DVFS lowers voltage and frequency together for light phases. Power falls roughly with V squared times f, while delay grows roughly with 1 over f.
Energy is power times delay. DVFS can cut energy while stretching delay, which is the tradeoff this lesson quantifies.
Take the lesson-1 block: 0.288 W switching at 0.8 V and 3 GHz, running a task in 1 s. Scale to 0.7 V and 2 GHz.
Power ratio = (V_new^2 x f_new) / (V_old^2 x f_old)
= (0.49 x 2e9) / (0.64 x 3e9)
= 0.98 / 1.92
= 0.5104
P_new = 0.288 W x 0.5104 = 0.147 W
Delay_new = 1 s x (3 / 2) = 1.5 s
E_new = 0.147 W x 1.5 s = 0.2205 J (was 0.288 J)Energy fell about 23 percent while delay rose 50 percent. The energy-delay product tells whether that trade was worth it.
EDP_old = 0.288 J x 1 s = 0.288 J-s
EDP_new = 0.2205 J x 1.5 s = 0.331 J-s
EDP rose, so this point buys energy at a steep delay cost.Static power does not scale with frequency. Stretching delay burns leakage longer, and past some point total energy rises.
Add the lesson-2 leakage of 0.4 W to both points above. The slow point then uses more total energy, which favors racing to halt.
E_total_old = (0.288 W + 0.4 W) x 1 s = 0.688 J
E_total_new = (0.147 W + 0.4 W) x 1.5 s = 0.547 W x 1.5 s = 0.8205 J
The slow point costs more energy once leakage dominates.The faster one. Equal energy times smaller delay gives a smaller product.
No. Power halves but delay doubles, so energy is unchanged. Only voltage cuts lower the energy.