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 192.168.135.0/24 is divided into /26 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 4?
192.168.135.0/24/26subnet 4 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
The network 192.168.137.0/24 is split with VLSM for these LANs: A: 15 hosts, B: 126 hosts, C: 60 hosts, D: 10 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: 15B: 126C: 60D: 10
x =
3.Computer networksMedium
What is the main purpose of a VLAN?
- ATo split one physical switched network into separate broadcast domains
- BTo give each device a public IP address
- CTo increase the speed of every switch port
- DTo encrypt all traffic between switches so other departments cannot read it
4.Computer networksHard
Why does TCP choose an unpredictable initial sequence number for each connection?
- ATo compress the data more efficiently
- BTo encrypt the payload of each segment
- CTo avoid confusion with old segments and make spoofed segments harder to inject
- DTo select the best route to the destination for the segments of that connection
5.Number systemsMedium
Write −84 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −848 bit
U =
6.Number systemsHard
What is the largest unsigned integer N that fits in 20 bits?
20 bit
N =
3Solutions
1.IPv4 subnetting
Answer: x = 255
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
26 = 64
Subnet number 4
network: (4 − 1)·64 = 192
broadcast: 192 + 64 − 1 = 255
Answer
x = 255
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
B: 126 hosts → 27 = 128 (/25)
C: 60 hosts → 26 = 64 (/26)
A: 15 hosts → 25 = 32 (/27)
D: 10 hosts → 24 = 16 (/28)
Allocation, largest first
B: .0 – .127
C: .128 – .191
A: .192 – .223
D: .224 – .239
Answer
x = 224
LAN D: 192.168.137.224/28.
3.Computer networks
Answer: A · To split one physical switched network into separate broadcast domains
Why this is the answer
VLANs group ports logically, so broadcasts stay within each VLAN. This improves security and limits broadcast traffic.
Answer
A · To split one physical switched network into separate broadcast domains
4.Computer networks
Answer: C · To avoid confusion with old segments and make spoofed segments harder to inject
Why this is the answer
Random initial sequence numbers stop delayed segments from old connections being accepted and make it hard for off-path attackers to forge valid segments.
Answer
C · To avoid confusion with old segments and make spoofed segments harder to inject
5.Number systems
Answer: U = 172
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
84 = 0101 0100
inverted: 1010 1011
+1 → 1010 1100
Read it as unsigned
1010 1100₂ = 256 − 84 = 172
Answer
U = 172
6.Number systems
Answer: N = 1.049 × 10⁶
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.
Largest value
220 − 1 = 1048575
Answer
N = 1048575
ExamProblems · examproblems.com/courses/computer-networks