samples/texts/1228241/page_15.md

is $d_i$, then $s_i^{(vdd)} = w_i/d_i$; $E_{vdd}$ is the optimal energy consumption obtained with these speeds. Let $s_i^{(algo)} = \min_i{s_{min} + u \times \delta \mid s_{min} + u \times \delta \ge s_i^{(vdd)}}$ be the smallest speed in $\mathcal{I}{inc}$ that is larger than $s_i^{(vdd)}$. There exists such a speed since, because of the values chosen for $\mathcal{I}{vdd}$, $s_i^{(vdd)} \le s_{max}$. The values $s_i^{(algo)}$ can be computed in time polynomial in the size of $\mathcal{I}{inc}$ and $K$. Let $E{algo}$ be the energy consumption obtained with these values.

In order to prove that this algorithm is an approximation of the optimal solution, we need to prove that $E_{algo} \le (1 + \frac{\delta}{s_{min}})^2(1 + \frac{1}{K})^2 \times E_{inc}$. For each task $T_i$, $s_i^{(algo)} - \delta \le s_i^{(vdd)} \le s_i^{(algo)}$. Since $s_{min} \le s_i^{(vdd)}$, we derive that $s_i^{(algo)} \le s_i^{(vdd)} \times (1 + \frac{\delta}{s_{min}})$. Summing over all tasks, we get

$$E_{algo}=\sum _i{w}_i(s_i^{(algo)}{)}^2\le \sum _i{w}_i(s_i^{(vdd)}\times (1+\frac{\delta }{s_{min}}){)}^2\le E_{vdd}\times (1+\frac{\delta }{s_{min}}{)}^2\mathrm{.}E\_\{algo\}\; =\; \backslash sum\_i\; w\_i\; (s\_i^\{(algo)\})^2\; \backslash le\; \backslash sum\_i\; w\_i\; (s\_i^\{(vdd)\}\; \backslash times\; (1\; +\; \backslash frac\{\backslash delta\}\{s\_\{min\}\}))^2\; \backslash le\; E\_\{vdd\}\; \backslash times\; (1\; +\; \backslash frac\{\backslash delta\}\{s\_\{min\}\})^2.$$

Next, we bound $E_{vdd}$ thanks to the optimal solution with the CONTINUOUS model, $E_{con}$. Let $\mathcal{I}{con}$ be the instance where the execution graph $G$, the deadline $D$, the speeds $s{min}$ and $s_{max}$ are the same as in instance $\mathcal{I}{inc}$, but now admissible speeds take any value between $s{min}$ and $s_{max}$. Let $s_i^{(con)}$ be the optimal continuous speed for task $T_i$, and let $0 \le u \le N$ be the value such that:

$$s_{min}\times {(1+\frac{1}{K})}^u\le s_i^{(con)}\le s_{min}\times {(1+\frac{1}{K})}^{u+1}=s_i^{\ast }\mathrm{.}s\_\{min\}\; \backslash times\; \backslash left(1\; +\; \backslash frac\{1\}\{K\}\backslash right)^u\; \backslash le\; s\_i^\{(con)\}\; \backslash le\; s\_\{min\}\; \backslash times\; \backslash left(1\; +\; \backslash frac\{1\}\{K\}\backslash right)^\{u+1\}\; =\; s\_i^*.$$

In order to bound the energy consumption for $I_{vdd}$, we assume that $T_i$ runs at speed $s_i^*$, instead of $s_i^{(vdd)}$. The solution with these speeds is a solution to $I_{vdd}$, and its energy consumption is $E^* \ge E_{vdd}$. From the previous inequalities, we deduce that $s_i^* \le s_i^{(con)} \times (1 + \frac{1}{K})$, and by summing over all tasks,

Proposition 3.

• For any integer $\delta > 0$, any instance of MINENERGY$(G, D)$ with the CONTINUOUS model can be approximated within a factor $(1 + \frac{\delta}{s_{min}})^2$ in the INCREMENTAL model with speed increment $\delta$.

• For any integer $K > 0$, any instance of MINENERGY$(G, D)$ with the DISCRETE model can be approximated within a factor $(1 + \frac{\alpha}{s_1})^2(1 + \frac{1}{K})^2$, with $\alpha = \max_{1 \le i < m}{s_{i+1} - s_i}$, in a time polynomial in the size of the instance and in $K$.

Proof. For the first part, let $s_i^{(con)}$ be the optimal continuous speed for task $T_i$ in instance $\mathcal{I}{con}$; $E{con}$ is the optimal energy consumption. For any task $T_i$, let $s_i$ be the speed of $\mathcal{I}{inc}$ such that $s_i - \delta < s_i^{con} \le s_i$. Then, $s_i^{(con)} \le s_i \times (1 + \frac{\delta}{s{min}})$. Let $E$ be the energy with speeds $s_i$. $E_{con} \le E \times (1 + \frac{\delta}{s_{min}})^2$. Let $E_{inc}$ be the optimal energy of $\mathcal{I}{inc}$. Then, $E{con} \le E_{inc} \times (1 + \frac{\delta}{s_{min}})^2$.

For the second part, we use the same algorithm as in Theorem 7. The same proof leads to the approximation ratio with $\alpha$ instead of $\delta$. □