Section 3.5.2: Additional Notes - Reducing the number of equations

Partitions
Incidence Matrix
Tearing
Solution by Tearing
Summarising
Equation set as a Graph
Bubble Point Calculation Revisited

Consider:

$\begin{eqnarray*}f_1(x_1) & = & 0 \\ f_2(x_1,x_2) & = & 0 \\ f_3(x_2,x_3,x_5... ... & 0 \\ f_5(x_1,x_4,x_5) & = & 0 \\ f_6(x_5,x_6) & = & 0 \\ \end{eqnarray*}$

We note that:

f₁ and f₂ can be rearranged and solved sequentially for x₁ and x₂
Even when x₁ and x₂ are known, f₃ to f₅ cannot be reordered to enable x₃, x₄ and x₅to be obtained sequentially.
However, x₆ only appears in f₆ , so that after we have solved f₃ to f₅, then x₅ will be known, and we can solve f₆ on its own for x₆.

We can call the subset (f₁,f₂) the head of the equation set, andf₆ the tail. (In general, of course, the equations need not have been numbered and ordered so that these appeared at the top and bottom of the list of equations!)

Partitions

The group (f₃,f₄,f₆) is called a partition of the set of equations and represents a subset that must be solved simultaneously. If we delete x₁ andx₂, which will be known after solvingf₁ andf₂, the partition is:

$\begin{eqnarray*}f_3(x_3,x_5) & = & 0 \\ f_4(x_3,x_4) & = & 0 \\ f_5(x_4,x_5) & = & 0 \end{eqnarray*}$

There are three equations in three unknowns, so our `work factor' is 3² = 9, only a quarter of that for the complete set.

Incidence Matrix

$\begin{displaymath}\left[ \begin{array}{c c c c c c} 1 & & & & & \\ 1 & 1 & & &... ... \\ 1 & & & 1 & 1 & \\ & & & & 1 & 1 \\ \end{array}\right] \end{displaymath}$

The partition is identifiable as the the group of equations with above diagonal elements.

Tearing

The partition alone is 3 equations, f₃, f₄, f₅ in $\{x_3, x_4,x_5\}$

These can be solved as follows:

Rearrange f₃ to give x₅ , in terms of x₃ , i.e:

$\begin{displaymath}x_5 = \phi_3(x_3) \end{displaymath}$

Similarly f₅ to give x₄ in terms of x₅ :

$\begin{displaymath}x_4 = \phi_5(x_5) \end{displaymath}$

And finally either: Evaluate f₄ ; if $f_4 \approx 0$ terminate; otherwise adjust x₃

This procedure is referred to as tearing the set of equations, reducing them here to solution for a single unknown x₃ , called the tear variable.

The incidence matrix:

$\begin{displaymath}\left[ \begin{array}{c c c} 1 & & 1 \\ 1 & 1 & \\ & 1 & 1 \\ \end{array}\right] \end{displaymath}$

has one variable with a supra-diagonal element; this tells us that we can reduce the equations to solving for a single variable.

Solution by Tearing

The procedure can be represented:

$\begin{displaymath}\begin{array}{ccccc} \phi_3 & \rightarrow & x_5 & \rightarrow... ...ow \\ x_3 & \leftarrow & \phi_4 & \leftarrow & x_4 \end{array}\end{displaymath}$

Clearly any one of x₃ , x₄ or x₅ could have been chosen as the unknown.

Summarizing ...

Solve essentially by rearranging the equations from the form:

$\begin{displaymath}{\bf f}({\bf x}) = {\bf0} \end{displaymath}$

into:

$\begin{displaymath}\tilde{{\bf x}} = {\bf\phi}({\bf x}) \end{displaymath}$

$\tilde{{\bf x}}$ , the output set of the equations contains all the x_i from ${\bf x}$ but not necessarily in the same order. For the example above ... Head:

$\begin{eqnarray*}x_1 & = & \phi_1 \\ x_2 & = & \phi_2(x_1) \\ \end{eqnarray*}$

Partition:

$\begin{eqnarray*}x_5 & = & \phi_3(x_2,x_3) \\ x_4 & = & \phi_5(x_1,x_5) \\ f_4(x_3,x_4) & = & 0 \\ \end{eqnarray*}$

Tail:

$\begin{eqnarray*}x_6 & = & \phi_6(x_5) \\ \end{eqnarray*}$

Equation set as a Graph

One and only one edge may leave an equation node. One and only one edge may enter a variable node. Partition cycle shown bold.

Bubble Point Calculation Revisited

Given a liquid mixture of n components with mol fraction composition $x_i, \ i=1 \ldots n$ , at pressure P , determine the temperature T , and the composition y_i of the vapour in equilibrium with the liquid.

P^_i - P^_i(T)	=	0	(1)
$\displaystyle k_i - \frac{P^*_i}{P}$	= 0		(2)
y_i - k_i x_i	= 0		(3)
$\displaystyle \Sigma y_i - 1$	= 0		(4)

The range of i and the summation is from 1 to n , the number of components.

With the equations in the above sequence and the output set, say for a binary, chosen to be:

$\begin{displaymath}\{P^*_1, k_1, y_1, P^*_2, k_2, y_2, T \}\end{displaymath}$

The incidence matrix is:

$\begin{displaymath}\left[ \begin{array}{c c c c c c c} 1 & & & & & & 1 \\ 1 & 1... ...1 \\ & & & & 1 & 1 \\ & & 1 & & & 1 \\ \end{array}\right] \end{displaymath}$

The final, summation, equation does not involve T explicitly. However, for any value of T the value of the function can be determined, and will be zero when T satisfies the equations.

Next - Section 3.5.3: Example Questions
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