Numericals Tx, Tp, RTT, MTU, MSS, MLS

All Formulas Overview

1. Transmission Delay $\boxed{T_x = \frac{{L}{R}$

$T\_x$ : Time to serialize all bits of a packet onto the link:
$L$ = Packet / segment size (bits)
$R$ = Link bandwidth (bits/sec)
$T\_{x,\min} = \frac{L}{R\_{\max\}}$
Effect of segment size:
- Larger segments → higher $T\_x$ , more efficient ❓
- Smaller segments → lower $T\_x$ , higher header overhead ❓

2. Propagation Delay $\boxed{T\_p = \frac{d}{v\}}$

$T\_p$ : Time for one bit to travel from sender to receiver:
$d$ = Distance (m)
$v$ = Propagation speed (~ $2 \times 10^8$ m/s in fiber/copper, $3 \times 10^8$ m/s in air)
$T\_{p,\min} = \frac{d\_{\min\}}{v\_{\max\}}$

3. Total Packet Delay $\boxed{T\_{\text{total\}} = T\_x + T\_p}$

First bit arrival: $T_{\text{first bit}} = T_p$
Last bit arrival: $T_{\text{last bit}} = T_x + T_p$

4. Round Trip Time (RTT)

Ignoring ACK transmission: $\boxed{RTT = 2T\_p}$ ⭐
Including ACK transmission: $\boxed{RTT = 2T\_p + T\_{x,\text{data\}} + T\_{x,\text{ack\}}}$

5. Bandwidth-Delay Product (BDP) $\boxed{\text{BDP} = R \times T\_p}$ ⭐

$BDP$ : Max number of bits in-flight
Minimum TCP window to fully utilize link: $\text{TCP window size} \geq \text{BDP}$
Number of segments in-flight: $n = \frac{\text{BDP\}}{\text{MSS\}}$

6. MTU / MLS / MSS

$MTU$ (Maximum Transmission Unit): Max frame size a link can carry (bytes)
$MLS$ (Maximum Link/Segment Size): Largest segment/frame link allows $\boxed{ \leq MTU}$
$MSS$ (Maximum Segment Size) $\boxed{\text{MSS} = \text{MLS} - \text{IP Header} - \text{TCP Header\}}$
Impact on $T\_x = \frac{\text{Segment Size (including headers)\}}{R}$
CSMA/CD minimum frame size: $\boxed{L\_{\min} = R \times 2T\_p}$ ⭐

7. Store-and-Forward / Multiple Links

1 packet across k links: $\boxed{T = k(T\_x + T\_p)}$
n packets across k links: $\boxed{T = k(T\_x + T\_p) + (n-1)T\_x}$
Cut-through switching (first packet delay): $T\_{\text{first\}} \approx T\_p + \text{header transmission}$

8. CSMA/CD & Stop-and-Wait

Collision detection requirement: $\boxed{T\_x \geq 2T\_p}$
Stop-and-Wait utilization: $\boxed{U = \frac{T\_x}{T\_x + 2T\_p\}}$ ⭐
Minimum frame size for Ethernet / CSMA/CD: $L\_{\min} = R \times 2T\_p$

9. End-to-End Delay Components $\boxed{T\_{\text{end-to-end\}} = \sum T\_x + \sum T\_p + \sum T\_q + \sum T\_{\text{process\}}}$

$T\_x$ = Transmission delay
$T\_p$ = Propagation delay
$T\_q$ = Queuing delay
$T\_{\text{process\}}$ = Processing delay

Shortcut / GATE-Friendly Formulas ⭐

$T\_x$ in ms: $\boxed{T\_x (\text{ms}) = \frac{L(\text{Mb})}{R(\text{Mbps})} \times 1000}$
$T\_p$ in ms (d in km, v ≈ $2 \times 10^8$ m/s): $\boxed{T\_p (\text{ms}) \approx 5 \times d (\text{1000 km})}$
Minimum frame size for collision detection: $\boxed{L\_{\min} = R \times 2T\_p}$
Number of in-flight segments: $\boxed{n = \frac{\text{BDP\}}{\text{MSS\}}}$

Transmission & Propagation Delay

Transmission and propagation delays belong to the Physical Layer, but CSMA/CD and Stop-and-Wait depend on them for correctness and performance — that’s why questions appear under those protocols.

