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 subnet uses the mask 255.255.240.0. How many usable host addresses H does it have?
255.255.240.0
H =
2.IPv4 subnettingHard
The network 192.168.149.0/24 is split with VLSM for these LANs: A: 36 hosts, B: 11 hosts, C: 4 hosts, D: 20 hosts, E: 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 C?
A: 36B: 11C: 4D: 20E: 10
x =
3.Computer networksMedium
Which IPv4 header field stops packets from circulating forever in a routing loop?
- AType of service
- BHeader checksum
- CIdentification
- DTime to live (TTL)
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?
- AA missing or wrong default gateway
- BThe TCP window is too small
- CA faulty DNS server setting
- DA duplicate MAC address on the internet
5.Number systemsMedium
The 8-bit pattern 1011 0011 stores a signed integer in two's complement. What is its value N in decimal?
1011 00118 bit
N =
6.Number systemsHard
A system must give every one of 3694 items its own binary code of the same length. What is the smallest number of bits b needed?
3694
b = bit
3Solutions
1.IPv4 subnetting
Answer: H = 4094
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).
Prefix length
255.255.240.0 = /20
Usable hosts
212 − 2 = 4094
Answer
H = 4094
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
A: 36 hosts → 26 = 64 (/26)
D: 20 hosts → 25 = 32 (/27)
B: 11 hosts → 24 = 16 (/28)
E: 10 hosts → 24 = 16 (/28)
C: 4 hosts → 23 = 8 (/29)
Allocation, largest first
A: .0 – .63
D: .64 – .95
B: .96 – .111
E: .112 – .127
C: .128 – .135
Answer
x = 128
LAN C: 192.168.149.128/29.
3.Computer networks
Answer: D · 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
D · Time to live (TTL)
4.Computer networks
Answer: A · 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
A · A missing or wrong default gateway
5.Number systems
Answer: N = −77
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.
−128 + 32 + 16 + 2 + 1 = −77
Answer
N = −77
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 < 3694
212 = 4096 ≥ 3694
Answer
b = 12
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