Sliding-Window ARQ: Minimum Distinct Sequence Numbers

Verified Sources

Nov 4, 2026

In sliding window ARQ, the number of distinct sequence numbers (typically the size of the sequence-number space modulo size) must be large enough to prevent ambiguity when packets are delayed, reordered, or retransmitted. A standard design rule is that the sequence-number space must exceed the maximum number of “in-flight” (and thus potentially ambiguous) packets between the sender and receiver windows. This is why protocols like Selective Repeat require a larger sequence-number space than Stop-and-Wait.

For a transmitter window size $N$ and a receiver window size $M$ , the minimum number of distinct sequence numbers required to ensure correct operation is: $\boxed{N+M}$
So among the options given, the correct answer is (iii) $M+N$ .

Key idea: the sequence-number space must be big enough so that a sequence number cannot be reused while it could still refer to an earlier (not-yet-finished) transmission that overlaps with the receiver’s acceptance window.

This course section focuses on deriving and validating the $N+M$ rule, and mapping it to the multiple-choice options.

Sliding Window ARQ (Selective Repeat & Sequence Number Space) — Learning Video

Core concepts and vocabulary

ARQ: sender retransmits lost/corrupted frames.
Sliding window: limits outstanding frames.
Sequence-number space: must avoid reuse ambiguity.
Selective Repeat: stricter sequence-number-space needs.

For Selective Repeat, the classic result for equal window sizes is that the sequence-number space must be at least twice the window size, commonly stated as $2W$ when sender and receiver windows are both $W$ . This prevents the receiver from confusing an old delayed frame with a newer one that has reused its sequence number. Such constraints are grounded in the ambiguity window caused by both sender and receiver windows overlapping in time.

Wikipedia - Selective Repeat ARQ - Describes Selective Repeat and the need for sufficient sequence number space to avoid ambiguity (commonly expressed as $2W$ for equal window sizes). ↩

Why the minimum distinct sequence numbers equal $M+N$

1
Step 1
With sender window $N$ , up to $N$ frames may be outstanding (sent but not yet definitively resolved). With receiver window $M$ , the receiver can accept/store up to $M$ frames that fall within its current acceptance range. Delays can cause a receiver to still be “concerned” about previously seen sequence numbers.
2
Step 2
If sequence numbers are reused too early, a late frame from an earlier transmission attempt could be mistaken for a frame in the receiver’s current acceptance window. Correct operation requires that these two ranges (old and new) never collide in sequence-number values.
3
Step 3
In the worst case, the sequence-number reuse would happen after the sender has advanced through its range ( $N$ ) while the receiver’s acceptance window ( $M$ ) still may overlap with delayed frames. Therefore, the maximum number of potentially ambiguous sequence numbers is $N+M$ .
4
Step 4
To prevent collision, the sequence-number space must have at least $N+M$ distinct values. Hence the minimum number of distinct sequence numbers required is $M+N$ .
5
Step 5
Therefore, the correct option is (iii) $M+N$ .

Practical rule of thumb

For sliding-window ARQ with independent sender/receiver windows, provision a sequence-number space large enough to cover the combined uncertainty due to both windows. That yields the $N+M$ minimum distinct sequence numbers rule.

Why smaller sequence number spaces break correctness

If the sequence-number space is too small, an old (delayed) frame can reappear with the same sequence number as a currently valid frame, causing the receiver to accept the wrong frame or discard a correct one—violating ARQ correctness.

Mapping to the provided options

Given transmitter window size $N$ and receiver window size $M$ , the minimum number of distinct sequence numbers required is:

(i) $\min(M,N)$ → too small: doesn’t cover worst-case overlap ambiguity.
(ii) $\max(M,N)$ → still too small for the same reason.
(iii) $M+N$ → matches the required no-collision condition.
(iv) $M\times N$ → far larger than necessary; ARQ ambiguity grows additively in this model, not multiplicatively.

Thus the correct answer is (iii) $M + N$ .

To connect with the well-known equal-window special case: if $M=N=W$ , then $M+N=2W$ , aligning with the classic “ $2W$ sequence space” requirement seen in Selective Repeat descriptions.

Wikipedia - Selective Repeat ARQ - Describes Selective Repeat and the need for sufficient sequence number space to avoid ambiguity (commonly expressed as $2W$ for equal window sizes). ↩

Which option matches the required sequence-number space?

Compare candidate formulas against the derived requirement $M+N$ .

Sequence numbers across time (conceptual lifecycle)

Window advancement

Sender sends up to N frames

Sender’s sequence numbers progress; multiple frames can be outstanding simultaneously."

Ambiguity window opens

Network delay can reorder/hold frames

Some old frames may arrive after the sender has advanced."

Receiver buffering/acceptance

Receiver acceptance limited to M

Receiver can only accept/store within its current window; wrong reuse breaks correctness."

Correct operation requires $M+N$

Sequence numbers reused safely only with enough distinct values

Sequence-number reuse must not collide with potentially late frames."

Common edge-case checks

Knowledge Check

Question 1 of 4

Q1Single choice

In a sliding window ARQ scheme with transmitter window size $N$ and receiver window size $M$ , the minimum number of distinct sequence numbers required is:

min(M, N)

max(M, N)

M + N

M × N

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