An Illustrative Example

Let us try to solve the problem which was presented in Section 5.2. According to HEALERS (Algorithm 15), we first use deadline partitioning to compute the current time-slice. Since, all tasks are ready for execution, the first time-slice T S₁ = [0,10).

5.5 An Illustrative Example

(a) Schedule for non-migrating tasks)

(b) Schedule for all tasks (c) Final energy-aware schedule for all tasks

Figure 5.1: An example to illustrate our proposed algorithm HEALERS

In order to compute the schedule matrixSM1 atT S1, HEALERS calls Algorithm 16.

First, it creates the lists L₁ and L₂ and initializes them to ∅. Then, it computes the required shares for each task T_i ∈T atT S₁. The computed shares are as follows:

sh[7×4] sh_i,1,1 sh_i,2,1 sh_i,3,1 sh_i,4,1

T₁ 5 13 8 12

T₂ 8 7 10 11

T₃ 11 8 4 9

T4 9 9 8 1

T₅ 9 11 11 14

T₆ 9 6 1 8

T₇ 12 4 7 2

Now, Algorithm 16 sortsL₁ based on computed shares innon-decreasing order, i.e., The sorted listL₁ at time-sliceT S₁ is (hi, j, sh_i,j,1i): h4,4,1i,{h6,3,1i,h7,4,2i,h3,3,4i, h7,2,4i, h1,1,5i, h6,2,6i, h2,2,7i, h7,3,7i, h1,3,8i, h2,1,8i, h3,2,8i, h4,3,8i, h6,4,8i, h3,4,9i, h4,1,9i, h4,2,9i, h5,1,9i, h6,1,9i, h2,3,10i}, h2,4,11i, h3,1,11i, h5,2,11i, h5,3,11i, h1,4,12i, h7,1,12i, h1,2,13i}. Then, Algorithm 16 computes the schedule for non-migrating tasks.

Scheduling Non-migrating Tasks: Algorithm 17 extracts the first element from the list L₁, i.e., h4,4,1i (taskT₄ requires a share of 1 unit on V₄). The taskT₄ can be fully allocated on V₄ and hence, the schedule matrix SM₁[4][4] is updated ash0,1i. Since, T₄ has been completely scheduled on V₄, all entries of T₄, i.e. h4,3,8i, h4,1,9i and h4,2,9i have been removed from L₁. Now, the above process is repeated by extracting the first

element fromL₁, i.e.,h6,3,1i. Since,V₃ can fully accommodateT₆, it has been scheduled onV₃ withSM₁[6][3] =h0,1i. Similarly,T₇ is scheduled (according toh7,4,2i) onV₄,T₃ is scheduled (according to h3,3,4i) on V₃, T₁ is scheduled (according to h1,1,5i) onV₁ andT₂ is scheduled (according toh2,2,7i) onV₂. Now,h5,1,9ibecomes the first element of the list L1. According to this, if T5 is allocated on V1, it requires a share of 9 units.

However, V₁ does not have the capacity to accommodate T₁ and hence, the element h5,1,9i has been inserted into L₂. Similarly, the other three elements corresponding to the task T₅ has been moved to L₂, i.e., h5,2,11i, h5,3,11i and h5,4,14i. Since, the list L₁ becomes empty, the execution moves from Algorithm 17 to Algorithm 18. The schedule matrix SM1[7×4] for non-migrating tasks at T S₁, is as follows:

SM_1[7×4] V₁ V₂ V₃ V₄

T₁ h0,5i 0 0 0

T2 0 h0,7i 0 0

T₃ 0 0 h0,4i 0

T₄ 0 0 0 h0,1i

T₅ 0 0 0 0

T₆ 0 0 h0,1i 0

T₇ 0 0 0 h1,3i

Scheduling of Migrating Tasks: It may be observed thatT₅could not be allocated in the earlier phase and therefore it must be partitioned into multiple chunks and scheduled on more than one core using the SCHEDULE-MIGRATE algorithm (Algorithm 18). It first moves all entries corresponding to T₅ from list L₂ to L₃. Then, it extracts the first element from L₃, i.e. h5,1,9i, and computes the unused capacity of V₁. Since, there is a residual spare capacity uc₁ of 5 on V₁, T₅ has been partially allocated on V₁, SM₁[5][1] =h5,10i. The unallocated share of T₅ with respect to V₁ us_5,1 becomes 9−5

= 4.

