TCP/IP Transmission Efficiency for a 16-Byte Application Payload

Verified Sources

Nov 4, 2026

To compute “efficiency” at each layer, we use the same core idea:
efficiency = $\dfrac{\text{useful bytes}}{\text{useful bytes}+\text{protocol overhead at that layer}}$ .

Here, the client sends 16 bytes of application data using TCP to a server. We assume:

TCP header has no options (minimum size).
IP header has no options.
Data link uses Ethernet with no extra options beyond the standard Ethernet header/trailer framing overhead.

We will compute efficiencies at:

TCP layer
IP layer
Data link (Ethernet) layer

Key bytes to track

TCP header size: TCP segment includes a fixed header size when options are absent.
IPv4 header size: IPv4 datagrams include a fixed header when options are absent.
Ethernet frame overhead: Ethernet adds a fixed header and trailer (FCS), and the transmitted frame must be considered as the bytes on the wire at the data link layer.

⚠️ Important modeling note: Different instructors/textbooks define “total bytes” slightly differently at Ethernet (e.g., whether to include preamble/IFG, or only header+payload+FCS). This section follows the common “frame bytes excluding preamble/IFG” convention: Ethernet header (14) + payload + FCS (4). (No options are assumed for IP or TCP.)

Compute TCP-layer transmission efficiency (useful bytes vs TCP total bytes)

1
Step 1
Useful bytes = 16 bytes (the application payload carried by TCP).
2
Step 2
TCP overhead = TCP header size with no options (minimum header).
3
Step 3
Efficiency $_{TCP}$ = $\dfrac{16}{16+\text{TCP\_header}}$ .

Let the TCP header length (no options) be 20 bytes (minimum TCP header).
Then:

\text{Efficiency}_{TCP}=\frac{16}{16+20}=\frac{16}{36}\approx 0.4444 \; (44.44\%)

Compute IP-layer transmission efficiency (useful bytes vs IP total bytes)

1
Step 1
Useful bytes at the IP layer = the entire TCP segment bytes carried by IP.
2
Step 2
TCP segment bytes = TCP header + payload = 20 + 16.
3
Step 3
IP overhead = IPv4 header size with no options (minimum IPv4 header).
4
Step 4
Efficiency $_{IP}$ = $\dfrac{\text{TCP segment bytes}}{\text{TCP segment bytes} + \text{IP\_header}}$ .

IPv4 header (no options) is 20 bytes.

TCP segment size:

\text{TCP segment} = 20 + 16 = 36 \text{ bytes}

IP datagram size:

\text{IP datagram} = 20 + 36 = 56 \text{ bytes}

Then:

\text{Efficiency}_{IP}=\frac{36}{56}=\frac{9}{14}\approx 0.6429 \; (64.29\%)

Compute Ethernet data link-layer transmission efficiency (useful bytes vs Ethernet frame bytes)

1
Step 1
Useful bytes = the payload handed to Ethernet from IP, i.e., the IP datagram bytes.
2
Step 2
Ethernet header = 14 bytes; FCS trailer = 4 bytes.
3
Step 3
Frame bytes = 14 + IP datagram + 4.
4
Step 4
Efficiency $_{DL}$ = $\dfrac{\text{IP datagram bytes}}{\text{14 + IP datagram + 4}}$ .

We already computed the IP datagram size as 56 bytes.

Ethernet overhead (typical modeling, excluding preamble/IFG):

Ethernet header: 14 bytes
FCS: 4 bytes

So Ethernet frame total bytes:

\text{Ethernet frame} = 14 + 56 + 4 = 74 \text{ bytes}

Then:

\text{Efficiency}_{DL}=\frac{56}{74}=\frac{28}{37}\approx 0.7568 \; (75.68\%)

Ethernet minimum frame size caveat

Some Ethernet physical-layer models enforce a minimum frame size, which can reduce efficiency when payloads are small. This course section assumes efficiency based on standard header+payload+FCS frame bytes (no minimum-frame padding).

Consistency tip across layers

When comparing TCP/IP vs Ethernet efficiencies, keep the definition of “useful bytes” consistent: at each layer, “useful” should be the exact byte block that layer passes downward/upward, and “total” should include that layer’s header/trailer overhead.

Transmission efficiency by protocol layer (16-byte payload)

Assumes: TCP no options (20B header), IPv4 no options (20B header), Ethernet header 14B + FCS 4B (excluding preamble/IFG and padding).

Definitions used in this calculation

Byte accounting path

TCP encapsulation

Step A

16 bytes payload + 20-byte TCP header => 36 bytes TCP segment."

IP encapsulation

Step B

36-byte TCP segment + 20-byte IPv4 header => 56-byte IP datagram."

Ethernet encapsulation

Step C

56-byte IP datagram + 14-byte Ethernet header + 4-byte FCS => 74-byte Ethernet frame."

Knowledge Check

Question 1 of 3

Q1Single choice

At the TCP layer, what is the transmission efficiency for a 16-byte payload if the TCP header is 20 bytes and there are no TCP options?

$\frac{16}{16+20}=\frac{4}{9}\approx 44.44\%$

$\frac{20}{16+20}=\frac{5}{9}\approx 55.56\%$

$\frac{16}{20}=80\%$

$\frac{20}{36}\approx 55.56\%$

Explore Related Topics

TCP Characterization: Layer, Connection Semantics, and Reliability

Protocol Field (Network Layer) vs Port Numbers (Transport Layer): Purpose, Sizing, and Design Tradeoffs

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

Browse all research articles