I trying to use matlab to solve a polynomial with variable parameters. I am using fsolve. The code is given below:
function my_fsolve3()
% Constant Parameters
A = 6.24;
B = 5.68e-5;
C = 1.7e-3;
D = 6.55e-8;
E = 5.3e-8;
F = 9.46e-1;
t = [0;600;1200;1800;2400;3000;10200;17400;24600;31800;...
39000;46200;53400;60600;67800;75000;82200;89400;96600;103800;...
111000;118200;125400;132600;139800;147000;154200;161400;168600;175800;183000];
G = [0.00;4.88E-06;4.88E-06;5.11E-06;5.11E-06;5.35E-06;5.35E-06;5.36E-06;5.36E-06;5.07E-06;....
5.07E-06;4.93E-06;5.30E-06;5.30E-06;5.30E-06;9.46E-06;9.46E-06;1.13E-05;1.15E-05;1.10E-05;...
1.10E-05;1.04E-05;1.04E-05;1.04E-05;1.03E-05;1.04E-05;1.06E-05;1.13E-05;1.13E-05;1.13E-05;1.03E-05];
H = [0.00;3.34E-01;6.79E-01;1.04E+00;1.41E+00;6.00E+00;1.07E+01;1.56E+01;2.07E+01;2.59E+01;...
3.14E+01;3.67E+01;4.18E+01;4.66E+01;5.09E+01;5.51E+01;5.90E+01;6.23E+01;6.56E+01;6.87E+01;...
7.12E+01;7.36E+01;7.59E+01;7.78E+01;7.95E+01;8.11E+01;8.24E+01;8.24E+01;8.23E+01;8.21E+01;8.20E+01];
I = [0.00;4.88E-06;4.88E-06;5.11E-06;5.11E-06;5.35E-06;5.35E-06;5.36E-06;5.36E-06;5.07E-06;...
5.07E-06;4.93E-06;5.30E-06;5.30E-06;5.30E-06;9.46E-06;9.46E-06;1.13E-05;1.15E-05;1.10E-05;...
1.10E-05;1.04E-05;1.04E-05;1.04E-05;1.03E-05;1.04E-05;1.06E-05;1.13E-05;1.13E-05;1.13E-05;1.03E-05];
J = [1.78E-07;7.41E-06;9.33E-06;1.20E-05;1.05E-05;1.74E-05;3.72E-05;3.55E-05;1.00E-04;4.07E-02;...
2.45E-01;6.17E-01;1.32E+00;2.29E+00;2.34E+00;2.40E+00;1.82E+00;1.38E+00;2.09E+00;1.82E+00;...
1.58E+00;2.29E+00;1.62E+00;1.12E+00;8.91E-01;8.51E-01;7.59E-01;8.71E-01;1.12E+00;1.00E+00;8.51E-01];
K = [9.75;8.13;8.03;7.92;7.98;7.76;7.43;7.45;7.00;4.39;...
3.61;3.21;2.88;2.64;2.63;2.62;2.74;2.86;2.68;2.74;...
2.8;2.64;2.79;2.95;3.05;3.07;3.12;3.06;2.95;3.00;3.07];
for i = 1:length(t)
s(i) = fsolve(#(x) x(i) + 2.* G(i) - ((H(i).* A.*x(i))/(x(i).^2 + A.*x(i) + A.*B))- 2.*((H(i).*A.*B)/(x(i).^2 ...
+ A.*x(i) + A.*B))-((I(i).*C.*x(i))/(x(i).^2 + C.*x(i) + C.*D))- 2.*((I(i).*C.*D)/(x(i).^2 ...
+ C.*x(i) + C.*D))- E/x(i), F);
end
L = 6 - log10(s);
plot(t,L)
hold on
plot (t,K,'v')
xlabel('Time (sec)')
ylabel('pH')
hold off
legend('pH-Mod','pH-Exp')
end
I do not know how can I ensure that the index value does not exceed the number of the array elements. In line 34, I indexed to the end of the independent variable, thinking that it will prevent the error. The error message is
Index exceeds the number of array elements (1).
Error in
my_fsolve3>#(x)x(i)+2.*G(i)-((H(i).*A.*x(i))/(x(i).^2+A.*x(i)+A.*B))-2.*((H(i).*A.*B)/(x(i).^2+A.*x(i)+A.*B))-((I(i).*C.*x(i))/(x(i).^2+C.*x(i)+C.*D))-2.*((I(i).*C.*D)/(x(i).^2+C.*x(i)+C.*D))-E/x(i)
(line 36)
s(i) = fsolve(#(x) x(i) + 2.* G(i) - ((H(i).* A.*x(i))/(x(i).^2 + A.*x(i) + A.*B))- 2.*((H(i).*A.*B)/(x(i).^2 ...
