circle the network portion of these addresses:
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1) a. circle the network portion of these addresses:
i. 209.240.80.78
ii. 199.155.77.56
iii. 117.89.56.45
iv. 215.45.45.0
v. 192.200.15.0
vi. 209.240.80.78
b. circle the host portion of these addresses
i. 198.125.87.177
ii. 223.250.200.222
iii. 17.45.222.45
iv. 126.201.54.231
v. 191.41.35.112
vi. 198.125.87.177
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- 52- A network for a local office requires 4 LAN subnets as follows: · Subnet A - North Office with 31 hosts; · Subnet B - East Office with 15 hosts; and · Subnet C - South Office with 7 hosts. · Subnet D – West Office with 4 hosts. Given the IP Address 192.168.15.0 /24 and above described network requirements: 1. Design an optimal address scheme using Variable Length Subnet Mask (VLSM) method; What will be the subnet mask of Subnet A? Select one: A. 255.255.255.192 B. 255.255.128.255 C. 255.255.255.128 D. 255.255.192.192 E. 255.255.192.05.1a)Define subnetting in the context of computer networks. Explain its purpose and benefits. b) Given the IP address 192.168.0.0 and subnet mask 255.255.255.0, calculate the number of subnets and the number of hosts per subnet. Show your calculations. c) Explain the concept of a subnet mask and its role in subnetting1-In a network based on the bus topology, the bus is a non-shareable resource for which the machines must compete in order to transmit messages. How is deadlock controlled in this context? 2-Using 32-bit Internet addresses was originally thought to provide ample room for expansion, but that conjecture is not proving to be accurate. IPV6 uses 128-bit addressing. Will that prove to be adequate? Justify your answer’ (for example, you might compare the number of possible addresses to the population of the world).
- 57- A network for a local office requires 4 LAN subnets as follows: · Subnet A - North Office with 31 hosts; · Subnet B - East Office with 15 hosts; and · Subnet C - South Office with 7 hosts. · Subnet D – West Office with 4 hosts. Given the IP Address 192.168.15.0 /24 and above described network requirements: 1. Design an optimal address scheme using Variable Length Subnet Mask (VLSM) method; What will be the range of usable (assignable) host addresses in Subnet C? Select one: A. 192.168.15.64 to 192.168.15.128 B. 192.168.15.97 to 192.168.15.110 C. 192.168.15.140 to 192.168.15.191 D. 192.168.15.128 to 192.168.15.139 E. 192.168.15.193 to 192.168.15.254Short answer27. A company is assigned to an address block 136.23.12.64/26, now it needs to be further divided into four subnets of the same size. Question: (1) How long is the network prefix of each subnet? (2) How many addresses are there in each subnet? (3) What is the address block of each subnet? (4) What is the minimum address that each subnet can be assigned to a host? (5) What is the maximum address that each subnet can be assigned to a host?58- A network for a local office requires 4 LAN subnets as follows: · Subnet A - North Office with 31 hosts; · Subnet B - East Office with 15 hosts; and · Subnet C - South Office with 7 hosts. · Subnet D – West Office with 4 hosts. Given the IP Address 192.168.15.0 /24 and above described network requirements: 1. Design an optimal address scheme using Variable Length Subnet Mask (VLSM) method What will be the network address of Subnet D? Select one: A. 192.168.15.232 B. 192.168.15.191 C. 192.168.15.112 D. 192.168.15.240
- 53- A network for a local office requires 4 LAN subnets as follows: · Subnet A - North Office with 31 hosts; · Subnet B - East Office with 15 hosts; and · Subnet C - South Office with 7 hosts. · Subnet D – West Office with 4 hosts. Given the IP Address 192.168.15.0 /24 and above described network requirements: 1. Design an optimal address scheme using Variable Length Subnet Mask (VLSM) method; and What will be the network address of Subnet B? Select one: A. 192.168.15.64 B. 192.168.15.0 C. 192.168.15.32 D. 192.168.15.1A 1-km-long, 10-Mbps CSMA/CD LAN (i.e., Classic Ethernet)has a propagation speed of 200 m/μsec (i.e., how fast the radio signal moves on the link). Thereare no repeaters in this system. Data frames are 512 bits long, including 480 bits of payload (i.e.,effective data) and 32 bits of overhead (e.g., header, checksum, and other overhead fields). Everydata frame should be confirmed by a 64-bit ACK frame. We assume the sender needs to use onecontention slot to seize the channel before transmitting the data frame, and the receiver needsto use one contention slot to seize the channel before transmitting the ACK frame. Answer thefollowing questions: a. What is the minimum length of a contention slot in μsec? b. What is the effective data rate for the sender to send one data frame? Assume that there areno collisions. Show the calculation c. What would be the impact on the total delay and the effective data rate if we used a longerdata payload size in the data frame? Given one application…42. Though there are practical reasons why we don't complicate our lives this way in the real world, from a theoretical perspective we could borrow non-consecutive bits for subnetting. So, when borrowing the first host bit, the second host bit and the last two host bits (counting from the left of the host bit field) from 172.1.0.0 can you determine the following: a. subnet maskb. network address for 0th subnet, 1st subnet and 2nd subnet?c. total number of subnets and number of useable subnets?
- Given a network address of 218.35.50.0, 5 needed subnets, determine the following and Fill up the table below. show the step by step soluIdentify the two wrong statements from the following. 1- CO iPvd uses 32bit address and IPv6 uses 64bit address.2- GO iPvd uses 32bit address and IPV6 uses 64bit hexadecimal address.3- CO iPvd uses 32bit address and IPV6 uses 64 bit binary addressG 4- Pv. uses 32bit hexadecimal address and IPV6 uses 64 bit address.42. Though there are practical reasons why we don't complicate our lives this way in the real world, from a theoretical perspective we could borrow non-consecutive bits for subnetting. So, when borrowing the first host bit, the second host bit and the last two host bits (counting from the left of the host bit field) from 172.1.0.0 can you determine the following: g. 1,203rd useable host address?