1. Transmission Delay $(Tx)$

Definition: Time required to serialize (push) all bits of a packet onto the link.

$\boxed{T_x = \frac{L}{R}}$

$L$ = Packet size (bits)
$R$ = Link bandwidth / transmission rate (bits/sec)

Key Properties:

Depends on packet size (L) and link (Bandwidth) rate (R)
- Larger packet → higher $(T\_x)$
- Higher bandwidth → lower $(T\_x)$
Independent of distance
Also called serialization delay

Minimum Transmission Time:

Occurs when bandwidth is maximum for a given packet: $T\_{x, \min} = \frac{L}{R\_\text{max\}}$

Example:

$L = 8\text{ Mb},\ R = 4\text{ Mbps} \implies T\_x = \frac{8 \times 10^6}{4 \times 10^6} = 2\ \text{sec}$

2. Propagation Delay $(Tp)$

Definition: Time taken by one bit to propagate from sender to receiver or Time taken for a bit to travel from sender to receiver.

$\boxed{T_p = \frac{d}{v}}$

$d$ = distance (meters)
$v$ = propagation speed in medium $\~2\times10^8$ m/s in copper/fiber, $3\times10^8$ m/s in air

Key Properties:

Depends on distance and medium, not on packet size.
- Long-distance links → $(T\_p)$ dominates
- Short links → $(T\_x)$ dominates
Independent of packet size and bandwidth
Typical values: Fiber/Copper: $( v \approx 2 \times 10^8 ) m/s$ Air/Vacuum: $( 3 \times 10^8 ) m/s$

Minimum Propagation Time:

Occurs when distance is minimum and speed is maximum: $T\_{p, \min} = \frac{d\_\text{min\}}{v\_\text{max\}}$

Example:

$d = 1000\text{ km},\ v = 2 \times 10^8 \text{ m/s} \implies T\_p = \frac{1000 \times 10^3}{2 \times 10^8} = 0.005\ \text{sec}$

Total Time to Send a Packet $(T\_\text{total})$

$\boxed{T_\text{total} = T_x + T_p}$

Observation: ⭐

LAN (short distance, high bandwidth)
→ $( T\_p )$ is very small, $( T\_x )$ may dominate Ex: For LAN: ( $T\_x >> T\_p)$
WAN / Satellite (long distance)
→ $( T\_p )$ dominates Ex: For satellite: $(T\_p >> T\_x)$

Time for First Bit vs Last Bit

First bit arrival time: $T\_{\text{first bit\}} = T\_p$
Last bit arrival time: $T\_{\text{last bit\}} = T\_x + T\_p$ ⭐

This is frequently tested in GATE. GATE often asks: “When does the receiver get the complete packet?”

Bandwidth–Delay Product $(BDP)$ ⭐⭐

Definition: Number of bits that can be present “in-flight” in the link at any instant.

$\boxed{\text{BDP} = R \times T_p}$

Number of bits present in the link simultaneously

Rate at which source sending bits (bits/sec)
   x
Time for one bit to reach destination  (sec)
   =
bits not yet reached destination but midway (bits)

Equals minimum TCP window size to fully utilize the link ⭐

TCP window size ≥ BDP (to fully utilize link)

Reason:
- TCP window size = max bits sender can send without ACK
- If ( TCP window size < BDP ) -> sender stops sending before the “in-flight” capacity of the link is full -> Some link capacity remains idle -> Throughput < R (link rate)

Significance:

Helps design buffer size in routers
Indicates how many bits are “on the wire”

Example:

$R = 10 \text{ Mbps},\ T\_p = 0.01\text{ sec} \implies \text{BDP} = 10^7 \times 0.01 = 10^5 \text{ bits}$

If window < BDP → link underutilized (very common GATE concept)

6. Effect of Packet Segmentation (Pipelining) / Multiple Packets

Case 1: Single link, no store-and-forward

For multiple packets sent back-to-back, total transmission time for (n) packets: $T = nT\_x + T\_p$
If pipelining (like TCP windowing), $(T\_p)$ can be overlapped, so effective total time for large (n): $(T ≈ n \cdot T\_x)$

Case 2: Store-and-Forward, k links (VERY IMPORTANT)

For 1 packet across k links: $T = k(T\_x + T\_p)$
For n packets:
$T = k(T\_x + T\_p) + (n-1)T\_x$

This is one of the most asked GATE formulas.

