WIMAX HARQ

HARQ is supported in WiMAX that uses OFDMA physical layer. As mentioned before, WiMAX uses a basic stop-and-wait HARQ protocol. Using multiple HARQ channels can compensate the propagation delay of the stop-and-wait scheme, that is, one channel transmits data while others are waiting for feedbacks. Therefore, using a small number of HARQ channels (e.g., using 6 channels), multichannel stop-and-wait HARQ is a simple, efficient protocol that simultaneously achieves high throughput and low memory requirement.

In this scheme, each HARQ channel is independent of each other; that is, a data burst can only be retransmitted by the HARQ channel that initially sent the data burst. Each HARQ channel is distinguished by a HARQ channel identifier (ACID – ARQ channel identifier). Data reordering in an SS is done by referring to Protocol Data Unit (PDU) sequence numbers that are enabled when the HARQ operation is used. The HARQ Downlink Map Information Element (DL MAP IE) contains the information about the DL HARQ Chase sub-burst IE. It specifies the location of HARQ sub bursts, the ACID, the HARQ Identifier Sequence Number (AI_SN), etc. By referring to MAP IE, an SS can correctly retrieve a given data burst. HARQ feedbacks are sent by the SS after a fixed delay (this is called synchronous feedbacks). To specify the start of a new transmission on each HARQ channel, one-bit HARQ AI_SN is toggled on each successful transmission.

When CC is used in WiMAX HARQ, the total buffer capability needed in a specific SS is determined by the maximum data size and may be transmitted by a single channel at a time, and the total number of HARQ channels provided for the SS (i.e., the product of multiplying these two values). In the IEEE 802.16 Standard, this buffer capacity is used by the BS as a rate control mechanism when sending data to an SS.

1. A Simple Example for WiMAX HARQ

In this section, the multiple-channel stop-and-wait HARQ used by WiMAX is illustrated by a simple example. As mentioned above, these HARQ channels are independent of each other; retransmissions of a data burst can only be done by its initial sending HARQ channel. Thus, a negative acknowledged (NAKed) data burst can only be resent via the initial sending channel until it is successfully received.

Figure 1 shows the channel activities versus time frames of a HARQ scheme, from a BS to a particular SS, assuming that four channels are available for the SS. The first row shows the consecutive time frames. The next four rows show the data bursts sent and the ACK or NAK received from/by each channel at the corresponding time frame. The last row shows the data bursts storing in the SS buffer at the corresponding time frame (waiting to be forwarded to the upper layer). It is assumed that data bursts have either been received correctly or erroneously. Erroneously received data bursts are placed in the receiver’s buffer waiting for the correct copies. On the other hand, correctly received data bursts are placed in the receiver’s buffer waiting to be forwarded further (either to the next hop or to the upper layer) in the right sequence, that is, the receiver is still waiting for a correct copy of some earlier data bursts.

 

Figure 1: A simple example of the WiMAX HARQ scheme.

Figure 1 may be explained below. Data bursts 1 through 4 are sent via channels 1 through 4, at time frames 1 through 4, respectively. A corresponding ACK or NAK will be received by the BS by the end of the 4th time frame after the BS sent out the data burst. So, at the end of 4th time frame, a NAK for data burst 1 is received, and thus the second copy of data burst 1 is sent at time frame 5, via channel 1 (the same channel it was initially sent). Similarly the second copy of data burst 2 is sent at time frame 6 via channel 2. Since an ACK for data burst 3 is received at time frame 6, a new data burst (5) is sent via channel 3 at time frame 7.

The data bursts stored at the receiver’s buffer may be explained as follows. At time frame 4, data bursts 1, 2, and 3 are in the receiver’s buffer. Data bursts 1 and 2 have been erroneously received (and are waiting for correct copies) while data burst 3, even though has been correctly received, is also waiting so that it may be forwarded further in the correct order. Similarly, at time frame 5, data bursts 1 through 4 are in the buffer. At time frame 6, only data bursts 2 through 4 are left in the buffer since data burst 1 has been correctly forwarded. At time 7, after data burst 2 is correctly forwarded, data bursts 3 and 4 are also forwarded, so nothing is left in the buffer. The rest of the time frames may be explained similarly.