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 192.168.10.0/24 must be divided into at least 5 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/245 subnets
H =
2.IPv4 subnettingHard
The network 192.168.40.0/24 is split with VLSM for these LANs: A: 6 hosts, B: 15 hosts, C: 14 hosts, D: 11 hosts, E: 34 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: 6B: 15C: 14D: 11E: 34
x =
3.Computer networksMedium
What uniquely identifies one end of a TCP connection on the internet?
- AA domain name alone
- BA MAC address alone
- CAn IP address together with a port number
- DA subnet mask and gateway
4.Computer networksHard
A firewall blocks all ICMP. Small web requests work, but large downloads over some paths hang. What is the likely cause?
- ADNS needs ICMP to resolve names
- BTCP cannot open connections without ICMP
- CTLS certificates are sent over ICMP
- DPath MTU discovery fails because "fragmentation needed" messages are blocked
5.Number systemsMedium
The 8-bit pattern 1110 0000 stores a signed integer in two's complement. What is its value N in decimal?
1110 00008 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 0111 0000 << 3. What is the result's value R in decimal?
0111 0000<< 3
R =
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).
Bits to borrow
23 = 8 ≥ 5 → b = 3, /27
Hosts per subnet
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 112
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
E: 34 hosts → 26 = 64 (/26)
B: 15 hosts → 25 = 32 (/27)
C: 14 hosts → 24 = 16 (/28)
D: 11 hosts → 24 = 16 (/28)
A: 6 hosts → 23 = 8 (/29)
Allocation, largest first
E: .0 – .63
B: .64 – .95
C: .96 – .111
D: .112 – .127
A: .128 – .135
Answer
x = 112
LAN D: 192.168.40.112/28.
3.Computer networks
Answer: C · An IP address together with a port number
Why this is the answer
A TCP connection is identified by source IP and port plus destination IP and port. Ports separate the many applications on one host.
Answer
C · An IP address together with a port number
4.Computer networks
Answer: D · Path MTU discovery fails because "fragmentation needed" messages are blocked
Why this is the answer
PMTUD relies on ICMP messages to learn the smaller MTU. If they are dropped, large packets vanish silently, creating a "black hole" connection.
Answer
D · Path MTU discovery fails because "fragmentation needed" messages are blocked
5.Number systems
Answer: N = −32
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 + 32 = −32
Answer
N = −32
6.Number systems
Answer: R = 128
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.
0111 0000 << 3
= 1000 0000
Value of the result
1000 0000₂ = 128
Answer
R = 128
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