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 192.168.247.0/24 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 network address of subnet 3?
192.168.247.0/24/26subnet 3 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 172.20.195.213/21. What is the third octet x of its network address?
172.20.195.213/21
x =
3.Computer networksMedium
Which IPv4 header field stops packets from circulating forever in a routing loop?
- ATime to live (TTL)
- BIdentification
- CHeader checksum
- DType of service
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 default gateway
- CThe network card driver
- DThe physical cable
5.Number systemsMedium
Set the bits so they store −112 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1128 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0001 1100 << 3. What is the result's value R in decimal?
0001 1100<< 3
R =
3Solutions
1.IPv4 subnetting
Answer: x = 128
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 3
network: (3 − 1)·64 = 128
broadcast: 128 + 64 − 1 = 191
Answer
x = 128
2.IPv4 subnetting
Answer: x = 192
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: 172.20.192.0
broadcast: 172.20.199.255
Answer
x = 192
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: 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: U = 144
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.
Invert and add one
112 = 0111 0000
inverted: 1000 1111
+1 → 1001 0000
Read it as unsigned
1001 0000₂ = 256 − 112 = 144
Answer
U = 144
6.Number systems
Answer: R = 224
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.
0001 1100 << 3
= 1110 0000
Value of the result
1110 0000₂ = 224
Answer
R = 224
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