Why does Go-Back-N use 2^n minus 1 as window size but Selective Repeat uses 2^(n-1)?

In Selective Repeat, if the window were 2^n-1, new and old sequence numbers would overlap in the receiver's buffer causing ambiguity. Halving the window to 2^(n-1) ensures the sender's and receiver's windows never overlap.

Why do we subtract 2 when calculating usable hosts in a subnet?

Every subnet reserves the network address (all host bits 0) and broadcast address (all host bits 1). These cannot be assigned to devices, so usable hosts = 2^h - 2.

Computer Networks Formulas – Complete Quick Reference Sheet

Q: What is the difference between transmission delay and propagation delay?

Transmission delay is the time to push all bits of a packet onto the wire — it depends on packet size and link bandwidth. Propagation delay is the time for a single bit to physically travel from sender to receiver — it depends on distance and medium speed.

Q: How do I convert SNR from dB to linear for Shannon's formula?

SNR_linear = 10^(SNR_dB / 10). For example, 30 dB = 10^3 = 1000. Then use log2(1 + 1000) in Shannon's formula.

Q: What does bandwidth-delay product tell you in practice?

BDP tells you how much data can be in flight in the network at once. If the TCP receive window is smaller than the BDP, the sender must wait for ACKs, wasting link capacity. Fully utilising a high-speed long-distance link requires a TCP window at least equal to the BDP.

What You Will Find Here: Every key formula used in Computer Networks — bandwidth-delay product, propagation delay, throughput, subnetting, TCP window size, Shannon capacity, and more. Each formula is explained with a brief note on when to use it and a worked example.

1. Transmission & Propagation Delay

When you send a file across a network, two independent delays add up before the last bit reaches the destination.

Transmission Delay

Time to push all bits of the packet onto the wire.

Transmission Delay (T_t) = L / B
L = packet size in bits | B = link bandwidth in bps

Example: A 1 MB file (8 × 10⁶ bits) on a 100 Mbps link:
T_t = 8,000,000 / 100,000,000 = 0.08 s = 80 ms

Propagation Delay

Time for one bit to travel from sender to receiver through the medium.

Propagation Delay (T_p) = D / V
D = distance in metres | V = propagation speed (~2 × 10⁸ m/s in copper, ~3 × 10⁸ m/s in fibre/air)

Example: 3000 km fibre link: T_p = 3,000,000 / 3×10⁸ = 10 ms

Total End-to-End Delay (store-and-forward)

Total Delay = n × T_t + T_p
n = number of links (each router stores the full packet before forwarding)

Bandwidth-Delay Product (BDP)

BDP tells you how many bits are “in flight” in the network at any moment — useful for sizing TCP buffers.

BDP = Bandwidth × Round-Trip Time

Example: 100 Mbps link, RTT = 100 ms: BDP = 10⁸ × 0.1 = 10⁷ bits = 10 Mb

2. Throughput & Efficiency

Efficiency

Efficiency = Throughput / Bandwidth
Efficiency = T_t / (T_t + 2T_p) (for stop-and-wait)

Stop-and-Wait Efficiency

With stop-and-wait, the sender waits for an ACK before sending the next packet. Let a = T_p / T_t:

Efficiency_SW = 1 / (1 + 2a)

When propagation delay is large relative to transmission delay (a >> 1), efficiency collapses — this is why sliding window protocols exist.

3. Channel Capacity — Shannon & Nyquist

Shannon’s Theorem (noisy channel)

The theoretical maximum data rate for a channel with noise:

C = B × log₂(1 + SNR)
C = capacity in bps | B = bandwidth in Hz | SNR = signal-to-noise ratio (linear, not dB)

SNR conversion: SNR_dB = 10 × log₁₀(SNR_linear)

Example: 3 kHz telephone channel, SNR = 1000:
C = 3000 × log₂(1001) ≈ 3000 × 10 = 30 kbps

Nyquist’s Theorem (noiseless channel)

C = 2B × log₂(M)
M = number of signal levels

Example: 3 kHz channel, 8 signal levels: C = 2 × 3000 × log₂(8) = 6000 × 3 = 18 kbps

