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
A department needs 100 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
100 hosts
p =
2.IPv4 subnettingHard
The network 192.168.57.0/24 is split with VLSM for these LANs: A: 93 hosts, B: 19 hosts, C: 53 hosts, D: 9 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 D?
A: 93B: 19C: 53D: 9
x =
3.Computer networksMedium
Why is UDP often preferred for live video calls?
- AUDP encrypts data automatically
- BUDP guarantees every packet arrives, so the picture never freezes
- CLow delay matters more than retransmitting every lost packet
- DUDP is required for all video
4.Computer networksHard
A user can ping 8.8.8.8 but cannot open any website by name. Where is the problem most likely?
- AName resolution (DNS)
- BThe network card driver
- CThe default gateway
- DThe physical cable
5.Number systemsMedium
The 8-bit pattern 0011 0111 stores a signed integer in two's complement. What is its value N in decimal?
0011 01118 bit
N =
6.Number systemsHard
A system must give every one of 2641 items its own binary code of the same length. What is the smallest number of bits b needed?
2641
b = bit
3Solutions
1.IPv4 subnetting
Answer: p = 25
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).
Host bits needed
26 − 2 = 62 < 100
27 − 2 = 126 ≥ 100 → h = 7
Prefix length
p = 32 − 7 = 25
Answer
p = /25
2.IPv4 subnetting
Answer: x = 224
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
A: 93 hosts → 27 = 128 (/25)
C: 53 hosts → 26 = 64 (/26)
B: 19 hosts → 25 = 32 (/27)
D: 9 hosts → 24 = 16 (/28)
Allocation, largest first
A: .0 – .127
C: .128 – .191
B: .192 – .223
D: .224 – .239
Answer
x = 224
LAN D: 192.168.57.224/28.
3.Computer networks
Answer: C · Low delay matters more than retransmitting every lost packet
Why this is the answer
Retransmitting lost packets would add delay; real-time media tolerates some loss, so UDP-based protocols like RTP are used.
Answer
C · Low delay matters more than retransmitting every lost packet
4.Computer networks
Answer: A · Name resolution (DNS)
Why this is the answer
A successful ping to an external IP proves cabling, addressing and routing work. Failing names point to a DNS server or resolver setting problem.
Answer
A · Name resolution (DNS)
5.Number systems
Answer: N = 55
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.
Add the place values
The other bits keep their normal positive values.
32 + 16 + 4 + 2 + 1 = 55
Answer
N = 55
6.Number systems
Answer: b = 12 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
211 = 2048 < 2641
212 = 4096 ≥ 2641
Answer
b = 12
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