CS4500 Test 4 Study Guide
CS4500 Test 4 Study Guide
Note: This is a study guide, answers may be incorrect, double check and best of luck.
- A ________ network is totally isolated from the global Internet. Private
- On a network that uses NAT, the router can use ______ global address(es). A pool of
- In the sending computer, UDP sends a data unit from the _____ layer. Application
- TCP uses ____ for error detection. Checksum, Acknowledge and time-out
- _____ is an IP layer security protocol that only provides integrity and authentication. AH
- The ______ timer keeps track of the time between the sending of a segment and the receipt of an acknowledgement. Retransmission
- The definition of reliable delivery includes ______. Error-Free Delivery
- Which of the following is not part of the UDP user datagram header? Length of Header
- The _____ defines the server program. Well-Known Port Number
- If the incoming queue of a UDP client overflows, _____ The user datagrams are discarded and a port unreachable message is sent
- TCP lies between the ______ and the ______ layers of the TCP/IP protocol suite. Application; Transport
- Which of the following does UDP guarantee? None of the Above
- IP is responsible for _____ communication while TCP is responsible for _____ communications.Host-to-host ; Process-to-process
- The foreign network address of a mobile host with a care-of address of 47.1.2.3./8 is ______. N/A
- A host can be identified by _____ while a program running on the host can be identified by _______. AN IP address ; a port number
- A network has addresses that range from 192.168.4.0 to 192.168.5.0. This is probably a(n) _____ network. Private
- In _____, data are sent or processed at a very inefficient rate, such as 1 byte at a time. Silly window syndrome
- The _____ address uniquely identifies a running application program. Socket
- If a segment carries data along with an acknowledgment, this is called ______. Piggybanking
- The ____ field is used to order packets of a message. Sequence number
- If the mobile host is on the same network as the remote host, an inefficient situation called _____ occurs when the remote host sends a message to the mobile host. Triangle Routing
- Which of the following is not a valid acknowledgement number? 2^32
- Which field indicates the length of the TCP header? Header Length
- The source port address on the UDP user datagram header defines _____. The application program of the sending computer
- The _____ defines the client program. Ephemeral Port Number
- What is the maximum number of no-operation options in one 32-bit word? 32/8=4
- If the outgoing queue of a UDP client overflows, ______. The operating system asks the server to wait before any more messages are sent
- What is the maximum number of the end-of-option options in the entire TCP header? 320/32 = 10
- To prevent silly window syndrome created by a receiver that processes data at a very slow rate, _____ can be used. Clark’s Solution and Delayed Acknowledgment
- An ACK number of 1,000 always means that _____. None of the Above
- The _____ timer prevents a long idle connection between two TPCs. Keep-Alive
- The _____ timer is needed to handle the zero window-size advertisement. Persistence
- Karin’s algorithm is used in calculations by the _____ timer. Retransmission
- Multiply the header length field by ____ to find the total number of bytes in the TCP header. Four
- A VPN uses ____ to guarantee privacy. IPSec and Tunneling
- A server issues ____ open while a client issues ____ open. Passive; Active
- ______ is a collection of protocols that provide security at the IP layer level. IPSec
- Urgent data requires the urgent pointer field as well as the URG bit in the ____ field. Control
- Connection establishment involves a ______ handshake; connection termination involves a _____ handshake. Three-way
- Tunneling is a technique in which the IP datagram is first ____ and then ____ . Encrypted; encapsulated in another datagram
- On a network that uses NAT, the _____ has a translation table. Router
- To prevent silly window syndrome created by a sender that sends data at a very slow rate, _____ can be used. Nagle’s Algorithm
- When the IP layer of a receiving host receives a datagram, _____. Transport Layer protocol takes over
- If the incoming queue of a UDP servers overflows, ______. The user datagrams are discarded and a port unreachable message is sent
- _____ is a protocol that provides security at the transport layer level. TLS
- Which option defines the maximum number of bytes in a TCP segment? Maximum Segment Size
- An _____ is a private network with limited external access that uses the TCP/IP protocol suite. Extranet
- The handshake protocol and data exchange protocol are part of _____. SSH
Short Problem
- Calculate the UDP checksum for the following binary numbers. Give the results in binary. Use the 16-bit format.
01000111 11100001
00000111 10001001
00000111 10001011
00000111 10001111
01000111 11100001
+00000111 10001001
01001000 01101010
01001000 01101010
+00000111 10001011
01001111 11110101
01001111 11110101
+00000111 10001111
01010000 01110000 <- UDP Checksum
- The following is dump of a TCP header in hexadecimal format.
05320017 00000001 0000000 500207FF 0000000
a.) What is the source port number (in decimal form)?
05 32 = 1330
b.) What is the destination port number (in decimal form)?
00 17 = 23
c.) What is the sequence number (in decimal form)?
00 00 00 01 = 1
d.) What is the acknowledgement number (in decimal form)?
00 00 00 00 = 0
e.) What is the window size (in decimal form)?
07 FF = 2047 bytes
- Given the class-A address 16.17.18.20 and mask 255.255.0.0, do starting and broadcast addresses exist for the block? If so, what are they? No
- A supernet mask is 255.255.248.0. How many class C networks are combined to make this supernet? 8
- In a class C subnet, we know the IP address of one of the hosts and the mask as given below:
IP address: 182.44.82.16
Mask: 255.255.255.192
What is the first (network address) and the last address (broadcast address) in this subnet? First 182.44.82.64 - last 182.44.82.127
- The subnet mask for a class C network is 255.255.255.248. How many subnetworks are available? How many assignable hosts ids per subnet?
2^3-2= 6 subnets ; 2^4= 16 hosts ids per subnet
Problems
- An ISP is granted a block of classless address starting with 150.80.0.0/16. The ISP wants to distribute blocks of addresses to different customer groups as follows:
- Customer Group 1: The number of customers in this group is 100. The type of customer is a medium-size business that needs 128 addresses.
- Customer Group 2: The number of customers in this group is 200. The type of customer is a small-size business that needs 32 addresses.
- Customer Group 3: The number of customer in this group is 1024. The type of customer is a household that needs 8 addresses.
Design the blocks (ranges) for each customer in slash notation. For a,b and c, only show blocks for the 1st, 2nd and last customer.
d.) After a, b, and c are satisfied, how many addresses remain for the ISP?
A)
1st Customer
|
150.80.0.0/16
|
150.80.0.127/16
|
2nd Customer
|
150.80.0.128/16
|
150.80.0.255/16
|
100th
|
150.80.49.128/16
|
150.80.49.255/16
|
B
2^5 = 32; 32-5 = 27 prefixes
| ||
1st Customer
|
150.80.50.0/27
|
150.80.50.31/27
|
2nd Customer
|
150.80.50.32/27
|
150.80.50.63/27
|
200th Customer
|
150.80.74.96/16
|
150.80.74.127/27
|
C
2^3=8; 32-3=29
| ||
1st Customer
|
150.80.75.0/29
|
150.80.75.3/29
|
2nd Customer
|
150.80.75.4/29
|
150.80.75.7/29
|
1024th Customer
|
150.80.75.251./29
|
150.80.75.255/29
|
D
8^16 = 65536 Total available
100*128 = 12800
200*32 = 6400
1024*4 = 4096
65,536 - 12800 - 6400 - 4096 = 42,240 left
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