4. Error Detection — Hamming Distance

Hamming Distance

The number of bit positions where two codewords differ.

d(x, y) = number of bit positions where x and y differ

Error Detection vs Correction

To detect up to e errors: d_min ≥ e + 1
To correct up to e errors: d_min ≥ 2e + 1

Hamming Code — Number of Parity Bits

For a message of m data bits, the number of parity bits r must satisfy:

2^r ≥ m + r + 1

Example: m = 4 data bits: try r = 3 → 2³ = 8 ≥ 4 + 3 + 1 = 8. So r = 3 parity bits, total codeword = 7 bits.

CRC — Key Relationship

Remainder = (Message × x^r) mod Generator
r = degree of generator polynomial (number of parity bits appended)

5. Sliding Window Protocols

Sliding window allows the sender to have multiple unacknowledged frames in transit at once.

Window Size for 100% Efficiency

Window Size W ≥ 1 + 2a where a = T_p/T_t

Efficiency with Window Size W

If W ≥ 1 + 2a: Efficiency = 1 (100%)
If W < 1 + 2a: Efficiency = W / (1 + 2a)

Sequence Number Bits

Protocol	Window Size	Sequence Numbers Needed	Bits Required
Stop-and-Wait	1	2	1
Go-Back-N	W_s	W_s + 1 (at minimum)	n bits → W_s = 2ⁿ − 1
Selective Repeat	W_s	2 × W_s	n bits → W_s = 2ⁿ⁻¹

Throughput

Throughput = Efficiency × Bandwidth

6. ALOHA Throughput

Pure ALOHA

S = G × e^−2G
S = throughput (fraction of capacity) | G = offered load

Maximum throughput = 1/(2e) ≈ 18.4% at G = 0.5

Slotted ALOHA

S = G × e^−G

Maximum throughput = 1/e ≈ 36.8% at G = 1

7. IP Subnetting & CIDR

Subnet Mask Basics

Number of hosts per subnet = 2^h − 2
h = number of host bits | subtract 2 for network address and broadcast address

CIDR Notation

192.168.1.0/24 means 24 network bits, 8 host bits
Hosts = 2³²⁻²⁴ − 2 = 2⁸ − 2 = 254 usable hosts

Number of Subnets

Subnets = 2^s where s = subnet bits borrowed from host portion

Common Subnet Quick Reference

CIDR	Subnet Mask	Hosts	Block Size
/30	255.255.255.252	2	4
/29	255.255.255.248	6	8
/28	255.255.255.240	14	16
/27	255.255.255.224	30	32
/26	255.255.255.192	62	64
/25	255.255.255.128	126	128
/24	255.255.255.0	254	256
/16	255.255.0.0	65,534	65,536

Routing — Longest Prefix Match

When a router has multiple matching routes, it always picks the one with the longest (most specific) prefix.

8. TCP — Window, RTT & Throughput

TCP Throughput

TCP Throughput = Window Size / RTT
Window Size in bytes, RTT in seconds → throughput in bytes/sec

RTT Estimation (Jacobson’s Algorithm)

SRTT = (1 − α) × SRTT + α × RTT_sample
DevRTT = (1 − β) × DevRTT + β × |RTT_sample − SRTT|
TimeoutInterval = SRTT + 4 × DevRTT
Typical: α = 0.125, β = 0.25

TCP Slow Start & Congestion Avoidance

Phase	cwnd Growth	Trigger
Slow Start	Doubles each RTT (exponential)	Start / after timeout
Congestion Avoidance	+1 MSS per RTT (linear)	cwnd ≥ ssthresh
Fast Recovery	Linear after triple duplicate ACK	3 duplicate ACKs

TCP Maximum Segment Size (MSS)

MSS = MTU − IP header − TCP header = 1500 − 20 − 20 = 1460 bytes
(for standard Ethernet MTU of 1500 bytes)

