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 10.128.184.0/22 is divided into equal subnets with prefix /24. How many subnets N are there?
10.128.184.0/22/24
N =
2.IPv4 subnettingHard
A router has routes to these networks: 10.161.68.0/24, 10.161.69.0/24, 10.161.70.0/24, 10.161.71.0/24, 10.161.72.0/24, 10.161.73.0/24, 10.161.74.0/24, 10.161.75.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.161.68.0/24 … 10.161.75.0/24
p =
3.Computer networksMedium
What is the usual order of messages when a client obtains an address from DHCP?
- ADiscover, Request, Offer, Acknowledge
- BDiscover, Offer, Request, Acknowledge
- CRequest, Offer, Discover, Acknowledge
- DOffer, Discover, Acknowledge, Request
4.Computer networksHard
Hosts in VLAN 10 and VLAN 20 on the same switch cannot communicate. What is needed?
- AThe same MAC address on both hosts
- BA router or layer 3 switch to route between the VLANs
- CA longer cable between the hosts
- DA hub connected to both VLANs
5.Number systemsMedium
Write −34 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −348 bit
U =
6.Number systemsHard
A system must give every one of 3195 items its own binary code of the same length. What is the smallest number of bits b needed?
3195
b = bit
3Solutions
1.IPv4 subnetting
Answer: N = 4
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).
Borrowed bits
24 − 22 = 2
Number of subnets
N = 22 = 4
Answer
N = 4
Each /24 subnet has 28 − 2 = 254 usable hosts.
2.IPv4 subnetting
Answer: p = 20
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Compare the bits
68 = 01000100
75 = 01001011
common leading bits: 4
Prefix length
p = 16 + 4 = 20
Answer
p = /20
Summary route: 10.161.64.0/20 (it also covers 8 network(s) that are not in the list).
3.Computer networks
Answer: B · 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
B · Discover, Offer, Request, Acknowledge
4.Computer networks
Answer: B · A router or layer 3 switch to route between the VLANs
Why this is the answer
VLANs are isolated at layer 2, so traffic between them must be routed, for example by a router on a trunk ("router on a stick") or a layer 3 switch.
Answer
B · A router or layer 3 switch to route between the VLANs
5.Number systems
Answer: U = 222
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.
Invert and add one
34 = 0010 0010
inverted: 1101 1101
+1 → 1101 1110
Read it as unsigned
1101 1110₂ = 256 − 34 = 222
Answer
U = 222
6.Number systems
Answer: b = 12 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
211 = 2048 < 3195
212 = 4096 ≥ 3195
Answer
b = 12
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