Chapter 3 Background On WiMedia MAC
3.2 DRP Mechanism
3.2.3 Retransmit procedures in DRP reservations:
In a hard DRP reservation period, if the reservation owner transmit a packet using I-mm ACK or B-ACK but does not receive the acknowledgement frame, it may retransmit the packet within the same reservation period. In a soft DRP reservation period, devices retransmitting a packet shall follow the PCA mechanism with back-off procedure except the reservation owner.
CHAPTER 4
Figure 4 System architecture based on WiMedia MAC
In this paper, the objective is how to obtain the minimal number of reserved medium access slots while achieving the quality of service for video transmission. To achieve the design objective, we propose a system architecture that shows the communication between the WiMedia MAC and applications, depict in Figure 4. We add additional optimization mechanism supporting the minimal reservation based on the WiMedia MAC. To provide maximal throughput in the optimization process, we propose a method to calculate the optimal payload length according to the bit error rate (BER) produced by the modulation scheme and the signal to noise ratio (SNR) from the lower MAC. The method to obtain the optimal payload length is to calculate the effective throughput by selecting the appropriate packet error rate consisted of the bit error rate and the payload length. Hence, we can obtain the optimal payload length from the maximal effective throughput.
Furthermore, we propose a reservation mechanism supporting the video
transmission from the video stream server. This mechanism integrating the payload length optimization, the MAC layer acknowledgement policy and the video frame packetization provides the minimal number of reserved medium access slots for achieving the target frame error rate from the requirement of MAC layer. The packetization could fragment or pack the frame from the video frame buffer depending on the expected payload length. In this mechanism, we reduce the frame error rate through the retransmission and reduce the overhead in the reserved medium access slots by the optimization combining with the packetization. Finally, we can derive the number of reserved medium access slots from estimating the necessary frame transactions including the retransmission to satisfy the target frame error rate.
In the system architecture, we use the Imm-ACK for our acknowledgement policy. The reason to use Imm-ACK rather than B-ACK is that B-ACK may not guarantee the delay requirement of real-time traffic in the reserved medium access slots each superframe when the interval of retransmission is not chosen correctly. The requirement of using B-ACK for high speed transmission is that the packet error rate needs to be small. Otherwise, the transmitted packets may be seriously dropped due to the high packet error rate. On the other side, we may use the Imm-ACK to protect the transmitting packet even when the channel condition is bad or the packet error rate is large.
CHAPTER 5
PROTOCOL OVERHEAD IN WIMEDIA
In the WiMeida MAC, we find the variation of the payload length and the frame transaction mostly affects the throughput in wireless environment. Hence, this section will describe an overhead of frame exchange in a reservation period.
5.1 Overhead in DRP reservation period
In the hard reservation, no other device except for the reservation target and owner can access the channel, so the overhead is mostly contributed by frame transactions. At the beginning of reserved MASs, the owner should initiate the frame transaction by transmitting request to send (RTS) to the reservation target. While the reservation target receives RTS, it should respond clear to send (CTS) after a period of time. The period of time between RTS and CTS is called the Short Inter-frame Space (SIFS). If the reservation owner receives the CTS correctly, it starts to transmit data packet according to different acknowledgment policy. There are three kinds of acknowledgment policies. The first is the No-ACK policy in which a packet should not be acknowledged by the recipient. The second is the Imm-ACK policy in which the reservation target should respond with an Imm-ACK frame after a period of time, SIFS, while receiving the packet which is shown in Figure 3. The third is the B-ACK in which allows the reservation owner to transmit multiple packet with receiving a single acknowledgement packet from the recipient indicating which packets need to be retransmitted. And the interval between each frame transmitted by reservation owner is minimum inter-frame space (MIFS). All the related parameters are shown in Table III.
Table III WiMedia MAC timing parameters
Parameter Value Description
mGuardTime 12us Guard Time
mMasLength 256us MAS Period
mMaxFragmentCount 8 Maximum Fragment number
pSIFS 10us SIFS Time
6 bits 10 octets 2 octets 6 bits 6 octets 4 bits 4 octets 6 bits
Figure 5 Standard PPDU structure
The physical layer convergence procedure (PLCP) is a frame format for MAC layer controlling the PHY layer. The physical protocol data unit (PPDU) is composed of three components: the PLCP preamble, the PLCP header, and the physical service data unit (PSDU). These components are shown in Figure 5. The PLCP preamble formed by a frame synchronization sequence and a channel estimation sequence is to assist the receiver in timing synchronization, carrier-offset recovery, and channel estimation. The PLCP header is to convey necessary information about the PHY and the MAC to aid in decoding the PSDU at the receiver. The PSDU is formed by
concatenating the payload with the frame check sequence (FCS), tail bits, and pad bits, which are inserted in order to align the data stream on the boundary.
