SME
Computer Networks · Problem type
Stop-and-wait utilization
Use this for “A frame takes 0.12 ms to send and the round trip is 20 ms. How busy is a stop-and-wait link?”
Diagram
0.6% of 20.12 ms is 0.12 ms.
This is an example. Type your own values to start a new problem.
U
%
=
100
×
dₜ (ms)
RTT (ms)
+
dₜ (ms)
T
Cycle time
worked out (not typed) · ms
Formulas
U
=
100
×
dₜ
RTT + dₜ
0.5964
=
100
×
0.12
20 + 0.12
T
=
RTT
+
dₜ
20.12 = 20 + 0.12
Assumptions
•
The sender sends one frame, then waits a full round trip for its acknowledgment before the next.
•
No frame is lost, and the acknowledgment is short enough that its sending time is ignored.
•
Utilization U is the share of each cycle spent sending, as a percent.
Step-by-step
WE KNOW
dₜ = 0.12 ms, RTT = 20 ms
FIND
utilization (U), cycle time (T)
Step 1 · Find utilization (U)
Use U = 100 × dₜ ÷ (RTT + dₜ)
Each cycle is one frame’s sending time plus the wait for its acknowledgment; 100 makes the share a percent.
U
=
100
×
0.12
20 + 0.12
U
=
100
×
0.12
20.12
U
=
12
20.12
U
=
0.5964%
Step 2 · Find cycle time (T)
Use T = RTT + dₜ
One cycle sends the frame, then waits the round trip for its acknowledgment.
T
=
20
+
0.12
T
=
20.12
ms
Check
Put the numbers back in. Both sides should match.
0.5964
=
100
×
0.12
20 + 0.12
✓
20.12 = 20 + 0.12 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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