Case 3: Cut-through switching (rare but asked)

Transmission overlaps
Delay < store-and-forward
First packet delay ≈ $( T\_p + \text{header transmission} )$

7. Delay Components in Networks / End-to-End Delay

Transmission delay: $(T\_x = \frac{L}{R})$
Propagation delay: $(T\_p = \frac{d}{v})$
Queuing delay: Time spent waiting in router queues
Processing delay: Time routers take to process headers

Total end-to-end delay:

$T\_{\text{end-to-end\}} = \sum T\_x + \sum T\_p + \sum T\_q + \sum T\_{\text{process\}}$
$T\_\text{end-to-end} = T\_x + T\_p + T\_\text{q} + T\_\text{process}$

Where:

$( T\_q )$ : Queuing delay (variable, load-dependent)
$( T\_{\text{process\}} )$ : Header processing (small but non-zero)

GATE usually ignores queue & processing unless specified.

8. Round Trip Time (RTT)

Definition : It is the total time taken for a data packet to travel from the sender to the receiver and for the corresponding acknowledgment (ACK) to travel back to the sender.

Ignoring ACK Tx

$\boxed{RTT = 2T_p}$

With ACK transmission:
$\boxed{RTT = 2T_p + T_{x,\text{data}} + T_{x,\text{ack}}}$

Satellite links → RTT dominates TCP performance.

9. Summary

Transmission delay

If L in Mb, R in Mbps:
$T\_x(\text{ms}) = \frac{L}{R} \times 1000$

Propagation delay

At ( $v = 2 \times 10^8$ ) m/s:
$T\_p(\text{ms}) = 5 \times d(\text{in 1000 km})$

Core Difference

Aspect	Transmission Delay ( $T\_x$ )	Propagation Delay ( $T\_p$ )
Depends on	Packet size, bandwidth	Distance, medium
Independent of	Distance	Packet size
Physical meaning	Time to send bits	Time for bits to travel
Controlled by	Sender’s data rate	Speed of signal
Dominates in	High-bandwidth links	Long-distance links

GATE-Style Numericals and Tips

Given: L, R, d, v

Step 1: Convert all units to bits, seconds, meters
Step 2: Compute ( $T\_x = L / R$ )
Step 3: Compute ( $T\_p = d / v$ )
Step 4: Total delay = ( $T\_x + T\_p$ )

Shortcut Formulas for GATE:

( $T\_x$ ) in milliseconds if L in Mb, R in Mbps: $T\_x (\text{ms}) = \frac{L (\text{Mb})}{R (\text{Mbps})} \times 1000$
( $T\_p$ ) in milliseconds if d in km, v in 10^8 m/s: $T\_p (\text{ms}) = \frac{d (\text{km})}{v (10^8 \text{ m/s})} \times 5$ (Approximation commonly used in GATE calculations)

Important Tips:

For minimum $T\_x$ → maximize $R$
For minimum $T\_p$ → minimize $d$ or increase $v$
Bandwidth-delay product tells you how much data can be “in-flight” before acknowledgment

Critical Condition: ( $\boldsymbol{T\_x \ge 2T\_p}$ ) ⭐

This condition appears repeatedly in CSMA/CD and Stop-and-Wait analysis, even though $(T\_x)$ and $(T\_p)$ are Physical Layer delays.

$\boxed{T_x \ge RTT = 2T_p}$

Meaning of the Condition -> The sender must keep transmitting long enough so that a signal (or collision) from the farthest node can travel to the receiver and back before transmission ends.

Why does ( $2T\_p$ ) appear?

( $T\_p$ ): time for signal to travel one way
Worst case:
- Collision occurs at the farthest node
- Collision information must return to sender
  ->Round-trip propagation delay = $2T\_p$

A. Role in CSMA/CD (MOST IMPORTANT USE)

Collision Detection Requirement

For CSMA/CD to detect a collision, the sender must still be transmitting when the collision signal returns.

