CS计算机代考程序代写 DHCP algorithm Microsoft Word – quiz3sol.docx

Microsoft Word – quiz3sol.docx

CMPT 471 QUIZ 3 SOLUTION

1. [2 points] What type of ICMPv4 message is not used in the IPv4 implementation of
traceroute in the virtual lab?
A. destination unreachable: port unreachable
B. destination unreachable: host unreachable
C. destination unreachable: network unreachable
D. time to live exceeded message

B and D
A and B
A, C and D
A and D
A and C

NO CORRECT ANSWER, POINTS GIVEN TO ALL STUDENTS

2. [2 points] QUESTION REMOVED (DUPLICATE)

3. [2 points] QUESTION REMOVED (DUPLICATE)

4. [2 points] Which of the following statements is false?

A DHCPDISCOVER message is used to discover if the lease on a packet is still valid
A DHCPRELEASE message is used to release a lease the host no longer needs
When in the rebind time expires, then any further DHCPREQUEST messages sent will

be broadcast so they reach all DHCP servers
A DHCP client does not send DHCP offer messages
A packet that has passed through one dhcprelay will have the IP address of the host

running the dhcprelay in the DHCP message it is carrying

5. [5 points] Give a detailed step by step description of the algorithm used to generate
EUI-64 addresses for a host given an IPv6 prefix. The given prefix may be a global prefix,
a link local prefix or a unique local address range prefix. Each step should be no more
than 2 short sentences and may be as short as one phrase. You should not have more
than 6 steps. Be concise, clear and complete. Number your steps.

1) Take the 48 bit Ethernet address of the interface and break it into two parts
2) Insert ff:fe between the two parts
3) Convert the first two hexadecimal digits to binary
4) Flip the seventh bit of the resulting binary
5) Convert the binary back to hexadecimal
6) Append the resulting 64 bit EUI 64 address to the IPv6 prefix

1 point for steps 1,2,4,6, ½ point for steps 3 and 5

6. [5 points] Consider SLAAC ( IPv6 automatic address configuration)

An IPv6 address is made up of three parts, the routing prefix, the interface ID and the
subnet.
 The interface ID is the portion of the address that is unique to each host on the

network. T
 The global routing prefix is extracted from the router announcement.
 The routing prefix for the link local address is fe80::/48 for the link local address.
Consider the ifconfig output shown below.

What is the length of each part of the address for this case?
Interface ID 64 bits (1 point)
Prefix 64 bits (1 point)
Subnet 0 bits (1 point)
What is in each of the three parts for the link local address? Include all zeros in the
addresses.
Prefix fdd0:8184:d967:0019 (1 point)
Interface ID b6a8:475b:4448:fff9 (1 point)

7. [5 points] A source host sends an IPv4 packet with an MTU of 1500 octets. The MTU or
maximum transmission unit indicates the length of the data field in the Ethernet frame
(the length of the IP packet in the Ethernet frame). The IP packet in the Ethernet frame
includes the IP header (32 bytes), the UDP header (8 bytes) and the UDP data.

If the IP header is 32 bytes long are there any options in the header?

On its way to the destination the Ethernet frame passes through a network with a MTU
of 650 octets. Explain how the IP packet is fragmented using the diagram below. The
diagram shows the original Ethernet frame and an Ethernet frame containing each of
the resulting fragments. For each of the numbers on the diagram give the value that
corresponds to it.

1) 1500 bytes 6) 0
2) 0 bytes 7) 650
3) 0 bytes 8) 77
4) 650 9) 650
5) 0 10) 1232

½ point for each of the 10 items

8. [6 points] Two networks belonging to company A have CIDR address blocks
of 202.78.52.0/22, and 202.78.64.0/20. Company B has CIDR address block
202.78.32.0/20 Company A’s router is directly connected ( through a point to point link)
to eth0 to router Y. Company B’s router is directly connected ( through a point to point
link) to eth1 on router Y.

1) [1 points] For Company B’s network how many addresses are used by the
network?
202.78.32.0/20 32 -20=12 212 = 4096 addresses (1 point) 4094 for hosts

2) [2 points] Which of the addresses in 1 , if any, cannot be assigned to network
hosts? Why? Give the addresses and the reason each address is not used
202.78.56.0, 202.78.52.0, 202.78.64.0, (½ point)
Network IDs not used as host addresses(½ point)
202.78.63.255, 202.78.55.255, 202.78.79.255 (½ point)
Broadcast addresses that should not be used as host addresses.

(½ point)
3) [3 points] Given the diagram and table below. What would b\e the aggregated

CIDR network (netid and mask) that would be used in the ISP’s routing table for
company A?
202.78.0.0 (1 point) 255.255.128.0 or /17 (1 point)
Is the network allocation for company B within company A’s aggregated
network? Why?
Yes. Because 202.78.32 in within the range of addresses for company A’s
network(½ point) from 202.78.0.0 to 202.78.127.0(½ point)

arrow points to ISP router(which connects to Internet)