Class-Based Scheduling Algorithm

In real applications, it is likely that there are only a few possible periods to schedule TSs.

Therefore, one can partition TSs into classes such that two TSs are in the same class if and only if they have identical periods of schedule. Assume that there are C classes, called Class 1, Class 2, ..., and Class C . Let p_i represent the period of Class i and n_i the number of TSs in Class i. If n_i = 0, then Class i is considered not exist. For ease of description, we assume that n_i > 0 for all i, 1 ≤ i ≤ C. The notations and their definitions used in this section can also be referred in Table 3.2.

Consider Class i and let e₁, e₂, ..., and e_ni be the TSs in the class. A TS, say, e₁, is selected as the representative of Class i. Let X_i = {x_i,m}^∞_m=−∞ be a sequence of period p_i such that x_i,m= p_i× m for all m. We shall use the representative TS as reference within the class and, therefore, assign X_i as the scheduled instants of TS e₁. Let o_i,s be the intra-class offset of TS e_swith respect to TS e₁. As a result, the scheduled instants for TS e_sis X_i+o_i,s. Let

Xi,s= Xi+ oi,s (3.20)

and

X_i = ∪

1≤s≤ni

X_i,s. (3.21)

Assume that a new TS in Class j is to be scheduled. The impact of Xi to the new TS can be analyzed as follows.

Let Y = {y_m}^∞_m=−∞ be a periodic sequence of period p_j with y_m = p_j × m for all m.

According to the results presented in the previous section, we need to construct a scheduling matrix of size n_i× G_i,j, where G_i,j = gcd(p_i, p_j). The s^th row of the scheduling matrix is [ d(X_i, Y + 0) d(X_i, Y + 1) ... d(X_i, Y + G_i,j − 1) ] circularly shifted to the right by o_i,s positions. By taking the minimum element in each column, we obtain

R(X_i, Y ) = [ d(X_i, Y + 0) d(X_i, Y + 1) ... d(X_i, Y + G_i,j− 1) ]. (3.22)

Note that the optimal SST of the new TS cannot be determined solely by R(Xi, Y ) because there are still TSs in other classes.

Assume that R(X_i, Y ), 1 ≤ i ≤ C, are obtained. We shall use the representative TS of Class 1 as reference of the overall system. Let

X = ∪

1≤i≤CXi. (3.23)

Further, let O_i, 1 ≤ i ≤ C, be the inter-class offset of the Class i representative TS with respect to the Class 1 representative TS. To determine the optimal SST of the new TS, we construct a global scheduling matrix M of size C × GL_j ,where

GL_j = lcm{G_1,j, G_2,j, ..., G_C,j}. (3.24)

The i^th row of M is R(Xi, Y ) repeated for GLj/Gi,j times and then circularly shifted to the right by Oi positions. By taking the minimum element of each column, we obtain

R(X, Y ) = [ d(X, Y + 0) d(X, Y + 1) ... d(X, Y + GL_j− 1) ]. (3.25)

Finally, the optimal scheduled instants of the new TS is given by Y + k^∗, where k^∗ satisfies

k^∗ = arg max

0≤k≤GLj−1d(X, Y + k). (3.26)

If the new TS is considered as the (n + 1)^th TS of Class j, then we update the intra-class offset

o_j,nj+1 = k^∗− O_j. (3.27)

3.3.1 Suggested Implementation Method

To reduce online scheduling complexity, we allocate C pairs of arrays for each class. Again, consider Class i . Denote the k^th element of the j^th pair of arrays by , Ai,j[k] and , Bi,j[k], 0 ≤ k ≤ Gi,j− 1. The array Ai,j stores [ d(Xi, Y + 0) d(Xi, Y + 1) ... d(Xi, Y + Gi,j− 1) ] for Y = {y_m}^∞_m=−∞ with y_m = p_j× m for all m. Note that A_i,j represents the impact of the

representative TS e1 to a new TS of Class j. The array Bi,j stores R(Xi, Y ) and represents the impact of all TSs in Class i to a new TS of Class j. When a new TS in Class j is to be scheduled, the arrays B_i,j, 1 ≤ i ≤ C, are used to construct the global scheduling matrix M as illustrated in Fig. 3.4. All we need to do is taking the minimum of each column and then find the maximum among the minima. The complexity is only C × GL_j comparisons. After the new TS is scheduled, we need to update B_j,l[k] , 1 ≤ l ≤ C and, 0 ≤ k ≤ G_j,l − 1, as

