Using The IP Number Mask Is 2192023 Please Find The Answers

Using The Ip Number Mask Is 2192023 Please Find The Answers To

Using The Ip Number Mask Is 2192023 Please Find The Answers To

The assignment involves analyzing two IP address ranges and subnetting schemes. First, the task concerns the IP address 21.9.2.0/23, requiring calculations of host bits before and after subnetting, number of borrowed bits, usable hosts per subnet, total subnets, and identifying specific subnet and broadcast addresses for subnet 8190. The second part involves the IP address 158.20.1.0/27, with similar subnetting calculations for subnet 2012. Finally, the task includes creating a VLSM addressing scheme for the class C network 223.201.10.0, assigning subnets with different subnet masks, shading, and color-coding for visualization.

Paper For Above instruction

Subnetting Calculations for 21.9.2.0/23

Address 21.9.2.0/23 provides a subnet mask of 255.255.254.0, indicating that the first 23 bits are allocated for network identification, and the remaining bits are available for host addresses. Initially, an IP network with a /23 prefix has 32 - 23 = 9 host bits, as the total IPv4 address space comprises 32 bits.

1. Host Bits Before Subnetting

Before subnetting, the subnet mask is /8 (class A default) or /23 as given, but in terms of host bits, a class C network traditionally has 8 host bits. However, as per the provided mask /23, the total host bits available per subnet initially are 32 - 23 = 9 bits (since all bits are host bits in a default /23). Thus, the number of host bits before subnetting is 9.

2. Host Bits After Subnetting

After subnetting, the subnet mask might borrow bits from the host portion. Given the mask remains /23, the number of host bits after subnetting stays at 9 because no additional bits are borrowed unless specified. Therefore, the number of host bits after subnetting is 9.

3. Borrowed Host Bits to Subnet

The question appears to evaluate if subnetting involves borrowing bits. Since starting with /23, if subnetting is performed further, the borrowed bits depend on the subnetting process itself. Without extra subnetting, no bits are borrowed from the host portion. Therefore, the number of borrowed bits is 0 in this scenario, unless additional subnetting is specified.

4. Usable Hosts per Subnet

The formula for usable hosts is 2host bits - 2. For 9 host bits, usable hosts per subnet = 29 - 2 = 512 - 2 = 510. Thus, each subnet can support approximately 510 usable hosts.

5. Usable Subnets

The total number of subnets depends on the subnetting process. Since the starting mask is /23, and assuming no further subnetting, the number of subnets varies according to the bits borrowed from the network address. However, typically, with /23, the total subnets in a Class A or Class B network can be computed based on the bits borrowed. For this case, assuming default Class B network with 16 network bits, possibly 2 bits borrowed to make /23, resulting in 22 = 4 subnets. This can be refined once the specific network class is confirmed.

6. Subnet Mask in Dotted Decimal

The subnet mask for /23 is 255.255.254.0.

7. Subnet Mask in Hexadecimal

Converting 255.255.254.0 to hex: 255 = FF, 255 = FF, 254 = FE, 0 = 00, so the mask is FF.FF.FE.00 in hex.

8. Subnet IP Number for Subnet 8190

To find subnet 8190, assuming subnet numbering begins at 0, with each subnet increment representing a block of addresses determined by the subnet mask. Since with /23, each subnet has 512 addresses, subnet 8190 corresponds to subnet number 8190 in decimal. The subnet IP is then calculated: starting from 21.9.2.0, the actual subnet address for subnet 8190 is obtained by multiplying subnet number with the size of each subnet (512 addresses). This calculation indicates the subnet IP for subnet 8190 can be pinpointed as 21.9.0.0 or a similar value depending on the network segmentation, but in this case, considering the total subnets and their possible numbering, subnet 8190 points outside typical range and indicates larger addressing schemes. For demonstration, if subnet 0 is 21.9.2.0, then subnet 8190 is beyond standard limits, but assuming sequential numbering, the subnet IP number is 21.9.0.0.

9. Broadcast IP Number for Subnet 8190

The broadcast IP address for a subnet is the last IP in its range. For a subnet with network address 21.9.x.x/23, the broadcast address is network address + 511 (the total number of hosts per subnet minus 1). Therefore, assuming subnet 8190 corresponds to 21.9.0.0/23, the broadcast address is 21.9.1.255.

10. Usable Host IP Numbers of Subnet 8190

Usable host IPs are the addresses from network address + 1 up to broadcast address - 1. For example, if the network address is 21.9.0.0 and broadcast is 21.9.1.255, then usable hosts are from 21.9.0.1 to 21.9.1.254.

Subnetting Calculations for 158.20.1.0/27

The IP range 158.20.1.0/27 has a subnet mask of 255.255.255.224, providing 32 addresses per subnet.

11. Host Bits Before Subnetting

Class C default mask is /24, giving 8 host bits. Thus, before subnetting, there are 8 host bits.

12. Host Bits After Subnetting

With /27 mask, the number of host bits reduces to 32 - 27 = 5 bits.

13. Borrowed Host Bits to Subnet

Host bits borrowed are 27 - 24 = 3 bits.

14. Usable Hosts per Subnet

Number of usable hosts per subnet = 25 - 2 = 32 - 2 = 30 hosts.

15. Usable Subnets

Number of subnets in this scheme if starting from /24 is 23 = 8.

16. Subnet Mask in Dotted Decimal

255.255.255.224.

17. Subnet Mask in Hexadecimal

FF.FF.FF.E0.

18. Subnet IP for Subnet 2012

Calculating the subnet ID for subnet 2012 involves determining the block size: 32 addresses per subnet. The network starting at 158.20.1.0, with subnetting, the subnet 2012 corresponds to 158.20.1.0 + (2012 * 32 addresses), which equates to a subnet network address of 158.20.48.0.

19. Broadcast IP for Subnet 2012

Broadcast address = network address + 31 = 158.20.48.31.

20. Usable host IPs for Subnet 2012

Range from 158.20.48.1 to 158.20.48.30 is the set of usable host IP addresses.

VLSM Addressing Scheme for 223.201.10.0

The network 223.201.10.0 is a Class C address. To optimize the address space for various department sizes, variable-length subnet masks (VLSM) are used. Starting with the largest subnet, assign appropriate subnets with mask sizes tailored to each group's size, shading, and coloring as instructed.

For example, a large department requiring 60 hosts could use /26 (255.255.255.192), which provides 62 usable addresses, shaded in light blue. Smaller departments needing fewer hosts might use /28 (255.255.255.240), shaded in purple, or /30 (255.255.255.252) for point-to-point links shaded in red. The entire IP scheme involves sequential subnet allocation, with the largest subnets first, then smaller ones, ensuring no overlaps and efficient utilization.

Conclusion

Proper understanding of subnetting, including calculating host bits, subnet masks, subnet and broadcast addresses, along with visual representation via VLSM, enhances efficient IP address management in various network architectures. Accurate calculations ensure optimized address space and network scalability, fundamental skills for network administrators and architects.

References

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