dynamic system model

Philip Wong

1. The problem statement, all variables and given/known data
I'm trying to learn the dynamic system model, I was given the following problem to work with. Can someone give me a walk through of did I did right or wrong?

Situation:
A conservationist is studying a colony of endangered variable oyster-catchers. Birds that are chicks one year will form adult breeding pairs the following year. The probability of a chick surviving to become an adult is 0.3. The probability an adult will survive to breed another year is 0.5. Each year a breeding pair produce on average 2.2 chicks (that is a breeding rate of 1.1 per adult). Initially there are 80 chicks and 20 adult birds.

a) Represent this situation by a dynamic system model of the form Xn = A^n X0. You must find matrix A and the initial state vector X0.

b) Find X1, the state vector representing the oyster-catcher colony 1 year later.

c) Find all eigenvalues of matrix A in part (a)

d) By finding the eigenvector corresponding to the numerically largest eigenvalue, estimate the long term ratio of chicks:adults. Hint: If landa is an eigenvalue of matrix A with eigenvector v then A^n v = landa^n v

e) Long-term, is this colony of oyster-catchers likely to survive?
2. Known equation
A - landa I (to find eigenvalue and eigenvector)
|A-landa I| = 0 (to find eigenvector and eigenvalue)
AX0=c1Av1+c2Av2 (to find the rate of change after x time)

3. The attempt at a solution
a) Chicks n+1 = 0 chicks n + 1.1 adults n
Adults n+1 = 0.3 chicks n + 0.5 adults n
where X0 = (80; 20)
So A= Xn+1 = (0, 1.1; 0.3, 0.5)(chicks n; adult n)
therefore Xn = A^n X0 = (0, 1.1; 0.3, 0.5)(80; 20)

b)
At first I tried to find the eigenvector and eigenvalue by hand, then implied it to the equation A = VDV^-1 to find the value for the long term run. Below is my steps:
A-landa I:
(0, 1.1; 0.3, 0.5) - landa (1, 0; 0 1) = (0-landa, 1.1; 0.3, 0.1-landa]

det (A-landa I) = 0:
(-landa)(0.1-landa)-0.33 = 0
expand it: landa^2 - 0.1landa - 0.33 =0
factories: I have got into some trouble factoring with decimal points, so I used matlab to get the eigenvector and eigenvalue which gives:
a =

0 1.1000
0.3000 0.5000

>> [v,d]=eig(a)

v =

-0.9461 -0.7821
0.3238 -0.6232


d =

-0.3765 0
0 0.8765

So landa1 = -0.3765, landa 2 = 0.8765

there I put these numbers into the following equation to work out X1 (after one year):
X0 = c1*landa 1*v1 + c2*landa 2*v2
= c1 *(-0.9491; 0.3238) + c2 *(-0.7821; -0.6232)
= c1 * -0.3765 * (-0.9491; 0.3238) + c2 * 0.8765 *(-0.7821; -0.6232)
Since X0 = (80;20)
So X1 = 80 * -0.3765 * (-0.9491; 0.3238) + 20 * 0.8765 *(-0.7821; -0.6232)

This is the part where I got completely stuck, I knew I did something wrong, I've read the textbook over and over again. But I can't find a solution of where I did wrong, and so I couldn't do the next part. can somebody please help?!

Filip Larsen

In b) you only need to calculate the population after one year, i.e. with n = 1 calculate x₁ = A¹ x₀ = A x₀. No need for eigenvalues yet.

Also, your characteristic equation (before you start using matlab) is not entirely correct.

Philip Wong

what do you mean by "Also, your characteristic equation (before you start using matlab) is not entirely correct."?

Filip Larsen

Just before you turn to matlab, you write the equation "landa^2 - 0.1landa - 0.33 =0", but the 0.1 should have been 0.5. I cannot see if this is a typo so I thought I would just mention it in case you went back to solving for the eigenvalues manually (its good to check with matlab, but you should have no trouble calculating the values manually too by solving the characteristic equation).

And, just for your information, the Greek letter is called lambda, not landa.

Philip Wong

yes whenever I'm doing excerise or pratice I always do the question first manually before I check it with matlab, the only time I don't do manually it anymore is when I got so used to working with those type of questions that I could mentally think the answer out, and then I check it with matlab. And yes it was I typo, ahaha! It is 0.5 instead of 0.1

Philip Wong

x₁ = A¹ x₀ = A x₀
So x₁= C₁(-0.3765)^1(-0.8481; 0.3238) + C₂(0.8765)^1(-0.7821;-0.6232)

but since lambda 2 is the domain value it will over-run lambda 1 in the long term, therefore the state vector after 1 year will be 0.8765x_k?

or did I think too much, the state vector is simply :
x₁= C₁(-0.3765)^1(-0.8481; 0.3238) + C₂(0.8765)^1(-0.7821;-0.6232)

Filip Larsen

If this is for b) then you are still getting ahead of yourself. You only need to perform a simple matrix multiplication of A and x₀ to get x₁. In this model this is, by definition, how the state for one generation is advanced to the state for the next generation. If you were to calculate, say, the 10th generation you would have to do the multiplication 10 times to get to x₁₀. By the way, it can be instructive to plot how the state develops over time, especially in order to prepare yourself for part d) and e).

Philip Wong

aha! so x₁ = [0 1.1; 0.3 0.5] *[80 20]
so say if I want to work out ₁₀, would I do this:
[0^10,1.1; 0.3,0.5^10]*[80,20]

Filip Larsen

Yes.

No. The definition of your model really says x_i+1 = A x_i, that is, to get state for generation i+1 you matrix multiply A with the state vector for generation i, i.e. the previous generation. So, if you know x₉ you can calculate x₁₀ = A x₉, but since x₉ likewise can be written based on the state vector for the previous generation you also have x₉ = A x₈ which can be inserted into the equation for x₁₀ to give x₁₀ = A (A x₈) = (A A) x₈, the last step because matrix multiplication is commutative. If you continue replacing x in this fashion until you get to to x₀ you arrive at x₁₀ = (A A ... 10 times ... A) x₀ = A¹⁰ x₀. Just like with scalar numbers, you can think of the (integer) power of a matrix as a shorthand notation of simply multiplying out the matrix as many times as the power says.

Your questions seem to indicate you have missed something fundamentally somewhere in your textbook. If the above explanation is all news to you, I recommend that you take some time to read up on the parts in your textbook that are relevant for this problem or you will most likely have a very tough time understanding the eigenvalues and eigenvectors.

Philip Wong

ar for the x₁₀ I think I skip a bit too much steps along the way. When I think about it again now, if I want to get the x₁₀ I would use diagnolisation and symmetry.
so for x₁₀ would I do the following?
A = VDV^-1
[0 1.1; 0.3 0.5]^10 =
[-0.9461,-0.7821; 0.3238,-0.6232]*[-0.3765^10, 0; 0,0.8765^10]*[-0.7394, 0.9279; -0.3842, -1.1225]

which gives:
[0.0805, 0.2349; 0.0641, 0.1872]

since X₁₀ = A^10 x₀, then:
[0.0805, 0.2349; 0.0641, 0.1872]*[80;20]

which gives:
11.1340
8.8692

is this correct?

Filip Larsen

That is correct.

Philip Wong

yay! thanks for your help