Evaluating the Performance Impact of No-Wait Approach to Resolving Write Conflicts in Databases

Abstract

A model of a large distributed database where write conflicts are handled by wait-free aborting of transactions, is introduced and analysed in the steady state. Performance measures, including the rate of aborts, are evaluated. Numerical results are presented, illustrating the behaviour of the system when various parameters are varied. The accuracy of certain approximations is assessed by means of simulations.

Appendix

Proof of Lemma. We invoke Rouché’s theorem, which states that if two holomorphic functions, $\phi (z)$ and $\psi (z)$, satisfy $|\phi (z)|>|\psi (z)|$ on a simple closed contour, then $\phi (z)$ and $\phi (z)+\psi (z)$ have the same number of zeros inside that contour. Each zero is counted according to its multiplicity.

We represent P(z) as $P(z)= \phi (z)+\psi (z)$, where

$$ \phi (z)= -(\lambda +N\mu )z $$

and

$$ \psi (z)= \lambda + \sum _{i=1}^N \xi _i z^{i+1} $$

The closed contour is the unit circle. When $|z|=1$, $|\phi (z)|=\lambda +N\mu$. Applying the triangle inequality to $\psi (z)$, we find

$$ |\psi (z)| \le \lambda + \sum _{i=1}^N \xi _i = \lambda +N\mu \;. $$

Moreover, the inequality is strict everywhere on the contour, except at $z=1$, where it is an equality.

Note that the derivative of P(z) is positive at $z=1$. This is because

$$ P^\prime (1) = -\lambda + \sum _{i=1}^N i\xi _i >0\;, $$

according to (12). Hence, we can choose a sufficiently small number, $\epsilon$, such that $P(1-\epsilon )<0$. Modifying the contour slightly in the vicinity of $z=1$, by making it pass through the point $z=1-\epsilon$, would ensure that the inequality $|\phi (z)|>|\psi (z)|$ is strict on the entire modified contour.

The function $\phi (z)$ is linear and has a single zero, $z=0$, inside the contour. Therefore, by Rouché’s theorem, P(z) also has a single zero inside the contour. That zero, $z_0$, must be in the interval $(0,1-\epsilon )$, because $P(0)=\lambda >0$ and $P(1-\epsilon )<0$. This completes the proof.