k = 0.019;
Pstar = 100;
H = 33;
h = 0.1;
X = 36;
N = round(X/h);
t = zeros(1,N+1);
P = zeros(1,N+1);
P(1) = 84;
t(1) = 0;
yHeun = zeros(1,N+1);
yHeun(1)=84;
a = 1; b = 100;
while b-a >0.5
c = (a+b)/2;
for n = 1:N
t(n+1) = t(n) + h;
Inside = nthroot(sin(2*pi*t/12),15);
Harvest = c*0.5*(Inside+1);
P(n+1) = P(n) + h*(k*P(n)*(Pstar-P(n))-Harvest(n));
if P < 0
P = 0;
end
yHeun(n+1) = yHeun(n) + h*0.5*((k*P(n)*(Pstar-P(n))-Harvest(n))+(k*P(n+1)*(Pstar-P(n+1))-Harvest(n+1)));
end
if sign(yHeun(c)) == sign(yHeun(a))
c = a;
else
c = b;
end
end
disp(['The root is between ' num2str(a) ' and ' num2str(b) '.'])
This is the code i'm trying to run and i know it probably sucks but im terrible at coding and every time i try to run the code, it says:
Attempted to access yHeun(50.5); index must be a positive integer or
logical.
Error in Matlab3Q4 (line 30)
if sign(yHeun(c)) == sign(yHeun(a))
I don't have ANY idea how to make yHeun(c or a or whatever) return anything that would be an integer. I dont think i did the while+for loop right either.
Question: "Begin with the higher bound of H being 100 (the high value results in a population of 0 after 36 months), and the lower bound being 1. Put the solver from Problem #3 above in the middle of a while loop and keep bisecting the higher and lower bounds of H until the difference between the higher and lower bound is less than 0.5."
I guess that line 30 (with the error) is this one:
if sign(yHeun(c)) == sign(yHeun(a))
Here, I guess c is equal to 50.5, as a result of c = (a+b)/2 above (BTW you can discover whether I guessed right by debugging - try adding disp(c) before line 30).
To force a number to be an integer, use floor:
c = floor((a+b)/2);
It seems you are trying to use some sort of divide-and-conquer algorithm; it should be enough to stop when b - a is equal to 1.
Related
I don't know where to go with this. I think I have the right stuff down but I don't understand.
https://imgur.com/a/V6gdDdr
It keeps running the loop forever and I don't know why.
n=9;
r =0;
p = 0;
syms x
v=1.7;
while abs(v-r) > 10^(-5)
n=n+1;
r = 0;
a = 0;
b= 1/n;
for i = 1:n
r = r + exp(((i+1)+exp(i))/2)*(b-a)
end
['done']
end
The output should be 85. But I'm getting like a couple thousand. I have tried changing the equation in the for loop but I do not know why it is giving me symbolic errors.
Several problems with the posted code:
The parentheses are misaligned in the r=r + ... statement
Instead of exp(i) and exp(i+1), you should use exp(bi) and exp(b(i+1)) in order to account for the given spacing.
Your value of "v" which is supposed be the exact answer should equal e-1, which is 1.71828. Using the approximate 1.7 is going to be very problematic when you are trying to converge to the exact solution within 5 decimal places.
Your for loop should go from 0 to n-1, this way you don't add up any values past your integral range.
n=9;
r =0;
p = 0;
v=e-1;
while abs(v-r) > 10^(-5)
n=n+1;
r = 0;
a = 0;
b= 1/n;
for i = 0:n-1
r = r + ((exp(b*(i+1))+exp(b*i))/2)*(b-a);
end
end
I want to find the smallest integer l that satisfies l^2 >= x, and mod(l,2)=0.
In the following example x=75, and hence l=10, since the previous even number doesn't fulfil the inequation: 8^2 <= 75 <= 10^2
I have tried this (ignoring the even-number requirement, which I can't to work):
syms l integer
eqn1 = l^2 >= 75;
% eqn2 = mod(l,2) == 0;
[sol_l, param, cond] = solve(eqn1, l, 'ReturnConditions', true);
But this does not give me anything helpful directly:
sol_l =
k
param =
k
cond =
(75^(1/2) <= k | k <= -75^(1/2)) & in(k, 'integer')
I would like to evaluate the conditions on the parameter and find the smallest value that satisfies the conditions.
