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

The data for the code below is:

Time (seconds) (t)
21.6
24
25.8
28.2
30
31.8
34.2
36
37.8
40.2
42
43.8
46.2
48
49.8
52.2
54
55.8
58.2
60
61.8
64.2
66
67.8
70.2
72
73.8
76.2
78
79.8
82.2
84
85.8
88.2
90
91.8
94.2
96
97.8
100.2
102
103.8
106.2

DOT (%)
10.7590349
12.33302268
13.71026199
15.63632598
17.14818267
18.59790825
19.99585793
21.27990059
22.83317801
24.50036243
26.0225743
27.94863829
29.84363674
31.4279797
32.77415346
34.15139277
35.63218391
37.05084395
38.61447655
39.97100549
41.67961065
43.15684995
44.14414414
45.69742156
46.88826758
48.29657243
50.17086052
51.24779952
52.78036657
54.02298851
55.22418971
56.63249456
57.76120949
58.98312105
60.07041524
61.30268199
62.65921094
63.96396396
64.80273377
65.69327949
66.67702185
67.62969866
68.4477581

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.

[Roger Stafford]

The code you give looks as though you are trying to carry out a Newton-Raphson procedure to find the values of x(i) for which f(x(i)) is equal to zero, except that you are permitting only one iteration for each value of i. That is not likely to get an accurate result, particulary since the time intervals change from one point to the next in t.

Even more strange is that upon exiting the for-loop, the value of KLAh depends only on the last x(i) value, namely x(43), that was computed in the loop. All previous values are ignored.

You should explain in words, not code, what your computation is supposed to accomplish. Only then can useful advice be given.

[Abhishek Jani]

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]

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