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02-28-2002, 12:05 PM
I keep running into a problem with this one. I can see that the problem requires solving a 3x3 linear system, but I can't get the coefficients right. I keep getting a negative value for one of the weights. Can anyone help me? :emb: <:)

02-28-2002, 12:14 PM
The system I get is:

-.02x+.02y+.03z=.008
.08x+.07y+.18z=.093
.008x-.007y+.054z=-.0305

Does anyone else get that system?

New at pd
02-28-2002, 01:03 PM
It's not a 3-variable system, as x + y + z =1, or z = 1 - x - y.

<font size=-1>[ This Message was edited by: New at c6 on 2002-02-28 13:04 ]</font>

02-28-2002, 01:12 PM
D'OH!!!

:wall:

New at pd
02-28-2002, 01:29 PM
The system of equations is as follows

Equation 1, using currency selection
6=(x-30)*1.1+(y-10)*0.9+(z-60)*1.3

Equation 2, using country selection
-1.5=(x-30)*0.1+(y-10)*0.05+(z-60)*0.15

Equation 3, using stock selection
0.8=x*-0.02+y*0.02+z*0.03

02-28-2002, 01:44 PM
OK...that explains it. Thanks, New.

Anybody know the phone number to a good tractor-trailer school...I think I'm going to need it.

<font size=-1>[ This Message was edited by: Enough Exams Already on 2002-02-28 14:12 ]</font>

New at pd
02-28-2002, 01:57 PM
Now, if I only knew the other 1,500 pages of material this well. d*mn.

<font size=-1>[ This Message was edited by: New at c6 on 2002-02-28 13:57 ]</font>

john6
03-28-2002, 12:17 PM
the first equation on currency doesn't use
1+i i.e i=-10%, 1+i = .9
it uses i --> -10%

I am not getting the so called solutions that the society released. They give us nothing for question 5b.
My matrix solutions keeps on solving for y> 1

Does anyone out there know how the society came up with .4,.2,.4 ???????
For x,y,z

New at pd
03-28-2002, 12:39 PM
On 2002-03-28 12:17, john6 wrote:
the first equation on currency doesn't use
1+i i.e i=-10%, 1+i = .9
it uses i --> -10%

I am not getting the so called solutions that the society released. They give us nothing for question 5b.
My matrix solutions keeps on solving for y> 1

Does anyone out there know how the society came up with .4,.2,.4 ???????
For x,y,z

Currency selection is defined as the sum of
(w_port - w_index)*E1/E0, where Ei is the exchange rate in \$/Foreign unit.

So, if there is -10% appreciation, E1/E0 = .9

Songbird
03-28-2002, 12:50 PM
[(10%)X+(-10%)Y+(30%)Z] - [(10%)0.3+(-10%)0.1+(30%)0.6] = -6.0%

[(10%)X+(5%)Y+(15%)Z] - [(10%)0.3+(5%)0.1+(15%)0.6] = -1.5%

(8%-10%)X + (7%-5%)Y + (18%-15%)Z = +0.8%

Solving gives you (X,Y,Z) = (0.4,0.2,0.4)

New at pd
03-28-2002, 12:57 PM
As far as currency selection goes, it's just a matter of preference if you prefer to use just -.1 or .9. If you use .9, you are effectively adding a term of x + y + z - (.1 + .3 + .6) to the equation. Since, by constraints, x + y + z = 1, the net effect is 0.

03-29-2002, 12:39 AM
song bird please refer to the question no. 11 of study note 2001. Why is 0.505 is added to the stock prices and how it is calculated?

<font size=-1>[ This Message was edited by: secada on 2002-03-29 00:40 ]</font>

<font size=-1>[ This Message was edited by: secada on 2002-03-30 03:34 ]</font>

actuary boy
03-30-2002, 10:15 AM
The .505 is the dividend paid in 60 days.