Does anyone know how to do question 8 ? D:

I understand that the last equation is the one that is redundant because it ends up getting you the parameter, and so then we need to change the a(t+1) equation... but other than that I don't know what to do next .. .

can anyone help out with questions 7 and 8?

Originally Posted by Crisht

can anyone help out with questions 7 and 8?

For Q7, you can find the determinant of your matrix A and you end up with p.
So from this ... you can figure out the conclusion to this right ?

just to confirm
eigenvectors are

x1 [0.02, .3, 1]^T
x2 [-.1, -.9. 1]^T
x3 [-.2, -.8, 1]^T

where X1 = 50
X2 = -10
X3 = -5

ignore this post

Hey can anyone help me with question number 4?

I think I'm on the right track but just need to confirm.

4.a) {-30 -150 500}^T
4.b) not so sure
4.c) [46 74 180] ^T

I have very different values then you. I don't think your values are correct because you have negative's.. and you can't have negative machinery o__O. Furthermore, your values don't add up to 13200 (recall there are 13200 components)

I got the following (could be wrong though)

4a) [500, 6200, 6500]^T
4b) [130, 2320, 10750]^T
4c) [130, 2320, 10750]^T
4d) 2 months: [215, 3140, 9845]^T
6 months: approximately [200, 3000, 10000]^T

Originally Posted by Ush.s

I have very different values then you. I don't think your values are correct because you have negative's.. and you can't have negative machinery o__O. Furthermore, your values don't add up to 13200 (recall there are 13200 components)

I got the following (could be wrong though)

4a) [500, 6200, 6500]^T
4b) [130, 2320, 10750]^T
4c) [130, 2320, 10750]^T
4d) 2 months: [215, 3140, 9845]^T
6 months: approximately [200, 3000, 10000]^T

^I'm getting the same. Also, I've been staring at question 8 for the past half hour and I still don't understand. The TA posted this on facebook:

In question 5 we found the eigenvectors who had eigenvalue 1 by solving the system of equations given by (A-I)x=0 (or some equivalent method). Naturally we get a whole family of eigenvectors, because any scalar multiple of an eigenvector is still an eigenvector for the same eigenvalue. Luckily, in our case the family only has one free parameter, so the "eigenspace" associated to 1 is just one-dimensional, they are all just a scalar times one fixed vector. If we wanted, we could choose that vector to be the eigenvector whose coordinates add to 1 (i.e. 100%), so that the coordinates are equal to the proportions of each type of part.

In question 8 we DO want to pick the eigenvector with that property, and we do so by adding an extra constraint to our system of equations. When we row reduce the system [(A-I) | 0] we get a row of zeros on the bottom, indicating that any one of the original equations can be written in terms of the other two. This is where the free parameter in question 5 comes from. To eliminate the free parameter we must add an independent equation, and we must choose the equation so that the unique solution to the new system has the desired property (that the coordinates sum to 1).

So you do in this question is:
- start with the system of equations [(A-I) | 0]
- replace one of the equations with a new equation, carefully chosen so that if x=(x_1,x_2,x_3) is the unique solution then x_1 + x_2 + x_3 = 1 (there are few equations you could choose, some more obvious than others)

And that's it. The question states that you don't even need to solve the new system that you get.

I sorta get what he's trying to say, but I'm still not sure of how to do this. Can someone please explain? Thanks.

for question 6 which steady state vector do we use? the costs would be different for the first few months!

Originally Posted by Bug324

^I'm getting the same. Also, I've been staring at question 8 for the past half hour and I still don't understand. The TA posted this on facebook:

In question 5 we found the eigenvectors who had eigenvalue 1 by solving the system of equations given by (A-I)x=0 (or some equivalent method). Naturally we get a whole family of eigenvectors, because any scalar multiple of an eigenvector is still an eigenvector for the same eigenvalue. Luckily, in our case the family only has one free parameter, so the "eigenspace" associated to 1 is just one-dimensional, they are all just a scalar times one fixed vector. If we wanted, we could choose that vector to be the eigenvector whose coordinates add to 1 (i.e. 100%), so that the coordinates are equal to the proportions of each type of part.

In question 8 we DO want to pick the eigenvector with that property, and we do so by adding an extra constraint to our system of equations. When we row reduce the system [(A-I) | 0] we get a row of zeros on the bottom, indicating that any one of the original equations can be written in terms of the other two. This is where the free parameter in question 5 comes from. To eliminate the free parameter we must add an independent equation, and we must choose the equation so that the unique solution to the new system has the desired property (that the coordinates sum to 1).

So you do in this question is:
- start with the system of equations [(A-I) | 0]
- replace one of the equations with a new equation, carefully chosen so that if x=(x_1,x_2,x_3) is the unique solution then x_1 + x_2 + x_3 = 1 (there are few equations you could choose, some more obvious than others)

And that's it. The question states that you don't even need to solve the new system that you get.

I sorta get what he's trying to say, but I'm still not sure of how to do this. Can someone please explain? Thanks.

replace 3rd equation of (lambdaI-A)x=0( the matrix which u used to find steady state vector in Q5)
with x1+x2+x3=1

Originally Posted by Differential

replace 3rd equation of (lambdaI-A)x=0( the matrix which u used to find steady state vector in Q5)
with x1+x2+x3=1

OH, okay, thanks! And we can choose any values for x1, x2, and x3, as long as they add to 1? For example, 0,0,1? Or do we keep the general equation?

Originally Posted by Bug324

OH, okay, thanks! And we can choose any values for x1, x2, and x3, as long as they add to 1? For example, 0,0,1? Or do we keep the general equation?

yes but u have to get unique solution and sum of 100% of components so..

can you pleasee give me a hint for number six

Originally Posted by Ush.s

can you pleasee give me a hint for number six

redo question 5 with new values p and q

for number 5 I got
V = x[p, p+q, 1]^T =S