Transmission Delay

In the previous page, we explored propagation delay—the irreducible time governed by physics and geography. Now we turn to transmission delay, the second critical timing component. Unlike propagation delay, transmission delay is controllable through engineering choices.

Transmission delay represents the time required to push all the bits of a frame from the sender's network interface onto the transmission medium. Think of it like pouring water from a bucket onto a conveyor belt: the propagation delay is how long the belt takes to carry the water across the room, but the transmission delay is how long it takes to empty the bucket onto the belt.

Here's the key insight: You can empty the bucket faster with a wider spout (more bandwidth), or you can use a smaller bucket (smaller frame). Both approaches reduce transmission delay. This controllability makes transmission delay a primary design variable in network engineering.

For Stop-and-Wait ARQ, transmission delay is the only time the channel is productively used. The ratio of transmission delay to total cycle time directly determines efficiency. Understanding how to calculate and optimize transmission delay is essential for protocol analysis.

Definition:

Transmission delay (also called emission delay or store-and-forward delay) is the time required to push all bits of a frame onto the transmission medium:

$$T_t = \frac{L}{B}$$

Where:

• Tt = Transmission delay (seconds)
• L = Frame size (bits)
• B = Bandwidth/data rate (bits per second)

Physical Interpretation:

A network interface card (NIC) transmits bits sequentially at the link's data rate. If the bandwidth is 100 Mbps (100 million bits per second), the NIC can push 100 million bits onto the wire every second—or equivalently, one bit every 10 nanoseconds.

For a 12,000-bit frame at 100 Mbps:

$$T_t = \frac{12,000}{100 \times 10^6} = 120 \times 10^{-6} \text{ s} = 120 \text{ μs}$$

The NIC spends 120 microseconds pushing all 12,000 bits onto the wire, one after another.

The Bit Emission Process:

When a sender transmits a frame:

1. t = 0: First bit is placed on the medium
2. t = Tt: Last bit is placed on the medium
3. t = Tt + Tp: First bit arrives at receiver
4. t = Tt + Tp (+ negligible): Last bit arrives at receiver

Note that the last bit arrives only slightly after the first bit because both travel at the same propagation velocity—the "length" of the frame on the wire is Tt × v meters.

Frame Length on the Wire:

At any instant, a frame in transit occupies a physical length on the transmission medium:

$$\text{Frame length on wire} = T_t \times v = \frac{L}{B} \times v = \frac{L \times v}{B}$$

Example:

• 1500-byte frame (L = 12,000 bits)
• 1 Gbps link (B = 10⁹ bps)
• Fiber optic (v = 2 × 10⁸ m/s)

$$\text{Wire length} = \frac{12,000 \times 2 \times 10^8}{10^9} = 2,400 \text{ meters} = 2.4 \text{ km}$$

The entire 1500-byte frame, while in transit, occupies 2.4 km of fiber!

The Inverse Relationship:

Transmission delay is inversely proportional to bandwidth:

$$T_t = \frac{L}{B} \propto \frac{1}{B}$$

Doubling the bandwidth halves the transmission time. This relationship drives the constant pursuit of higher bandwidth in network technology.

Bandwidth Evolution and Impact:

The history of Ethernet illustrates this relationship dramatically:

The Efficiency Paradox Revisited:

Counter-intuitively, increasing bandwidth can decrease Stop-and-Wait efficiency! Here's why:

As bandwidth increases:

• Transmission time (Tt) decreases
• Propagation delay (Tp) remains constant
• The ratio a = Tp/Tt increases
• Utilization U = 1/(1+2a) decreases

Example: 1000 km link, 1500-byte frame

Increasing bandwidth from 1 Mbps to 10 Gbps improves raw capacity 10,000×, but Stop-and-Wait utilization drops from 54.5% to 0.012%—a 4,542× drop in efficiency!

The Direct Relationship:

Transmission delay is directly proportional to frame size:

$$T_t = \frac{L}{B} \propto L$$

Larger frames mean longer transmission times. However, for Stop-and-Wait, larger frames actually improve efficiency:

$$a = \frac{T_p}{T_t} = \frac{T_p \cdot B}{L} \propto \frac{1}{L}$$

Since a decreases with larger L, utilization U = 1/(1+2a) increases.

