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IP Subnetting VLSMs And Troubleshooting IP

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Instructions:

Answer 50 questions in 15 minutes.
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Match each statement with the correct term.
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This is a study tool. The 3 wrong answers for each question are randomly chosen from answers to other questions. So, you might find at times the answers obvious, but you will see it re-enforces your understanding as you take the test each time.

1. If an Ethernet port on a router were assigned an IP address of 172.16.112.1/25 - what would be the valid subnet address of this host?
A /25 mask is 255.255.255.128. Used with a Class B network - the third and fourth octets are used for subnetting with a total of 9 subnet bits - 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet -
11000000 = 192
A CIDR address of /19 is 255.255.224.0. This is a Class B address - so that is only 3 subnet bits but provides 13 host bits - or 8 subnets - each with 8 -190 hosts.
/29 is 255.255.255.248. The fourth octet is a block size of 8. 0 - 8 - 16 - 24. The host is in the 16 subnet - broadcast of 23. Valid host 17-22

2. To test the IP stack on your local host - which IP address would you ping?
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
0.0 - 64.0 - 128.0 - and 192.0
To test the local stack on your host - ping the loopback interface of 127.0.0.1
11110000 = 240

3. Variable Length Subnet Masks (VLSMs)
11111000 = 248
Create many networks using subnet masks of different lengths from one network
Tracert - Show ip arp
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s

4. /27
Possible DNS problem
RIPv2 - EIGRP - or OSPF
/23 is 255.255.254.0. The third octet is a block size of 2. 0 - 2 - 4. The subnet is in the 16.2.0 subnet - the broadcast address is 16.3.255
11100000 = 224

5. 4 Troubleshooting Steps
A Class B network ID with a /22 mask is 255.255.252.0 - with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask
/30 is 255.255.255.252. The valid subnet is 192.168.100.24 - broadcast is 192.168.100.27 - and valid hosts are 192.168.100.25 and 26
11111000 = 248
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination

6. You have an interface on a router with the IP address of 192.168.192.10/29. What is the broadcast address the hosts will use on this LAN?
.1 - .126 & .129 - .254
RIPv2 - EIGRP - or OSPF
A Class B network ID with a /22 mask is 255.255.252.0 - with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask
A /29 (255.255.255.248) has a block size of 8 in the fourth octet. This means the subnets are 0 - 8 - 16 - 24 - etc. 10 is in the 8 subnet. The next subnet is 16 - so 15 is the broadcast address.

7. Given 192.168.10.0/28 - How many subnets?
A /25 mask is 255.255.255.128. Used with a Class B network - the third and fourth octets are used for subnetting with a total of 9 subnet bits - 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet -
2
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
Discard it; by default will discard any broadcast packets

8. You have an interface on a router with the IP address of 192.168.192.10/29. Including the router interface - how many hosts can have IP addresses on the LAN attached to router interface?
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
A /29 (255.255.255.248) - regardless of the class of address - has only three hosts bits. Six hosts is the maximum amount of hosts on this LAN - including the router interface.
/26 is 255.255.255.192. The fourth octet has a block size of 64. 0 - 64 - 128. The host is in the 64 subnet - broadcast of 127. Valid host 65-126
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.

9. /25
10000000 = 128
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
A Class B network ID with a /22 mask is 255.255.252.0 - with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask
/28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0 - 16 - 32 - 48. The host is in the 32 subnet - with a broadcast address of 47. Valid hosts 33-46

10. How many hosts are available with a Class C /29 mask?
/26 is 255.255.255.192. The fourth octet has a block size of 64. 0 - 64 - 128. The host is in the 64 subnet - broadcast of 127. Valid host 65-126
10000000 = 128
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
11000000 = 192

11. What are the valid hosts in each subnet?
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
11111000 = 248
Traceroute - Arp -a - Ipconfig /all

12. Given 192.168.10.0/28 - What are the valid hosts?
First - if you have two hosts directly connected - as shown in the graphic - then you need a crossover cable. A straight-through cable won't work. Second - the hosts have different masks - which puts them in different subnets. The easily solution is
.1 - .126 & .129 - .254
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
Discard it; by default will discard any broadcast packets

13. Unable to Ping Remote Destination
A /25 mask is 255.255.255.128. Used with a Class B network - the third and fourth octets are used for subnetting with a total of 9 subnet bits - 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet -
/28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0 - 16 - 32 - 48. The host is in the 32 subnet - with a broadcast address of 47. Valid hosts 33-46
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
The routers IP address on the E0 interface is 172.16.2.1/23 - which is a 255.255.254.0. This makes the third octet a block size of 2. The routers interface is in the 2.0 subnet - the broadcast address is 3.255 because the next subnet is 4.0. The vali

