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Thread: Trying to find an exponential decay factor for a project.

  1. #1
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    Trying to find an exponential decay factor for a project.

    Hello! Got a huge project a couple days ago, buying cars, houses, etc.

    I'm trying to find the exponential decay factor of the value of my vehicle.

    It all seems somewhat easy, however I came to a place where I appear to be stuck.

    Up to this point, I've had to choose a 2009 model car that's been around since at least 2006 (for those who are curious, I chose a 2009 GMC Yukon 4-Door SLE 4WD). I picked accessories and whatnot and added them to the price of the vehicle.

    You will need to make a table (years from purchase starting at 0 years old, total mileage, and car value at the beginning of the next year) and a graph (comparing age to car value) that has at the minimum four years of data.
    So I made a table using the percents given from the website that my teacher told me to use.

    http://www.kbb.com/kbb/NewCars/ResaleVa ... 37053.aspx

    I couldn't be lazy though, I had to take the price of my vehicle with the accessories and multiply it by the percents given instead of the numbers already given.

    Note: The total mileage is 12000 per year, as per what my teacher instructed.

    Code:
    Year	            Retained MSRP %	Total Mileage	Resale Value
    2009	              100%	                  0	   $43,728.00
    2010	               42%	              12000	   $18,365.76
    2011	               33%	              24000	   $14.430.24
    2012	               27%	              36000	   $11,806.56
    2013	               23%	              48000	   $10,057.44

    Then the next part is where I'm having trouble.

    From the list of data, you will need to develop an equation in which you can input the age of the car and the output would be the value of the car. The form of your equation should be exponential and you will probably have to use exponential regress on the TI calculators to find your decay factor. Your decay factor should be rounded to the nearest thousandth.
    I really haven't made any progress, but I did doodle some stuff down that I guess I can post since its already typed up.

    Equation I can put the age of the car and output be value?

    Y=ab^x?

    a and b are constants, x is age.

    If so…
    Y=ab^0
    Y=a(1)
    Y=a

    ???

    a=43728??? b=.42???
    a is resale value, b is percent retained…

    AGH!
    I don’t know what the equation is…
    Please note that the above work is most likely complete garbage.

    If anybody could please assist me in any way on finding this evasive equation, it would be greatly appreciated. Thanks!
    Take whatever advice I give you with a grain of salt, I'm just here to learn more about my favorite subject =]

    Hush my dear, don't say a word. Utter but not a scream, nor quiver or writhe in the face of this overwhelming madness.

  2. #2
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    Re: Trying to find an exponential decay factor for a project.

    You need to input the data (year, resale value) into you calculator and do the required regression analysis. The calculator will tell you the a and b values. You are correct, however, in assuming that "a" should equal the initial value of the vehicle.

  3. #3
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    Re: Trying to find an exponential decay factor for a project.

    Wow! That was a quick reply! Anyway, I have a TI89+ if it matters, how do I perform the required regression analysis so that it tells me my a and b? Thanks!
    Take whatever advice I give you with a grain of salt, I'm just here to learn more about my favorite subject =]

    Hush my dear, don't say a word. Utter but not a scream, nor quiver or writhe in the face of this overwhelming madness.

  4. #4
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    Re: Trying to find an exponential decay factor for a project.

    Err, my bad. TI 83+. Got the numbers mixed up.

    Anywho, just bumping this up a little.
    Take whatever advice I give you with a grain of salt, I'm just here to learn more about my favorite subject =]

    Hush my dear, don't say a word. Utter but not a scream, nor quiver or writhe in the face of this overwhelming madness.

  5. #5
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    Re: Trying to find an exponential decay factor for a project.

    Here's a tutorial on how to do an exponential regression with the TI-83:

    http://online.redwoods.cc.ca.us/instruc ... sld006.htm

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    Re: Trying to find an exponential decay factor for a project.

    Thanks for the quick reply!

    I followed that exactly, and it made sense, but I think it goes over my head at like slide 30-31ish. Im not sure how to relate that to my problem. I have everything else up to that point however.
    Take whatever advice I give you with a grain of salt, I'm just here to learn more about my favorite subject =]

    Hush my dear, don't say a word. Utter but not a scream, nor quiver or writhe in the face of this overwhelming madness.

  7. #7
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by Go^3
    … it goes over my head at like slide 30-31ish. Im not sure how to relate that to my problem …


    I'm not sure how to relate that phrase to your exercise, either. I do not even know what the phrase "slide 30-31ish" means.

    Your data does not look like exponential decay, to me. The initial drop from 100% to 42% is too large compared with the subsequent drops. Are you sure that your data is correct?

    The example that Galactus provided (via link) uses the Least Squares method to find the best-fit line, after using a theory to change from exponential modeling to linear modeling and converting the population data to logarithms.

    I would not have gone down that road because (1) that road is very convoluted, and (2) I'm thinking that the TI-83 comes with base-e exponential regression built-in. (See slide #22. If you scroll down that menu, then you should find it. If not, then I'm wrong.)

    However, since you stated that you understand that example, and since it is the method suggested by Galactus, I'll try to explain how that example relates to your exercise.

    In that example, they are looking for a model that gives some elk population as a function of the number of elapsed years since 1978.

    y = elk population after t elapsed years

    t = number of elapsed years since 1978

    Some of their data:

    (6, 70)
    (7, 82)
    (8, 96)

    These data points tell us that there were 70 elk 6 years after 1978. In other word, there were 70 elk in 1984.