When transmitting a packet, the PLCP preamble is sent first, followed by the PLCP header and then the PSDU. The interval of PLCP preamble defined as a Standard PLCP preamble or a burst PLCP preamble is 9.375us or 5.625us which depends on the frame transaction type. The PLCP header must be sent at a data rate of 39.4Mb/s. The PSDU can be sent at the data rate chosen from 53.3Mb/s, 80Mb/s, 106.7Mb/s, 160Mb/s, 200Mb/s, 320Mb/s, 400Mb/s, and 480Mb/s. The length of PLCP Header mainly composed of PHY header, MAC header, and header check sequence (HCS) is 26 bytes. The length of PSDU varies according to the frame payload from 0-4095 bytes.
CHAPTER 6
OPTIMIZE THE PAYLOAD LENGTH IN DRP
In this section, we formulate a mathematical model to analyze the throughput variation with the payload length changed in DRP and find the optimal payload length to reach the maximal throughput.
6.1 Packet Success Rate (PSR)
If the locations of the transmitter and the receiver are fixed, the channel condition is relative stable. The packet success rate is constant and depends on the payload length, PHY mode, and SNR. The packet success rate is given by [6][7]:
Ps γ, L = PPLCPi γ, LPLCP ∗ PPSDUj γ, L (1) where
L: payload length in bytes,
LPLCP: length of PLCP header in bytes,
Ps(): packet success rate (PSR) defined as the probability of receiving a packet correctly at receiver,
PPLCPi (): the probability of receiving a correct PLCP header corresponding to PHY mode i, PPSDUj (): the probability of receiving a right PSDU corresponding to PHY mode j,
γ: signal to noise ratio (SNR).
In AWGN channel, the PSR in WiMedia is
Ps γ, L = 1 − beri γ 8∗LPLCP ∗ 1 − berj γ 8∗LPSDU (2)
where berk() is bit error rate corresponding to PHY mode k.
Because the PLCP header is transmitted at lower data rate (39.6 Mbps) than the rate of PSDU, the error of the PLCP header is small, and the packet success rate is dominated by the probability of receiving a right PSDU. That is
Ps γ, L ≈ PPSDUj γ, L (3) and
Ps γ, L ≈ 1 − berj γ 8∗LPSDU (4)
in AWGN channel, where L=LPLCP.
6.2 Mathematical Model for throughput analysis in DRP with Imm-ACK
Figure 6 Frame transaction analysis with Imm-ACK
We define the throughput as the average number of payload bits per second received correctly in a successive reserved time. The transmission mechanism using Imm-ACK policy in a contiguous reserved time is shown in Figure 6. Using Imm-ACK, the transmission time of each packet contains not only the transmission time of PPDU but also the transmission time of Imm-ACK and two SIFS time. For convenience, we define that T2 is the total transmission time of a packet, as shown in Figure 6. If the payload length of each packet is constant in a continuous time, the number of packets during the time is
N =M ∗ TMAS − T1 − Tguard
T2 (5)
where
M: the number of reserved MAS, TMAS: the interval of a MAS,
TRTS: the time for transmitting a RTS frame, TCTS: the time for transmitting a CTS frame, TPre: the time of preamble,
TPacket : the time consumed by a packet, TAck: the time consumed by Imm-ACK frame, Tguard: guard time,
T1: the transmission time of RTS an CTS, which is equal to 2 ∗ TPre +RLPLCP
PLCP + SIFS T2: 2∗SIFS+2∗ TPre+8∗LPLCPRPLCP)+LPSDURPSDU
RPSDU: data rate corresponding to PHY mode, RPLCP: data rate (39.6Mbs) to transmit PLCP header
For the environment where the transmitter and the receiver’s locations are fixed, the PSR of each packet is the same, so the number of correct received packets X could be considered a random variable with the binomial distribution. The probability mass function of X, B(X, N, Ps γ, L ) is
B(X, N, Ps γ, L ) = N
X Ps γ, L x 1 − P γ, L 𝑁−𝑋 (6) The throughput in a successive reserved time is given by:
Th =N ∗ Ps γ, L ∗ (8 ∗ L)
M ∗ TMAS (7)
To reach the maximal throughput, the optimal payload length could be obtained by solving the following optimization problem:
Max N ∗ Ps γ, L ∗ (8 ∗ L) M ∗ TMAS subject to 0 ≤ L ≤ 4095,
L ∈ N
(8)
This optimization problem is an integer programming problem, and it is hard to solved in MAC. In order to get analytical solution, we approximate it as:
Max Ps γ, L′ ∗ L′ T2
subject to 0 ≤ L′ ≤ 4095
(9)
where L’ is a approximate real number of L.
