IPv4 subnetting
- Hosts per subnet
- 2(32 − prefix) − 2 (/31: 2, /32: 1)
- Block size
- 256 − (mask octet)
- Network address
- IP AND mask
- Broadcast address
- IP OR (NOT mask): all host bits set to 1
- Subnets from n borrowed bits
- 2n
Computer Networks: cram pack
Formula sheets, the key method steps for every topic and 6 practice problems, with full solutions at the back.
1Formulas and methods
Number systems
- Place values
- binary 2k, octal 8k, hex 16k
- To decimal
- Σ dk·bk
- From decimal
- divide by the base, read the remainders bottom-up
- Hex ↔ binary
- one hex digit = 4 bits
- Two’s complement
- invert the bits and add 1
2Practice problems
1.IPv4 subnettingMedium
The network 192.168.166.0/24 is divided into /27 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 8?
192.168.166.0/24/27subnet 8 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A router has routes to these networks: 10.252.137.0/24, 10.252.138.0/24, 10.252.139.0/24, 10.252.140.0/24, 10.252.141.0/24, 10.252.142.0/24, 10.252.143.0/24, 10.252.144.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.252.137.0/24 … 10.252.144.0/24
p =
3.Computer networksMedium
Using even parity, which parity bit is added to the data bits 1011001?
- A0
- B4
- C1
- D2
4.Computer networksHard
Why was the TCP window scale option introduced?
- AUDP needed a way to control its sending rate on congested paths
- BThe 32-bit window field was too large and wasted header space in every TCP segment sent
- CThe 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
- DWithout it, TCP could not detect lost segments on long-distance links
5.Number systemsMedium
Set the bits so they store −95 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −958 bit
U =
6.Number systemsHard
A system must give every one of 4571 items its own binary code of the same length. What is the smallest number of bits b needed?
4571
b = bit
3Solutions
1.IPv4 subnetting
Answer: x = 224
IPv4 subnetting: an address has 32 bits. The prefix /p says how many leading bits identify the network; the remaining 32 − p bits number the hosts. A subnet has 2³²⁻ᵖ addresses, but the first (network address) and last (broadcast address) cannot be given to hosts, so 2³²⁻ᵖ − 2 hosts fit. Exceptions: a /31 (point-to-point link, RFC 3021) uses both of its 2 addresses for hosts, and a /32 is one single address (a single host or a loopback).
Block size
25 = 32
Subnet number 8
network: (8 − 1)·32 = 224
broadcast: 224 + 32 − 1 = 255
Answer
x = 224
2.IPv4 subnetting
Answer: p = 19
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Compare the bits
137 = 10001001
144 = 10010000
common leading bits: 3
Prefix length
p = 16 + 3 = 19
Answer
p = /19
Summary route: 10.252.128.0/19 (it also covers 24 network(s) that are not in the list).
3.Computer networks
Answer: A · 0
Why this is the answer
1011001 contains four 1-bits, already an even number, so the parity bit is 0. With odd parity it would be 1.
Answer
A · 0
4.Computer networks
Answer: C · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
Why this is the answer
The original header allows at most 65,535 bytes of unacknowledged data. Window scaling multiplies this by a power of two so high bandwidth-delay paths can be filled.
Answer
C · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
5.Number systems
Answer: U = 161
Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.
Invert and add one
95 = 0101 1111
inverted: 1010 0000
+1 → 1010 0001
Read it as unsigned
1010 0001₂ = 256 − 95 = 161
Answer
U = 161
6.Number systems
Answer: b = 13 bit
Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.
Powers of two
212 = 4096 < 4571
213 = 8192 ≥ 4571
Answer
b = 13
ExamProblems · examproblems.com/courses/computer-networks