An Energy-Efficient Scheduling Model

Assume that there are n STAs with uplink traffic and the required transmission time for traffic arrivals to the STAs in one SI are independent, identically distributed (i.i.d.). The derivation is not limited to traffic of identical distributions while i.i.d. assumption yields more concise results.

Problem Formulation

Without knowledge of exact buffer statuses of STAs, the AP can only estimate the trans-mission time required by each STA. As a result, the system BU might have to be sacrificed if STAs are put to sleep for energy saving because it is possible that the designated wake-up time of an STA is larger than the actual total transmission time of STAs assigned to transmit before it in some SI. As such, we define the scheduling problem of EE-Multipoll as follows. The notations and their definitions used in this section are listed in Table 5.1. In the

following description, CP-Multipoll represents the ordered-contention multi-polling scheme.

Determine WT₁ ≤ WT₂ ≤ ... ≤ WT_n to maximize energy saving subject to at most x% degradation of bandwidth utilization compared with the CP-Multipoll scheme.

Note that the original CP-Multipoll did not take PM into account. The access scheme, however, can function normally without modifications if the shortest job first policy is adopted. Therefore, to compare the performance on energy saving, we assume that SJF policy is applied and thus an STA keeps awake till the end of its transmission and then enters the Doze state. As a result, the CP-Multipoll corresponds to W Ti = 0 for all i, 1 ≤ i ≤ n, and x = 0. It seems difficult to solve the above optimization problem. In the following, we present a feasible solution with an additional constraint of degrading exactly x% BU for the first i STAs for all i. Since the first STA does not have the overhearing problem, it has W T_i = 0 and can start to transmit after SIFS of the EE-Multipoll frame.

The i^th STA is assigned backoff value of (i − 1) slots.

Consider first the BU of the CP-Multipoll scheme. To simplify analysis, we assume that all traffic arrivals in one SI are served in the following polling period. To compute the BU for the first i STAs, we assume that there are i scheduled STAs and the AP is considered as STA (i + 1) with an assigned backoff value i. The BU for the first i STAs, which is defined as the average time used for transmission by the first i STAs over the access start time of STA (i + 1), can be obtained as

i(1 − p)L

tMP(i) + i(1 − p)L + i · Slot + (i(1 − p) + 1)SIFS . (5.1) Let us now evaluate the BU of our proposed feasible solution. Since the required trans-mission time (with pdf h(t) = p · δ(t) + (1 − p)l(t)) is independent of the transtrans-mission start time (with pdf si(t)), it is clear that ui(t) = p · ui−1(t) + (1 − p)(l(t) ⊗ si(t)), where ⊗ represents the convolution operation and u₀(t) = δ(t). The BU for the first i STAs is also

defined by _t ^i(1−p)L

MP(i)+Si+1 to agree with that of CP-Multipoll.

Solution for the Defined Problem

The process to calculate W T_i can be depicted in two steps.

Step 1. Find the target S_i satisfying the requirement of BU.

Step 2. Solve for W T_i making the S_i as we expected by their relationship.

The derivation, which is shown below, is done iteratively from i = 1 to i = n.

u i-1 (t ) id l b t im e

W T

id le b u sy

= ?

W T i

= ?

Figure 5.5: Illustration of the calculation of WTS.

Consider the first STA with W T₁ = 0 . We have s₁(t) = δ(t − SIFS ) , u₁(t) =p · δ(t)+(1 − p)l(t−SIFS ) , and S₁= SIFS is the same as that for the CP-Multipoll scheme. Now consider the second STA. The BU for the first STA is given by _t ^(1−p)L

MP(1)+S2. Since the BU is allowed to degrade by x% as compared with the CP-Multipoll scheme, we have

tMP(1) + (1 − p)L + Slot + (2 − p)SIFS tMP(1) + S2

= (100 − x)%. (5.2)

Therefore, the target average transmission start time for STA 2, i.e. S2, can be derived. The next step is to find a suitable W T₂ resulting in the target S₂.

Since STA 2 wakes up at time W T₂, we consider the duration spent by STA 1 as W T₂ if its actual finish time is shorter than W T2. Therefore, the average transmission start time

for STA 2 is given by

With the obtained W T₂ and u₁(t), the pdf of the transmission start time for STA 2 if it has data to transmit, s₂(t), is then given by

s₂(t) =

Note that STA 2 spends SIFS + Slot to check the channel status before its transmission.