Also, I would like to enforce the mod(l,2)=0 condition somehow, but I don't seem to get that work.
Using the solve for this task is like using a cannon to kill a mosquito. Actually, the answer of Lidia Parrilla is good and fast, although it can be simplified as follows:
l = ceil(sqrt(x));
if (mod(x,2) ~= 0)
l = l + 1;
end
% if x = 75, then l = 10
But I would like to point out something that no one else noticed. The condition provided by the solve function for l^2 >= 75 is:
75^(1/2) <= k | k <= -75^(1/2)
and it's absolutely correct. Since l is being raised to the power of 2, and since a negative number raised to the power of 2 produces a positive number, the equation will always have two distinct solutions: a negative one and a positive one.
For x = 75, the solutions will be l = 10 and l = -10. So, if you want to find the smallest number (and a negative number is always smaller than a positive one), the right solution will be:
l = ceil(sqrt(x));
if (mod(x,2) ~= 0)
l = l + 1;
end
l = l * -1;
If you want to return both solutions, the result will be:
l_pos = ceil(sqrt(x));
if (mod(x,2) ~= 0)
l_pos = l_pos + 1;
end
l_neg = l_pos * -1;
l = [l_neg l_pos];
I guess the easiest solution without employing the inequality and solve function would be to find the exact solution to your equation l^2 >= x, and then finding the next even integer. The code would look like this:
x = 75;
y = ceil(sqrt(x)); %Ceil finds the next bigger integer
if(~mod(y,2)) %If it's even, we got the solution
sol = y;
else %If not, get the next integer
sol = y+1;
end
The previous code gives the correct solution to the provided example (x = 75; sol = 10)
I had written the following matlab code to optimize the following LP
max b'x
s.t A'x <= 0;
x <= d;
Also d is
d = {1,2..m}
and A is defined in the code. I am getting the error:
Maximum number of iterations exceeded; increase options.MaxIter.
Upon googling, someone said it is not very good that the error is occuring. and the problem has to reformulated. Any idea how to reformulate it.
The solution is very simple as A > 0, b> 0 and d>0 therefore x = 0
m = 10;
d = [1:1:m];
for j = 1:m,
for i = 1:m,
A(i,j) = 1/(i+j-1);
end
end
for i = 1:m,
b(i)=0;
end
for i = 1:m
lb(i) = -inf;
end
b;
lb = lb';
f = A*d';
[x,fval,exitflag,output] = linprog(-f,A,b,[],[],lb,d); %minimzation problem. Hence -f, A = A'
I'd used the optimist = ('MaxIter', 10000). To stop that error message.
I'm trying to multiply (element wise) a vector V of length N by a randomly generated number in the range (a,b), while keeping the sum of the vector equal to a total amount, E. I want to do this in MATLAB, but I'm not sure how. Getting random numbers between a certain range I know how to do:
minrand = 0;
maxrand = 1;
randfac = (maxrand-minrand).*rand(1,N) + minrand;
But yeah, beyond that I'm pretty clueless. I guess the random numbers can't really be generated like this, because if we call the random numbers the vector R, then I want that
R_1*V1 + R_2*V2 .... + R_N*V_N = E. So I guess it's a big equation. Is there any way to solve it, while putting constraints on the max and min values of R?
You can pick pairs of two elements (in all combinations) and add and subtract an equal random number.
% Make up a random vector
N=10;
randfac = 10*rand(1,N);
%OP Answer here: Given randfac with sum E re-randomize it
E = sum(randfac);
minrand = 0;
maxrand = 2;
disp(randfac)
% v = [6.4685 2.9652 6.6567 1.6153 7.3581 0.0237 7.1025
% 3.2381 1.9176 1.3561]
disp(sum(randfac))
% E = 38.7019
r = minrand + (maxrand-minrand)*rand(N*N,1);
k = 1;
for i=1:N
for j=1:N
randfac(i) = randfac(i)-r(k);
randfac(j) = randfac(j)+r(k);
k = k + 1;
end
end
disp(randfac)
% v = [5.4905 0.7051 4.7646 1.3479 9.3722 -1.4222 7.9275
% 7.5777 1.7549 1.1836]
disp(sum(randfac))
% E = 38.7019
Just divide the vector with the sum and multiply with the target E.
randfac = (maxrand-minrand).*rand(1,N) + minrand;
randfac = E*randfac/sum(randfac);
as long as the operator is linear, the result is going to retain it's randomness. Below is some sample code:
minrand = 0;
maxrand = 1;
N = 1000; %size
v = (maxrand-minrand).*rand(1,N) + minrand;
E = 100; %Target sum
A = sum(v);
randfac = (E/A)*v;
disp(sum(randfac))
% 100.0000
First of all with random numbers in the interval of [a b] you can't guarantee that you will have the same summation (same E). For example if [a b]=[1 2] of course the E will increase.
