Q6. Define the type of the following destination addresses:
a. 4A:30:10:21:10:1A b. 47:20:1B:2E:08:EE
c. FF:FF:FF:FF:FF:FF Q7. An Ethernet MAC sublayer receives 1510 bytes of data from the upper layer. Can the data be encapsulated in one frame? If not, how many frames need to be sent? What is the size of the data in each frame?
7. The maximum data size in the Standard Ethernet is 1500 bytes. The data of 1510 bytes, therefore, must be split between two frames.
Data size for the first frame: 1500 bytes
Data size for the second frame: 46 bytes (with padding).
Q8. What are the common Standard Ethernet implementations?
Q9. What are the common Fast Ethernet implementations?
Q10. An Ethernet MAC sublayer receives 42 bytes of data from the upper layer. How many bytes of padding must be added to the data?
CHAPTER 15 Connecting LANs Q1 1. How is a repeater different from an amplifier?
An amplifier amplifies the signal, as well as noise that may come with the signal,whereas a repeater regenerates the signal, bit for bit, at the original strength.
Q2. How is a hub related to a repeater?
A hub is a multiport repeater.
Q3 Complete the Five categories of connecting devices.
CHAPTER 19 Network Layer: Logical Addressing Q1. What is the number of bits in an IPv4 address? What is the number of bits in an IPv6 address?
1. An IPv4 address is 32 bits long. An IPv6 address is 128 bits long.
Q2. What are the differences between classful addressing and classless addressing in IPv4?
2. Classful addressing assigns an organization a Class A, Class B, or Class C block of addresses. Classless addressing assigns an organization a block of contiguous addresses based on its needs.
Q3. List the classes in classful addressing and define the application of each class (unicast , multicast, broadcast, or reserve). Q4. What is the network address in a block of addresses? How can we find the network address if one of the addresses in a block is given? 4. The network address in a block of addresses is the first address. The mask can be ANDed with any address in the block to find the network address.
5 - Explain the type of Classes with example?
س 1- اشرح انواع الكلاسات مع اعطاء امثلة ؟
الحل: (ص 552)
IP : 126.96.36.199
IP : 188.8.131.52
IP : 184.108.40.206
6) Write the following address in binary digits?
حول العنوان التالي من النظام العشرى الى النظام الثنائي:
الحل (لمزيد من المعلومات ص 555) :
Q7) What is the address space in each of the following systems?
a. A system with 8-bit addresses
b. A system with 16-bit addresses 7
Q8). Find the class of the following IP addresses.
a. 11110111 11110011 10000111 11011101
b. 10101111 11000000 11110000 00011101
c. 11011111 10110000 00011111 01011101
d. 11101111 11110111 11000111 00011101
a. Class E (first four bits are 1s)
b. Class B (first bit is 1 and second bit is 0
c. Class C (first two bits are 1s and the third bit is 0)
d. Class D (first three bits are 1s and the fourth bit is 0) Q9) In a block of addresses, we know the IP address of one host is 220.127.116.11/16. What are the first address (network address) and the last address (limited broadcast address) in this block? 9. With the information given, the first address is found by ANDing the host address with the mask 255.255.0.0 (/16).
The last address can be found by ORing the host address with the mask complement 0.0.255.255.
The first address 18.104.22.168
The last address 22.214.171.124
Q10) An organization is granted the block 192.168.1.0/8. The administrator wants to create 32 fixed-length subnets.
a. Find the subnet mask.
b. Find the number of addresses in each subnet.
c. Find the first and last addresses in subnet 192.168.1.8.
d. Find all the host addresses in subnet 192.168.1. .
: ( نستخدم هذه الطريقة في حالة عدد الشبكات الفرعية متساوي)
From formula : No of Subnets = 2n , n is the number bits that borrowed from the host
32 = 2n ==>n = 5,
From formula : No of hosts = 2x -2, n is the number bits that remained for the host
Then in the host port of class C Class is:
128 64 32 16 8 4 2 1
Network (حدود الشبكة) host
The network will be: ( عدد الشبكات الفرعية هي)
Subnet Mask ; 255.255.255.248
اذا فتحنا احد هذه الشبكات الفرعية فان
(كل شيكة تحوى على 6 عناوين )
192.168.1.8 192.168.1.8 Network ID
192.168.1.15 Broadcast ID
Q11) An organization is granted the block 126.96.36.199/8. The administrator wants to create 500 fixed-length subnets.
a. Find the subnet mask.
b. Find the number of addresses in each subnet.
c. Find the first and last addresses in subnet 1.
d. Find the first and last addresses in subnet 500.
a. If we calculate the checksum, we get 0x0000. The packet is not corrupted.
b. Since the length of the header is 20 bytes, there are no options.
c. Since M = 0 and offset = 0, the packet is not fragmented.
d. The total length is 84. Data size is 64 bytes (84 20).
e. Since the value of time to live = 32, the packet may visit up to 32 more routers.
f. The identification number of the packet is 3.
g. The type of service is normal. Q5. What is the difference between the delivery of a frame in the data link layer and the delivery of a packet in the network layer?
The delivery of a frame in the data link layer is node-to-node. The delivery of a packet at the network layer is host-to-host.
Q6 The value of HLEN in an IPv4 datagram is 7. How many option bytes are present?
Q7. The size of the option field of an IPv4 datagram is 20 bytes. What is the value of HLEN? What is the value in binary? If the size of the option field is 20 bytes, then the total length of the header is 40 bytes (20 byte base header plus 20 bytes of options). The HLEN field will be the total number of bytes in the header divided by 4, in this case ten (1010 in binary). Q8. An IPv4 datagram is carrying 1024 bytes of data. If there is no option information, what is the value of the header length field? What is the value of the total length field? Since there is no option information, the header length is 20, which means that the value of HLEN field is 5 or 0101 in binary. The value of total length is 1024 + 20 or 1044 (00000100 00010100 in binary).
Q9. Table 20.5 lists the MTUs for many different protocols. The MTUs range from 296 to 65,535. What would be the advantages of having a large MTU? What would be the advantages of having a small MTU? Advantages of a large MTU:
■ Good for transferring large amounts of data over long distances
■ No fragmentation necessary; faster delivery and no reassembly
■ Fewer lost datagrams
■ More efficient (less overhead)
Advantages of a small MTU:
■ Good for transferring time-sensitive data such as audio or video
■ Better suited for multiplexing Q10. An IPv4 datagram arrives with fragmentation offset of 0 and anM bit (more fragment bit) of O. Is this a first fragment, middle fragment, or last fragment? If the M (more) bit is zero, this means that the datagram is either the last fragment or the it is not fragmented at all. Since the offset is 0, it cannot be the last fragment of a fragmented datagram.The datagram is not fragmented
CHAPTER 21 Network Layer: Address Mapping And Error Reporting Q1. Is the size of the ARP packet fixed? Explain.
The size of an ARP packet is variable, depending on the length of the logical and physical addresses used. Q2 What is the size of an Ethernet frame carrying an ARP packet if The size of the ARP packet is 28 bytes?
Since the size of the ARP is 28 bytes. We need to pad the data to have the minimum size of 46. The size of the packet in the Ethernet frame is then calculated as 6 + 6 + 2 + 46 + 4 = 64 bytes (without preamble and SFD). Q3. What is the minimum size of an IPv4 packet that carries an ICMPv4 packet? What is the maximum size?