Error in fsolve (line 242)
fuser = feval(funfcn{3},x,varargin{:});
Error in my_fsolve3 (line 36)
s(i) = fsolve(#(x) x(i) + 2.* G(i) - ((H(i).* A.*x(i))/(x(i).^2 + A.*x(i) + A.*B))- 2.*((H(i).*A.*B)/(x(i).^2 ...
Caused by:
Failure in initial objective function evaluation. FSOLVE cannot continue.
Your anonymous function has x as an argument which is initialized to the scalar F.
You try to access x(i) which fails for i>1 because x is scalar. Solution: replace x(i) by x.
s(i) = fsolve(#(x) x + 2.* G(i) - ((H(i).* A.*x)/(x.^2 + A.*x + A.*B))- 2.*((H(i).*A.*B)/(x.^2 ...
+ A.*x + A.*B))-((I(i).*C.*x)/(x.^2 + C.*x + C.*D))- 2.*((I(i).*C.*D)/(x.^2 ...
+ C.*x + C.*D))- E/x, F);
By the way, you do not need the dot operator in multiplications of scalars, a simple * will do in your code.
Related
I saw some textbook about the worst-case space complexity of Fibonacci Sequence. However, I have the following question:
You can start with a concrete example and generalize. Start with n = 5.
S(5) = S(4) + c
= (S(3) + c) + c
= ((S(2) + c) + c) + c
= (((S(1) + c) + c) + c) + c
= S(1) + 4c
There are 4 c's when n = 5. In general, there are n-1 c's.
This is my code and when I run it appears index was outside the bounds of the array in line b = Asc(y(j + m)). I have tried Try and Catch and it didn't work out.
Public Function SMITH(x, y, SX, SY)
Dim a, b, j As Integer
result = 0
m = x.Length
n = y.Length
preBmBc(x)
preQsBc(x)
j = 0
While (j <= (n - m))
If (SX = SY.ToString.Substring(j, m)) Then
result = 1
End If
a = Asc(y(j + (m - 1)))
b = Asc(y(j + m))
j = j + Math.Max(bmBc(a), qsBc(b))
End While
Return result
End Function
Have you tried making m = "m-1" for b also? You really need to step through the code using breakpoints and the debugger to figure out when the program throws the OutOfRangeException.
I use vba on excel 2007, OS: windows vista, to make calculation using kinematic wave equation in finite difference scheme. But, when it runs the run-time 5 (invalid procedure call or arguments) message appears. I really don't what is going wrong. Anyone can help?
Sub kwave()
Dim u(500, 500), yy(500, 500), alpha, dt, dx, m, n, so, r, f, X, L, K As Single
Dim i, j As Integer
dx = 0.1
dt = 0.01
L = 10
m = 5 / 3
r = 1
f = 0.5
n = 0.025
so = 0.1 'this is slope
alpha = 1 / n * so ^ 0.5
X = 0
For i = 0 To 100
Cells(i + 1, 1) = X
u(i, 1) = L - so * X
X = X + dx
Cells(i + 1, 2) = u(i, 1)
Next i
For j = 0 To 100
For i = 1 To 100
'predictor step
u(i, j + 1) = u(i, j) - alpha * dt / dx * (u(i + 1, j) ^ m - u(i, j) ^ m) + (r - f) * dt
'corrector step
K = u(i, j + 1) ^ m - u(i - 1, j + 1) ^ m '<<<<----- RUNTIME ERROR 5 HAPPENS AT THIS LINE
yy(i, j + 1) = 0.5 * ((yy(i, j) + u(i, j + 1)) - alpha * dt / dx * K + (r - f) * dt)
Next i
Next j
End Sub
You are declaring the variables wrong- the array should store a double/single but it is defaulting to a variant. See this article.
http://www.cpearson.com/excel/declaringvariables.aspx -
"Pay Attention To Variables Declared With One Dim Statement
VBA allows declaring more than one variable with a single Dim
statement. I don't like this for stylistic reasons, but others do
prefer it. However, it is important to remember how variables will be
typed. Consider the following code:
Dim J, K, L As Long You may think that all three variables are
declared as Long types. This is not the case. Only L is typed as a
Long. The variables J and K are typed as Variant. This declaration is
functionally equivalent to the following:
Dim J As Variant, K As Variant, L As Long You should use the As Type
modifier for each variable declared with the Dim statement:
Dim J As Long, K As Long, L As Long "
Additionally, when i = 99 and j = 10, u(99,11), which is j+1, produces a negative number. Note that this does not fully cause the problem though, because you can raise negative numbers to exponents. Ex, -5^3 = -125
I'm new to VBA, and I have a question, i.e I have a mathematical function 1 + 2x¹ + 3x² + 4x³ + ... +10x⁹ and I need to resolve it into two ways:
I can use raising operations(analog pow in Pascal) and IF statement;
without rising operations and IF statement...
