电子工程代写|并行计算代写Parallel Computing代考|CS525

电子工程代写|并行计算代写Parallel Computing代考|Round-Off Error Analysis

With a formal system at hand, we can analyse longer calculations. This analysis describes how a computer realises a particular source code. Our activity consists of two steps: We first write down the calculation as done on the machine and replace all arithmetic operations by their machine counterpart. In a second step, we use the rules from (6.1) and thereafter to map the machine calculations onto “real” mathematical operators including error thresholds. The first step is purely mechanical. The second step requires us to think about potential variable content.

I demonstrate this agenda via the summation of three variables. Technically, each compute step would require us to work with an $\epsilon$ of its own:
\begin{aligned} x+M y+M z &=(x+y)\left(1+\epsilon_{1}\right)+M z \ &=\left((x+y)\left(1+\epsilon_{1}\right)+z\right)\left(1+\epsilon_{2}\right) \ &=(x+y)+(x+y) \epsilon_{1}+(x+y) \epsilon_{2}+(x+y) \epsilon_{1} \epsilon_{2}+z+z \epsilon_{2} \end{aligned}

For each calculation, we have another error. As discussed before, we use a pessimistic upper bound and hence use a generic $\epsilon$ as worst case, i.e. we set $\epsilon_{1}=\epsilon_{2}=\epsilon$. Furthermore, we eliminate higher order terms and end up with
\begin{aligned} x++{M} y++{M} z &=((x+y)(1+\epsilon)+z)(1+\epsilon) \ &=(x+y+z+x \epsilon+y \epsilon)(1+\epsilon) \ &=x+y+z+x \epsilon+y \epsilon+x \epsilon+y \epsilon+z \epsilon+x \epsilon^{2}+y \epsilon^{2} \ &=x+y+z+x \epsilon+y \epsilon+x \epsilon+y \epsilon+z \epsilon \ &=x(1+2 \epsilon)+y(1+2 \epsilon)+z(1+\epsilon) . \end{aligned}
We reiterate that this is not a precise formula in a mathematical sense: Per term, the $\epsilon$ accepts the role of a bad guy and switches sign and magnitude just as appropriate. The equals sign also is really a $\approx$ as we drop the higher order terms. All of this has been known before.

电子工程代写|并行计算代写Parallel Computing代考|Associativity and worst-case analysis

Both the worst-case character of our analysis and the lack of associativity become apparent from a simple example: Let a machine work with a significand of two bits (the leading 1 is not stored). We compute the aforementioned $f=x+y+z$ for $x=0.875, y=-0.75$ and $z=0.1875$. If we evaluate the formula left to right and normalise the data after each step, we obtain a result of $f=0.3125$. This is exact. If we however evaluate the second addition first, we obtain $f=0.375$.

For a machine, the second result is not wrong. It just suffers from an amplification of the individual round-off errors. Indeed, $2 x+2 y+z=0.4375<1$ is the “amplification” term of the left-to-right evaluation. Our machine errors are not amplified. For the second variant, we obtain $2 y+2 z+x=1.125+0.875>1$. Here, round-off errors are amplified.

On a machine, it makes a difference in which order we evaluate the individual terms. Although our round-off formalism yields worst-case estimates, it does give us some hints what we should do in practice: It tells us that some operations are potentially more harmful than others. Multiplying a number by two for example is safe-see our previous discussion on Sect. 4.2.1. Dividing by three might be tricky.

Now, we can dive into the details of our calculation. From the intermediate step (6.2), we see that the error introduced by the first calculation is eventually increased by the next operation.

电子工程代写|并行计算代写Parallel Computing代考|Round-Off Error Analysis

$$x+M y+M z=(x+y)\left(1+\epsilon_{1}\right)+M z \quad=\left((x+y)\left(1+\epsilon_{1}\right)+z\right)\left(1+\epsilon_{2}\right)=(x+y)+(x+y) \epsilon_{1}$$

$$x++M y++M z=((x+y)(1+\epsilon)+z)(1+\epsilon) \quad=(x+y+z+x \epsilon+y \epsilon)(1+\epsilon)=x+y+z+$$

电子工程代写|并行计算代写Parallel Computing代考|Associativity and worst-case analysis

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