Intuitive Explanation:

With larger frames:

• The fixed overhead of waiting for ACKs (2Tp) is amortized over more data
• Each round-trip delivers more bytes
• The ratio of productive time to waiting time improves

Example: 100 Mbps link, 10 km distance

Tp = 10,000 / (2 × 10⁸) = 50 μs

Increasing frame size from 100 bytes to 9,000 bytes (jumbo frame) improves throughput from 7.4 Mbps to 87.7 Mbps—an 11.8× improvement!

Frame Size Constraints:

1. Maximum Transmission Unit (MTU):

2. Error Recovery Cost:

If a frame is corrupted, the entire frame must be retransmitted. For larger frames:

• More data is lost per error
• Retransmission takes longer
• The expected cost of errors increases

Error cost analysis:

Expected overhead due to errors = Frame size × Error probability × Retransmission cost

3. Latency Sensitivity:

Large frames monopolize the link during transmission. For real-time traffic:

• A 9,000-byte jumbo frame on a 100 Mbps link takes 720 μs to transmit
• Voice packets waiting behind it experience this as delay
• Interactive applications become sluggish

4. Fragmentation:

When frames exceed path MTU, they must be fragmented:

• Each fragment needs its own headers
• Loss of any fragment requires retransmission of entire original
• Modern protocols (IPv6, path MTU discovery) avoid fragmentation

Bit Time:

The bit time is the duration of a single bit on the transmission medium:

$$T_{\text{bit}} = \frac{1}{B}$$

Transmission delay is simply frame size multiplied by bit time:

$$T_t = L \times T_{\text{bit}} = L \times \frac{1}{B} = \frac{L}{B}$$

Example Bit Times:

Physical Significance:

At 10 Gbps (100 picosecond bit time), light travels only 2 centimeters during one bit time! This illustrates the incredible precision required in high-speed network equipment.

Slot Time in Ethernet:

In CSMA/CD Ethernet, the slot time is critical for collision detection. It represents the maximum time to detect a collision—the time for a signal to propagate to the farthest point and back, plus jam signal:

$$T_{\text{slot}} = \text{Maximum RTT in collision domain} + \text{Jam time}$$

For 10 Mbps Ethernet:

• Slot time = 512 bit times = 51.2 μs
• Maximum network diameter: 2,500 meters

For 100 Mbps Fast Ethernet:

• Slot time = 512 bit times = 5.12 μs
• Maximum network diameter: 250 meters (shrunk 10×!)

For Gigabit Ethernet:

• Slot time would be 512 bit times = 0.512 μs → too small!
• Solution: Carrier extension pads small frames to 512 bytes = 4,096 bits
• Extended slot time = 4.096 μs

Why Slot Time Matters:

For collision detection to work reliably, the sender must still be transmitting when a collision signal returns. The minimum frame size must be at least the slot time:

$$L_{\text{min}} = B \times T_{\text{slot}}$$

For 10 Mbps Ethernet: L_min = 10 × 10⁶ × 51.2 × 10⁻⁶ = 512 bits = 64 bytes

This is why Ethernet has a 64-byte minimum frame size!

When calculating transmission delay, we must consider the total frame size, not just the payload. Every layer adds overhead:

Ethernet Frame Structure:

True Transmission Time:

For a 1500-byte payload:

$$L_{\text{total}} = 8 + 6 + 6 + 2 + 1500 + 4 + 12 = 1538 \text{ bytes} = 12,304 \text{ bits}$$

(Including preamble and IFG)

Overhead percentage: 38/1538 = 2.5%

Layer Encapsulation:

Real network traffic has multiple layers of headers:

Example: HTTPS Web Request Overhead

For a small 100-byte HTTP response payload:

Overhead ratio: 133/233 = 57%

For small payloads, overhead can exceed the actual data!

Efficiency with Overhead:

When analyzing Stop-and-Wait efficiency, consider:

Payload Efficiency:

$$\eta_{\text{payload}} = \frac{L_{\text{payload}}}{L_{\text{total}}}$$

Combined Efficiency (utilization × payload efficiency):

$$\eta_{\text{combined}} = U \times \eta_{\text{payload}} = \frac{T_t}{T_t + 2T_p} \times \frac{L_{\text{payload}}}{L_{\text{total}}}$$

Example:

• 100 Mbps link, 10 km
• 1000-byte payload, 1038 bytes total (with overhead)
• Tp = 50 μs, Tt = 83.04 μs
• Utilization: 45.4%
• Payload efficiency: 96.3%
• Combined efficiency: 43.7%

Both utilization losses and overhead losses compound!

