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I have an optimisation problem.

I wish to maximise a function subject to a constraint. It is the constraint that is causing me problems. I am using an addin in Matlab which does the optimisation however the constraints that I have used before have been in the format of the line below.

 b_l <= Ax <= b_u

The constraint is,

 Sum(x .* stock)*BetaBM - 0.1 <= Sum(x .* stock.*BetaSK) <= Sum(x .*stock)*BetaBM + 0.1

where,

 x is 2000 by 1 vector
 stock is 2000 by 1 vector
 BetaBM is a scalar
 BetaSK is 2000 by 1 vector

 x - is the weight of each stock in the fund. It cannot be more than 100% but can be less.
 stock - I am looking at M&A deals. The stock variable is a number between 0 and 1 which represents how much of the deal is being paid for in the acquires stock. 0 would mean the deal is purely cash. If there is part of the deal being paid in stock I will hedge the beta exposure against the S&P Index.

 BetaBM - is the S&P beta.
 BetaSK - contains all the individual beta for all the stocks in the fund

I need to get the constraint in the format b_l <= Ax <= b_u if at all possible?

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    $\begingroup$ please, check your question: something is missing BetaSK is. $\endgroup$ – Arrigo Sep 29 '14 at 14:40
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    $\begingroup$ I think there is something to correct: in MATLAB, x*stock will not work if both vectors have the same dimensions (2000x1) because it is not a correct matrix product. Do you mean matrix product or dot product when you use *? $\endgroup$ – Arrigo Sep 29 '14 at 14:56
  • $\begingroup$ sorry if I was using matlab your right they would not work. I meant to use .* $\endgroup$ – mHelpMe Sep 29 '14 at 15:16
  • $\begingroup$ is this a programming question? If yes .. off-topic. If no: then please write something about what the variables mean and what the constraint means. $\endgroup$ – Richard Sep 29 '14 at 15:28
  • $\begingroup$ @Richard - please see my revised post. If it is not clear please let me know $\endgroup$ – mHelpMe Sep 29 '14 at 15:46
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If you have a vector of weights $w=(w_1,\ldots,w_n)^T$ then $(1,\ldots,1)* w = \sum_{i=1}^n w_i$ thus a sum condtion can be formulated by multiplication with a row of ones. A $\le$ can be put into an $\ge$ by multiplying with $(-1)$ and if you have to put all your constraints into on $A$ then you usually stack all the row vectors together. In your case the matrix $A$ will consist of rows of ones. As long as you don't explain your constraint I can not add more.

I will try: If you want to put a constraint on your porfolio beta then you should use a constraint: $$(\beta_1,\ldots,\beta_n) * w \le \beta_{BM}+0.1$$ and $$(\beta_1,\ldots,\beta_n)*w \ge \beta_{BM}-0.1$$ which the same as $$ -(\beta_1,\ldots,\beta_n)*w \le 0.1-\beta_{BM}$$. So you could use a matrix $$A = (\beta_1,\ldots,\beta_n;-\beta_1,\ldots,-\beta_n)$$ and right hand side $b = (\beta_{BM}+0.1;0.1-\beta_{BM})$ then your condition is: $$ A w \le b. $$

EDIT after a remark by PO: If you have $$ (\sum_{i=1}^n w_i stock_i) \beta_{BM} -0.1 \le \sum_{i=1}^n w_i stock_i \beta_i $$ then this is equivalent to $$ \sum_{i=1}^n w_i stock_i (\beta_{BM} - \beta_i) \le 0.1 $$ then you define scalars $k_i = stock_i (\beta_{BM} - \beta_i)$ and the same as above holds for $\sum_{i=1}^n w_i k_i \le 0.1$. For the right hand side you get $$ \sum_{i=1}^n w_i (-k_i) \le 0.1. $$ Your matrix $A$ has two rows, one $(k_1,\ldots,k_n)$ and one $(-k_1,\ldots,-k_n)$ the rhs is $(0.1, 0.1)$.

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  • $\begingroup$ Firstly thanks for such a clear answer. I have a silly question (maths is not primary skill which you can probably tell). Doing it this way have we broken the contraint b_l <= Ax <= b_u into two seperate contraints which will achieve the same goal? I.e we Ax <= b where b is (0.1, 0.1). I believe in Matlab I can just put b_l equal to [ ]. $\endgroup$ – mHelpMe Sep 30 '14 at 8:02
  • $\begingroup$ It is quite some time ago that I used Matlab. But it should work that you only define b_u and leave b_l empty. Or you use a formulation where you still stack to rows together for A but keep a b_l. Give it a try and tell us. $\endgroup$ – Richard Sep 30 '14 at 8:25
  • $\begingroup$ I'm pretty sure that setting b_l to [ ] will work. I will try that now. Sorry I do not quite follow where you say 'keep a b_l'. I understand that A will be stacked (i.e. two rows, k & -k) that you mentioned in your post $\endgroup$ – mHelpMe Sep 30 '14 at 8:30
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    $\begingroup$ I though that maybe there is a way to keep the $B_l \le A \le b_u$ formulation. But please forget it for now. It is quite usualy to formulate the constraint as $A x \le b_u$. This is fully general as you can always multiply rows and rhs by (-1). $\endgroup$ – Richard Sep 30 '14 at 8:39
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    $\begingroup$ I guess actually if needs be in matlab I can put B_l to -inf. I will mark the answer as correct, thank-you again. $\endgroup$ – mHelpMe Sep 30 '14 at 9:15

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