Therefore, we have s2(t) = 0 for t < W T2+SIFS + Slot because STA 2 cannot start its trans-mission before waking up and sensing the channel being idle for a duration of SIFS + Slot.

For t = W T₂ + SIFS + Slot, we have s₂(t) = U₁(W T₂)δ(t − (W T₂ + SIF S + Slot)) where U₁(W T₂) represents the probability that the time used by STA 1 is less than or equal to W T₂. Finally, for t > W T₂+ SIFS + Slot, it holds that s₂(t) = u₁(t − (SIFS + Slot)) because STA 2 starts to transmit SIFS + Slot after STA 1 finishes its transmission. After s₂(t) and S₂ are obtained, the pdf of average total duration for STAs 1 and 2 to finish their transmissions can be derived as u₂(t) = p · u₁(t) + (1 − p)(l(t) ⊗ s₂(t)). The acquired u₂(t) can be used in

Similar to the derivation of W T₂, we can obtain the average transmission start time for STA

3 as

Similarly, we can solve for W T3 in (5.6) by using the obtained S3 from (5.5).

Likewise, after the W T₃ is derived, the s₃(t) is given by start to transmit before waking up and sensing the channel for SIFS + Slot. The second region, i.e., W T₃ + SIFS + Slot < t < W T₃ + SIFS + 2Slot, represents the situation that the total duration spent by STAs 1 and 2 is in between W T₃ and W T₃+ Slot. For the case that the medium is still busy after STA 3 wakes up, the sensing time for STA 3 actually depends on whether or not STA 2 has data to transmit. If it does not, then the ongoing one is STA 1 and STA 3 has to sense the channel for SIFS + 2Slot, the assigned backoff value plus the duration of one SIFS . On the other hand, if STA 2 has data to transmit, then STA 3 only needs to sense the channel for SIFS + Slot before transmission. For simplicity, we assume that STA 2 has data to transmit in our analysis. For t = W T₃+ SIFS + 2Slot, we have s₃(t) = U₂(W T₃)δ(t − (W T₃+ SIFS + 2Slot)) where U₂(W T₃) denotes the probability that the total duration spent by STAs 1 and 2 is less than or equal to W T₃. Finally, for t > W T₃ + SIFS + 2Slot, we have s₃(t) = u₂(t − (SIFS + Slot)) because STA 3 starts its transmission SIFS +Slot after STAs 1 and 2 finish their transmissions. Again, here we assume

that STA 2 has data to transmit to simplify the analysis. We make similar assumption in deriving sk(t) for k ≥ 4. In other words, we assume that STA (k − 1) has data to transmit if the medium is busy when STA k wakes up. In a real system, the sensing time of STA k, if medium is busy when it wakes up, is equal to SIFS + (k − j)Slot if STA j is the last one to transmit before STA k does.

In general, to derive W T_k, we should first compute the target average transmission start time S_k for STA k by solving

tMP(k−1)+(k−1)(1−p)L+(k−1)Slot+(k−(k−1)p)SIFS

tMP(k−1)+Sk = (100 − x)%. (5.8)

The obtained value of S_k is then used to derive W T_k by solving

Sk = Uk−1(WTk)(WTk + SIFS + (k − 1)Slot))

The bisection method can be adopted to solve for W T_i. The initial values for the lower bound and the upper bound can be selected as W T_i−1 and S_i − SIFS − Slot, respectively.

In real implementation, the W Ti can only be represented by finite precisions. Therefore, the solving process of W Ti is stopped when both the precision has been met and the resulted degradation of BU is smaller than the constraint of our defined optimization problem.

Retransmission caused by channel error can be easily incorporated in the computation of WTS. Let A be the random variable for the transmission time of an STA without channel error. Assume that frame error probability is q. All we need to do is to compute the WTS with the modified random variable B = (1 + q)A. Moreover, since the hardware delay should be taken into account in a real system, W T_i will be compared with the hardware delay D once it is derived:

W T_i =

½W T_i, if W T_i > D.

0, if W T_i ≤ D. (5.11)