Selective Repeat ARQ: Window Rules, Worked Example and Exam Question Patterns

Selective Repeat ARQ sounds simple: retransmit only the missing frame. It gets confusing when sender and receiver windows move independently, frames are buffered, and sequence numbers wrap. Selective Repeat uses independent sender and receiver windows, a half-space rule, and efficiency calculations.

Related reading: Selective Repeat MCQs and Go-Back-N ARQ.

Selective Repeat ARQ: what the protocol is solving

An unreliable data link may lose or damage a frame or ACK. Automatic Repeat reQuest (ARQ) uses numbered frames, ACKs, and retransmission timers. An m-bit sequence field repeats modulo 2^m.

Selective Repeat buffers valid out-of-order frames. The sender tracks ACKs and keeps a timer per outstanding frame. An expiry retransmits only that frame.

Keep three states separate:

• Received: the receiver obtained a valid frame.

• Acknowledged: its ACK reached the sender.

• Delivered: the receiver passed it upward in order.

The Computer Networks guide places these states in the wider data-link context.

Selective Repeat window size and sequence-number rule

An m-bit field provides sequence numbers 0 through 2^m - 1. For equal sender and receiver windows:

W <= 2^(m-1)

More generally, W_s + W_r <= 2^m. Equal windows reduce this to the half-space rule.

For m = 3, the sequence space is 0, 1, 2, 3, 4, 5, 6, 7. Therefore, the largest equal window is W = 2^(3-1) = 4.

Why is W = 5 illegal? After old frames 0 to 4 are accepted, the receive base becomes 5 and its new window is [5,6,7,0,1]. Positions 0 and 1 overlap the old window [0,1,2,3,4]. A delayed old frame 0 now looks like new frame 0 after wraparound.

Selective Repeat ARQ worked example: frame 1 is lost

Let m = 3 and W = 4. Both windows start at [0,1,2,3]. Frames 0,1,2,3 are sent in order. Frame 1 is lost; every other frame and ACK arrives. Each outstanding frame has a timer.

ACKs 2, 3, and 4 cannot move the sender past base 1. ACK 1 closes the gap and exposes the contiguous ACKed run.

Selective Repeat efficiency: a complete numerical

Assume no loss, negligible ACK transmission and processing time, frame transmission time T_t, one-way propagation T_p, and a = T_p/T_t. Ideal utilisation is:

U = min(1, W/(1 + 2a))

A 1,000-byte frame contains 1,000 x 8 = 8,000 bits. On a 1 Mbps link:

T_t = 8,000/1,000,000 = 0.008 s = 8 ms

With T_p = 20 ms, a = 20/8 = 2.5, so 1 + 2a = 1 + 5 = 6. For W = 4:

U = 4/6 = 0.6667, or about 66.7%.

Under the same assumptions, stop-and-wait gives 1/6 = 0.1667, or about 16.7%.

Full ideal utilisation needs W >= 6. A 3-bit field caps legal equal windows at 4. The smallest usable field is m = 4, since 6 <= 2^(4-1) = 8. Loss, ACK overhead, and processing reduce throughput.

Selective Repeat ARQ versus Go-Back-N

With frame 1 lost, Go-Back-N discards later frames and retransmits the suffix from 1. Selective Repeat buffers 2,3,4 and retransmits only 1.

For transport-layer reliability and ordering, continue with TCP vs UDP: Transport Layer Explained. TCP must not be equated mechanically with classroom Selective Repeat.

Selective Repeat ARQ traps that change the answer

• ACK is not delivery. ACK 2 says frame 2 was received, not that it reached the upper layer.

• A gap blocks the sender base. ACKed frames above base 1 cannot slide the sender past unacknowledged frame 1.

• Buffering is not reordering delivery. Frames can arrive out of order, but delivery remains in order.

• Range checks are modulo checks. A window such as [5,6,7,0] is contiguous in a modulo-8 space.

If ACK 1 is lost after the receiver reaches base 5, a duplicate frame 1 must be recognised, ACKed again where required, and never delivered twice.

After every event, write S_base, R_base, the sender ACK bitmap, and receiver buffer. Move a base only across a contiguous run, then choose the timer that expires.

Selective Repeat ARQ exam question patterns

Practise four forms: maximum W from m; minimum m from W; buffering, ACK, delivery, and retransmission traces; and ideal utilisation from frame size, rate, propagation, and window.

Two rapid checks:

1. For m = 4, maximum equal SR window is 2^(4-1) = 8.

2. For required W = 9, minimum m is 5: 9 > 2^(4-1) = 8, while 9 <= 2^(5-1) = 16.

Now lose frame 3 instead. Frames 0,1,2 deliver, frame 4 may be buffered after the window opens, and only 3 is retransmitted. KnowledgeGate has over 20 live practice questions on this topic; use Computer Networks MCQs after reproducing the trace unaided. is retransmitted. KnowledgeGate has over 20 live practice questions on this topic; use Computer Networks MCQs after reproducing the trace unaided.

Selective Repeat ARQ: the short version and next step

• Buffer valid out-of-order frames.

• ACK frames individually.

• Retransmit only what is missing.

• Keep equal windows within half the sequence space.

Delivery remains in order even when arrival and ACK order do not. Use GATE Guidance by Sanchit Sir for a structured route through the wider CS syllabus. For timed, topic-wise practice, use the GATE Test Series.

Redraw the frame-1-loss timeline from memory. Then change m, W, and the lost-frame number, and recompute every window instead of memorising one trace.