14. 192.168.100.37/28
.1 - .126 & .129 - .254
/23 is 255.255.254.0. The third octet is a block size of 2. 0 - 2 - 4. The subnet is in the 16.2.0 subnet - the broadcast address is 16.3.255
/28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0 - 16 - 32 - 48. The host is in the 32 subnet - with a broadcast address of 47. Valid hosts 33-46
RIPv2 - EIGRP - or OSPF

15. /29
Allows you to use the first and last subnet in your network design; turned this command on by default
This is a pretty simple question. A /28 is 255.255.255.240 - which means that our block size is 16 in the fourth octet. 0 - 16 - 32 - 48 - 64 - 80 - etc. The host is in the 64 subnet.
All interfaces within the classful address space have the same subnet mask
11111000 = 248

16. Given 172.16.0.0/18 - How many hosts per subnet?
A Class C subnet mask of 255.255.255.224 is 3 bits on and 5 bits off (11100000) and provides 8 subnets - each with 30 hosts. However - if the command ip subnet-zero is not used - then only 6 subnets would be available for use.
0.0 - 64.0 - 128.0 - and 192.0
/28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0 - 16 - 32 - 48. The host is in the 32 subnet - with a broadcast address of 47. Valid hosts 33-46
16 -382

17. Given 192.168.10.0/28 - What are the valid subnets?
127 & 255
Each network segment can use a different subnet mask
0 & 128
A /30 - regardless of the class of address - has a 252 in the fourth octet. This means we have a block size of 4 and our subnets are 0 - 4 - 8 - 12 - 16 - etc. Address 14 is obviously in the 12 subnet.

18. Which two statements describe the IP address 10.16.3.65/23?
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
The mask 255.255.254.0 (/23) used with a Class A means that there are 15 subnet bits and 9 hosts bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in the interesting octet 0 - 2 - 4 - 6 - etc. - all the way to 254. T
The 10.32 subnet; The broadcast is 10.63
/23 is 255.255.254.0. The third octet is a block size of 2. 0 - 2 - 4. The subnet is in the 16.2.0 subnet - the broadcast address is 16.3.255

19. /26
0 & 128
/27 is 255.255.255.224. The fourth octet is a block size of 32. Count by 32s until you pass the host address of 66. 0 - 32 - 64. The host is in the 32 subnet - broadcast address of 63. Valid host range of 33-62.
.1 - .126 & .129 - .254
11000000 = 192

20. What is the subnetwork address for a host with the IP address 200.10.5.68/28?
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
Tracert - Show ip arp
The numbers between the subnets and the broadcasts omitting the all 0s and all 1s
This is a pretty simple question. A /28 is 255.255.255.240 - which means that our block size is 16 in the fourth octet. 0 - 16 - 32 - 48 - 64 - 80 - etc. The host is in the 64 subnet.

21. Using the following illustration - what would be the IP address of E0 if you were using the eighth subnet? The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range. The zero subnet should not be considered vali
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
A 240 mask is 4 subnet bits and provides 16 subnets - each with 14 hosts. We need more subnets - so let's add subnet bits. One more subnet bit would be a 248 mask. This provides 5 subnet bits (32 subnets) with 3 hosts bits (6 host per subnet).
A /28 is a 255.255.255.240 mask. We need to count to the eighth subnet - not starting at subnet-zero. 16 - 32 - 48 - 64 - 80 - 96 - 112 - 128. The ninth subnet is 144 (we need this to help us find the 128 subnet broadcast address - which is 143).The

22. A router receives a packet on an interface with a destination address of 172.16.46.191/26. What will the router do with this packet?
This is a pretty simple question. A /28 is 255.255.255.240 - which means that our block size is 16 in the fourth octet. 0 - 16 - 32 - 48 - 64 - 80 - etc. The host is in the 64 subnet.
/27 is 255.255.255.224. The fourth octet is a block size of 32. Count by 32s until you pass the host address of 66. 0 - 32 - 64. The host is in the 32 subnet - broadcast address of 63. Valid host range of 33-62.
Discard it; by default will discard any broadcast packets
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination

23. Given 192.168.10.0/28 - How many hosts per subnet?
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
126
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
/26 is 255.255.255.192. The fourth octet has a block size of 64. 0 - 64 - 128. The host is in the 64 subnet - broadcast of 127. Valid host 65-126

