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 10.103.174.0/24 is divided into equal subnets with prefix /29. How many subnets N are there?
10.103.174.0/24/29
N =
2.IPv4 subnettingHard
The network 192.168.133.0/24 is split with VLSM for these LANs: A: 11 hosts, B: 10 hosts, C: 45 hosts, D: 24 hosts, E: 3 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN E?
A: 11B: 10C: 45D: 24E: 3
x =
3.Computer networksMedium
A browser receives HTTP status code 404. What does it mean?
- AThe requested resource was not found
- BThe request succeeded
- CThe server had an internal error
- DThe client must log in first
4.Computer networksHard
How does fragmentation differ in IPv6 compared with IPv4?
- AIPv6 does not allow packets larger than 576 bytes
- BIPv6 fragments at the transport layer inside TCP
- CIPv6 routers fragment every packet larger than 64 bytes before forwarding it on the path
- DRouters never fragment IPv6 packets; only the source can, using path MTU discovery
5.Number systemsMedium
Write −2 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −28 bit
U =
6.Number systemsHard
An uncompressed image is 800 × 600 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
800 × 600 px32 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: N = 32
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).
Borrowed bits
29 − 24 = 5
Number of subnets
N = 25 = 32
Answer
N = 32
Each /29 subnet has 23 − 2 = 6 usable hosts.
2.IPv4 subnetting
Answer: x = 128
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
Block size per LAN
C: 45 hosts → 26 = 64 (/26)
D: 24 hosts → 25 = 32 (/27)
A: 11 hosts → 24 = 16 (/28)
B: 10 hosts → 24 = 16 (/28)
E: 3 hosts → 23 = 8 (/29)
Allocation, largest first
C: .0 – .63
D: .64 – .95
A: .96 – .111
B: .112 – .127
E: .128 – .135
Answer
x = 128
LAN E: 192.168.133.128/29.
3.Computer networks
Answer: A · The requested resource was not found
Why this is the answer
404 Not Found means the server has no resource at that URL. 200 is OK, 401 is Unauthorized and 500 is Internal Server Error.
Answer
A · The requested resource was not found
4.Computer networks
Answer: D · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
Why this is the answer
An IPv6 router drops an oversized packet and returns ICMPv6 Packet Too Big. The source then sends smaller packets or uses a fragment extension header.
Answer
D · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
5.Number systems
Answer: U = 254
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
2 = 0000 0010
inverted: 1111 1101
+1 → 1111 1110
Read it as unsigned
1111 1110₂ = 256 − 2 = 254
Answer
U = 254
6.Number systems
Answer: M = 1875 KiB
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.
Bits and bytes
800·600·32 = 15360000 bit = 1920000 B
Convert to KiB
1920000/1024 = 1875 KiB
Answer
M = 1875 KiB
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