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 255.255.255.224. How many usable host addresses H does it have?
255.255.255.224
H =
2.IPv4 subnettingHard
The network 192.168.55.0/24 is split with VLSM for these LANs: A: 26 hosts, B: 21 hosts, C: 6 hosts, D: 5 hosts, E: 11 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: 26B: 21C: 6D: 5E: 11
x =
3.Computer networksMedium
What is the IPv6 loopback address?
- Afe80::1
- B::1
- C::
- Dff02::1
4.Computer networksHard
A student runs a game server on a PC behind a home NAT router. Friends on the internet cannot connect. What is the usual fix?
- AConfigure port forwarding on the router to the PC's private address
- BTurn off the PC's DNS client
- CChange the PC's subnet mask to /32
- DGive the PC a different MAC address so the router can find it from outside
5.Number systemsMedium
The 8-bit pattern 1101 0111 stores a signed integer in two's complement. What is its value N in decimal?
1101 01118 bit
N =
6.Number systemsHard
A system must give every one of 4996 items its own binary code of the same length. What is the smallest number of bits b needed?
4996
b = bit
3Solutions
1.IPv4 subnetting
Answer: H = 30
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
255.255.255.224 = /27
Usable hosts
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 64
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: 26 hosts → 25 = 32 (/27)
B: 21 hosts → 25 = 32 (/27)
E: 11 hosts → 24 = 16 (/28)
C: 6 hosts → 23 = 8 (/29)
D: 5 hosts → 23 = 8 (/29)
Allocation, largest first
A: .0 – .31
B: .32 – .63
E: .64 – .79
C: .80 – .87
D: .88 – .95
Answer
x = 64
LAN E: 192.168.55.64/28.
3.Computer networks
Answer: B · ::1
Why this is the answer
::1 is the loopback address. :: is the unspecified address, fe80::/10 is link-local and ff02::1 is the all-nodes multicast group.
Answer
B · ::1
4.Computer networks
Answer: A · Configure port forwarding on the router to the PC's private address
Why this is the answer
NAT creates mappings only for connections started from inside. A port-forwarding rule maps a public port to the internal host so inbound connections can reach it.
Answer
A · Configure port forwarding on the router to the PC's private address
5.Number systems
Answer: N = −41
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 + 64 + 16 + 4 + 2 + 1 = −41
Answer
N = −41
6.Number systems
Answer: b = 13 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
212 = 4096 < 4996
213 = 8192 ≥ 4996
Answer
b = 13
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