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 192.168.157.0/24 is divided into /25 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 2?
192.168.157.0/24/25subnet 2 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A router has routes to these networks: 10.54.124.0/24, 10.54.125.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.54.124.0/24 … 10.54.125.0/24
p =
3.Computer networksMedium
Which IPv4 header field stops packets from circulating forever in a routing loop?
- AType of service
- BTime to live (TTL)
- CIdentification
- DHeader checksum
4.Computer networksHard
A PC can reach printers on its own subnet but nothing beyond it. Its IP address and mask are correct. What is the most likely fault?
- AThe TCP window is too small
- BA duplicate MAC address on the internet
- CA missing or wrong default gateway
- DA faulty DNS server setting
5.Number systemsMedium
Write −5 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −58 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0110 0101 << 1. What is the result's value R in decimal?
0110 0101<< 1
R =
3Solutions
1.IPv4 subnetting
Answer: x = 128
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
27 = 128
Subnet number 2
network: (2 − 1)·128 = 128
broadcast: 128 + 128 − 1 = 255
Answer
x = 128
2.IPv4 subnetting
Answer: p = 23
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
124 = 01111100
125 = 01111101
common leading bits: 7
Prefix length
p = 16 + 7 = 23
Answer
p = /23
Summary route: 10.54.124.0/23
3.Computer networks
Answer: B · Time to live (TTL)
Why this is the answer
Every router decrements the TTL; at zero the packet is discarded and an ICMP Time Exceeded message is usually sent. IPv6 calls it hop limit.
Answer
B · Time to live (TTL)
4.Computer networks
Answer: C · A missing or wrong default gateway
Why this is the answer
Hosts send off-subnet packets to the default gateway. If it is wrong or missing, local traffic still works but nothing outside is reachable, even by IP address.
Answer
C · A missing or wrong default gateway
5.Number systems
Answer: U = 251
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.
Invert and add one
5 = 0000 0101
inverted: 1111 1010
+1 → 1111 1011
Read it as unsigned
1111 1011₂ = 256 − 5 = 251
Answer
U = 251
6.Number systems
Answer: R = 202
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.
Bit by bit
Shifting left by 1 moves every bit 1 places to the left; bits that fall off the left end are lost and zeros come in on the right.
0110 0101 << 1
= 1100 1010
Value of the result
1100 1010₂ = 202
Answer
R = 202
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