16

I’ve been following the CCENT official certification book(100-105) and came upon this question in the “do I know this already?” quiz. The books only covered /24 subnetting only so far.

Which of the following is a network broadcast address?
a. 10.1.255.255
b. 192.168.255.1
c. 224.1.1.255
d. 172.30.255.255

As no subnetting notation has been included, I’ll stick with .255 ending as being broadcast.

  • a = seems correct.
  • b = incorrect. ends with .1
  • c = incorrect, it’s a class d multicast.
  • d = seems corrects too.

The answer says ONLY D is correct.

So why is A incorrect? My understanding:

  • 10.1.255.0 as the network ID
  • 10.1.255.1 to 10.1.255.254 as valid IP addresses.
  • 10.1.255.255 as the broadcast

Class a = 8 Bits network ID. 24 Bits Host ID. (0 subnet bits as theirs only 1 whole subnet? is this correct?)
subnet mask = 255.0.0.0
I believe it's my lack of understanding of how subnet masks correlate to Ip addresses

  • 33
    The correct answer to the certification question would be: "This is a stupid question. Classes are long dead." Or, to be more polite: "It is impossible to tell without knowing the network size." Or, a counter-question: "Why are testing for decades-obsolete technology, I want my money back!" – Jörg W Mittag Oct 10 at 17:55
  • 2
    "As no subnetting notation has been included". This question has nothing to do with subnetting. It would be a valid question even if no subnetting existed. This questino is entirely about network addresses and network broadcast addresses. It has nothing to do with subnets. VLSM predated CIDR and they are two different things. "I believe it's my lack of understanding of how subnet masks correlate to Ip addresses". Nope. Question is not about subnets, it's about networks. A subnet is a portion of a network. – David Schwartz Oct 10 at 20:46
  • 8
    This question has been made obsolete in the last millennium: Classes are long dead and IPv4 is obsolete. – Reinstate Monica - M. Schröder Oct 10 at 21:08
  • 8
    @DavidSchwartz You missed your coffee today. It's been decades since a network could be defined without its prefix length. Class A, B or C networks are no longer a thing. – jcaron Oct 10 at 23:19
  • 1
    @jcaron I agree. But that has nothing to do with subnets or subnetworking. VLSM and CIDR are two different things. – David Schwartz Oct 11 at 15:30
40

I believe the book wrongly assumes network classes are still in effect. So a) would be a "Class A" network, where 10.255.255.255 would be the broadcast address. Another hint: There is no explicit network size specified (/24, /27, ..) so it is implied you know about network classes. Classical example of outdated literature.

  • 10
    This question is indeed incomplete without stating the given subnet size resp. subnet mask length. Even 192.168.255.1 could be considered the broadcast address of 192.168.255.0/31 (but then again.. maybe not, since /31 are a bit peculiar in that context). – Marc 'netztier' Luethi Oct 10 at 9:13
  • 1
    Thanks for the great answer :) I've just found out that because the default subnet mask of class A is 255.0.0.0, then the broadcast can only be 10.255.255.255(like you've specified) – Bilal Baroudi Oct 10 at 9:16
  • 2
    @Marc'netztier'Luethi /31 subnets have no broadcast address (no use for p2p), so 192.168.255.1 can't be one, regardless of prefix length. ;-) – Zac67 Oct 10 at 9:25
  • 6
    @DavidSchwartz How exactly are "networks" and "subnets" different, ever since CIDR? – Marc 'netztier' Luethi Oct 10 at 20:55
  • 3
    "Classical example of outdated literature". I see what you did there. – Jonas Schäfer Oct 12 at 11:20
7

This is a "traditional" exam. question which contains:

  1. Missing information
  2. A trap
  3. A hint

The missing information is the subnet mask or CIDR number of bits.

The trap is answer (a) : 10.1.255.255/16 is a broadcast address, as is 10.1.255.255/24, but without mask information, we must assume classful addressing, and 10.1.255.255/8 is a unicast address.

The hint is the word network in the question. Being pedantic, 10.0.0.0/8 is a network and 10.1.0.0/16 is a subnet.

Those who voted for Sebastian Wiesinger's answer would be marked wrong in the exam.

This sort of trap existed in real life 20 years ago. I hope nowadays no one is using 10.0.0.0/8 for small networks any more.

  • Knowing your stuff and being good at exams are two different skill sets... Unfortunately sometimes the latter matters more – FreeSoftwareServers Oct 13 at 3:56
  • Noone should assume classful addressing in 2019. That is just wrong. Any documentation stating or assuming otherwise is wrong as well. This exam would not be worth the money payed to take it. – Sebastian Wiesinger Oct 22 at 15:31
  • No, both are networks. The later is also a subnet of the former. – Ricky Beam Oct 25 at 16:07
3

TL;DR

10.1.255.255 is the valid directed broadcast address for the networks

10.0.0.0/15
10.1.0.0/16
10.1.128/17
10.1.192/18
10.1.224.0/19
10.1.240.0/20
10.1.248.0/21
10.1.252.0/22
10.1.254.0/23
10.1.255.0/24
10.1.255.128/25
10.1.255.192/26
10.1.255.224/27
10.1.255.240/28
10.1.255.248/29
10.1.255.252/30

The directed broadcast address for a network has all bits in the address's host part set to 1, so in binary form

00001010.00000001.11111111.11111111

works for

00001010.00000000.00000000.00000000/15
\-network-addr-/\----host-part----/

through

00001010.00000001.11111111.11111100/30
\--------network-address--------/\/host-part

10.1.255.255 is not a valid broadcast address for network prefixed 10.0.0.0/7 to 10.0.0.0/14.

For people still living in the early 1990s, 10.1.255.255 may imply class A, or /8 in CIDR notation, making it a host address. Network classes were obsoleted in 1993 by RFCs 1518 and 1519 and replaced by Classless Inter-Domain Routing (CIDR).

-1

The address shown are defined in RFCs 1918 and 5771 as reserved addresses for specific use:

a. 10.1.255.255     is part of 10.0.0.0/8 (rfc 1918)
b. 192.168.255.1    is part of 192.168.0.0/16 (rfc 1918)
c. 224.1.1.255      is part of 224.1.0.0/16 (rfc 5771)
d. 172.30.255.255   is part of 172.16.0.0/12 (rfc 1918)

In particular, RFC 1918 states that the network 10.0.0.0/8 is a single number, 172.16.0.0/12 are 16 contiguous class B (/16) network numbers, and 192.168.0.0/16 are 256 class C (/24) network numbers. Following that definition, answers (a) and (b) cannot be broadcast addresses, because for (a) should be 10.255.255.255 and for (b) should be 192.168.255.255. Answer (d) is in fact the broadcast address of the network 172.30.0.0/16, always taking as basis the document RFC 1918. As of answer (b), RFC 5771 explicitly states that the network 224.1.0.0/16 is "RESERVED" and its broadcast address is then 224.1.255.255, so (c) is also wrong. Therefore, the only valid broadcast address present in the list is in answer (d).

Yes, CIDR is the modern way, and Classes are obsolete, but the IETF documentation is the agreed reference for network implementations, and these addresses are defined there.

  • 5
    Even RFC 1918, old as it is, is aware of CIDR and presents 10/8 and friends as "blocks", noting that in pre-CIDR notation, they're just 1x class A network, 16x class B and 256x class C. That doesn't seem to imply that they should only be used as classful networks, not when the document was written, let alone 23 years later. – ilkkachu Oct 11 at 21:11

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