Now, Algorithm 18 extracts the next element h5,2,11i fromL3 and computes unal- located shares of T₅ with respect to V₂ (i.e. us_5,2 = 4×1.1/0.9 = 4.8 (approx)). Then, it checks the spare capacity of V₂. Since, the spare capacity ofV₂ (i.e. uc₂ = 3), is not sufficient enough to accommodate the unallocated share ofτ₅ (i.e. 4.8), it has been par- tially allocated on V₂. That is, SM₁[5][2] = h0,3i. The unallocated share of T₅ become

5.5 An Illustrative Example

us_5,2 = 4.8−3 = 1.8. It may be noted that T₂ has already been scheduled at V₂ from 0 to 7. To avoid the overlap withT₅, the schedule ofT₂ is updated as: SM₁[2][2] =h3,10i.

Next, Algorithm 18 extracts the element h5,3,11i from L₃ and checks the spare capacity of V₃. Since, the spare capacity of V₃ (i.e. uc₃ = 5) is sufficient enough to accommodate the unallocated share of T₅ (i.e. us_5,3 = 1.8×1.1/1.1 = 1.8), it has been allocated on V₃. That is, SM₁[5][3] = h3,5i. The tasks that are already scheduled on V₃ are then adjusted to avoid overlaps. The Gantt chart representation of the schedule matrixSM1[7×4](including migrating tasks) for time-sliceT S₁, is depicted in Figure 5.1b.

Energy-aware scheduling: As we can observe from the schedule matrix SM₁ (in Figure 5.1a),V₁ and V₂ do not have any spare capacity in the current time-slice. Hence, COMPUTE-EA-SCHEDULE allowsV₁andV₂to run at their highest available frequency and we are not able to reduce any power consumption at these cores. Hence, total normalized power consumption on V₁ and V₂ (refer Table 5.2): (P_d+P_s +P_on)×20

= (1327 + 714 + 276)×20 = 46.34W. On the other hand, V3 has a spare capacity of 3 time slots. This can be utilized to lower the operating frequency of V3. However, the execution window of T₅ will overlap with its own execution on other cores, if the operating frequency is decreased during its execution. Hence, the spare capacity of 3 units is shared only among the non-migrating tasks T₆ and T₃ which together require 5 time units at frequency f_max. The required frequency (refer Table 5.2) to execute the tasks T₆ and T₃ is f_opt =d(1 + 4)/8e = 0.68. For simplicity of explanation, we are using only df_optein this example but the algorithm uses combination of two frequencies df_opte and bf_optc practically. The modified execution shares of T₆ and T₃ becomes 2 (= 2/0.68) and 6 (= 4/0.68), respectively. Therefore, normalized power consumption on V3: = (655 + 461 + 276)×8 + (1327 + 714 + 276)×2 = 15.77W. In core V4, the required frequency to execute the tasks T₄ and T₇ is f_opt = (1 + 2)/10 = 0.33. Since, f_opt < f_cr (0.41 in our system) , we setf_opt = 0.41. The modified execution shares ofT₄ and T₇ becomes 3 (= 1/0.41) and 5 (= 2/0.41), respectively. V₄ is kept in suspension mode for the remaining 2 slots in the time-slice. Hence, normalized power consumption

on V₄: = (267 + 289 + 276)×8 + (80µW)×2 = 6.65W. The power consumption in the system using energy oblivious strategy would have been (1327 + 714 + 276)×40

= 92.68W. Therefore, the overall percentage of fractional power saved in a system is:

P = (92.68−(46.34 + 15.77 + 6.65))/92.68×100 = 25.80%. The final schedule for the given task set for T S1 is shown in Figure 5.1c. Similarly, the schedule for subsequent time-slices can be computed.