Hence,

$\boxed{T\_x \ge 2T\_p}$

Consequence:

Determines minimum frame size
$L\_{\min} = R \cdot 2T\_p$

If $(T\_x < 2T\_p)$ :

Sender finishes transmission early
Collision goes undetected ❌
CSMA/CD fails

B. Role in Stop-and-Wait

In Stop-and-Wait, utilization depends on the relation between $(T\_x)$ and $(2T\_p)$ :

$U = \frac{T\_x}{T\_x + 2T\_p}$

Cases:

Relation	Effect
$(T\_x \ll 2T\_p)$	Very low utilization (long idle wait)
$(T\_x = 2T\_p)$	Boundary condition
$(T\_x \gg 2T\_p)$	High utilization

-> Satellite links → ( $2T\_p$ ) dominates → poor Stop-and-Wait performance

Where GATE Uses This Condition

Minimum frame size
Ethernet design
CSMA/CD correctness
Stop-and-Wait utilization
RTT-dominated links

Transmission & MTU / MSS / MLS

MTU (Maximum Transmission Unit)

Maximum frame size that a link layer protocol can carry.
Example: Ethernet → MTU = 1500 bytes.
Limits the maximum IP packet size, which in turn limits TCP segment size / MLS.

MLS (Maximum Link / Segment Size)

Largest allowed segment or frame a link can transmit in a single transmission.
Often determined by MTU.
Ensures minimum transmission time for CSMA/CD / Stop-and-Wait:

$\text{MLS} \geq L_{\text{min}} = R \times 2T_p$

TCP segment size must not exceed MLS; otherwise, fragmentation occurs.

MSS (Maximum Segment Size / Largest Segment Size)

Maximum TCP payload that can fit inside IP packet without fragmentation:

$\text{MSS} = \text{MLS} - \text{IP Header} - \text{TCP Header}$

Example: Ethernet MTU = 1500B → MLS = 1500B, IP header = 20B, TCP header = 20B → MSS = 1460B

Impact on Transmission Delay ( $T\_x$ )

Transmission delay is proportional to segment size ( $L$ ):

$T_x = \frac{L}{R} \text{ , } L = \text{TCP Segment Size (including TCP/IP headers)}$

Smaller segments → smaller $T\_x$ , more header overhead
Larger segments (near MLS) → higher $T\_x$ , fewer headers → better efficiency
Minimum frame size / minimum $T\_x$ in CSMA/CD:

$L_{\text{min}} = R \times 2T_p$

If MTU / MLS < $L\_{\text{min\}}$ , padding is required to detect collisions.

Bandwidth-Delay Product (BDP) & MSS

BDP tells how many bits can be “in-flight” in the link.
TCP window must accommodate multiple segments if MSS < BDP:

$\text{Number of segments in-flight} = \frac{\text{BDP}}{\text{MSS}}$

Large MLS / MSS → fewer segments → better pipelining efficiency.

Propagation Delay vs MTU / Segment Size

Large segments (near MLS / MTU) → higher $T\_x$ → can overlap with $T\_p$ → better link utilization.
Small segments → low $T\_x$ → $T\_p$ dominates → poor Stop-and-Wait or CSMA/CD utilization.

Example Including MTU / MLS

Link: R = 10 Mbps, $T\_p$ = 0.01 s, MTU / MLS = 1500B (12,000 bits)

$T_x = \frac{L}{R} = \frac{12,000}{10,000,000} = 0.0012 \text{ s}$

$T_p = 0.01 \text{ s}$

$\text{BDP} = R \times T_p = 10,000,000 \times 0.01 = 100,000 \text{ bits}$

Number of segments in-flight to fully utilize link:

$n = \frac{\text{BDP}}{L} = \frac{100,000}{12,000} \approx 9 \text{ segments}$

Integration with CSMA/CD / Stop-and-Wait

Minimum frame size: $L\_{\text{min\}} = R \times 2T\_p$
MLS / MTU must allow minimum $L\_{\text{min\}}$ , otherwise padding is required.
Stop-and-Wait utilization improves if segment size ≈ MLS / MTU to maximize $T\_x$ .

Numericals Tx, Tp, RTT, MTU, MSS, MLS

All Formulas Overview

Transmission & Propagation Delay

Critical Condition: ( T_x≥2T_p\boldsymbol{T\_x \ge 2T\_p}T_x≥2T_p ) ⭐

Transmission & MTU / MSS / MLS

Critical Condition: ( $\boldsymbol{T\_x \ge 2T\_p}$ ) ⭐