Efficiency of a 16-Byte TCP Transmission

Given: useful data = 16 bytes; TCP header = 20 bytes (no options); IP header = 20 bytes (no options); Ethernet header = 6 (destination MAC) + 6 (source MAC) + 2 (type) = 14 bytes and trailer (FCS) = 4 bytes.

Efficiency = useful data ÷ total bytes transmitted on that layer.

1. Efficiency at the TCP level

$$\eta_{TCP} = \frac{16}{16 + 20} = \frac{16}{36} = 0.4444$$

η_TCP = 44.4 % (overhead 20 bytes of TCP header).

2. Efficiency at the IP level

The IP packet carries the TCP header + data:

$$\eta_{IP} = \frac{16}{16 + 20 + 20} = \frac{16}{56} = 0.2857$$

η_IP = 28.6 % (overhead = 20-byte IP header + 20-byte TCP header = 40 bytes).

3. Efficiency at the data link level (Ethernet)

The Ethernet frame = 14-byte header + (IP packet = 56 bytes) + 4-byte FCS:

$$\eta_{DLL} = \frac{16}{16 + 20 + 20 + 14 + 4} = \frac{16}{74} = 0.2162$$

η_DLL = 21.6 %

Note: the frame payload is 16 + 20 + 20 = 56 bytes, which is above Ethernet's 46-byte minimum payload, so no padding is added — 74 bytes is the true frame size.

If the 8-byte preamble and the 12-byte interframe gap on the wire are also counted: $\eta = \frac{16}{74+20} = \frac{16}{94} = 17.0\%$.

Summary

Level	Useful bytes	Total bytes	Efficiency
TCP	16	36	44.4 %
IP	16	56	28.6 %
Data link (Ethernet)	16	74	21.6 %

As expected, efficiency falls monotonically as each layer adds its own header to the same 16 bytes of useful data.