SME
Computer Networks · Problem type
Throughput limited by the window
Use this for “A TCP connection has a 64 KiB window and a 50 ms round-trip time. What is its highest throughput?”
This is an example. Type your own values to start a new problem.
T (Mb/s)
=
8
×
W (KiB)
RTT (ms)
Formulas
T
=
8
×
W
RTT
10,485,760
=
8
×
65,536
0.05
Formulas are worked in B, s, b/s. The step-by-step shows the conversions.
Assumptions
•
The sender keeps at most one window of unacknowledged bytes in flight, then waits for an acknowledgment.
•
So one window crosses each round trip at best: no losses, and the link is faster than this.
•
The window is in bytes with 1 KiB = 1024 bytes; rates are in bits with 1 Mb/s = 10⁶ bits/s.
Step-by-step
WE KNOW
W = 64 KiB, RTT = 50 ms
FIND
throughput (T)
Step 1 · Convert to the formula’s units
The formulas don’t work directly in the units you chose, so convert first (to B, s, b/s).
W = 64 KiB = 65,536 B (1 KiB = 1024 B)
RTT = 50 ms = 0.05 s (1 s = 1000 ms)
Step 2 · Find throughput (T)
Use T = 8 × W ÷ RTT
One window’s bits arrive each round trip, so divide them by the round-trip time.
T
=
8
×
65,536
0.05
T
=
524,288
0.05
T
=
10,485,760
b/s
Step 3 · Convert the answers
Change each answer back to the unit you chose.
T = 10,485,760 b/s = 10.4858 Mb/s (1 Mb/s = 10⁶ b/s)
Check
Put the numbers back in. Both sides should match.
10,485,760
=
8
×
65,536
0.05
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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