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Ill Conditioning

In the vast majority of cases the solver will give a correct solution to the problem. However, some problems are by nature ill-conditioned in which case small changes in the input data can lead to more significant changes in the results.

Taking a simple example to look at ill-conditioning; consider a simple two spring system, where the springs are connected in series. The stiffness of the first spring is k1k_{1} and that of the second is k2k_{2} and we assume that k2k_{2} is much greater than k1k_1 (k2>>k1)\left( k_{2} > > k_{1} \right).

In this case the equations describing the system is

{0f2}=[k1+k2k2k2k2]{u1u2}\begin{Bmatrix} 0 \\ f_{2} \\ \end{Bmatrix} = \begin{bmatrix} k_{1} + k_{2} & - k_{2} \\ - k_{2} & k_{2} \\ \end{bmatrix}\begin{Bmatrix} u_{1} \\ u_{2} \\ \end{Bmatrix}

As in a solver based on a Gaussian elimination technique, we use these equations to arrive at a relationship between u2u_{2} and u1u_{1}

u2=f2+k2u1k2u_{2} = \frac{f_{2} + k_{2}u_{1}}{k_{2}}

which when substituted in the other equation gives:

(k1+k2)u1k2(f2+k2u1k2)=0\left( k_{1} + k_{2} \right)u_{1} - k_{2}\left( \frac{f_{2} + k_{2}u_{1}}{k_{2}} \right) = 0


[k1+(k2k2)]u1=f2\left\lbrack k_{1} + \left( k_{2} - k_{2} \right) \right\rbrack u_{1} = f_{2}

With exact arithmetic the term (k2k2)\left( k_{2} - k_{2} \right)would be zero, however, if k2k_{2} is large compared with k1k_{1} and due to limited precision, some error will be introduced in the calculation. If this error is denoted by ee, then the equation we have is

[k1+e]u1=f2\left\lbrack k_{1} + e \right\rbrack u_{1} = f_{2}

We have then a system as shown below where the error is like adding a third spring, which acts in parallel with k1k_{1}.

The expected reaction is f2f_{2}, but the reaction that is calculated is

r=k1u1=k1f2k1+er = k_{1}u_{1} = \frac{k_{1}f_{2}}{k_{1} + e}

Thus the reaction is in error by a factor

k1k1+e\frac{k_{1}}{k_{1} + e}