24. Classful Routing
All interfaces within the classful address space have the same subnet mask
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination
/28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0 - 16 - 32 - 48. The host is in the 32 subnet - with a broadcast address of 47. Valid hosts 33-46
Use different size masks on each router interface

25. Subnet Mask
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
32-bit value that allows the recipient of IP packets to distinguish the network ID portion of the IP address from the host ID portion of the IP address
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255
The mask 255.255.254.0 (/23) used with a Class A means that there are 15 subnet bits and 9 hosts bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in the interesting octet 0 - 2 - 4 - 6 - etc. - all the way to 254. T

26. Which configuration command must be in effect to allow the use of 8 subnets if the Class C subnet mask is 255.255.255.224?
A Class C subnet mask of 255.255.255.224 is 3 bits on and 5 bits off (11100000) and provides 8 subnets - each with 30 hosts. However - if the command ip subnet-zero is not used - then only 6 subnets would be available for use.
10000000 = 128
RIPv2 - EIGRP - or OSPF
11111000 = 248

27. What is the subnetwork number of a host with an IP address of 172.16.66.0/21?
/29 is 255.255.255.248 - which is 5 subnet bits and 3 hosts bits. This is only 6 hosts per subnet
/28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0 - 16 - 32 - 48. The host is in the 32 subnet - with a broadcast address of 47. Valid hosts 33-46
A /21 is 255.255.248.0 - which means we have a block size of 8 in the third octet - so we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet is 72.0 - so the broadcast address of the 64 subnet is 71.255.
A /28 is a 255.255.255.240 mask. We need to count to the eighth subnet - not starting at subnet-zero. 16 - 32 - 48 - 64 - 80 - 96 - 112 - 128. The ninth subnet is 144 (we need this to help us find the 128 subnet broadcast address - which is 143).The

28. Unable to Ping Loopback
Tracert - Show ip arp
IP stack failure; Reinstall TCP/IP
.1 - .126 & .129 - .254
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21

29. You have a network with a subnet of 172.16.17.0/22. Which are valid host addresses?
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
A /29 is 255.255.255.248 - which is a block size of 8 in the fourth octet. The subnets are 0 - 8 - 16 - 24 - 32 - 40 - etc. 192.168.19.24 is the 24 subnet - and since 32 is the next subnet - the broadcast address for the 24 subnet is 31. 192.168.19.2
A /27 (255.255.255.224) is 3 bits on and 5 bits off. This provides 8 subnets - each with 30 hosts. Does it matter if this mask is used with a Class A - B or C network address? Not at all. The amount of hosts bits would never change.
A Class B network ID with a /22 mask is 255.255.252.0 - with a block size of 4 in the third octet. The network address in the question is in subnet 172.16.16.0 with a broadcast address of 172.16.19.255. Only option E even has the correct subnet mask

30. What subnet and broadcast address is the IP address 172.16.10.33 255.255.255.224 (/27) a member of?
/30 is 255.255.255.252. The valid subnet is 192.168.100.24 - broadcast is 192.168.100.27 - and valid hosts are 192.168.100.25 and 26
A /30 - regardless of the class of address - has a 252 in the fourth octet. This means we have a block size of 4 and our subnets are 0 - 4 - 8 - 12 - 16 - etc. Address 14 is obviously in the 12 subnet.
The 10.32 subnet; The broadcast is 10.63
RIPv2 - EIGRP - or OSPF

31. What subnet and broadcast address is the IP address 172.16.66.10 255.255.192.0 (/18) a member of?
Traceroute - Arp -a - Ipconfig /all
The subnet is 172.16.64.0. The broadcast must be 172.16.127.255
RIPv2 - EIGRP - or OSPF
/27 is 255.255.255.224. The fourth octet is a block size of 32. Count by 32s until you pass the host address of 66. 0 - 32 - 64. The host is in the 32 subnet - broadcast address of 63. Valid host range of 33-62.

