Exponential Decay?

Forums: Science And Math
Email this Topic • Print this Page

This may be more appropiate in the Computer Programming forum but. . .

A book I'm working on has me generating a table for an equation.

I decided to do the old "Exponential Growth/Decay" formula

N=N(0)*e^(k*t)

Where e = natural log
k= rate of growth or decay
t= time.

I wrote a C++ function:

Code:



double calculateExp(double timeZero, int duration, double rate){ 

       //Sub variable 

       double growthFactor; 

       

       //determine growthFactor 

       growthFactor = rate * duration; 

       //Complete the equation  

       return timeZero*(exp(growthFactor)); 

}

Here's what I came up as a problem:

Starting with 1000 bacteria, with a rate of decay of .112 how many bacteria are left at the end of each hour.

So my inputs were:

Initial Value=1000
rate =-.112 (decay)
Time = 10.

The results were:

Quote:

1 hour = 894
2 '' = 799
3 '' = 714
4 '' = 638
5 '' = 570
6 '' = 509
7 '' = 455
8 '' = 406
9 '' = 362
10 = 323

The code that generated this table was

Code:



for (int length = 1; length <= time ; length++){ 

        cout << "Time: "; 

        result = calculateExp(value, 1, rateOfChange); 

        value = (int)result; 

        cout << value << endl;   

    }

Topic Stats
Top Replies
Link to this Topic

Type: Discussion • Score: 1 • Views: 3,070 • Replies: 4

No top replies

Whats the problem?
Hey,

What is the problem? The method seems correct.

But the ROUNDING of double result to an integer value may loose precision

Try this java code similar to your C++ for() loop but without loss of precision

Code:

for (int length = 1; length <= time ; length++) { 

    result = calculateExp(1000, length, rateOfChange); 

    value = (int)result; 

    System.out.println("Hour "+length+":"+value); 

}

This gives following output

Quote:

Hour 1:894
Hour 2:799
Hour 3:714
Hour 4:638
Hour 5:571
Hour 6:510
Hour 7:456
Hour 8:408
Hour 9:364
Hour 10:326

In the above output, Bold results show difference with your output series.

It is due to LOSS of double precision.

0 Replies

Basically have the variable 'value' be a double?

0 Replies

Answer

NeoGuin wrote:

Basically have the variable 'value' be a double?

Ok I mean you have my way of calling calculateExp() or you can pass value as double in you method

But then (int)result is required becoz number of bacteria is in whole number.

Smile

0 Replies

OK.

But other than loss precision the ROUTINES are correct!

0 Replies

Exponential Decay?

Related Topics

Quick Links

My Account

able2know