TCP Sequence Number Space

Sequence numbers: 0 to 2³² − 1 (32-bit field)
ACK number = next expected byte

9. Application Layer — Key Numbers

Protocol	Port	Transport	Key Detail
HTTP	80	TCP	Stateless; persistent vs non-persistent connections
HTTPS	443	TCP	HTTP over TLS/SSL
DNS	53	UDP (queries) / TCP (zone transfers)	Hierarchical; TTL controls caching
FTP	21 (control), 20 (data)	TCP	Two separate connections
SMTP	25	TCP	Sending mail; push protocol
POP3	110	TCP	Download and delete from server
IMAP	143	TCP	Sync, keep on server
DHCP	67 (server), 68 (client)	UDP	DORA process
Telnet	23	TCP	Unencrypted remote login
SSH	22	TCP	Encrypted remote login
SNMP	161	UDP	Network management

HTTP RTT Formula (Non-Persistent)

Total time = 2 RTT + T_t(file) per object
1 RTT for TCP handshake + 1 RTT for HTTP request/response + file transmission time

HTTP RTT Formula (Persistent with Pipelining)

Total time = 2 RTT + T_t(all objects) (one handshake, then pipeline)

10. Quick Reference Table

Formula	Expression	Unit
Transmission delay	L / B	seconds
Propagation delay	D / V	seconds
Total delay (n links)	n × T_t + T_p	seconds
BDP	BW × RTT	bits
Stop-and-Wait efficiency	1 / (1 + 2a)	—
Sliding window efficiency (W < 1+2a)	W / (1 + 2a)	—
Go-Back-N window (n bits)	2ⁿ − 1	frames
Selective Repeat window (n bits)	2ⁿ⁻¹	frames
Pure ALOHA max throughput	1 / 2e ≈ 18.4%	—
Slotted ALOHA max throughput	1 / e ≈ 36.8%	—
Shannon capacity	B log₂(1 + SNR)	bps
Nyquist capacity	2B log₂(M)	bps
Hamming parity bits	2^r ≥ m + r + 1	bits
Hosts per subnet	2^h − 2	hosts
TCP throughput	W / RTT	bytes/sec
MSS (Ethernet)	1460 bytes	bytes

11. Frequently Asked Questions

What is the difference between transmission delay and propagation delay?

Transmission delay is the time to push all bits of a packet onto the wire — it depends on packet size and link bandwidth. Propagation delay is the time for a single bit to physically travel from sender to receiver — it depends on distance and the speed of the medium. A large file on a slow link has high transmission delay; a short packet across a continent has high propagation delay.

Why does Go-Back-N use 2^n − 1 as window size but Selective Repeat uses 2^(n-1)?

In Selective Repeat, the receiver also maintains a window. If the window size were 2ⁿ−1, new and old sequence numbers would overlap in the receiver’s buffer, causing ambiguity — the receiver cannot tell if an arriving frame is a retransmission of an old frame or a new one. Halving the window to 2^n-1 ensures the sender’s and receiver’s windows never overlap.

How do I convert SNR from dB to linear for Shannon’s formula?

Shannon’s formula needs linear SNR (not dB). Convert using: SNR_linear = 10^{(SNR_dB / 10)}. For example, 30 dB SNR = 10³ = 1000 linear. Then plug 1000 into log₂(1 + 1000) ≈ 10.

Why do we subtract 2 when calculating the number of usable hosts in a subnet?

Every subnet reserves two addresses: the network address (all host bits = 0) and the broadcast address (all host bits = 1). These cannot be assigned to devices. So for a /24 subnet with 256 total addresses, only 254 are usable for hosts.

What does bandwidth-delay product tell you in practice?

BDP tells you how much data can be “in flight” in the network at once. If your TCP receive window is smaller than the BDP, the sender will have to stop and wait for ACKs, wasting link capacity. To fully utilise a high-speed long-distance link, you need a large TCP window — at least equal to the BDP.

What is the efficiency of stop-and-wait at a = 10?

Efficiency = 1 / (1 + 2 × 10) = 1/21 ≈ 4.76%. This shows how poor stop-and-wait is on high-latency links — the sender is idle for 95% of the time waiting for ACKs. Sliding window protocols fix this by allowing multiple frames in flight.