32. What's the broadcast address of each subnet?
The subnet is 172.16.45.12 - with a broadcast of 172.16.45.15
A /29 (255.255.255.248) - regardless of the class of address - has only three hosts bits. Six hosts is the maximum amount of hosts on this LAN - including the router interface.
The routers IP address on the E0 interface is 172.16.2.1/23 - which is a 255.255.254.0. This makes the third octet a block size of 2. The routers interface is in the 2.0 subnet - the broadcast address is 3.255 because the next subnet is 4.0. The vali
The number just before the next subnet; the broadcast of the last subnet is always 255

33. Additional Windows Troubleshooting
Discard it; by default will discard any broadcast packets
A Class C subnet mask of 255.255.255.224 is 3 bits on and 5 bits off (11100000) and provides 8 subnets - each with 30 hosts. However - if the command ip subnet-zero is not used - then only 6 subnets would be available for use.
A /29 (255.255.255.248) has a block size of 8 in the fourth octet. This means the subnets are 0 - 8 - 16 - 24 - etc. 10 is in the 8 subnet. The next subnet is 16 - so 15 is the broadcast address.
Traceroute - Arp -a - Ipconfig /all

34. How many valid hosts per subnet?
All nodes in the network use the same subnet mask
2y - 2 where y is the number of unmasked bits (or 0's)
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255

35. IP Subnet-Zero
Allows you to use the first and last subnet in your network design; turned this command on by default
10000000 = 128
A CIDR address of /19 is 255.255.224.0. This is a Class B address - so that is only 3 subnet bits but provides 13 host bits - or 8 subnets - each with 8 -190 hosts.
A /25 mask is 255.255.255.128. Used with a Class B network - the third and fourth octets are used for subnetting with a total of 9 subnet bits - 8 bits in the third octet and 1 bit in the fourth octet. Since there is only 1 bit in the fourth octet -

36. Given 192.168.10.0/28 - What's the broadcast address for each subnet?
A Class C subnet mask of 255.255.255.224 is 3 bits on and 5 bits off (11100000) and provides 8 subnets - each with 30 hosts. However - if the command ip subnet-zero is not used - then only 6 subnets would be available for use.
255.255.255.128
127 & 255
All interfaces within the classful address space have the same subnet mask

37. 192.168.100.25/30
This is a pretty simple question. A /28 is 255.255.255.240 - which means that our block size is 16 in the fourth octet. 0 - 16 - 32 - 48 - 64 - 80 - etc. The host is in the 64 subnet.
A CIDR address of /19 is 255.255.224.0. This is a Class B address - so that is only 3 subnet bits but provides 13 host bits - or 8 subnets - each with 8 -190 hosts.
/30 is 255.255.255.252. The valid subnet is 192.168.100.24 - broadcast is 192.168.100.27 - and valid hosts are 192.168.100.25 and 26
Local physical network problem between NIC and router

38. You have a Class B network and need 29 subnets. What is your mask?
0.0 - 64.0 - 128.0 - and 192.0
Allows you to use the first and last subnet in your network design; turned this command on by default
This is 5 bits of subnetting - which provides 32 subnets. This is our best answer - a /21
A /28 is a 255.255.255.240 mask. We need to count to the eighth subnet - not starting at subnet-zero. 16 - 32 - 48 - 64 - 80 - 96 - 112 - 128. The ninth subnet is 144 (we need this to help us find the 128 subnet broadcast address - which is 143).The

39. Variable Length Subnet Masks (VLSMs)
A /28 is a 255.255.255.240 mask. The first subnet is 16 (remember that the question stated not to use subnet zero) and the next subnet is 32 - so our broadcast address is 31. This makes our host range 17-30. 30 is the last valid host.
Each network segment can use a different subnet mask
A /21 is 255.255.248.0 - which means we have a block size of 8 in the third octet - so we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet is 72.0 - so the broadcast address of the 64 subnet is 71.255.
The 10.32 subnet; The broadcast is 10.63

40. Classless Routing Protocols
RIPv2 - EIGRP - or OSPF
256 - subnet mask = block size; start with 0 and add the block size until the mask value is reached
126
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255

41. Class C Subnet Masks
A /28 is a 255.255.255.240 mask. We need to count to the eighth subnet - not starting at subnet-zero. 16 - 32 - 48 - 64 - 80 - 96 - 112 - 128. The ninth subnet is 144 (we need this to help us find the 128 subnet broadcast address - which is 143).The
/25 (with ip subnet-zero) - /26 - /27 - /28 - /29 - /30
The routers IP address on the E0 interface is 172.16.2.1/23 - which is a 255.255.254.0. This makes the third octet a block size of 2. The routers interface is in the 2.0 subnet - the broadcast address is 3.255 because the next subnet is 4.0. The vali
This subnet address must be in the 172.16.32.0 subnet - and the broadcast must be 172.16.47.255