Let's work through several transmission delay calculations with varying complexity.

Example 1: Basic Calculation

A host transmits a 500-byte frame over a 10 Mbps link.

$$T_t = \frac{L}{B} = \frac{500 \times 8}{10 \times 10^6} = \frac{4000}{10^7} = 400 \times 10^{-6} = 400 \text{ μs}$$

Example 2: Unit Conversion Challenge

A 2.5 KB file is transmitted over a 45 Mbps T3 link.

$$L = 2.5 \times 1024 \times 8 = 20,480 \text{ bits}$$ $$T_t = \frac{20,480}{45 \times 10^6} = 455 \times 10^{-6} \approx 455 \text{ μs}$$

Example 3: High-Speed Link

A standard 1500-byte Ethernet frame on a 40 Gbps backbone.

$$T_t = \frac{1500 \times 8}{40 \times 10^9} = \frac{12,000}{4 \times 10^{10}} = 3 \times 10^{-7} = 300 \text{ ns}$$

Example 4: Including Overhead

Calculate the transmission time for a 1460-byte TCP payload over 100 Mbps Ethernet.

Total frame size:

• Ethernet preamble + SFD: 8 bytes
• Ethernet header: 14 bytes
• IP header: 20 bytes
• TCP header: 20 bytes
• Payload: 1460 bytes
• Ethernet FCS: 4 bytes
• IFG: 12 bytes
• Total: 1538 bytes

$$T_t = \frac{1538 \times 8}{100 \times 10^6} = \frac{12,304}{10^8} = 123.04 \text{ μs}$$

Vs. payload-only calculation: $$T_t^{\text{payload}} = \frac{1460 \times 8}{10^8} = 116.8 \text{ μs}$$

Overhead adds 5.3% to transmission time.

Example 5: Comparative Analysis

Compare transmission times for the same 1500-byte frame across different technologies:

Transmission time spans 7 orders of magnitude!

In multi-hop networks, frames are transmitted multiple times—once on each link. This significantly affects end-to-end delay.

Store-and-Forward Networks:

Most network devices (routers, switches) use store-and-forward processing:

1. Receive the entire frame
2. Check for errors (FCS verification)
3. Make forwarding decision
4. Transmit on output port

Each hop adds a complete transmission delay:

$$T_{\text{total}} = \sum_{i=1}^{n} T_t^{(i)} + \sum_{i=1}^{n} T_p^{(i)}$$

For n hops with identical links:

$$T_{\text{total}} = n \cdot T_t + \sum_{i=1}^{n} T_p^{(i)}$$

Example: 3-Hop Path

A packet traverses three 1 Gbps links with distances 100 km, 500 km, and 50 km.

Frame size: 1500 bytes

Transmission delays: $$T_t = 3 \times \frac{12,000}{10^9} = 3 \times 12 \text{ μs} = 36 \text{ μs}$$

Propagation delays: $$T_p = \frac{100,000 + 500,000 + 50,000}{2 \times 10^8} = \frac{650,000}{2 \times 10^8} = 3.25 \text{ ms}$$

Total: 3.25 ms + 36 μs ≈ 3.29 ms

Propagation dominates!

Heterogeneous Networks:

Real networks have links with different bandwidths. Each link contributes its own transmission delay:

$$T_t^{\text{total}} = \frac{L}{B_1} + \frac{L}{B_2} + ... + \frac{L}{B_n}$$

Example: Campus to Internet

Path: Workstation → Building Switch → Core Router → ISP Router → Internet

Total Tt: 12 + 1.2 + 1.2 + 0.12 = 14.52 μs Total Tp: 0.5 + 5 + 250 + 1000 = 1,255.5 μs = 1.26 ms

The slowest link (1 Gbps) dominates transmission delay despite being the shortest!

The Bottleneck Effect:

Overall throughput is limited by the slowest link—the bottleneck:

$$\text{Throughput}_{\text{max}} = \min(B_1, B_2, ..., B_n)$$

Upgrading non-bottleneck links provides no throughput benefit.