Stop-and-Wait Minimum Frame Size for 50% Link Utilization

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Nov 4, 2026

In the stop-and-wait ARQ model, the sender transmits one frame of size $L$ bits and then waits for its acknowledgment (ACK) before sending the next frame. With negligible ACK transmission time and negligible processing time at nodes, the main cost is the propagation delay over the link.

Let:

keyword $T_{\text{tx}}$ : time to push a frame onto the wire
keyword $a$ : one-way propagation delay
keyword $RTT$ : time for an entire exchange (frame propagation + ACK propagation)
keyword $U$ : fraction of time the sender is actively transmitting useful frame bits

For stop-and-wait, during each cycle:

Sender transmits for $T_{\text{tx}}$
Then it must wait until the ACK arrives, which (with negligible ACK transmission time) is $2a$

So the cycle time is:

T_{\text{cycle}} = T_{\text{tx}} + 2a

The sender is busy transmitting for $T_{\text{tx}}$ , therefore:

U = \frac{T_{\text{tx}}}{T_{\text{tx}} + 2a}

We need $U \ge 0.5$ .

Key takeaway: larger frames increase $T_{\text{tx}}$ and improve utilization by reducing the relative impact of idle waiting ( $2a$ ).

Compute minimum frame size for $U \ge 0.5$

1
Step 1
Bit rate $R_b = 64\,\text{kbps} = 64{,}000\,\text{b/s}$ . One-way propagation delay $a = 20\,\text{ms} = 0.02\,\text{s}$ . ACK transmission time and processing time are negligible.
2
Step 2
Use $U = \frac{T_{\text{tx}}}{T_{\text{tx}} + 2a}$ . Require $\frac{T_{\text{tx}}}{T_{\text{tx}} + 2a} \ge 0.5$ .
3
Step 3
From $\frac{T_{\text{tx}}}{T_{\text{tx}} + 2a} \ge 0.5$ : $2T_{\text{tx}} \ge T_{\text{tx}} + 2a \Rightarrow T_{\text{tx}} \ge 2a$ .
4
Step 4
Transmission time $T_{\text{tx}} = \frac{L}{R_b}$ (where $L$ is frame size in bits). So $\frac{L}{R_b} \ge 2a \Rightarrow L \ge 2aR_b$ .
5
Step 5
$L_{\min} \ge 2(0.02)(64{,}000) = 2{,}560\,\text{bits}$ .
6
Step 6
$2{,}560\,\text{bits} / 8 = 320\,\text{bytes}$ . The minimum frame size to achieve at least 50% utilization is therefore 320 bytes (or larger).

Utilization threshold for stop-and-wait (ACK transmission negligible)

At 50% utilization, transmission time must be at least twice the one-way propagation delay ( $T_tx \ge 2a$ ).

Interpretation of the 50% condition

When $U \ge 0.5$ , the sender must be transmitting for at least as long as it is waiting for the ACK. Under the stated assumptions, waiting is essentially $2a$ , so the condition becomes $T_{\text{tx}} \ge 2a$ .

Common variations and edge cases

Stop-and-Wait ARQ and Throughput/Utilization Intuition

Knowledge Check

Question 1 of 4

Q1Single choice

In stop-and-wait with negligible ACK transmission and negligible processing, the sender’s idle time after sending a frame is approximately:

T_tx

a + T_tx

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