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 1513 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1513 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.209.174/25. What is the fourth octet x of its network address?
10.20.209.174/25
x =
3.Computer networksMedium
How does a stateful firewall differ from a simple packet filter?
- AIt only checks the MAC address of each frame
- BIt encrypts every packet it forwards
- CIt blocks all incoming traffic without exceptions, including replies to outgoing requests
- DIt tracks connections and allows replies that belong to established sessions
4.Computer networksHard
The count-to-infinity problem occurs in which kind of routing protocol, and what helps against it?
- ALink-state; flooding link updates
- BPath-vector; longer AS paths
- CStatic routing; adding default routes
- DDistance-vector; split horizon and poison reverse
5.Number systemsMedium
Set the bits so they store −109 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1098 bit
U =
6.Number systemsHard
What is the largest unsigned integer N that fits in 12 bits?
12 bit
N =
3Solutions
1.IPv4 subnetting
Answer: p = 21
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
210 − 2 = 1022 < 1513
211 − 2 = 2046 ≥ 1513 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.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
27 = 128
Subnet of this host
network: ⌊174/128⌋·128 = 128
broadcast: 128 + 128 − 1 = 255
usable hosts: 129 … 254
Answer
x = 128
3.Computer networks
Answer: D · It tracks connections and allows replies that belong to established sessions
Why this is the answer
A stateful firewall keeps a table of active connections, so return traffic is allowed automatically. A stateless filter judges each packet in isolation.
Answer
D · It tracks connections and allows replies that belong to established sessions
4.Computer networks
Answer: D · Distance-vector; split horizon and poison reverse
Why this is the answer
After a failure, distance-vector routers can keep advertising stale routes to each other, raising the metric slowly. Split horizon, poison reverse and a maximum metric limit this.
Answer
D · Distance-vector; split horizon and poison reverse
5.Number systems
Answer: U = 147
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
109 = 0110 1101
inverted: 1001 0010
+1 → 1001 0011
Read it as unsigned
1001 0011₂ = 256 − 109 = 147
Answer
U = 147
6.Number systems
Answer: N = 4095
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.
Largest value
212 − 1 = 4095
Answer
N = 4095
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