MedVision ad

QMA finals help please (1 Viewer)

angelfromabuv

New Member
Joined
Apr 22, 2006
Messages
9
Gender
Female
HSC
2006
Hey guys i need a bit of direction with a couple QMA questoins if anyones up for it.
First is the whole Ax=B thing,
We were trying to find a local maximum, (i.e. fx=fy=0)
We have:

fx= -2(x-1)
fy= -2 (y-1)
Then solve for any (x,y) that satisfy fx=fy=0

Then we had to solve with the matrix thing, and the correct matrix is
|2 0| 2|
|0 2| 2|

The 2, 0, 0, 2 is the A matrix, then the [2, 2] is the solution matrix but i don't understand how htey got that? It was something about 2 of x + 0 of y=2 but... :S


2nd question is about '06 finals paper
-1. For Q3 how do we find the Iso-objective line? Do we just stick in any point and solve?

-2. (e) Down to what level of investment will the change you found in (d) for every decrease of one dollar apply? -->What does this even mean?

-3. What's the deal with the Hessian matrix thing? Coz its a fair blurr to me :S!

I know that sounds like a lot sorry but any nudge in teh right direction is much appreciated thanks!
 

shinji

Is in A State Of Trance
Joined
Feb 2, 2005
Messages
2,733
Location
Syd-ney
Gender
Male
HSC
2006
For the first question, i don't really understand the context of it. Would you be able to provide a reference to the actual question?

2nd questions;
- 1
Yeah, to find the Iso-objective line yuo would just plug in any number to be the revenue. Then you would make Y equal to something and then solve for x. and then graph.

-2 Basically, saying as you decrease $1 of investment, how much would maximum return change. What you found in (d), (decrease in $1 -> decrease in max. return of 6cents), up to what point would this be applicable?
I.e, you would have to refer to you graph that you drew.

-3. : hessian matrix. i don't think we've learnt it, but i did some research on it and did the question now i can't find it. lol
 

Loz_metalhead

Member
Joined
Feb 5, 2005
Messages
800
Gender
Female
HSC
2006
We learn't the hessian matrix and its pretty important

Have a look over the lecture notes. The hessian matrix is the second order condition for questions with a constraint-it will tell you if its a minimum and maximum. gx is the derivative of x in the constraint, gy is the derivative of the y in the constraint.

Lxx-its just d2y/dx2
Lxy-d2y/dxdy
 

angelfromabuv

New Member
Joined
Apr 22, 2006
Messages
9
Gender
Female
HSC
2006
Ah ur a champ thanks heaps. Which lecturer do u have? QMA not my fave subject haha, i got Fox, i feel bad for him tho coz everyone likes Angus. That thing was from Lecture 24, if it's too much trouble dw bout it. I just have no idea how htey got the solutions vector as [2, 2] for example 24.2.

Goodluck tomorow! doesn't sound like u need it tho
haha fingers crossed we don't get any of that hessian stuff watever it is
 

Chris.

Member
Joined
Nov 27, 2006
Messages
538
Gender
Male
HSC
2006
hoping to get 30/60

is the exam out of 60 marks?
 

shinji

Is in A State Of Trance
Joined
Feb 2, 2005
Messages
2,733
Location
Syd-ney
Gender
Male
HSC
2006
Chris. said:
hoping to get 30/60

is the exam out of 60 marks?
65 marks :p

But in reply to the first question, i know how they get it i think.

