Exercises — IPv6 — address format, why needed, key differences

WHAT: multiply digits by bits-per-digit. WHY: each hex digit stands for exactly 4 bits ($16=2^4$), so counting digits then multiplying by 4 gives bits. $8\text{ hextets}\times 4\text{ digits}\times 4\text{ bits}=128\text{ bits}.$ Answer: 128 bits.

WHY $2^{128}$: 128 independent on/off choices give $2^{128}$ combinations (multiplication principle: 2 options per bit, 128 bits). $2^{128}\approx 3.4\times 10^{38}.$ Answer: $2^{128}\approx 3.4\times 10^{38}$ addresses.

Loopback = "talk to myself", the address 0000:0000:0000:0000:0000:0000:0000:0001. All 7 leading hextets are zero → collapse with ::, last hextet 0001 drops leading zeros to 1. Answer: ::1. (:: alone is the all-zeros unspecified address, not loopback.)

Go hextet by hextet, remove zeros on the left of each group; a group of all zeros becomes a single 0.

• 2001 → 2001
• 0db8 → db8
• 0000 → 0
• 0042 → 42
• 0000 → 0
• 8a2e → 8a2e
• 0370 → 370
• 7334 → 7334 Answer: 2001:db8:0:42:0:8a2e:370:7334.

Step 1 (leading zeros): fe80:0:0:0:204:61ff:fe9d:f156. Step 2 (::): hextets 2,3,4 are all zero — a single run of length 3. Replace that run with ::. Answer: fe80::204:61ff:fe9d:f156. This fe80::/10 is a link-local address (auto-assigned, not routable — see DHCP and SLAAC).

:: means "insert enough all-zero hextets to reach 8 groups". We currently see 2001, db8, and 1 = 3 non-zero hextets, so the gap holds $8-3=5$ zero hextets, placed where the :: sits. Then pad each group to 4 digits. Answer: 2001:0db8:0000:0000:0000:0000:0000:0001.

Leading zeros first: 2001:db8:0:0:ab:0:0:1234. Find zero-runs: run A = hextets 3–4 (length 2); run B = hextets 6–7 (length 2). A tie. Rule: :: may appear only once. On a tie, compress the first longest run; write the other explicitly. Answer: 2001:db8::ab:0:0:1234. 2001:db8::ab::1234 is illegal (two :: → the parser can't split the zeros between the gaps).

Leading zeros: ff02:0:0:0:0:0:0:1. Zero-runs: hextets 2–7, one run of length 6 (the longest possible non-trivial run here). Collapse all six with ::. Answer: ff02::1. (This is the "all-nodes" multicast group — see Multicast vs Broadcast vs Anycast. IPv6 uses this instead of IPv4 broadcast.)

• (a) Illegal — two ::. Ambiguous: the parser can't tell how many zero-hextets go in each gap.
• (b) Legal — the unspecified address, all 128 bits zero.
• (c) Legal but not fully compressed — those six trailing zeros should be ::, giving 2001:db8::. It's valid, just not the canonical short form.
• (d) Illegal — 12345 is 5 hex digits; a hextet is at most 4 hex digits (16 bits, max ffff). Answer: (a) and (d) are illegal.

WHAT the /64 means: the first 64 bits (4 hextets) are the network prefix; the last 64 bits (4 hextets) are the interface ID (see Subnetting).

• Prefix hextets: 2001 : db8 : acad : 1
• Interface hextets: 0000 : 0000 : 0000 : 00ff → 0 : 0 : 0 : ff Join: 2001:db8:acad:1:0:0:0:ff. Compress: the three zero hextets (positions 5–7) form the longest run → ::. Answer: 2001:db8:acad:1::ff.

IPv4 has $2^{32}$ addresses, IPv6 has $2^{128}$. Divide: $\frac{2^{128}}{2^{32}}=2^{128-32}=2^{96}.$ WHY subtract exponents: dividing equal bases subtracts powers ($a^m/a^n=a^{m-n}$). $2^{96}\approx 7.9\times 10^{28}.$ Answer: $2^{96}\approx 7.9\times 10^{28}$ times larger.

Per packet: header $=40$ B, data $=8$ B, total $=48$ B. Overhead fraction $=\frac{40}{48}=\frac{5}{6}\approx 0.8333$. Multiplying every packet by 1000 doesn't change the ratio. Answer: $\approx 83.33\%$ of bytes are header. (This is why the fixed header and extension-header design aim to keep the base header lean — see IP Header Structure.)

Compress: leading zeros → 2001:db8:0:0:abcd:0:0:1234. Two zero-runs of length 2 (hextets 3–4 and 6–7); tie → compress the first: 2001:db8::abcd:0:0:1234. Expand back: :: must supply $8-6=2$ zero hextets between db8 and abcd: 2001:db8:0000:0000:abcd:0000:0000:1234. Check: identical to the start. The two zero-hextets in positions 6–7 stayed explicit throughout — good. ✔ Answer: the round trip is lossless; full form 2001:0db8:0000:0000:abcd:0000:0000:1234.

• (a) Broadcast wakes all N hosts on the link; a targeted multicast (e.g. a routers-only group ff02::2) wakes only the subscribed subset (often far fewer). ff02::1 still reaches all, but most IPv6 traffic uses narrow groups.
• (b) Fewer interrupted hosts ⇒ fewer CPUs waking to process irrelevant frames ⇒ no cascading "everyone answers everyone" storms. Answer: broadcast = all N; targeted multicast = subscribers only; narrowing the audience removes the storm. (See Multicast vs Broadcast vs Anycast.)

• All zeros → single run of 8 zero-hextets → collapses entirely to :: (the unspecified address).
• Second: leading zeros give 2001:db8:0:0:0:0:0:0; hextets 3–8 are one run of length 6 → collapse the trailing run: 2001:db8::. The :: sits at the end because that's where the zero-run is. Answer: :: and 2001:db8::. Lesson: :: can sit at the start, middle, or end — wherever the longest zero-run actually is.