# A system of linear equations with linear constraints

Mathematical problem. Suppose we have $2n$ indeterminates $x_1,\dots,x_n$ and $y_1,\dots,y_n$ (which are denoted by $q$ with indices and called abundances below) and $m$ subsets $P_1,\dots,P_m$ of $\lbrace 1,2,\dots,n\rbrace$. Given real $r_1,\dots,r_m$, how to determine the indeterminates if they satisfy the system of linear homogeneous equations $$\sum_{p\in P_i}x_p-r_i\sum_{p\in P_i}y_p=0 \quad\text{for } i=1,\dots,m$$ and the additional linear constraints $$0\le x_p\le 1 \quad\text{and}\quad 0\le y_p\le 1 \quad\text{for } p=1,\dots,n \quad?$$

Background. Proteins are important molecules in living organisms. In molecular biology you often want to deduce the fold change of abundance of the proteins in one sample versus another.

Proteins are built up of a sequence of molecules (amino acids). Peptides are parts of proteins, $i.e.$ part of the sequence. Using current technology (Mass spectrometry), you can only detect the peptides and ratios of these peptides.

Motivation. Normally if you want to deduce the ratio of a protein in sample A versus B, you just take the geometric mean of the ratios of the peptides specific to that protein. . However, you will observe peptides which are shared by proteins. It is assumed that the ratio of a peptide which is shared by multiple proteins is the sum of the abundances of the proteins in sample A divided by the sum of the abundances of the proteins in sample B. My question targets how to still be able to use these observations, and deduce ratios for proteins having only shared proteins.

Given

• $P = \lbrace p_1,\...,p_n \rbrace$ ... Proteins
• $Q = \left( P_1, ..., P_m \right), P_i \subseteq P$ ...proteins which share peptide i
• $R = \left( r_1,\...,r_m \right)$ ... Observed ratios for peptides

find $q_{p,a},q_{p,b} \forall p \in P$ such that $$r_i=\frac{\sum_{p \in P_i}q_{p,a}}{\sum_{p \in P_i}q_{p,b}}$$

The real abundances $q_{p,a},q_{p,b}$ are not known and not of interest, just their relation to each other.

My approach

Solve a nonlinear system of equations minimizing $$\log(r_i)=\log\sum_{p \in P_i}q_{p,a} - \log\sum_{p \in P_i}q_{p,b}, \forall r_i \in R$$

Is there a way to approach this in a linear system? Of course I can try solve $\sum_{p \in P_i}q_{p,a} - r_i \times \sum_{p \in P_i}q_{p,b} = 0, \forall r_i \in R$, but the ratios are not dealt with correctly.

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The question (even I can't get it) has nothing to do with linear algebra and combinatorics. Your problem is really not defined: what are the sets $S_i$? why can't I take all $q_{s,a}$ to be the same and all $q_{s,b}$ to be the same (and differ one from the other by $r_i$)? You really have to put some effort in order to give more details. I've never heard about proteomics and peptides... –  Wadim Zudilin Jun 8 '10 at 12:20
If this is minimization subject to some constraints (for example various sums of variables being given), then it is probably accessible via the method of Lagrange multipliers. As it is, the question doesn't seem properly posed. –  Charles Matthews Jun 8 '10 at 12:40
This seems to be a duplicate of mathoverflow.net/questions/26817 asked by the same author week ago. –  Wadim Zudilin Jun 8 '10 at 13:53
I try to pose the question better this time and deleted the old question. Sorry for the mistake with $S_i$, it means $P_i$. $q_{p,a}$ and $q_{p,b}$ are the relative abundances of proteins rather than ratios of peptides $r_i$. I'll explain peptides and protein in the main text. –  Florian Breitwieser Jun 9 '10 at 8:14
There are still troubles with your formulation. If I denote $x_p=q_{p,a}$ and $y_p=q_{p,b}$, $p=1,2,\dots,n$, there are $2n$ indeterminates, $x_1,\dots,x_n,y_1,\dots,y_n$. Then there are $m$ homogeneous linear equations $\sum_{p\in P_i}(x_p-r_iy_p)=0$ --- note that these are linear equations, as $r_i$ are given numbers. The general solution of the system of $m$ equations in $2n>m$ variables forms a subspace of $\mathbb R^{2n}$ of dimension $\ge m-2n$. Without further constraints you can only pick one element from the subspace. Aren't you happy with it? –  Wadim Zudilin Jun 9 '10 at 11:34