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 255.255.255.248. How many usable host addresses H does it have?
255.255.255.248
H =
2.IPv4 subnettingHard
A host has the address 172.20.237.93/21. What is the third octet x of its network address?
172.20.237.93/21
x =
3.Computer networksMedium
Which IPv4 header field stops packets from circulating forever in a routing loop?
- ATime to live (TTL)
- BHeader checksum
- CType of service
- DIdentification
4.Computer networksHard
A user can ping 8.8.8.8 but cannot open any website by name. Where is the problem most likely?
- AThe default gateway
- BThe network card driver
- CThe physical cable
- DName resolution (DNS)
5.Number systemsMedium
The 8-bit pattern 1000 1010 stores a signed integer in two's complement. What is its value N in decimal?
1000 10108 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1100 0100 << 3. What is the result's value R in decimal?
1100 0100<< 3
R =
3Solutions
1.IPv4 subnetting
Answer: H = 6
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
255.255.255.248 = /29
Usable hosts
23 − 2 = 6
Answer
H = 6
2.IPv4 subnetting
Answer: x = 232
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 in the third octet
23 = 8
Subnet of this host
network: 172.20.232.0
broadcast: 172.20.239.255
Answer
x = 232
3.Computer networks
Answer: A · 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
A · Time to live (TTL)
4.Computer networks
Answer: D · 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
D · Name resolution (DNS)
5.Number systems
Answer: N = −118
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 + 8 + 2 = −118
Answer
N = −118
6.Number systems
Answer: R = 32
Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.
Bit by bit
Shifting left by 3 moves every bit 3 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1100 0100 << 3
= 0010 0000
Value of the result
0010 0000₂ = 32
Answer
R = 32
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