Module 8 Flashcards
Q18-2 Why is routing the routing the responsibility of the network layer? In other words, why can’t the routing be done at the transport layer or data-link layer?
Routing moves data between layer 3 devices. Routing requires the information contained in the IP header to make its decisions its decisions such as the source and destination IP address. This information does not exist at either the transport layer or the data-link layer.
Q18-1 Why does the network layer protocol need to provide packetizing service to the transport layer? Why can’t the transport layer send out the segments with out encapsulating then in datagrams?
. The transport layer communication is between two ports; the network layer communication is between two hosts. This means that each layer has a different source/destination address pair; each layer needs a different header to accommodate these pair of addresses. In addition, there are other pieces of information that need to be separately added to the corresponding header.
Q18-3 Distinguish between the process of routing a packet from the source to the destination and the process of forwarding a packet at each router.
Forwarding is delivery to the next node. A router uses its forwarding table to send a packet out of one of its interfaces and to make it to reach to the next node. In other words, forwarding is the decision a router makes to send a packet out of one of its interfaces. Routing, on the other hand, is an end-to-end delivery resulting in a path from the source to the destination for each packet. This means a routing process is a series of forwarding processes. To enable each router to perform its forwarding duty, routing protocols need to be running all of the time to provide updated information for forwarding tables. Although forwarding is something we can see in the foreground, in the background, routing provides help to the routers to do forwarding.
Q18-7 Do we have any of the following at the network layer of TCP/IP? If not why? a) flow control b) error control c) congestion control
None of these services are implemented for the IP protocol in order to make it simple.
Q18-11 In classless addressing, we know the first address and the number of addresses in the block. Can we find the prefix length?
Yes. We can find the prefix length using only the block size. The prefix length is directly related to the block size as shown below: n = 32 − log2N First = 192.168.0.1 Number of addresses = 64 255.255.255.192 2^bits borrowed
P18-1 What is the size of the address space in each of the following systems? a) A system in which each address is only 16 bits b)A system in which each address is made of six hexadecimal bits. c) A system in which each address is made of four octal digits.
The size of the address in each case is the base to the power of the number of digits: a. The size of the address space is 216 = 65,536. b. The size of the address space is 166 = 16,777,216. c. The size of the address space is 84 = 4096
P18-3 Rewrite the following IP addresses using dotted-decimal notation: a) 0101110 1011000 01110101 00010101 b)10001001 10001110 11010000 00110001 c) 01010111 10000100 00110111 00001111
We change each 8-bit section to the corresponding decimal value and insert dots between the bytes. a. 94.176.117.21 b. 137.142.208.49 c. 87.132.55.15
P18-4 Find the class of each of the following classful IP addresses. a) 130.34.54.12 b) 200.34.2.1 c) 145.34.2.8
a) 130.34.54.12 Class B b) 200.34.2.1 Class C c) 145.34.2.8 Class B
P18-7 In classless addressing what is the size of the block N if the value of the prefix length (n) is one of the following. a) n = 0 b) n = 14 c) n = 32
We can use the formula N = 2^32− n a. N = 232−0 = 4,294,967,296 b. N = 232−14 = 262,144 c. N = 232−32 = 1
P18-11 Which of the following cannot be a class in CIDR a) 255.225.0.0 b) 255.192.0.0 c) 255.255.255.6
We first write each potential mask in binary notation and then check if it has a contiguous number of 1s from the left followed by 0s. a. 11111111 11100001 00000000 00000000 Not a mask b. 11111111 11000000 00000000 00000000 A mask c. 11111111 11111111 11111111 00000110 Not a mask
P18-13 Show the n leftmost bits that can be used in a forwarding table. a. 170.40.11.0 /24 b. 110.40.240.0 /22 c. 70.14.0.0 /18
We write the address in binary and then keep only the leftmost n bits. a. 10101010 00101000 00001011 b. 01101110 00101000 111100 c. 01000110 00001110 00
P18-17 Assume we have an internet with a 12 bit address space. The addresses are divided between 8 networks (N0 - N7), with 64, 192, and 256 addresses respectively. The internetwork communication is done through a router with 8 interfaces (m0 to m7). Show the internet outline and the forwarding table (with two columns: prefix in binary and the interface number) for the only router that connects the networks. Assign a network address to each network.
The total number of addresses is 212 = 4096. This means that there are 512 addresses for each network. We can divide the whole address space into eight blocks (block 0 to block7), each of 512 addresses. The addresses in each block are allocated as (0 to 511), (512 to 1023), (1024 to 1535), (1536 to 2047), …, (3584 to 4095). It can be checked that each block is allocated according to the two restrictions needed for the proper operation of CIDR. First, the number of addresses in each block is a power of 2. Second, the first address is divisible by the number of addresses as shown below: Block 0: 0 / 512 = 0 Block 1: 512 / 512 = 1 Block 2: 1024 / 512 = 2
P18-19 Combine the following three blocks of addresses into a single block. a) 16.27.24.0/26 b) 16.27.24.64 /26 c) 16.27.24.128 /25
16.27.24.0 /24
P18-21 An ISP is granted the block 16.12.64.0 /20. The ISP needs to allocate addresses for 8 organizations, each with 256 addresses. a) Find the number and range of the addresses in the ISP block b) Find the range of addresses for each organization and the range of unallocated addresses. c) Show the outline of the address distribution and the forwarding table.
Q18-10 In classless addressing, we know the first and the last address in the block. Can we find the prefix length? If the answer is yes, show the process.
The answer is Yes, we can find the prefix length based on the first address and the number of addresses in the block. For example, say you the address 192.168.0.1 and you know there is 32 addresses in the block. You would determine how bits are required to have that many addresses in the block. 2^5 = 32 therefore 27 bits are required for the prefix. 192.168.0.1 /27
P18-2 Rewrite the following IP addresses in binary notation:
- 110.11.5.88
- 12.74.16.18
- 201.24.44.32
- 01110010.00001011.00000101.01011000
- 00001100.01001010.00010000.00010010
- 11001001.01001010.00010000.00100000
P18-16 Assume we have an internet with an 8 bit address space. The addresses are divided between four networks (n0 - n3). The internetnetwork communication is done through a router with four interfaces (m0 - m3). Show the internet outline and the forwarding table (with two columns: prefix in binary and the interface number) for only router that connects the networks. Assign a network address to each network.
