Q1.) You work for a large communications corporation named GlobeComm which has been assigned a Class A network address. Currently, the company has 1,000 subnets in offices around the world. You want to add 100 new subnets over the next three years, and you want to allow for the largest possible number of host addresses per subnet. Which subnet mask would you choose?
Ans: 255.255.224.0
Requirements: 1,000 subnets + 100 subnets = 1,100 subnets, and as many host addresses as possible
2^10-2 = 1022 we Know from memory that 2^10= 1024, and will yield ONLY 1022 usable subnets with 10 bits of subnetting Because our requirements ask for over 1024 subnets, we must borrow one more host bit for our subnetting:
2^11-2 = (1024 * 2) - 2 = 2046 unique usable subnets
11111111 00000000 00000000 00000000 = 255.0.0.0 Default class A subnet mask
11111111 11111111 11100000 00000000 = 255.255.224.0 Default Class A subnet mask with our 11 bits of subnetting.
Q2.) Which of the following is a valid IP host address given the network ID of 191.254.0.0 while using 11 bits for subnetting?
a. 191.254.0.32
b. 191.254.0.96
c. 191.254.1.29
d. 191.54.1.64
Sol: Answer: Choice C is correct: 191.254.1.29. The network ID 191.254.0.0 is a Class B network address with a default subnet mask of 255.255.0.0. Adding 11 subnetting bits to the default 16-bit subnet mask yields 27 subnet bits and makes the new subnet mask
255.255.255.224. Using 11 bits for subnetting provides 2046 subnets with a maximum of 30 hosts per subnet. To fin the subnet address interval, use the following procedure. Convert the subnet mask 255.255.255.224 into binary. Next, convert the lowest order subnet bit to decimal. In this item, the decimal value of the lowest order non-zero bit is 32. Use the decimal value of the lowest-order bit as the interval for calculating subnet IDs. The following table displays the host ID ranges for the first 10 subnets.
Subnet ID Host ID Range
191.254.0.0 191.254.0.1 -191.254.0.30
191.254.0.32 191.254.0.33 -191.254.0.62
191.254.0.64 191.254.0.65 -191.254.0.94
191.254.0.96 191.254.0.97 -191.254.0.126
191.254.0.128 191.254.0.129-191.254.0.158
191.254.0.160 191.254.0.161-191.254.0.190
191.254.0.192 191.254.0.193-191.254.0.222
191.254.0.224 191.254.0.225-191.254.0.254
191.254.1.0 191.254.1.1 -191.254.1.30
191.254.1.32 191.254.1.33 -191.254.1.62
...
As shown in the table, the IP addresses 191.254.0.32, 191.254.0.96 and 191.254.1.64 are subnet IDs in this scenario. Therefore, these IP addresses cannot be used as host IDs, given a network ID of 191.254.0.0 with 11 bits of subnetting.
Q3.) NuTex Corporation has been assigned the Class B network address 165.87.0.0. NuTex needs to divide the network into eight subnets. What subnet mask should be applied to the network to provide the most hosts per subnet?
a. 255.255.192.0
b. 255.255.224.0
c. 255.255.240.0
d. 255.255.248.0
*Answer: 255.255.240.0
2^3-2 = 6 subnets
2^4-2 = 14 subnets
(2^16 - 2^4) -2 = 2^12 - 2 = 4094 hosts possible per subnet
Because 3 bits of subnetting will only yield 6 usable subnets (7 if ip subnet-zero is enabled), you will need to borrow another bit. To meet your minimum requirement of 8 subnets, and the maximum number of hosts per subnet, this Subnet mask will be correct: 255.255.240.0
Q4.) What is the network ID portion of the IP address 191.154.25.66 if the default subnet mask is used? (Done)
Answer: 191.154 - If the default subnet mask is used, then the network ID portion of the IP address 191.154.25.66 is 191.154. The first octet, 191, indicates that this is a class B address. In a class B address, the first two octets of the address represent the network portion. The default subnet mask for a Class B network address is 255.255.0.0. If a network administrator had implemented a subet mask other than the default, then the network portion of the address would be different. For example, an address of 191.154.25.66 with a subnet mask of 255.255.255.0 would have a network ID of 191.154.25.
A Class A address has a default subnet mask of 255.0.0.0; therfore, the first octet of a Class A address represents the network portion of the address, if the default subnet mask is used. A Class C address has a default subnet mask of 255.255.255.0; therefore the first three octets of a Class C address represent the network portion of the network portion of the address, if the default mask is used. The bits to the right of the network portion of an IP address are the host portion of the address.