In AWGN channel, the problem becomes: optimization problem, the analytical solution of optimal L,opt in AWGN channel is
L′opt = −b − b2− 4ac
The approximated solution of optimal payload length in AWGN channel, Lopt is
where round() is a function which output is an nearest integer to the input.
6.3 Numerical Result
The variation of throughput with various bit error rates in AWGN is shown in Figure 7 and Figure 8. If the successive reservation time is fixed, the throughput using the payload length L′opt would be close to the maximal throughput using the payload length Lopt at higher tramission rate RPSDU. Because the packet transmission time T2 is shorter at the higher data rate, the number of successful packets be more, and N′ approximates N very well. With the same reason, N′ approximates N at longer reservation time. In Figure 9 and Figure 10, it shows the variation of packet error rate and L′opt with different RPSDU. If the bit error rate is small, the payload length could be larger, because the packet error rate increases slowly when the payload length gets longer. The number of packets in the continuous interval would becomes more at higher RPSDU, but the packet error rate increases more slowly than it, . Figure 11 and Figure 12 show the throughput and packet error rate with different bit error rate using the optimal payload length at RPSDU = 480Mbps and RPSDU = 200Mbps. In Table IV, we set the target packet error rate must be less than 0.05. It shows its throughput is much less than that with larger packet error rates, because the proportion of overhead is too large. We must consider both the proportion of overhead and the packet error rate to reach the maximal throughput.
Figure 7 Variation of throughput with bit error rate in AWGN using Lopt and L′opt at 480 Mbps and 200Mpswith M=4
Figure 8 Variation of throughput with bit error rate in AWGN using Lopt and L′opt at 480 Mbps and 200Mpswith M=8
Figure 9 Variation of L′opt with different RPSDU
Figure 10 Variation of packet error rate with different RPSDU
Figure 11 Throughput and packet error rate with different bit error rate using the optimal payload length at RPSDU = 480Mbps
Figure 12 Throughput and packet error rate with different bit error rate using the optimal payload length at RPSDU = 200Mbps
Table IV Throughput for different payload length
Reservation MAS Number 8
Bit Error Rate 10−5
RPLCP 480 Mbps 200 Mbps
Packet Error Rate 0.27 0.05 0.23 0.05
Payload Length in bytes 4007 (Lopt) 636 3261 (Lopt) 636
Throughput in Mbps 193 75.5 107.9 61.4
CHAPTER 7
MINIMAL RESERVED TIME FOR HD H.264 VIDEO
In this section, we propose a model to evaluate the minimal reservation time for the wireless HD H.264 video transmission under the error constraint. First, we formulate an optimization problem to calculate the minimal number of packets that we must reserved to satisfy the error constraint before the decoder is empty. Second, we will show that the ratio of the minimal number of reserved packets to the packets to be transmitted approximates the reciprocal of packet success rate, when the number of transmitted packets is large enough. Third, we consider the impact of the payload length on the reservation time and find the optimal payload length to reach the minimal reservation time in AWGN. We would prove that the payload length for minimal reservation time would converge to L,opt if the number of packets to be transmitted is large enough. The condition is always satisfied because the video frame size of the HD H.264 video is large and would be fragmented into a large number of packets.
7.1 Minimal Reservation Time under fixed Packet Error Rate
For the wireless HD H.264 video transmission, the error rate of the video frame per 120-minute HD movie should be less than one [15]. The error rate of the video frame 𝑃𝑒 of this requirement must be less than 1 120 ∗ 60 ∗ f at f fps, where f is the frame rate of the transmitted video. Suppose that a video frame is fragmented into NF packets, and the packets error rate must be less than 1 − 1 − PN F e to satisfy the constraint without retransmission. If NF becomes larger, the packets error rate gets smaller. Because the size of the H.264 HD video frame is large, it could be fragmented to a great number of packets, so the packet error rate must be very small, and the complexity of PHY would be high. We could use the retransmission policy to solve this problem, but we must calculate the exact time for retransmission and reserve it before the deadline of the video frame, or the latency would happen.
Assume that we reserve the time which is enough to transmit NR packet where NR is larger than NF, the error rate of the video frame is the probability that there is less than NF-1 packets transmitting correctly. From Eq. (6), the probability of error is:
PF = NF−1Pi∗ 1 − P NR−i
Although the transmission time of the video frame may be less than the time we reserved, the remaining time would not be wasted, because it could be released in WiMedia. To find the minimal reservation time under the fixed packet error rate, we could find smallest NR and then times the T2 shown in Figure 6. We suppose that
This optimization problem is an integer programming problem. We must solve it and create a table to look up ahead of implementation. If NF and the packet error rate is small, Nmin would be several times of NF . It takes much time to retransmit the error packet. When NF big enough, the ratio Nmin to NF converges to the reciprocal of packet success rate. It is shown below:
If NF → ∞, then Nmin → ∞.