That's essentially plotting the function of the first partial derivatives into a matrix.
it's in the form;

Fx = -2x + 2
Fy = -2y + 2

to obtain the maxima of things, we would normally set the derivative equal to zero.
I.e, -2x + 2 = 0
-2y + 2 = 0

Then, u just move the 2's over to the other side,
so it becomes
2x + 0 = 2
0 + 2y = 2
(note, in a matrix, the first column would correspon to the "x" variable, 2nd column would be "Y" variable, etc)

and then remove the variables and then put it into matrix form ='s

|2 0|2|
|0 2|2|

hope you understood that.
 

angelfromabuv

New Member
Joined
Apr 22, 2006
Messages
9
Gender
Female
HSC
2006
hey shinji that helps a lot thank u! much clearer now :) on the slim chance that u may need my help lol i'll be around if u need anything as well, thanks!
 

shinji

Is in A State Of Trance
Joined
Feb 2, 2005
Messages
2,733
Location
Syd-ney
Gender
Male
HSC
2006
angelfromabuv said:
hey shinji that helps a lot thank u! much clearer now :) on the slim chance that u may need my help lol i'll be around if u need anything as well, thanks!
Well if you could kindly explain Lagrange Multiplier method for me, i'd be forever greatful! :D

haha.
i kinda..skippedthat lecture. ^^"
 

Chris.

Member
Joined
Nov 27, 2006
Messages
538
Gender
Male
HSC
2006
shinji said:
Well if you could kindly explain Lagrange Multiplier method for me, i'd be forever greatful! :D

haha.
i kinda..skippedthat lecture. ^^"
i just learnt that shit man dont mention it lol

i need 25/60 to pass, and i dont even know if i can achieve that. the weather sucks ass too. hoping to gain the majority of marks from annuities, matrices, linear programming and marks here and there from calculus.
 

angelfromabuv

New Member
Joined
Apr 22, 2006
Messages
9
Gender
Female
HSC
2006
Lagrange Multiplier? aahh that things a bit messy and i think i just got the hang of it, i think the best way i can explain it is to go thru the '06 question number 5? if that helps here we go:

The lagrange format is on the formula sheet, so for the f(x,y) bit that we've found from teh sectoin above--> L= (what you want to maximise) + lambda (constraint)

We insert functions given so that gives you:

L= 240x -[FONT=CMBX12~18]x^2[/FONT] + [FONT=CMBX12~18]140y -[/FONT] [FONT=CMBX12~18]y^2[/FONT] - [FONT=CMBX12~18]200 [/FONT]+ lambda [(80-(x + y)]
The last bit after lambda is the cost function which is our constraint (i get a bit confused as to what happened to teh 10 in front of teh 10(x + y) and wateva but yea...)

AFter that you do the partial differentiation for x, y and lambda, then set taht all = 0
Lx=0
Ly=0
L(lambda)=0

Then solve both your x and y for lambda, so now they both equal lambda, they equal each other i.e. (240-2x=140-2y)
Solve for y
Sub Y into the L(lambda)=0 equation and solve for Y
Now you have x and Y, then sub both of these into either the Lx=0 or the Ly=0 to find lambda,
Then sub all of these (x, y, lambda) into the initial profit equation to find your profit.

And then we have to check for a max using teh Hessian....and this is where i'm up to lol
does that make any sense? hope it helps a tad!

Did anyone get that Hessian question in Quiz 4? had me completely stumped no idea how to do it :S
sub
 

angelfromabuv

New Member
Joined
Apr 22, 2006
Messages
9
Gender
Female
HSC
2006
um someone might wana check i'm doing all the right thigns yea? i have trouble figuring out how to put in teh constraints...any tips :S?
 

Lavenderpup

Member
Joined
Nov 8, 2004
Messages
234
Location
Sydney
Gender
Female
HSC
2006
in quiz four, you use the small hessian to check which is fxxfyy - fxy^2 aka it's determinant gets used and if its >0 then its a max, and if its <0 then its a min.

and with lagrangian you use the bigger hessian as indicated by its g and Ls.
.. well that's how I remember that anyway :/
 

shinji

Is in A State Of Trance
Joined
Feb 2, 2005
Messages
2,733
Location
Syd-ney
Gender
Male
HSC
2006
Wow. thanks for that angelfromabuv!!

Also just to check, we can also use Matrices to solve the Lx=Ly=Llambda=0 yeah?
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top