42. Using the illustration from the previous question - what would be the IP address of S0 if you were using the first subnet? The network ID is 192.168.10.0/28 and you need to use the last available IP address in the range. Again - the zero subnet shoul
A /28 is a 255.255.255.240 mask. The first subnet is 16 (remember that the question stated not to use subnet zero) and the next subnet is 32 - so our broadcast address is 31. This makes our host range 17-30. 30 is the last valid host.
A /30 - regardless of the class of address - has a 252 in the fourth octet. This means we have a block size of 4 and our subnets are 0 - 4 - 8 - 12 - 16 - etc. Address 14 is obviously in the 12 subnet.
.0.1 - .63.254 - .64.1 - .127.254 - .128.1 - .191.254 - .192.1 - .255.254
To test the local stack on your host - ping the loopback interface of 127.0.0.1

43. What is the maximum number of IP addresses that can be assigned to hosts on a local subnet that uses the 255.255.255.224 subnet mask?
A /27 (255.255.255.224) is 3 bits on and 5 bits off. This provides 8 subnets - each with 30 hosts. Does it matter if this mask is used with a Class A - B or C network address? Not at all. The amount of hosts bits would never change.
Traceroute - Arp -a - Ipconfig /all
11110000 = 240
Discard it; by default will discard any broadcast packets

44. Unable to Ping Default Gateway
Remote physical network problem between NIC and destination; Additional troubleshooting required at destination
Local physical network problem between NIC and router
The mask 255.255.254.0 (/23) used with a Class A means that there are 15 subnet bits and 9 hosts bits. The block size in the third octet is 2 (256 - 254). So this makes the subnets in the interesting octet 0 - 2 - 4 - 6 - etc. - all the way to 254. T
11111000 = 248

45. If a host on a network has the address 172.16.45.14/30 - what is the subnetwork this host belongs to?
The subnet is 172.16.32.0 - and the broadcast must be 172.16.63.25
A /30 - regardless of the class of address - has a 252 in the fourth octet. This means we have a block size of 4 and our subnets are 0 - 4 - 8 - 12 - 16 - etc. Address 14 is obviously in the 12 subnet.
RIPv1 and IGRP
126

46. Classful Routing Protocols
A /29 (255.255.255.248) - regardless of the class of address - has only three hosts bits. Six hosts is the maximum amount of hosts on this LAN - including the router interface.
Traceroute - Arp -a - Ipconfig /all
RIPv1 and IGRP
First - if you have two hosts directly connected - as shown in the graphic - then you need a crossover cable. A straight-through cable won't work. Second - the hosts have different masks - which puts them in different subnets. The easily solution is

47. You need to configure a server that is on the subnet 192.168.19.24/29. The router has the first available host address. Which of the following should you assign to the server?
/30 is 255.255.255.252. The valid subnet is 192.168.100.24 - broadcast is 192.168.100.27 - and valid hosts are 192.168.100.25 and 26
A /29 is 255.255.255.248 - which is a block size of 8 in the fourth octet. The subnets are 0 - 8 - 16 - 24 - 32 - 40 - etc. 192.168.19.24 is the 24 subnet - and since 32 is the next subnet - the broadcast address for the 24 subnet is 31. 192.168.19.2
Traceroute - Arp -a - Ipconfig /all
Ping 127.0.0.1 (loopback) - Ping the local host - Ping Default Gateway (router) - Ping remote destination

48. /28
2y - 2 where y is the number of unmasked bits (or 0's)
11110000 = 240
RIPv1 and IGRP
0.0 - 64.0 - 128.0 - and 192.0

49. 192.168.100.99/26
0.0 - 64.0 - 128.0 - and 192.0
/28 is 255.255.255.240. The fourth octet is a block size of 16. Just count by 16s until you pass 37. 0 - 16 - 32 - 48. The host is in the 32 subnet - with a broadcast address of 47. Valid hosts 33-46
/26 is 255.255.255.192. The fourth octet has a block size of 64. 0 - 64 - 128. The host is in the 64 subnet - broadcast of 127. Valid host 65-126
2y - 2 where y is the number of unmasked bits (or 0's)

50. On a VLSM network - which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?
A point-to-point link uses only two hosts. A /30 - or 255.255.255.252 - mask provides two hosts per subnet.
A /21 is 255.255.248.0 - which means we have a block size of 8 in the third octet - so we just count by 8 until we reach 66. The subnet in this question is 64.0. The next subnet is 72.0 - so the broadcast address of the 64 subnet is 71.255.
11111000 = 248
63.255 - 127.255 - 191.255 - 255.255