Here is an idea, I don't know how random is this!
For even N I randomize V then divide it in two rows and multiply one of them with random numbers in [a b] but the second column will be multiplied to a vector to hold the summation fixed.
N = 10;
V = randi(100,[1 N]);
E = sum(V);
idx = randperm(N);
Vr = V(idx);
[~,ridx] = sort(idx);
Vr = reshape(Vr,[2 N/2]);
a = 1;
b = 3;
r1 = (b - a).*rand(1,N/2) + a;
r2 = (sum(Vr) - r1.*Vr(1,:))./Vr(2,:);
r = reshape([r1;r2],1,[]);
r = r(ridx);
Enew = sum(V.*r);
The example results are,
V = [12 82 25 51 81 51 31 87 6 74];
r = [2.8018 0.7363 1.9281 0.5451 1.9387 -0.4909 1.3076 0.8904 2.9236 0.8440];
with E = 500 as well as Enew.
I'm simply assigning one random number to a pair (It can be considered as half random!).
Okay, I have found a way to somewhat do this, but it is not elegant and there are probably better solutions. Starting with an initial vector e, for which sum(e) = E, I can randomize its values and end up with an e for which sum(e) is in the range [(1-threshold)E,(1+thresholdE)]. It is computationally expensive, and not pretty.
The idea is to first multiply e by random numbers in a certain range. Then, I will check what the sum is. If it is too big, I will decrease the value of the random numbers smaller than half of the range until the sum is no longer too big. If it is too small, I do the converse, and iterate until the sum is within the desired range.
e = somepredefinedvector
minrand = 0;
maxrand = 2;
randfac = (maxrand-minrand).*rand(1,N) + minrand;
e = randfac.*e;
threshold = 0.001;
while sum(e) < (1-threshold)*E || sum(e) > (1+threshold)*E
if sum(e) > (1+threshold)*E
for j = 1:N
if randfac(j) > (maxrand-minrand)/2
e(j) = e(j)/randfac(j);
randfac(j) = ((maxrand-minrand)/2-minrand).*rand(1,1) + minrand;
e(j) = randfac(j)*e(j);
end
if sum(e) > (1-threshold)*E && sum(e) < (1+threshold)*E
break
end
end
elseif sum(e) < (1-threshold)*E
for j = 1:N
if randfac(j) < (maxrand-minrand)/2
e(j) = e(j)/randfac(j);
randfac(j) = (maxrand-(maxrand-minrand)/2).*rand(1,1) + (maxrand-minrand)/2;
e(j) = randfac(j)*e(j);
end
if sum(e) > (1-threshold)*E && sum(e) < (1+threshold)*E
break
end
end
end
end
If I have a matrix and I want to apply a function to each row of the matrix. This function has three possible outputs, either x = 0, x = 1, or x > 0. There's a couple things I'm running into trouble with...
1) The cases that output x = 1 or x > 0 are different and I'm not sure how to differentiate between the two when writing my script.
2) My function isn't counting correctly? I think this might be a problem with how I have my loop set up?
This is what I've come up with. Logically, I feel like this should work (except for the hiccup w/ the first problem I've stated)
[m n] = size(matrix);
a = 0; b = 0; c = 0;
for i = 1 : m
x(i) = function(matrix(m,:));
if x > 0
a = a + 1;
end
if x == 0
b = b + 1;
end
if x == 1
c = c + 1;
end
end
First you probably have an error in line 4. It probably should be i instead of m.
x(i) = function(matrix(i,:));
You can calculate a, b and c out of the loop:
a = sum(x>0);
b = sum(x==0);
c = sum(x==1);
If you want to distinguish x==1 and x>0 then may be with sum(xor(x==1,x>0)).
Also you may have problem with precision error when comparing double values with 0 and 1.