10111111.10011010.00011001.01000010 = 191.154.25.66 IP address
11111111.11111111.00000000.00000000 = 255.255.0.0 Default Class B Subnet mask
10111111.10011010.00000000.00000000 = 191.154.0.0 Network ID is the Result of ANDing the IP address and Subnet mask
Q5.) Your company has the network ID 165.121.0.0. You are responsible for creating subnets on the network, and each subnet must provide at least 900 host IDs.
What subnet mask meets the requirement for the minimum number of host IDs and provides the greatest number of subnets?
*Answer: A Class B address with a subnet mask of 255.255.252.0 creates 62 subnets with 1,022 hosts each. The next largest subnet creates 126 subnets, but each subnet contains only 510 hosts. In binary format, the subnet mask reads as: 11111111 11111111 11111100 00000000. The third octet designates 1s for the first 6 digits. The 1s mask the subnet IDs. The calculation of the number of host IDs available for each subnet is based on the number of digits remaining in the network address after all of the subnets are created. To calculate the number of host IDs, raise 2 to the tenth power and then subtract 2. Note that there are two digits remaining in the third octet and eight digits in the fourth octet (00 00000000); these ten digits can be used to create host IDs. The two host IDs that are subtracted are the two host IDs that contain all 0s and all 1s; those IDs identify the subnet itself and the broadcast address for all hosts on the subnet, respectively. The number of possible host IDs in each subnet ranges from 00 00000001 through 11 11111110. Converted to decimal notation, this range is 0.1 through 3.254. For example, in the network 165.121.0.0/22, host addresses can range from 165.121.0.1 through 165.121.3.254.
6. What is the broadcast address on subnet 32 given a prefix notation of 12.1.0.0/12?
a. 12.32.0.1
b. 12.32.0.255
c. 12.23.255.255
d. 12.47.255.255
*Answer: Choice d is correct. Given a prefix notation of 12.1.0.0/12, the broadcast address on subnet 32 is 12.47.255.255. A prefix notation of /12 indicates a binary subnet mask of 11111111 11110000 00000000 00000000 or 255.240.0.0 in dotted-decimal notation. The subnet mask 255.240.0.0 contains 12 masked bits. The address 12.1.0.0 belongs to a Class A network address. When only one octet is used for subnetting, the interval between valid subnets can be determined by subtracting the value of the octet from 256, such as 256 - 240 = 16 in this scenario. The number 16 is both the first valid address of the subnet ID and the interval between valid subnets. Therefore, the first valid subnet ID is 12.16.0.0. Adding 16 to the first subnet ID value yields the second valid subnet, which is subnet 32, and the corresponding IP address for this subnet ID is 12.32.0.0. Continuing this process, the third subnet ID is 12.48.0.0 and so on. The broadcast address for subnet 32 is represented by 00001100 00101111 11111111 11111111 in binary or 12.47.255.255 in dotted-decimal notation. In the second octet of the broadcast address for subnet 32, the first 4 bits are used for the subnet ID (0010), and 00100000 represents subnet 32. The broadcast address on subnet 32 is found by assigning 1s to each of the 20 rightmost bits that represent the host portion of the address. Exhibit 1 in this explanation shows the addresses and masks involved in this scenario.
The broadcast mask is formed by using 0s for the subnet bits and 1s for the host bits. In this example, network 12.1.0.0/12 uses 12 network bits. Therefore, the first 12 bits of the broadcast mask will be 0s. The remaining 20 bits from the total of 32 bits will be 1s. This
yields 00000000 00001111 11111111 11111111 in this example. The broadcast address is calculated by performing a Boolean XOR operation on subnet 32 and the broadcast mask. A Boolean XOR operation compares two bits.
A Boolean XOR operation evaluates to True(1) only when one bit is True and the other bit is False (0). All other combinations result in False, such as when both bits are True or both bits are False. To do the complete XOR operation in this item, first write down the subnet address in binary. On a second line, write down the broadcast mask. Finally, perform a Boolean XOR operation on all corresponding bits.
00001100.00000001.00000000.00000000 = 12.1.0.0 network address
11111111.11110000.00000000.00000000 = 255.240.0.0 subnet mask
00001100.00010000.00000000.00000000 = 12.16.0.0 subnet 16 00001100.00100000.00000000.00000000 = 12.32.0.0 subnet 32 00001100.00110000.00000000.00000000 = 12.48.0.0 subnet 48 00000000.00001111.11111111.11111111 = 0.15.255.255
broadcast mask
00001100.00100000.00000000.00000000 = 12.32.0.0 subnet 32 00000000.00001111.11111111.11111111 = 0.15.255.255 broadcast mask
00001100.00101111.11111111.11111111 = 12.47.255.255 subnet 32 Broadcast address
Thus, the XOR operation yields the subnet 32 broadcast address of 12.47.255.255. Note that the sum of the network subnet mask and the network broadcast mask is always 11111111 11111111 11111111 11111111 in binary, or 255.255.255.255 in dotted decimal notation.