B_j,l[k] = min{B_j,l[k], A_j,l[k − o_j,nj+1]} (3.28)

because the impact of TSs in Class j to a new TS of every class is changed. Here, the index k − oj,nj+1 is performed modulo Gi,j. The complexity of the update process is P_C

l=1Gj,l

comparisons, which is upper bounded by C × p_j comparisons.

When a TS in Class i finishes, we need to update B_i,j[k], 1 ≤ j ≤ C and , 0 ≤ k ≤ G_i,j−1, as follows. Remove the finished TS so that the updated ni, denoted by n⁰_i, becomes ni − 1.

Construct a scheduling matrix of size n⁰_i× Gi,j such that the m^throw is Ai,j circularly shifted to the right by o_i,m positions. The content of B_i,j[k] is updated as the minimum of the k^th column. Of course, a new representative TS is selected before the update process if the finished one was originally the representative TS of Class i. The complexity of the update process is n⁰_i×P_C

j=1G_i,j comparisons.

Again, we break a tie based on column sum of the global scheduling matrix which requires 2(C − 1) additions and one comparison. Note that one can pre-compute A_i,j[k], 1 ≤ i, j ≤ C and , 0 ≤ k ≤ G_i,j− 1. The initial content of B_i,j[k] is set to a sufficiently large value for all j and k if there is no Class i TS, i.e., n_i = 0.

Once the STA has negotiated with the AP a power saving delivery service according to the S-APSD mechanism, it should automatically switch to the Awake state at designated scheduled time every SI unless any further notification. Therefore, the signaling load is saved effectively when compared with legacy power save mode. The start of its established service, i.e., the SST, can be easily found by the AP according to the repetition pattern of

×

Figure 3.4: The illustration of constructing a global scheduling matrix.

the established periodic schedules and the current time since the notion of sequence can be extended to both directions of the time line as shown in our basic idea. Since the service delivered every SI, the delay for the starting of the established service is upper bounded by the value of its SI, rather than the period of the total repetition pattern of different periodic service schedules.

Example 3.2. Assume that C = 2, n1 = 2, n2 = 1, p1 = 6, p2 = 9, and a new TS in Class 2 is to be scheduled. Based on the assumptions, we have G_1,1 = 6, G_1,2 = 3, G_2,1 = 3, and G_2,2 = 9. Besides, the contents of A_i,j[k] are given by A_1,1 = [0 1 2 3 2 1], A_1,2 = [0 1 1]; and A_2,1 = [0 1 1], A_2,2 = [0 1 2 3 4 4 3 2 1]. Let e₁ and e₂ represent the TSs in Class 1 with e₁ being the representative. Also, let f₁ be the representative TS in Class 2. Assume that TS e₁ was scheduled first, followed by TS f₁, and then TS e₂. As a result, when the new TS f₂ is to be scheduled, we have O2 = 1 (which is randomly selected from 1 and 2) and o1,2 = 3

(which is randomly selected among 2, 3, and 5). At this moment, the contents of Bi,j[k] are given by B1,1 = [0 1 1 0 1 1], B1,2 = [0 1 1]; and B2,1 = [0 1 1], B2,2 = [0 1 2 3 4 4 3 2 1].

The global scheduling matrix M =

· 0 1 1 0 1 1 0 1 1 1 0 1 2 3 4 4 3 2

¸ .

Therefore, the value of k^∗ can be chosen as 2, 4, 5, 7, or 8. We select k^∗ = 5 because it has maximum column sum. Then we compute o2,2 = k^∗ − O2 = 4 and update B2,1 = [0 0 1], B_2,2 = [0 1 2 1 0 1 2 2 1].