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 department needs 72 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
72 hosts
p =
2.IPv4 subnettingHard
The network 192.168.125.0/24 is split with VLSM for these LANs: A: 12 hosts, B: 47 hosts, C: 9 hosts, D: 26 hosts, E: 6 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: 12B: 47C: 9D: 26E: 6
x =
3.Computer networksMedium
What is the usual order of messages when a client obtains an address from DHCP?
- ADiscover, Request, Offer, Acknowledge
- BOffer, Discover, Acknowledge, Request
- CRequest, Offer, Discover, Acknowledge
- DDiscover, Offer, Request, Acknowledge
4.Computer networksHard
What does TCP fast retransmit do?
- AResends every segment twice to be safe
- BSkips the three-way handshake when reconnecting to known hosts, so data is resent sooner
- CResends a segment after three duplicate ACKs, without waiting for the timeout
- DDoubles the window after every loss
5.Number systemsMedium
Set the bits so they store −119 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1198 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1001 1000 OR 1000 1001. What is the result's value R in decimal?
1001 1000OR1000 1001
R =
3Solutions
1.IPv4 subnetting
Answer: p = 25
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).
Host bits needed
26 − 2 = 62 < 72
27 − 2 = 126 ≥ 72 → h = 7
Prefix length
p = 32 − 7 = 25
Answer
p = /25
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
B: 47 hosts → 26 = 64 (/26)
D: 26 hosts → 25 = 32 (/27)
A: 12 hosts → 24 = 16 (/28)
C: 9 hosts → 24 = 16 (/28)
E: 6 hosts → 23 = 8 (/29)
Allocation, largest first
B: .0 – .63
D: .64 – .95
A: .96 – .111
C: .112 – .127
E: .128 – .135
Answer
x = 128
LAN E: 192.168.125.128/29.
3.Computer networks
Answer: D · Discover, Offer, Request, Acknowledge
Why this is the answer
The client broadcasts Discover, servers reply with Offers, the client Requests one offer and the chosen server confirms with an Acknowledge.
Answer
D · Discover, Offer, Request, Acknowledge
4.Computer networks
Answer: C · Resends a segment after three duplicate ACKs, without waiting for the timeout
Why this is the answer
Three duplicate ACKs strongly suggest a single lost segment, so TCP retransmits it at once. This recovers much faster than waiting for the retransmission timer.
Answer
C · Resends a segment after three duplicate ACKs, without waiting for the timeout
5.Number systems
Answer: U = 137
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
119 = 0111 0111
inverted: 1000 1000
+1 → 1000 1001
Read it as unsigned
1000 1001₂ = 256 − 119 = 137
Answer
U = 137
6.Number systems
Answer: R = 153
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
OR gives 1 where at least one bit is 1.
1001 1000
OR 1000 1001
= 1001 1001
Value of the result
1001 1001₂ = 153
Answer
R = 153
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