lot and dancing
with girls.

yo ddelphi,

don't get confused,

always check back with your original example i.e.
Example (1) if Con A (50 tons and 10 years) + Con B (50 tons and 20
years)

if you used websearcher's recommended method you get:
Average years by ton of cooling = (50*20 + 50*10)/100
                                = 15 years (which is what you
expected)

if you used socal's method you get:
Average ton of cooling per averaged year = (50/10 + 50/20)/2
                                         = 3.75 ton/year

in conclusion, stick to websearcher's answer for your needs.  Socal,
you gave an answer that is different from what ddelphi needs.

OK, you are right my answer was inverted.  You wanted years per ton
rather than the inverse.

First, lets go back to your example of "Unit A produces 50 tons of
cooling and is 10 years old, and Unit B produces 50 tons of cooling
and is 20 years old".  You gave an answer of 15 years for the average.
 Well, the answer to your example is 10 10 years divided by 50 tons
equals 0.2 years per ton.  Then 20 years divided by 50 tons equals 0.4
years per ton.  The average of the average is 0.3 years per ton.  The
weighted average and the average of the average is the same since the
tonnage is the same.

Now for your question.  You want an answer "average age of the units
(by TONS of cooling).  So the answer must have the units of
"years/ton" as per your request.

Therefore, divide Years by Tons for each unit from A to H.

A	65.5	10	0.152671756
B	5	12	2.4
C	3	3	1
D	6	20	3.333333333
E	4	5	1.25
F	4	20	5
G	5	5	1
H	7.5	3	0.4

Then take the sum of each average (years/ton) and divide by 8 the
number of units.  You will get an answer of 1.817000636 years/ton for
the averge age of the units by tonnage of cooling.

The weighted average (taking into account the age of each unit over
the total age of all untis combined).  Therefore, each average is
multiplied by the "units age" divided by 78 years (total for all
units).  Then add all these numbers together to get the weight average
-- 2.72363 years/ton.

Your example of "Unit A produces 50 tons of cooling and is 10 years
old, and Unit B produces 50 tons of cooling and is 20 years old" with
your answer of 15 years actually has the units of (Years-Tons)/Tons
which reduces to Years, not your requested units of Years/Tons.  The
answer to your example and the answer given by the researcher are not
AVERAGES of years per ton, but are the AVERAGE of Years-Tons per Tons
of cooling.

The answer to your question is given by my answer.  The answer, if you
assume your example as "the de facto question" then the researcher's
answer is correct.

I now see how not stating one's questions correctly can cause
confusion.  As I implied before, I'm no mathmatician.

You see, I'm in sales, and the only calculation I have to make on a
regular basis is how many gin and tonics I need to pour down the
throat of my clients to get them to sign contracts. :-)

All is well, however, as websearcher apprently read my mind and
divined the answer I was looking for, despite my poorly worded
question.

I appreciate the comments posted by both socal and unstable.  I will,
in the future, phrase my questions with more care.