Arrange the following stages of sliding-window flow control using this…

Concept. Sliding-window flow control lets a sender transmit a whole window of frames before it waits on an acknowledgment, instead of stopping after every single frame. The window represents a transmission allowance that may slide forward onto the next set of frames only after an acknowledgment confirms the earlier frames arrived correctly. Retransmission is not a rung of that cycle at all: it is the conditional recovery branch that fires only when an expected acknowledgment fails to arrive.

Application. Applying that principle to the four listed stages, the order in which they occur on a normal pass is:

1. B — Sender sends a window of frames: the cycle can only begin with a transmission, because before it there is nothing to acknowledge and no acknowledged transmission range to advance past.

2. C — Receiver acknowledges the received frames: acknowledgment is the receiver’s response to that transmission, so it is the next event once the frames are on the line.

3. A — Sender moves the window to the next set of frames: the window slides only on the strength of that acknowledgment, which is what makes this stage follow it and not precede it.

4. D — If acknowledgment is not received, sender retransmits the frame: the stem states this stage conditionally, so it is the alternate failure branch of the same acknowledgment check that drives the window forward, and it is listed last rather than inside the successful cycle.

Cross-check against the other orderings. Each of the remaining arrangements breaks the acknowledgment rule at a different point:

• B, A, C, D — slides the window immediately after sending, before any acknowledgment exists to authorise it.

• B, C, D, A — puts retransmission between the acknowledgment and the window move, which would make a recovery action run on every successful cycle even though the stem guards it with “if acknowledgment is not received”.

• B, D, C, A — retransmits before the acknowledgment step is even reached, so the sender would be recovering from a failure it has not yet detected.

Source check. This item is UGC-NET Computer Science (subject code 087), Paper 2, 6 January 2025 Shift 1. NTA’s official final answer key keys Question ID 3767491573 to the arrangement B, C, A, D, and reputable PYQ archives (Testbook, EduRev) reproduce the same stage list with the same key. A small number of secondary compilations print B, C, D, A instead; that reading contradicts the stem’s own conditional wording for stage D, and it is not what the official key records.

Therefore the normal sequence is B, C, A, D.