I have tried this one:
Public Function test(x)
test = 1 + 2*x^1 + 3*x^2 + 4*x^3 + 5*x^4 + 6*x^5 + 7*x^6 + 8*x^7 + 9*x^8 + 10*x^9
End Function
but I think it returns the wrong answer - 2441406 with calling =test(5)
So can anyone give any advice, or help with my problem?
If you can't use VBA for this, there is a formula solution. Assuming your variable x is in cell A1, you would use this formula in another cell (I used B1):
=SUMPRODUCT(ROW($1:$10)*A1^(ROW($1:$10)-1))
When A1 = 5, it returned 23803711 as expected.
You will need * as the multiper:
Public Function test(x)
test = _
1 _
+ 2 * (x ^ 1) _
+ 3 * (x ^ 2) _
+ 4 * (x ^ 3) _
+ 5 * (x ^ 4) _
+ 6 * (x ^ 5) _
+ 7 * (x ^ 6) _
+ 8 * (x ^ 7) _
+ 9 * (x ^ 8) _
+ 10 * (x ^ 9)
End Function
I was running tests on my software today and found that some of the values it was producing weren't correct.
I decided to step through the code and noticed that the variables I had assigned to textbox values on my userform when hovered over said empty, even though when hovering over the textbox assigned to it, the value inputted by the user showed.
For Example,
n = BiTimeSteps_TextBox.Value
when hovered over
n = empty
even though
BiTimeSteps_TextBox.Value = 2
when hovered over.
So say I have a formula shortly after that says
d = n*2 ,
n when hovered over says empty and d is made 0 when it shouldn't be.
Someone told me if I switch it around to
BiTimeSteps_TextBox.Value = n
it should be recognised but it is still not.
What could possibly be causing this?
See full code below: (it aims to price options using the binomial tree pricing method)
S = BiCurrentStockPrice_TextBox.Value
X = BiStrikePrice_TextBox.Value
r = BiRisk_Free_Rate_TextBox.Value
T = BiexpTime_TextBox.Value
Sigma = BiVolatility_TextBox.Value
n = BiTimeSteps_TextBox.Value
Dim i, j, k As Integer
Dim p, V, u, d, dt As Double
dt = T / n ' This finds the value of dt
u = Exp(Sigma * Sqr(dt)) 'formula for the up factor
d = 1 - u 'formula for the down factor
'V value of option
'array having the values
Dim bin() As Double 'is a binomial arrays, it stores the value of each node, there is a loop
'work out the risk free probability
p = (1 + r - d) / (u - d)
'probability of going up
ReDim bin(n + 1) As Double
'it redims the array, and n+1 is used because it starts from zero
'------------------------------------------------------------------------------------------------------------------------------
''European Call
If BiCall_CheckBox = True Then
For i = 0 To n 'payoffs = value of option at final time
bin(i + 1) = Application.WorksheetFunction.Max(0, (u ^ (n - i)) * (d ^ i) * S - X)
'It takes the max payoff or 0
Cells(i + 20, n + 2) = bin(i + 1) 'to view payoffs on the isolated column on the right
Next i
End If
'european put
If BiPut_CheckBox = True Then
For i = 0 To n 'payoffs = value of option at final time
bin(i + 1) = Application.WorksheetFunction.Max(0, X - (S * (u * (n - i)) * (d * i)))
' European Put- It takes the max payoff or 0
Cells(i + 20, n + 2) = bin(i + 1) 'to view payoffs on the isolated column on the right
Next i
End If
For k = 1 To n 'backward column loop
For j = 1 To (n - k + 1) 'loop down the column loop
bin(j) = (p * bin(j) + (1 - p) * bin(j + 1)) / (1 + r)
Cells(j + 19, n - k + 2) = bin(j)
'' print the values along the column, view of tree
Next j
Next k
Worksheets("Binomial").Cells(17, 2) = bin(1) ' print of the value V
BiOptionPrice_TextBox = bin(1)