Budget question

Hi everyone
was hoping you guys(and or ladies) could help me with this question.

If you cut your budget from 99 to 66
What is the budget cut in terms of %???
is it 5%, 33%, 50%, 25% ???
thks for your help
j

It is (99-66)/99=33/99=1/3
So correct answer should be 100%/3=33,333...% or 33,(3)%

but if you cut your budget from 99 to 49 1/2 isnt that a 100% cut in your budget? followed reasoning that 99 to 66 would be a 50% budget cut?

Sorry, I'm good at math not at "budget", so I don't know what you exactly ask. I don't know how it's possible that budget cut from 99 to 49,5 is not 50%.

It is reverse: 49,5 to 99 is 100% because (99-49,5)/49,5=49,5/49,5=100% for (99-49,5)/99=49,5/99=50%

but vstjrt(and btw ,ty for help) why wouldnt the reasoning stand going from 99 to 49 1/2 as 100% in terms of chopping your expenses?

Because you have 99 as input value, not 49,5. So decrease of 49,5 is only 50% for 99, but increase of 49,5 for 49,5 is 100%.

it may help if you think that reducing your budget from 99 to 0, it's a 100% budget cut

i know it seems so obvious , ty, i just was not willing to look at it the most obvious way.

Percentages are a bit strange in the sense they are in fact multiplications or divisions:

If you cut your income by 40% and then increase it by 50%, you end up with less than what you started by 10%...

N(1-0.4)x1.5 = Nx0.9

The formula I use for percentage moves is: Ending number divided by starting number, minus one. Thus:

66 / 99 = 0.666667, and 0.66667 - 1 = -0.33.

So vsjrt is correct... that's a 33% drop.

Similarly, an increase from 66 to 99 is: 99 / 66 = 1.5, and 1.5 - 1 = .50... or a 50% increase.

yes rex. that is what i was thinking. 66 to 99 = 50%. So why not the same when you are decreasing???!!! After all DECREASING YOUR BUDGET IS A POSITIVE FACTOR RIGHT???!

@jimO if you want to look equal, use logartithms.

The logarithm of the increase (99/66) is (about) +0.405465

The logarithm of the decrease (66/99) is (about) -0.405465

In any case (and in any base) they are exactly opposite numbers.

Let me pose a similar question. I think that the answer depends on the definition of "average".

A house is heated up by an outdoor ambience (at 25C) from 10C at 8 am to 20C at the noontime. What is the average temperature (T_avg) of the house during this 4-hour heating process?

(a) 15C; (b) higher than 15C; (c) lower than 15C.

I'll bite. It seems like the temperature will change faster when the difference is greater, implying the average would be > 15C.

Are you implying that the graph of the temperatures will be logarithmic? E.g., every hour the temperature changes by half the distance to 25C.

From Newton's law of cooling, the temperature will be an exponential:

T = To + (Ti - To) exp(-rt) with r = 0.27 s-1, Ti initial temp, To outside temp

But yes the temperature will be longer around 20 than around 10.

I have long known that LG people are bright and knowledgeable. Newton's law of cooling, my, yes. Logarithmic, Oh, yes, too.

If we take T_avg as Integral(T(t)*w(t))/t0, where T(t) is the exponential solution given by Carroll and t0 is 4 hours, then T_avg is indeed greater than 15C, if the weighting function w(t) is taken to be unity.

If we define T_avg simply as 0.5*(T_initial + T_final), then obviously T_avg = 15C.

If w(t) is some type of non-unity weighting functions, then it is possible T_avg can become lower than 15C.

Likewise, clearly, % = (a1 - a2)/a1 can be different from % = (a1 - a2)/a2.

Therefore, do not ask me, "Tim, are you happy these days?" Because if you do, my answer will be miles long, depending on the definition of "happy".

BTW, we seem to like to send flowers to beauties, don't we? :) Newton's law of cooling, George Washington's law of cherry-tree cutting,....

Your posts are always so poetic Tim, not sure I get it but it sends me in faraway lands full of cherry blossom trees and Gobans...

Why divide by t0? And what do you get for the integral?

T_avg IS defined as Integral(T(t))/t0, NOT 0.5*(T_initial + T_final). If you want to apply your "own" definition, why not simply "define" T_avg to be To+Ti? Or just "define" it to be, say, 20C? That is, at least when you are talking about the average temperature at a specific point over time. If you were to talk about the average temperature in the entire room at a specific point in time, I guess you should carefully consider if you really are interested in the average TEMPERATURE, or if you want to account for the heat capacity of all materials in the room, to try to calculate the theoretical "equilibrium temperature" of the room. Sounds fun!

Happiness, though, is a different matter. Not easily measured - nor defined. Yet, I suppose that if instead of asking you “Tim, are you happy these days?”, I were to ask you "Tim, are you happier today than you were this time last Friday?", it would be possible for you to answer with a simple "yes" or "no" :-)

Don't put money on that!

I meant for mean value from ti to tf, you need to divide the integral by (tf-ti), not t0...
Result is 15.89°C : see here

Carroll, your equation T(t) = To + (Ti - To)*exp(-rt) with r = 0.27 s^-1 is qualitatively wonderful, but quantitatively off. According to your equation, it takes only 5 sec for the house to reach 21.1C.

Based on Newton's cooling law (I meant to say that we might have given Mr. Newton too much credit), the original governing equation can be written as

m*cv*dT/dt = A*h*(25 - T), where

m: the mass of the house, which can be of Norwegian-, French-, or Estonian-style.

cv: heat capacity of the house, as Marius kindly mentioned

A: area of the house exposed to the air (assume that the floor is insulated)

h: so-called heat transfer coefficient

So, your r can be defined as r = A*h/(m*cv).

Let us try again to find an accurate value of r, such that, after exactly 4 hours, the house has reached exactly 20C. To unify our discussions, let us choose the following units: m (in kg), cv(in J/kg-C), A (in m^2), t (in sec), and h (in W/m^2-C. An appropriate value is about 10).

Only after having found the correct T(t), can we talk about T_avg and happiness.

Carrol got the NUMBERS right (ln(3)/4 = 0.2747), but he got the UNITS wrong. It should obviously be pr HOUR (h-1), not pr SECOND (s-1), as he wrote :-)

Fortunately, we don’t need to consider the areas and heat coefficients to find this – I think that could be messy...

Ah, Marius, thanks.

So, back to the original theme of this post, Carroll is only

(0.2747 - 0.27)/0.27 = 1.74% off. :)

Or 4(ln(3)/4-0.27)/ln(3) = 1.69% off ;)

But 0.27 is 0.2727, just expressed to fewer significant digits.

The question was posed with barely more than one significant digit of precision, expecting more in the answer is divisive. Had you expressed the starting and ending temperatures as 10.0C and 20.0C, then I'm sure you'd have been given a more precise anwer.

vstjrt   has it right :

If you  go from   99  to  66  ;

( decrease  the  fraction    ,  i. e .

it s like going  from  3  to 2    )

So,   33 %  is the  right answer  here !

Decreasing  a budget  ,   or    anything   by   50 %

means  you   slice  it  in half ! !

( i  e    100  to  50    ,      2   to  1  ,  etc      )

Try not  to make   %    or  fractions     or   Ratios      so  difficult     /

In  Math ,  find   the  simple  solution   first

( as  Ms.  Bagley  said  - -  'don 't make  a  mountain  out   of a  Mole Hill   '  )