The coding below gives me a final answer of 128, however I need a value around 124. Please help.

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Abhishek Jani 2015-3-11

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/182691-the-coding-below-gives-me-a-final-answer-of-128-however-i-need-a-value-around-124-please-help

评论： Roger Stafford 2015-3-11

The data for the code below is:

Time (seconds) (t)
6
24
8
2
30
8
2
36
8
2
42
8
2
48
8
2
54
8
2
60
8
2
66
8
2
72
8
2
78
8
2
84
8
2
90
8
2
96
8
2
102
8
2
DOT (%)
7590349
33302268
71026199
63632598
14818267
59790825
99585793
27990059
83317801
50036243
0225743
94863829
84363674
4279797
77415346
15139277
63218391
05084395
61447655
97100549
67961065
15684995
14414414
69742156
88826758
29657243
17086052
24779952
78036657
02298851
22418971
63249456
76120949
98312105
07041524
30268199
65921094
96396396
80273377
69327949
67702185
62969866
4477581

The code is as given below

x(1)=2;
x(2)=3;
error = 0.0001;
for i=3:43
    f=@(x) (1./(x-60)).*(x.*exp(-t(i)./x)-60.*exp(-t(i)/60)) - ((100-DOT(i))./100);
     x(i) = x(i-1) - (f(x(i-1)))*((x(i-1) - x(i-2))/(f(x(i-1)) - f(x(i-2))));
    disp (x(i))
    if abs(f(x(i)))< error
        break
    end
end
   tm = (x(i)); 
  disp('This is tm')
   disp(tm)
   kLa = (1/tm);
   disp('kLa = 1/tm')
 disp(kLa)
 KLAh = (kLa.*3600);
 disp('kLa per hour is')
 disp(KLAh)

KLAh is supposed to be around 124. Please help.

3 个评论
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Abhishek Jani 2015-3-11

It's a secant method and hence only the last value is used because once it falls below the error value, it terminates, hence converging.

Roger Stafford 2015-3-11

在 MATLAB Online 中打开

In the secant method, as with many other solver algorithms, any parameters that enter into the equation to be solved should always be held constant while attaining convergence. In your code the 't' and 'DOT' variables are acting as parameters, but you are varying them throughout your iteration. If you do manage to get

abs(f(x(i)))< error

at some point in this procedure it will be a solution that corresponds to the particular values of 't' and 'DOT' at that point, and you have no reason to suppose that this will be the value 1.24. It depends on where your procedure happens to break.

You apparently have a misunderstanding of the secant method. You had better do a more careful study of it. See:

http://en.wikipedia.org/wiki/Secant_method

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