Chapter 21- IP: Internet Addressing
21.1 Could IP be redesigned to use hardware addresses instead of the IP addresses it currently uses? Why or why not?
Answer: MAC addresses cannot be used, because the Internet can include multiple network technologies and each technology can define its own MAC addresses. But if the Internet were to be designed from the beginning, for sure so many things would be different, probably much easier. Backward compatibility and various tries by different companies made the system complicated.
21.2 What does the Internet address hierarchy allow a local manager to do?
Answer: It allows a local manager to divide the network into subnetworks for better management in terms of security and traffic segregation.
21.3 In the original IPv4 classful address scheme, was it possible to determine the class of an address from the address itself? Explain.
Answer: Yes, since in the classful addressing scheme initial bit(s) gives indication about the class being used.
21.4 Write a computer program that accepts a dotted decimal address as input and displays a string of 32 bits.
21.5 Write a computer program that accepts a colon-hex address as input and displays a string of 128 bits.
21.6 Write a computer program that reads an IPv4 address in dotted decimal form and determines whether the address is a multicast address.
21.7 Write a computer program that translates between CIDR slash notation and an equivalent dotted decimal value.
21.8 If an ISP assigned you a /28 IPv4 address block, how many computers could you assign an address?
Answer: When an organization is assigned /28 CIDR address, it means 28 bits out of 32 bits are fixed, so 32-28 = 4 bits available for user space. So number of users 24-2 = 4, since the all 0s and all 1s address are having special use and can’t be assigned to a user.
21.9 If an ISP assigned you a /28 IPv6 address block, how many computers could you assign an address?
21.10 If an ISP offers a / 17 address block for N dollars per month and a / 16 address block for 1.5 N dollars per month, which has the cheapest cost per computer?
Answer: Number of addresses in /17 block 232-17 = 215 Price per address: 215/N = 215/N
Number of addresses in /16 block 232-16 = 216 Price per address: 216/1.5N = 2×215/1.5N = 1.33×215/N
So /16 address block will be cheaper in comparison with the price given for /17 block.
21.11 Is the CIDR prefix 1.2.3.4/29 valid? Why or why not?
21.12 Suppose you are an ISP with a / 24 IPv4 address block. Explain whether you accommodate a request from a customer who needs addresses for 255 computers. (Hint: consider the special addresses.)
Answer: For /24 address block, number of available addresses will be 232-24 = 28 = 256. However, a suffix with all 0s address is reserved for network ID and a suffix with all 1s address is reserved for broadcast address, so number of addresses that can be assigned to computers/hosts will be 256 -2 = 254.
21.13 Suppose you are an ISP that owns a / 22 IPv4 address block. Show the CIDR allocation you would use to allocate address blocks to four customers who need addresses for 60 computers each.
Answer: The /22 address block can be assigned as follows:
ddd.ddd.ddd.00/26
ddd.ddd.ddd.01/26
ddd.ddd.ddd.10/26
ddd.ddd.ddd.11/26
21.14 Suppose you are an ISP that owns a / 22 IPv4 address block. Can you accommodate requests from six customers who need addresses for 9, 15, 20, 41, 128, and 260 computers, respectively? If so, how? If not, explain why.
Answer: If an ISP owns /22 address block, then number of addresses available will be 232-22 = 210 = 1024
We should look at the distribution of these available addresses in a very efficient way. For example, if we want to assign a block for 260 addresses, 28 = 256 is less and 29 =512 is a lot (lots of addresses are wasted). The reader is expected to analyze the situation accordingly and answer the remaining of the question.
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