    Likewise, there were 82 elk in 1985 and 96 elk in 1986.

    Your exercise is looking for a model that gives the value of a vehicle as a function of the number of elapsed years since 2009.

    y = value of the vehicle after x elapsed years

    x = number of elapsed years since 2009

    Some of your data points are:

    (0, 43728)
    (1, 18365.76)
    (2, 14430.24)

    These data points tell us that the vehicle is worth $43,728 after zero years have elapsed (i.e., 2009).

    Likewise, the value is $18,365.76 after 1 year has elapsed (2010) and $14,430.24 after two years have elapsed (2011).

    (I'm concerned that your data shows a huge drop in value after the first year, compared to the drops in subsequent years; in other words, I'm not sure how close an exponential-decay model will fit this data.)

    Slide #14 gives the exponential decay model:

    y = C * e^(kt)

    C is the initial elk population (i.e., in 1978 when t = 0 because no years have elapsed).

    k is the exponential-decay factor they are trying to find.

    The model that you typed is:

    y = a * b^x

    This is NOT the same form. The base in your model is b. The base in their model is the number e. You stated that you want to find the exponential-decay factor for this exercise. This is k.

    You need to use the form given in Galactus' example. That is exponential decay.

    If the base b is not the number e, then it's not an exponential-decay model.

    (If you were to use the exponential regression instead, then the TI-83 would give you the value of the base b above and the value of the coefficient a.)

    Slide #15 explains the theory they used to switch to a linear model (i.e., ln[y] is a linear function of t).

    ln(y) = ln(C) + ln(e^[kt]) leads to the linear equation:

    ln(y) = k*t + ln(C)

    On slide #19, they show how to convert the y-values to get a list of ln(y) values.

    On slide #24, they show how to use the Least Squares regression using this new data (i.e., using List1 and List3).

    The values for a and b in slide #25 are the Least Squares result.

    In other words:

    k is a = 0.2415…

    ln(C) is b = 2.6131…

    (Do not confuse these symbols with the a and b in a*b^x. They are NOT the same.)

    After they made these substitutions into the linear equation, they arrived at:

    ln(y) = 0.2415 t + 2.6131

    Then they switched from this logarithmic equation to an exponential equation.

    e^(ln[y]) = e^(0.2415 t + 2.6131)

    y = 13.6415 e^(0.2415 t)

    You need to obtain the natural logarithm of your vehicle values, and put them into a third list, as they did.

    You need to take the values for a and b given by the TI-83's Least Squares regression from List1 and List3 and substitute them the same way.

    ln(y) = a x + b

    Exponentiate both sides to obtain your exponential-decay function in the form:

    y = C e^(kx)

    I hope my explanation helps you to understand how the example provided by Galactus relates to your exercise.

    If I wrote anything that you do not understand, then please ask specific questions.

    If you would like more help with this exercise, then please show whatever work you're able to accomplish, and try to say something about WHY you're stuck, so that I might determine where to continue helping you.

    Using exponential regression instead (on the TI-89) gives me the following result:

    y = 33221.2945 e^(-0.7131108x)

    As I suspected, this best-fit does not match your data very well because the value drops a whopping 42% after one year, but only drops by small percentages after that.

    This model gives a value of $16,282.33 in 2010. It gives a value of $1,916.97 in 2013. These results are not very close.

    I strongly suggest that you verify your data before proceeding.

    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

  8. #8
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    Re: Trying to find an exponential decay factor for a project.

    I was referring to the guide he posted at about here.

    http://online.redwoods.cc.ca.us/instruc ... sld030.htm


    My data is indeed correct as far as I know. I used the information directly from kbb.com as my teacher instructed to achieve those numbers. Due to your concern, I will recheck them soon.

    I'm gonna look at everything you typed in detail, but I just wanted to squeeze in a quick post telling you how thankful I am for your assistance. It's obvious to me that you spent a LOT of time typing that up for me, so you have my deepest gratitude!
    Take whatever advice I give you with a grain of salt, I'm just here to learn more about my favorite subject =]

    Hush my dear, don't say a word. Utter but not a scream, nor quiver or writhe in the face of this overwhelming madness.

  9. #9
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    Re: Trying to find an exponential decay factor for a project.

    Upon further investigation of the prices, I made a theory: SUVs drop their value much quicker than other cars.

    Here is where I got my numbers from:

    http://www.kbb.com/kbb/NewCars/ResaleVa ... 37053.aspx

    I didn't use those numbers, I merely multiplied my own by their percentages.
    Take whatever advice I give you with a grain of salt, I'm just here to learn more about my favorite subject =]

    Hush my dear, don't say a word. Utter but not a scream, nor quiver or writhe in the face of this overwhelming madness.

  10. #10
    Elite Member mmm4444bot's Avatar
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    Quote Originally Posted by Go^3

    … My data is indeed correct as far as I know.


    Okay. Are you sure you were told to use exponential decay as a model for this exercise?

    It's not a very good model for this data.

    Perhaps, if you consider the initial value in 2009 to be an outlier (i.e., a value far from most others in a set of data), and thus ignore it, then the model might work for values of x >= 1.

    Also, what is the significance of the mileage? I do not see how mileage affects the percentages that you copied from that web page.

    PS: I just realized what you meant when you wrote "slide 30-31ish". DOH!

    "English is the most ambiguous language in the world." ~ Yours Truly, 1969

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