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Exam Code: 1z0-808

Exam Name: Java SE 8 Programmer I

Version: V13.25

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NEW QUESTION: 1
Which two GRE features are configured to prevent fragmentation? (Choose two.)
A. TCP window size
B. TCP MSS
C. IP MTU
D. PMTUD
E. DF bit Clear
F. MTU ignore
Answer: B,D
Explanation:
Explanation
The IP protocol was designed for use on a wide variety of transmission links. Although the maximum length of an IP datagram is 65535, most transmission links enforce a smaller maximum packet length limit, called an MTU. The value of the MTU depends on the type of the transmission link. The design of IP accommodates MTU differences since it allows routers to fragment IP datagrams as necessary. The receiving station is responsible for the reassembly of the fragments back into the original full size IP datagram.
Fragmentation and Path Maximum Transmission Unit Discovery (PMTUD) is a standardized technique to determine the maximum transmission unit (MTU) size on the network path between two hosts, usually with the goal of avoiding IP fragmentation. PMTUD was originally intended for routers in IPv4. However, all modern operating systems use it on endpoints.
The TCP Maximum Segment Size (TCP MSS) defines the maximum amount of data that a host is willing to accept in a single TCP/IP datagram. This TCP/IP datagram might be fragmented at the IP layer. The MSS value is sent as a TCP header option only in TCP SYN segments. Each side of a TCP connection reports its MSS value to the other side. Contrary to popular belief, the MSS value is not negotiated between hosts. The sending host is required to limit the size of data in a single TCP segment to a value less than or equal to the MSS reported by the receiving host.
TCP MSS takes care of fragmentation at the two endpoints of a TCP connection, but it does not handle the case where there is a smaller MTU link in the middle between these two endpoints.
PMTUD was developed in order to avoid fragmentation in the path between the endpoints. It is used to dynamically determine the lowest MTU along the path from a packet's source to its destination.
Reference:
25885-pmtud-ipfrag.html

NEW QUESTION: 2
Which of the following IP addresses are valid Class B host addresses if a default Class B mask is in use? (Choose two.)
A. 190.6.5.4
B. 192.168.5.9
C. 127.0.0.1
D. 133.6.5.4
E. 10.6.8.35
Answer: A,D
Explanation:
The IP addresses 133.6.5.4 and 190.6.5.4 are both valid Class B addresses when a default mask is in use.
The Class B default mask is 255.255.0.0 and the range of valid addresses is 128.0.0.0-191.255.255.255.
The IP address 10.6.8.35 is a Class A address. The Class A default mask is 255.0.0.0 and the range of valid addresses is
1.0.0.0 - 127.255.255.255, with the exception of the range 127.0.0.1 - 127.255.255.255, which is reserved and cannot
be assigned.
The IP address 192.168.5.9 is a Class C address. The Class C default mask is 255.255.255.0 and the range of valid
addresses is 192.0.0.0 - 223.255.255.255.
The IP address 127.0.0.1 is a Class A address, but it comes from a reserved portion that cannot be assigned.
The range 127.0.0.1 - 127.255.255.255 is used for diagnostics, and although any address in the range will work as a
diagnostic address, 127.0.0.1 is known as the loopback address. If you can ping this address, or any address in the
127.0.0.1 - 127.255.255.255 range, then the NIC is working and TCP/IP is installed. The Class A default mask is
255.0.0.0 and the range of valid addresses is 1.0.0.0 - 127.255.255.255, with the exception of the range 127.0.0.1 -
127.255.255.255, which is reserved and cannot be assigned.


NEW QUESTION: 3

A. /unattend
B. /generalize
C. /oobe
D. /audit
Answer: C
Explanation:
The /oobe option restarts the computer into Windows Welcome mode. Windows Welcome enables end users to customize their Windows operating system, create user accounts, name the computer, and other tasks. Any settings in the oobe System configuration pass in an answer file are processed immediately before Windows Welcome starts.

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