Exercises — Subnetting — subnet mask, network - host bits, VLSM

(a) Host bits $h=32-n=32-28=4$. Why? The prefix eats the network side; whatever is left is the host side, and $n+h=32$ always. (b) Usable hosts $=2^h-2=2^4-2=16-2=14$. Why $-2$? The all-0 host pattern is the network address (the nameplate) and the all-1 host pattern is the broadcast address (the loudspeaker). Neither is assignable to a machine. Answers: 4 host bits, 14 usable hosts.

$/28$ means 28 ones, then $32-28=4$ zeros: 11111111.11111111.11111111.11110000. The first three octets are all ones $=255$ each. The last octet is 11110000. Why is that 240? Sum the place-values of the ones: $128+64+32+16=240$. Answer: 255.255.255.240.

Block size $=256-\text{mask octet}=256-240=16$. Why? Block size is how far apart consecutive subnet start-addresses sit inside that octet, and it equals $2^{(\text{host bits inside this octet})}=2^4=16$. Look at the top number line in the figure: the ticks land at 0, 16, 32, ... exactly 16 apart. Answer: 16.

Step 1 — Interesting octet & block size. $/27\Rightarrow$ mask 255.255.255.224 (last octet 11100000 $=128+64+32=224$). Block size $=256-224=32$. Step 2 — Boundaries in the 4th octet: $0,32,64,96,128,\dots$ (multiples of 32). Step 3 — Which block holds 77? $64\le 77<96$, so the subnet is 10.0.5.64. Look at the middle number line in the figure: 77's red dot sits inside the [64, 96) band. Answer: 10.0.5.64/27.

Broadcast = next boundary − 1. Next boundary after 64 is $64+32=96$, so broadcast $=96-1=95\Rightarrow$ 10.0.5.95. Why? Broadcast is the all-1s host pattern — the very last address before the next subnet begins. Host range = strictly between network and broadcast $=$ .65 to .94 $\Rightarrow$ 10.0.5.65 – 10.0.5.94. Sanity check: usable count $=94-65+1=30=2^5-2$. ✓ (host bits $h=32-27=5$). Answers: broadcast 10.0.5.95, hosts 10.0.5.65–10.0.5.94.

Borrowed bits $b=29-24=5$. Number of subnets $=2^b=2^5=32$. Why? Each borrowed bit doubles the number of distinct subnet labels; 5 bits give $2^5$ patterns. Host bits $h=32-29=3$, so usable hosts $=2^3-2=6$ per subnet. Answers: 32 subnets, 6 usable hosts each.

Block size $=$ broadcast − network $+1=63-0+1=64$. Why $+1$? We are counting addresses inclusive of both endpoints. Block size $64=2^h\Rightarrow h=6$ host bits. Therefore $n=32-6=26$. Answer: /26 (mask 255.255.255.192, since $256-64=192$).

$/28\Rightarrow$ block size $256-240=16$. Boundaries: $0,16,32,48,\dots$

• .20 falls in $[16,32)$ → subnet 192.168.0.16.
• .35 falls in $[32,48)$ → subnet 192.168.0.32. Different subnets ⇒ NO, they are not on the same subnet. A router (not a direct link) is needed between them. See the bottom number line in the figure: the two red dots straddle the boundary at 32. Answer: No — .20 is in .16, .35 is in .32.

Incorrect. 255.255.255.192 is /26 (last octet 11000000, two 1-bits $\Rightarrow$ prefix $24+2=26$). Host bits $h=32-26=6$, usable $=2^6-2=62$. The mistake: they read the mask octet 192 as a host count. But 192 describes where the dividing line is, not how many hosts fit. The host count depends only on how many host bits exist. Answer: 62 usable hosts, not 192.

Golden rule: biggest subnet first, so each block lands on its own size boundary and nothing overlaps.

Sales (60 hosts). Need $2^h-2\ge 60$. $2^6-2=62\ge 60$ ✓, $2^5-2=30$ ✗. So $h=6$, prefix $/26$, block size 64. → 192.168.100.0/26, hosts .1–.62, broadcast .63.

Eng (28 hosts). $2^5-2=30\ge 28$ ✓, $2^4-2=14$ ✗. So $h=5$, prefix $/27$, block size 32. Next free address is .64. → 192.168.100.64/27, hosts .65–.94, broadcast .95.

Ops (12 hosts). $2^4-2=14\ge 12$ ✓, $2^3-2=6$ ✗. So $h=4$, prefix $/28$, block size 16. Next free address is .96. → 192.168.100.96/28, hosts .97–.110, broadcast .111.

Link1 (2 hosts). $2^2-2=2\ge 2$ ✓. So $h=2$, prefix $/30$, block size 4. Next free address is .112. → 192.168.100.112/30, hosts .113–.114, broadcast .115.

No overlaps, everything aligned — each block starts at a multiple of its own size (0, 64, 96, 112). Plenty of /24 space remains beyond .116. Answers: Sales .0/26, Eng .64/27, Ops .96/28, Link1 .112/30.

A /29 has $h=3$, so $2^3=8$ addresses (.8–.15). Each link needs /30 (block size 4). Borrowing $b=30-29=1$ bit gives $2^1=2$ subnets — exactly what we need.

• 10.1.1.8/30: network .8, hosts .9–.10, broadcast .11.
• 10.1.1.12/30: network .12, hosts .13–.14, broadcast .15. Answers: Link A 10.1.1.8/30 (.9–.10), Link B 10.1.1.12/30 (.13–.14).

$h=32-31=1$, so $2^h-2=2^1-2=0$ by the classic rule — zero usable hosts. That looks useless! But on a point-to-point link there is no broadcast to send (only one other endpoint), so RFC 3021 lets a /31 use both addresses as the two hosts. It gives you 2 usable addresses on exactly the links where you don't need a broadcast. This is a documented exception to $2^h-2$, not a violation of it — the subtraction assumes a broadcast is needed, and here it isn't. Answers: 0 by the classic formula; 2 usable in practice on point-to-point links (RFC 3021).

$h=32-32=0$, so $2^0-2=1-2=-1$. A negative host count is the formula screaming "this is not a normal subnet." A /32 has no host bits — it names one single exact address (a host route). It is used in routing tables to point at exactly one machine, and as a loopback address. There is no network/broadcast split because there is nothing to split. Answer: /32 = a single host address; the formula gives $-1$, signalling a degenerate (host-route) case, not a usable subnet.

$/20$ means 20 network bits. The mask is 11111111.11111111.11110000.00000000 $=$ 255.255.240.0. The interesting octet is the 3rd, not the 4th — because that's where the ones run out. Block size in the 3rd octet $=256-240=16$. Boundaries in octet 3: $0,16,32,\dots ,128,144,\dots$ The address's 3rd octet is 130. Which block? $128\le 130<144$, so octet 3 rounds down to 128, and the 4th octet is 0. Subnet = 172.20.128.0/20. Broadcast = next boundary − 1 in octet 3 = .143.255 → 172.20.143.255. Host range 172.20.128.1 – 172.20.143.254. Answer: subnet 172.20.128.0/20, broadcast 172.20.143.255.