Q7. Your ISP has provided you a T1 connection to their network. They have provided the interface at your end of the Serial connection an IP address of 16.32.96.109/30. For this particular situation, you must supply the IP address of the ISP's interface at the opposite end of the serial connection as your Gateway of Last Resort. What IP address would you specify as the Gateway of last resort? (done)
Hints(Please answer)
a. How many hosts are allowed in each subnet? Answer: 2 usable hosts
32 possible bits in an IP version 4 IP address.
32 - 30 = 2 possible bits left for our host
2^2-2 = 2 usable hosts
b. What is the Broadcast address for this network? (Please show the "interesting" octet in binary)
c. What is the Network address for this network? (Please show the "interesting" octet in binary)
128 64 32 16 8 4 2 1
0 1 1 0 1 1 0 1 = 109 decimal
16.32.96.01101101 = 16.32.96.109 -IP address of one of your hosts in this small subnet
255.255.255.11111100 = 255.255.255.252 -Subnet Mask given in CIDR /30
16.32.96.01101100 = 16.32.96.108 -Network ID for this subnetwork 16.32.96.01101111 = 16.32.96.111 -Subnet broadcast address 16.32.96.01101110 = 16.32.96.110 -The only usable host address left range between the network and broadcast
8. You are a network administrator, seated at a workstation at the remote end of a campus network. You are attempting to troubleshoot a communication problem between that client workstation and the server at the other end of campus. This workstation has a static IP address of 181.197.187.212, with a subnet mask of 255.255.224.0. Because a particularly thorough security administrator, Frank, has removed most extraneous applications, including the Calculator, you must use paper and pencil to verify that the workstation is on the same subnet as your server at 181.197.143.236, with a subnet mask of 255.255.224.0. The user reports that the computer "hasn't worked right since it was installed last week." You cannot ping the server from the workstation. Are these two computers on the same subnet?
Answer: NO, see solution below...
Hint- Convert the "interesting octet" of the Subnet mask, of the workstation and then the server to binary. What are the results with dotted-decimal from the boring octets, combined with Binary for the interesting octet? Line them up, how do they compare?
What is the network ID for the Workstation's IP address? (Please show the interesting octet in binary)
What is the broadcast ID for network? (Please show the interesting octet in binary)
How many usable host addresses are available for the network?
Interesting Octet = 187 Convert to binary:
>=128 1
187-128=59
>=64 0
>=32 1
59-32=27
>=16 1
27-16=11
>=8 1
11-8=3
>=4 0
>=2 1
3-2=1
>=1 1
181.197.10111011.212 = 181.197.187.212 -Workstation's configured IP Address
255.255.11100000.0 = 255.255.224.0 -Workstation's configured Subnet Mask
181.197.10100000.0 = 181.197.160.0 -Workstation's Network ID 181.197.10100000.1 = 181.197.160.1 -First Host in Workstation's Range
...
181.197.10111111.254 = 181.197.191.254 -Last Host in Workstation's Range
181.197.10111111.255 = 181.197.191.255 -Workstation's Broadcast ID
...96-128-160-192-224-256 -Network IDs for a 224 Subnet Mask
Interesting Octet = 143 Convert to binary:
>=128 1
143-128=15
>=64 0
>=32 0
>=16 0
>=8 1
15-8=7
>=4 1
7-4=3
>=2 1
3-2=1
>=1 1
181.197.10001111.236 = 181.197.143.236 -Server's configured IP address
255.255.11110000.0 = 255.255.240.0 -Server's configured Subnet Mask
181.197.10000000.0 = 181.197.128.0 -Server's Network ID 181.197.10000000.1 = 181.197.128.1 -First Host in Server's Range ...
181.197.10001111.254 = 181.197.143.254 -Last Host in Server's Range
181.197.10001111.255 = 181.197.143.255 -Server's Broadcast ID
...96-112-128-144-160-176-192-208-224... -Subnet ID's for 240 Subnet Masks
If you are at the Workstation, and attempt to ping the server, the workstation must determine whether the server's IP address is on the same subnet as it's own. If the workstation determines that the server is on it's subnet, then it will send an ARP request for the server's IP address, otherwise, the workstation will need to Arp it's configured Default Gateway, because it handles all requests for addresses outside the local subnet, and might have a route to those other networks.
