SME
Computer Networks
Layered models
Use this for “A full Ethernet frame carries 1460 bytes of data behind 20-byte TCP and IP headers and 18 bytes of Ethernet framing. What fraction of the frame is data?”
Diagram
frame
=
1,460
+
20
+
20
+
18
=
1,518
B.
efficiency
=
1,460
1,518
=
96.18%.
Each layer adds its header in front of what it carries; the headers are not to scale.
This is an example. Type your own values to start a new problem.
frame (B)
=
payload (B)
+
TCP (B)
+
IP (B)
+
link (B)
efficiency
%
=
100
×
payload (B)
frame (B)
Formulas
frame = payload + TCP + IP + link
1,518 = 1,460 + 20 + 20 + 18
efficiency
=
100
×
payload
frame
96.1792
=
100
×
1,460
1,518
Assumptions
•
TCP and IP headers without options are 20 bytes each; options add up to 40 bytes more.
•
Ethernet adds 18 bytes: a 14-byte header and a 4-byte check sequence (the preamble and gap are left out).
•
Sizes are in bytes, and the payload is the application’s data in one segment.
Step-by-step
WE KNOW
payload = 1,460 B, TCP = 20 B, IP = 20 B, link = 18 B
FIND
frame size (frame), efficiency (efficiency)
Step 1 · Find frame size (frame)
Use frame = payload + TCP + IP + link
Each layer adds its header to what it carries: TCP to the data, IP to the segment, Ethernet to the datagram.
frame = 1,460 + 20 + 20 + 18
frame = 1,518 B
Step 2 · Find efficiency (efficiency)
Use efficiency = 100 × payload ÷ frame
The share of the frame that is the application’s data; × 100 makes it a percent.
efficiency
=
100
×
1,460
1,518
efficiency
=
146,000
1,518
efficiency = 96.1792%
Check
Put the numbers back in. Both sides should match.
1,518 = 1,460 + 20 + 20 + 18 ✓
96.1792
=
100
×
1,460
1,518
✓
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