Simultaneous Equations for Chapter 14 Convexity

[stylz]

Hi all,

I am having a bit of trouble getting my answers to the convexity questions that require simultaneous equations to find P and Q (example on page 29, question 14.17, and the practice exam question on page 33).

Could someone please advise what steps they did to get me started. I couldn't do it without using v from the start. Although the book said it could do it without adding v in until later).

It's quite frustrating as I understand the concepts of this chapter, but this basic mathematical procedure is tripping me up!

When I put v in at the start (from the tables) my answers are close to the answers in the book, but not that close (say a couple of $1000 out).

Please let me know if you need further detail as I'm happy to write out my workings.

Cheers

Thanks

 

[DevonMatthews]

I havn't got access to a recent version of the notes for this so i don't have much idea what your on about. Are you trying to apply the reddington conditions to a portfolio? Put the question up so i can help.....

 

[Muppet]

if you follow the same method as that used in the example on p29 then you'd get, working in 000s:


20v^3.5 + 18v^6 = 24.4875 = Pv^4 + Qv^7 (equating PVs) (1)

Then taking the negative derivative:

3.5*20v^4.5 + 6*18v^7 = 101.00702 = 4Pv^5 + 7Qv^8
times this by (1+i) = 1.1 gives:

111.1077 = 4Pv^4 + 7Qv^7 (2)

Solving (1) and (2) simultaneously gives:

P = 29.431
Q = 8.547

which are consistent with their answers.

 

[stylz]

"20v^3.5 + 18v^6 = 24.4875 = Pv^4 + Qv^7 (equating PVs) (1)

111.1077 = 4Pv^4 + 7Qv^7 (2)

Solving (1) and (2) simultaneously gives:

P = 29.431
Q = 8.547"

Hi Guys,
Actually, the solutions in my book for the practice exam question that you have shown above says P = 20,103 and Q = 4,385.
But nevertheless, I'm sorry, I haven't made myself clear. It's just the solving of the simultaneous equations bit that I am having trouble with.

What were the steps you used to solve for P and Q from equations (1) and (2)?

The other example (from p29) would be to solve for P and Q again, the following:

Pv^5 + Qv^10 = 62,418

5Pv^6 + 100Qv11 = 404,398

I tried to solve but my answers were about $1000 out which doesn't seem close enough. Not sure if its rounding errors or what. That's why I want to see what process other people use.
Cheers

 

[John Lee]

But nevertheless, I'm sorry, I haven't made myself clear. It's just the solving of the simultaneous equations bit that I am having trouble with.

What were the steps you used to solve for P and Q from equations (1) and (2)?

The other example (from p29) would be to solve for P and Q again, the following:

Pv^5 + Qv^10 = 62,418 (1)

5Pv^6 + 100Qv11 = 404,398 (2)

There are two ways to solve simultaneous equations:

1. Elimination

If we multiply equation (1) by 5v we get:

5Pv^6 + 5Qv^11 = 2021990v (3)

Now subtract this eqn (3) from eqn (2) to eliminate the 5Pv^6 term.

Don't be tempted to calculate the v's - just makes it very messy

2. Substitution

Rearrange eqn (1) for P to give:

P = (62,418 - Qv^10)/v^5

Substitute this into eqn (2).

Hope this helps

 

[stylz]

Sorry, but could you complete the steps for both Elimination and Substitution methods for me please. How do we get rid of the v? Or at what point do we add the v calculation in?

Using elimination I end up with Q = (404,398 - 312,090v)/95v.

Using substitution I end up with Q = (404,398 - 2021990v)/95v.

Thanks.

 

[Muppet]

"Actually, the solutions in my book for the practice exam question that you have shown above says P = 20,103 and Q = 4,385.
Cheers

I think they have X = £20,103 and Y = £4,385.

eg: X = Pv^5 = 29,431 / 1.1^4 = £20,102 (near enough)
Y = Qv^7 = 8,547 / 1.1^7 = £4,386

 

[Muppet]

Pv^5 + Qv^10 = 62,418 (1)

5Pv^6 + 100Qv11 = 404,398 (2)

Equation (2) should say 10Q, not 100Q.

So you'll end up dividing by 5 instead of 95. Also think you have "^11" missing from the end of your equation for Q using elimination.

 

[stylz]

Thank you Muppet and John.

 

[stylz]

DMT or Negative Derivative?

Do you think it will matter if you use Discounted Meant Term or the negative derivative in the second step for these kinds of questions in the exam?

The example in the notes does it by taking the negative derivative, but the practice exam question uses DMT.

 

[John Lee]

Do you think it will matter if you use Discounted Meant Term or the negative derivative in the second step for these kinds of questions in the exam?

The example in the notes does it by taking the negative derivative, but the practice exam question uses DMT.

Makes no difference whatsoever unless the question asks explicitly for one of them.