samples/texts/1228241/page_4.md

Corollary 1. A linear chain with $n$ tasks is equivalent to a single task of cost $W = \sum_{i=1}^{n} w_i$.

Indeed, in the optimal solution, the $n$ tasks are executed at the same speed, and they can be replaced by a single task of cost $W$, which is executed at the same speed and consumes the same amount of energy.

4.2.3. Fork and join graphs

Let $V = {T_1, \dots, T_n}$. We consider either a fork graph $G = (V \cup {T_0}, E)$, with $E = {(T_0, T_i), T_i \in V}$, or a join graph $G = (V \cup {T_0}, E)$, with $E = {(T_i, T_0), T_i \in V}$. $T_0$ is either the source of the fork or the sink of the join.

Theorem 1 (fork and join graphs) When $G$ is a fork (resp. join) execution graph with $n+1$ tasks $T_0, T_1, \dots, T_n$, the optimal solution to MINENERGY$(G, D)$ is the following:

• the execution speed of the source (resp. sink) $T_0$ is $s_0 = \frac{(\sum_{i=1}^n w_i^3)^{\frac{1}{3}} + w_0}{D}$;

• for the other tasks $T_i$, $1 \le i \le n$, we have $s_i = s_0 \times \frac{w_i}{(\sum_{i=1}^n w_i^3)^{\frac{1}{3}}} \quad \text{if } s_0 \le s_{max}$.

Otherwise, $T_0$ should be executed at speed $s_0 = s_{max}$, and the other speeds are $s_i = \frac{w_i}{D}$, with $D' = D - \frac{w_0}{s_{max}}$, if they do not exceed $s_{max}$ (Proposition 1 for independent tasks). Otherwise there is no solution.

If no speed exceeds $s_{max}$, the corresponding energy consumption is

$$\min E(G,D)=\frac{{({\left(\sum _{i=1}^n{w}_i^3\right)}^{\frac{1}{3}}+w_0)}^3}{D^2}\mathrm{.}\backslash min\; E(G,\; D)\; =\; \backslash frac\{\backslash left(\; \backslash left(\; \backslash sum\_\{i=1\}^\{n\}\; w\_i^3\; \backslash right)^\{\backslash frac\{1\}\{3\}\}\; +\; w\_0\; \backslash right)^3\}\{D^2\}.$$

Proof. Let $t_0 = \frac{w_0}{s_0}$. Then, the source or the sink requires a time $t_0$ for execution. For $1 \le i \le n$, task $T_i$ must be executed within a time $D - t_0$ so that the deadline is respected. Given $t_0$, we can compute the speed $s_i$ for task $T_i$ using Theorem 1, since the tasks are independent: $s_i = \frac{w_i}{D-t_0} = w_i \cdot \frac{s_0}{s_0 D - w_0}$. The objective is therefore to minimize $\sum_{i=0}^n w_i s_i^2$, which is a function of $s_0$:

$$\sum _{i=0}^n{w}_i{s}_i^2=w_0{s}_0^2+\sum _{i=1}^n{w}_i^3\cdot \frac{s_0^2}{(s_{0}D-w_0{)}^2}=s_0^2(w_0+\frac{\sum _{i=1}^n{w}_i^3}{(s_{0}D-w_0{)}^2})=f(s_0)\mathrm{.}\backslash sum\_\{i=0\}^\{n\}\; w\_i\; s\_i^2\; =\; w\_0\; s\_0^2\; +\; \backslash sum\_\{i=1\}^\{n\}\; w\_i^3\; \backslash cdot\; \backslash frac\{s\_0^2\}\{(s\_0\; D\; -\; w\_0)^2\}\; =\; s\_0^2\; \backslash left(\; w\_0\; +\; \backslash frac\{\backslash sum\_\{i=1\}^\{n\}\; w\_i^3\}\{(s\_0\; D\; -\; w\_0)^2\}\; \backslash right)\; =\; f(s\_0).$$

Let $W_3 = \sum_{i=1}^{n} w_i^3$. In order to find the value of $s_0$ that minimizes this function, we study the function $f(x)$, for $x > 0$. $f'(x) = 2x(w_0 + \frac{W_3}{(xD-w_0)^2}) - 2D \cdot x^2 \cdot \frac{W_3}{(xD-w_0)^3}$, and therefore $f'(x) = 0$ for $x = (W_3^{1/3} + w_0)/D$. We conclude that the optimal speed for task $T_0$ is $s_0 = (\sum_{i=1}^{n} w_i^3)^{1/3} + w_0/D$, if $s_0 \le s_{max}$. Otherwise, $T_0$ should be executed at the maximum speed $s_0 = s_{max}$, since it is the bottleneck task. In any case, for $1 \le i \le n$, the optimal speed for task $T_i$ is $s_i = w_i s_0/(s_0 D - w_0)$.