The Workstation must determine if the resource is on the local network or not, and must perform an ANDing operation. Workstation's question: Is this destination network address local to my subnet?
Workstation knows that 181.197.160.0 is it's network ID, and must determine if the server is on that same subnet by ANDing the Server's IP address against it's own Subnet Mask.
181.197.10100000.0 = 181.197.160.0 -Workstation's Network ID 181.197.10000000.0 = 181.197.128.0 -Server's Network ID
* -These two network ID's not exactly the same, off by one bit (32) Therefore, the Workstation knows the server isn't on an attached network,
Send packet to my Default Gateway, let it handle packets destined for other networks.
Q9. Which of the following might correct a problem like this? Change the subnet address of the workstation to 255.255.248.0?
Answer: No, You know this would make the create more subnets, when you would need fewer subnets with larger ranges. 181.197.10111011.212 = 181.197.187.212 -Workstation's configured IP Address
255.255.11111000.0 = 255.255.248.0 -Testing 248 Subnet Mask 181.197.10111000.0 = 181.197.184.0 -New Subnet ID 181.197.10111111.255 = 181.197.191.255 -Same Broadcast ID as configured, but subnets are increments of 8
Q10. Change the entire network's subnet mask to 255.255.240.0?
Answer: No, You would still be on different subnets, but changing the network's subnet mask to 255.255.192.0 would allow these two IP addresses to be on the same subnetwork. This might be a great deal of work, but you can see that creating fewer subnets will make those usable subnets have a broader range.
181.197.10001111.236 = 181.197.143.236 -Server's configured IP address
255.255.11110000.0 = 255.255.240.0 -Testing 240 subnet mask 181.197.10000000.0 = 181.197.128.0 -Server's Network ID with 240 subnet mask
181.197.10111011.212 = 181.197.187.212 -Workstation's configured IP Address
255.255.11110000.0 = 255.255.240.0 -Testing 240 subnet mask 181.197.10110000.0 = 181.197.176.0 -Workstation's Network ID with 240 subnet mask
181.197.10000000.0 = 181.197.128.0 -Server's Network ID with 240 subnet mask
181.197.10110000.0 = 181.197.176.0 -Workstation's Network ID with 240 subnet mask
** -These two network ID's are still on different subnets, even if both are on 240 subnet mask
Just for an Example, look what happens if you change the network's Subnet mask to 255.255.192.0:
There are 4 Unique Subnets created, only 2 are Usable.
181.197.10000000.0 = 181.197.128.0 -Server's Network ID with 192 subnet mask
181.197.10000000.0 = 181.197.128.0 -Workstation's Network ID with 192 subnet mask
Q11. Change the IP address of the server?
Answer, this was a tricky question, and there were many ways to answer this one. If you changed the IP Address and Subnet Mask for the server, then you would be able to place the server on the same subnet as the Workstation. The question only mentioned two computers on the network, so changing the server's IP address might not impact anyone else. There is also the possibility that the Server is handling thousands of clients, and changing the IP address might cause the server to become unreachable.
If you gave the server an IP address of 181.197.187.213/20, then the two devices would be able to communicate.
181.197.10111011.212 = 181.197.187.212 -Workstation's
configured IP Address
255.255.11100000.0 = 255.255.224.0 -Workstation's configured Subnet Mask
181.197.10100000.0 = 181.197.160.0 -Workstation's Network ID 181.197.10111011.213 = 181.197.187.213 -Server's New Address 255.255.11110000.0 = 255.255.240.0 -Server's New Subnet Mask 181.197.10110000.0 = 181.197.160.0 -Server's New Network ID
Interestingly enough, these two machines should be able to communicate on the same physical network, even with different subnet masks.
This is due to the fact that the workstation will first check to see if the server is on it's subnet.
Workstation:
255.255.11100000.0 = 255.255.224.0 -Workstation's configured Subnet Mask
181.197.10111011.213 = 181.197.187.213 -DESTINATION IP Address
181.197.10100000.0 = 181.197.160.0 -Result of ANDing the Destination Address,
Workstation determines that this destination is on it's own network, and ARP's for the Server's MAC address.
Workstation then initiates an ICMP Echo Request to the server.
Server:
255.255.11110000.0 = 255.255.240.0 -Server's configured Subnet Mask
181.197.10111011.212 = 181.197.187.212 -DESTINATION IP Address
255.255.10110000.0 = 181.197.176.0 -Result of ANDing the Destination Address,
Server Determines that the Workstation is on it's same network, and sends an Echo Reply.
https://cis.temple.edu/~tug29203/18spring-3329/reading//hw2a.pdf
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