By central limit theroem, X − Nmin ∗ P L Nmin ∗ P L ∗ 1 − P L
~N 0,1 ,
where X is the number of packets receivied correctly.
P X − Nmin ∗ P L
Because the video frame size of HD H.264 video is large and we can hold some video frame in the MAC buffer, the amount of fragmented packets is large enough to make NNmin
F converges to the reciprocal of packet error rate.
7.2 Minimal Reserved Time Analysis
The minimum reservation time is to ensure the QoS of HD video transmission over the error-prone wireless environments. In home environment, the transmission rate is basically stable, in this case, the minimal reserved time is primarily affected by the payload length. To calculate the optimal payload length for achieving the minimal reservation time, following optimization problem is constructed:
Min Nmin(L) ∗ T2 L (15)
where Nmin(L) is minimal reserved number of packets fragmented by payload length L, and T2 L is the packet transmission and ACK response time using payload length L.
where NF L is the number of the fragmented packets of each video frame using payload length L.
The optimization problem (15) is equivalent to:
Max 1
Nmin(L) ∗ T2 L
⇒ Max P L ∗ L
L ∗ NF L ∗ T2 L
Because the video frame size to transmit is constant for this problem, L ∗ NF L is a constant. The optimization problem becomes:
Max P L ∗ L T2(L)
This problem is the same as equation (9) and also an optimization problem to maximize the throughput, so L′opt = Lrop. For HD H.264 video transmission, the condition is always satisfied. We can use L′opt to fragment the video frames and use the equation (15) to calculated the reservation time in AWGN. The reservation time is
TR = Nmin L′opt ∗ T2 L′opt (16)
7.3 Numerical Result
When the number of fragmented packets is getting large, the ratio of minimal reservation packet number t would converge to the reciprocal of packet success rate.
The result is shown in Figure 13. It means that average retransmission times of each packet are about PSR1 .
Figure 13 Variation of minimal reservation packet number to the number of fragmented packets with bit error rate
If we buffer NMB video frames in the MAC and send NMB video frames every NMB ∗ f second, and the decoder buffer holds the same number of video frame, the latency of video would not happen. It means that the deadline of the NMB video frames is NMB ∗ f second, it must reserve enough time including retransmission to transmit NMB video frames before the deadline. Table VI is an example of created table by solving Eq. (14), where NMB is 15 at f=30, and the bit rate of video is from 10 Mbps to 20Mbps, and the range of payload length is from 125 bytes to 4095 bytes.
If NF is not on the table, we can use the linear interpolation to calculate the value of
NMIN
NF , and it could also satisfy the error constraint, because the convexity of the curve in Figure 13. Suppose that the bit rate and the frame rate of the HD H.264 video is 10 Mbps and 30fps respectively and buffer fifteen video frames in the MAC for our following simulation. The detail of parameter of video in our simulation is shown in Table VII.
TABLE VI Table of NNmin
Number of Frames in the MAC Buffer 15
Dead Line of the Video Frames 0.5 second (≈ 7 superframes)
Data Size in the Buffer 5 Mb
In Figure 14, we find that the approximated payload length for maximal throughput L′opt and the payload length Lrop of minimal Reservation time are close, and there is a little difference at small bit error rate, because the number of fragmented packets of the video is small. Figure 15 shows that the number of reserved MAS per superframe. The maximal MAS number which we can use per superframe is 224. From Figure 15, it could also transmit the HD video when the error rate is small such as 4 ∗ 10−4. Figure 16 and Figure 17 show the reserved MAS number with different bit error rates of 480Mbps and 200 Mbps with the error constraints of P = 10−8, 10−12, and 10−14 . We would find the reservation time is almost the same
and we can reserve a little more time to reach much smaller error rate of video frame.
When the bit error is small enough, the reservation time is small, and could support several to transmit HD H.264 video. If we try to let the packet error rate get small, the payload length must be getting small, and the overhead of each packet becomes larger, so the minimal reservation time must be longer to satisfy the error criteria of video
When the bit error is small enough, the reservation time is small, and could support several to transmit HD H.264 video. If we try to let the packet error rate get small, the payload length must be getting small, and the overhead of each packet becomes larger, so the minimal reservation time must be longer to satisfy the error criteria of video