The number of Networks = 2^b; b =Borrow bits
The Number of Usable hosts/IP =2^n-2; n = subnet mask

Magic Value:

Magic value	128	64	32	16	8	4	2	1
Subnet Mask	1	2	3	4	5	6	7	8
	9	10	11	12	13	14	15	16
	17	18	19	20	21	22	23	24
	25	26	27	28	29	30	31	32

Question

🡺 Questions on IP Addressing (google.com)

Questions on IP Addressing

Total points 58

Professional Networking

1. Supernetting provides the following advantages.

Correct answer 🡺 All

It reduces the size of routing updates.

It provides a better overview of network.

It decreases the use of resources such as Memory and CPU.

It decreases the required time in rebuilding the routing tables.

2. What is a Single Network?

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3. Supernetting is the opposite of Subnetting. It is

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1. True

2. False

4. What are the network address, broadcast address, and the subnet mask for a host with the IP Address below?IP Address: 189. 162. 65. 42/ 27

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Network Address: 189.162.65.33 , Broadcast Address: 189.162.65.63 , Subnet Mask: 255.255.255.224

Network Address: 189.162.65.32 , Broadcast Address: 189.162.65.63 , Subnet Mask: 255.255.255.224

Network Address: 189.162.65.33 , Broadcast Address: 189.162.65.63 , Subnet Mask: 255.255.255.240

Network Address: 189.162.65.33 , Broadcast Address: 189.162.65.64 , Subnet Mask: 255.255.255.240

5. What are the network address, broadcast address, and the subnet mask for a host with the IP Address below?IP Address: 105. 105. 40. 81/ 29

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Network Address: 105.105.40.64, Broadcast Address: 105.105.40.88, Subnet Mask: 255.255.255.248

Network Address: 105.105.40.64, Broadcast Address: 105.105.40.87, Subnet Mask: 255.255.255.248

Network Address: 105.105.40.80, Broadcast Address: 105.105.40.87, Subnet Mask: 255.255.255.248

Network Address: 105.105.40.64, Broadcast Address: 105.105.40.87, Subnet Mask: 255.255.255.240

6. What are the network address, broadcast address, and the subnet mask for a host with the IP Address below?IP Address: 42. 155. 65. 64/ 23

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Network Address: 42.155.224.0, Broadcast Address: 42.155.65.255 , Subnet Mask: 255.255.254.192

Network Address: 42.155.64.0, Broadcast Address: 42.155.65.255, Subnet Mask: 255.255.254.0

Network Address: 42.155.192.0, Broadcast Address: 42.155.65.254, Subnet Mask: 255.255.254.0

Network Address: 42.155.128.0, Broadcast Address: 42.155.65.255, Subnet Mask: 255.255.254.0

7. Subnet the Network 203.10.93.0/24 into 25 Subnets. After Subnetting, is IP 203.10.93.30 a valid Host ID?

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1. Yes

2. No

Explanation:

203.10.93.0/24 into 25 Subnets 🡺

🡺 / 25 Mean, 2^0=1, 2^1=2, 2^2 = 4, 2^3 =8, 2^4= 16, 2^5 =32, 2^6=64, 2^7 =128

2^5 =32 ; 32 > 25

5 is borrow bit , because 25 subnet mean = 2^5, 25 is exist on 32 ,

24+5 (borrow of bit)

= /29 Magic value= 8

Network = 0, 8, 16,24,32,40

IP = find out this is valid or not valid 🡺 203.10.93.30

203.10.93.0/29

203.10.93.0.8

203.10.93.0.16

203.10.93.0.24🡺 Network

203.10.93.0.25 🡺 1st valid IP

203.10.93.0.30 🡺 last valid IP (203.10.93.30) 🡺 so this is valid IP

203.10.93.0.31🡺 Broadcast

203.10.93.0.32

203.10.93.0.30

8. Your company has been using the network of 193.56.7.0 /24. You want to put each of the 6 floors in your building on a different subnet. What is the range of the last available subnet after doing Subnetting?

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193.56.7.248 – 193.56.7.255

193.56.7.32 – 193.56.7.64

193.56.7.224 – 193.56.7.255

193.56.7.240 – 193.56.7.255

Explanation:

6 subnet means = 2^3 = 8 , we need 6 , so =6 is exist on 8 because (8>6).

So , borrows bit = 3, (2^3)

/24 + 3 = /27 Magic value = 32

Network = 0, 32, 64, 96, 128,… 224 (32*8 =224) (256 but we go to the 255)

Ip = 193.56.7.0 /27 🡺

193.26.7.224 - 193.26.7.255

9. Is below-given IP address is valid or not and give the reason for the same:17.25.39.127/26

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Yes Valid IP, as it is first hosting IP address of network 17.25.39.126/26.

