# 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

myassignments-help数学代考价格说明

1、客户需提供物理代考的网址，相关账户，以及课程名称，Textbook等相关资料~客服会根据作业数量和持续时间给您定价~使收费透明，让您清楚的知道您的钱花在什么地方。

2、数学代写一般每篇报价约为600—1000rmb，费用根据持续时间、周作业量、成绩要求有所浮动(持续时间越长约便宜、周作业量越多约贵、成绩要求越高越贵)，报价后价格觉得合适，可以先付一周的款，我们帮你试做，满意后再继续，遇到Fail全额退款。

3、myassignments-help公司所有MATH作业代写服务支持付半款，全款，周付款，周付款一方面方便大家查阅自己的分数，一方面也方便大家资金周转，注意:每周固定周一时先预付下周的定金，不付定金不予继续做。物理代写一次性付清打9.5折。

Math作业代写、数学代写常见问题

myassignments-help擅长领域包含但不是全部: