Dynamic voltage scaling based energy-minimized partial task offloading in fog networks

Abstract

With the dynamic voltage scaling (DVS) technology, the terminal node (TN) can dynamically adjust its computational speed, thus providing a new way to save energy during task offloading in fog computing. Focusing on the scenario of one TN and multiple fog nodes (FNs), this paper proposed an Energy-Minimized Partial Task Offloading (EMPTO) scheme for the first time to reduce the overall energy consumption based on DVS technology. Firstly, by modeling the energy consumption and processing delay of task offloading, we formulated the problem of minimizing energy consumption. Then, using the variable substitution method, we transformed this energy minimization problem into a univariate optimization problem about the TN’s computational speed. By solving this problem, EMPTO gets the optimal TN’s computational speed, task offloading size between each pair of TN and FN, and the overall energy consumption. Finally, EMPTO selects the offloading scheme with the lowest overall energy consumption as the final scheme. Theoretical proof and simulation results show that EMPTO can achieve the minimum energy consumption by DVS technology under delay constraint.

Appendices

Appendix A

The energy consumption of the TN to calculate a unit bit of data is expressed as:

$$E_{T}^{C} = k_{T} f_{T}^{ 2} \alpha$$

(23)

It is easy to know that $E_{T}^{C}$ is a monotonically increasing function of the TN’s computational speed f_T.

The delay generated by the TN calculating the unit bit of data is expressed as α/f_T, which is a monotonically decreasing function of f_T.

Therefore, when Q_T takes any value under the constraint of D <  = D_max, we can always reduce the local computing energy consumption by adjusting f_T so that D_T = D_max, thereby minimizing the overall energy consumption.

The above proves Proposition 1.

Appendix B

According to (2) and (10), the processing delay of Q_i is rewritten as:

$$D_{i} = Q_{i} \left( {\frac{1}{{r_{i} }} + \frac{\alpha }{{f_{i} }}} \right) = \left( {Q - \frac{{D_{\max } f_{T} }}{\alpha }} \right)\left( {\frac{1}{{r_{i} }} + \frac{\alpha }{{f_{i} }}} \right)$$

(24)

To ensure that the optimization problem (8) is solvable, it needs to satisfy D ≤ D_max. It has been proved in Proposition 1 that D_T = D_max, then D_i ≤ D_max must be satisfied:

$$D_{i} = \left( {Q - \frac{{D_{\max } f_{T} }}{\alpha }} \right)\left( {\frac{1}{{r_{i} }} + \frac{\alpha }{{f_{i} }}} \right) \le D_{\max }$$

(25)

$$\to \frac{Q}{{D_{\max } }} \le \frac{{f_{T} }}{\alpha } + \left( {\frac{1}{{r_{i} }} + \frac{\alpha }{{f_{i} }}} \right)^{ - 1} \le \frac{{f_{T\max } }}{\alpha } + \left( {\frac{1}{{r_{i} }} + \frac{\alpha }{{f_{i} }}} \right)^{ - 1}$$

(26)

The above proves Proposition 2.

Appendix C

To ensure that problem (8) is solvable, Q_T ≤ Q and D_i ≤ D_max should be satisfied.

According to Q_T ≤ Q, we have

$$Q_{T} = \frac{{D_{\max } f_{T} }}{\alpha } \le Q \to f_{T} \le \frac{\alpha Q}{{D_{\max } }}$$

(27)

According to D_i ≤ D_max, we have

$$D_{i} = \left( {Q - \frac{{D_{\max } f_{T} }}{\alpha }} \right)\left( {\frac{1}{{r_{i} }} + \frac{\alpha }{{f_{i} }}} \right) \le D_{\max }$$

$$\to f_{T} \ge \alpha \left( {\frac{Q}{{D_{\max } }} - \left( {\frac{1}{{r_{i} }} + \frac{\alpha }{{f_{i} }}} \right)^{ - 1} } \right)$$

(28)

We denote the upper and lower bounds of the effective value range of f_T as $\widetilde{{f_{T\max } }}$ and $\widetilde{{f_{T\min } }}$. According to (27), (28) and $0 \le f_{T} \le f_{T\max }$,$\widetilde{{f_{T\max } }}$ and $\widetilde{{f_{T\min } }}$ are expressed as:

$$\widetilde{{f_{T\max } }} = \min \left( {f_{T\max } ,\frac{\alpha Q}{{D_{\max } }}} \right)$$

(29)

$$\widetilde{{f_{T\min } }} = \max \left( {0,\alpha \left( {\frac{Q}{{D_{\max } }} - \left( {\frac{1}{{r_{i} }} + \frac{\alpha }{{f_{i} }}} \right)^{ - 1} } \right)} \right)$$

(30)

The above proves Proposition 3.