Yes Valid IP, as it is the last host IP address of network 17.25.39.64/26.

No Valid IP, as it is the network address of network 17.25.39.127/26.

No Valid IP, as it is broadcast address of network 17.25.39.64/26.

Explanation:

/26 = mean = Magic value= 64

Network = 0, 64,128,192, (256 but we go to 255)

17.25.39.64

17.25.39.65 🡺 1st valid IP

17.25.39.127 🡺 broadcast

17.25.39.128

10. You work in an enterprise, where 150.17.0.0/16 network is used. This network has divided into 450 subnets. In one office, you need to configure the 10th valid IP address of the 14th subnet to your router interface. What will be that IP address?

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150.17.13.10

150.17.6.138

150.17.5.10

150.17.7.138

Explanation:

450 subnet means = 2^9 =512 = we need 450 which is exist on 512 (because (512>450)

2^9 = borrow bits =9

/16 +9 = /25 magic value =128

Network: 0,128 (256 but we can go to 255)

Network SL. Network SL.

150.17.0.0 /25 1 150.17.3.128 8

150.17.0.128 2 150.17.4.0 9

150.17.1.0 3 150.17.4.128 10

150.17.1.128 4 150.17.5.0 11

150.17.2.0 5 150.17.5.128 12

150.17.2.128 6 150.17.6.0 13

150.17.3.0 7 150.17.6.128 14 🡺14th Subnet

And, 10th Ip means = 150.17.6.138 (10th of Ip address)

11. You have a class ‘C’ network that needs 28 subnets while maximizing the number of valid host addresses available on each subnet. How many bits need to borrow from the host field to provide the correct combination, what will be the subnet mask and how many maximum valid hosts we will have in each subnet?

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Borrowed bits=5 ; Subnet Mask = /29 ; Maximum Number of Valid Hosts = 6

Borrowed bits=6 ; Subnet Mask = /28 ; Maximum Number of Valid Hosts = 6

Borrowed bits=5 ; Subnet Mask = /27 ; Maximum Number of Valid Hosts = 14

Borrowed bits=4 ; Subnet Mask = /29 ; Maximum Number of Valid Hosts = 30

Explanation:

In class C means = /24

We needs 28 subnet means = 2^5 = 32, (28 is exist on 32 because (32>28)

/24 +5 (borrow bits)

/29 = Magic value = 8

Network = 0,8,16,24…..

Valid host = 2^n – 2= (n=32-29)=3

2^3-2=8-2 =6 ( 6 valid host)

12. Which of the subnet masks we can use for a corporate network with 300 sub-networks and a maximum of 50 host addresses per subnet and working with only Class B address.

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255.255.255.128 & 255.255.255.192

255.255.255.240 & 255.255.255.192

255.255.255.128 & 255.255.255.248

255.255.255.0 & 255.255.255.192

Explanation:

Given that,

300= subnetwork

50 = host (maximum)

Working with = Class B (Ip address)

Now,

300= subnetwork 🡺 2^9 = 512, which is exist on 300 because (500>300)

So, borrow of bits = 2^9 = 9

In the question said, Class B Ip = /16 +9 =/25 subnet = 255.255.255.128 (Ans)

50 host = means (2^n-2= 2^6 -2=62 ) 50 is exist on 62 .62>50.

Hence, 2^6 =32 (32-b) b=6 (borrow bits)

32-6 =/26 = 255.255.255.192 (Ans)

13. 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?

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Explanation:

255.255.255.224 means =/27

In , /27

(32-27)=5 🡺 32 total network

5 =2^5-2= 30 Ans. (-2 mean = 1 network and 1 Broadcast)

14. 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?

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A. 172.16.112.0

B. 172.16.0.0

C. 172.16.96.0

D. 172.16.255.0

15. The network address of 172.16.0.0/19 provides how many subnets and hosts?

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A. 7 subnets, 30 hosts each

B. 8 subnets, 8,190 hosts each

C. 8 subnets, 2,046 hosts each

D. 7 subnets, 2,046 hosts each

16. On a VLSM network, which mask should you use on point-to-point WAN links in order to reduce the waste of IP addresses?

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A. /27

B. /28

C. /29

D. /30

17. What is the subnetwork address for a host with the IP address 200.10.5.68/28?

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A. 200.10.5.56

B. 200.10.5.32

C. 200.10.5.64

D. 200.10.5.0

18. Which two statements describe the IP address 10.16.3.65/23?

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A. 1 and 3

B. 2 and 4

C. 1, 2 and 4

D. 2, 3 and 4

19. Using the following illustration, what would be the IP address of E0 if you were using the Seventh 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 valid for this question.

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