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NEW QUESTION: 1
On the basis of the definitions of different services in various RFCs, traffic with Expedited
Forwarding per-hop behavior should be marked as which of these?
A. IP ToS of 0xEF
B. DSCP decimal 5
C. IP experimental ECN
D. Binary value of 101110
Answer: D
Explanation:
The assured forwarding (AF) model is used to provide priority values to different data applications. The Expedited Forwarding (EF) model is used to provide resources to latency (delay) sensitive real-time, interactive traffic. The EF model uses one marking - DSCP 46. DSCP 46 is backward compatible with an IP Precedence value of 5 as seen in the following binary pattern: 101110 = DSCP 46 The EF marking of 46 does NOT follow the drop preference rules of the assured forwarding model. Please do NOT think that the 11 means high drop preference. The EF model is used for voice over IP media traffic (RTP) by default in most vendors phones. Cisco IP Phones mark signaling packets (SCCP or SIP) to CS3 (24), while media (RTP) is marked to EF (DSCP 46) by default. All EF traffic is normally mapped to the priority queue (PQ) on Cisco switches and routers. The priority queue guarantees three critical services: Packet Loss Delay Jitter (delay variation)
The three most significant bits of 101 are only considered if IP Precedence was being used. The binary digits of 4 2 1 are used to factor the 101 binary pattern when only three digits are under consideration. The DSCP binary pattern of 101110 (46) uses six digits or binary values-32 16 8 4 2 1.
It is good to know how to convert a DSCP decimal value to an entire ToS octet (byte) values as well. The ToS byte uses all eight bits, while the DSCP is only using the leading six digits. The EF pattern discussed above will become 10111000 when considering the entire octet. Notice the two least significant zeros that were added to the 101110 binary pattern. Many network management utilities will only allow administrators to configure or display the entire ToS byte. A ping V from a Microsoft operating system requires setting the entire ToS byte. An extended ping from a Cisco router will also allow administrators to see the entire ToS byte. Sniffer Pro LAN and Wire Shark sniffers show the entire ToS field as well. IP accounting shows the entire ToS byte, while Netflow shows the ToS byte in hexadecimal format. The ToS byte value for EF is as follows: 128 64 32 16 8 4 2 1 1 0 1 1 1 0 0 0
A DSCP value of 46 results in a ToS byte value of 184. Although you can mark a ping with a ToS value of 184, the ICMP (ping) traffic will probably not be mapped to the proper application class. In the next blog, we will learn QoS models for using markings for different application classes.

NEW QUESTION: 2
Your network contains an Active Directory domain named contoso.com. The domain has Remote Desktop Services (RDS) deployed. The domain contains six servers. The servers are configured as shown in the following table.

You plan to configure Server6 as an additional RD Connection Broker in the RDS deployment.
You need to identify which servers require the SQL Server Native Client installed.
Which servers should you identify? (Each correct answer presents part of the solution. Choose all that apply.)
A. Server6
B. Server5
C. Server2
D. Server1
E. Server3
F. Server4
Answer: A,E
Explanation:
You must have a SQL Server setup that can be used by the RD Connection Broker servers to store data. At least SQL Server 2008 R2 must be used, and the minimum recommended SQL Server SKU for this is Standard with at least 4GB of RAM. RD Connection Broker High Availability in Windows Server 2012

NEW QUESTION: 3
Fifteen buoys, numbered 1 to 15, are positioned equidistantly along a straight path for a sailboat race.
The race begins at buoy 1, and 45 minutes into the race the fastest boat reaches buoy 10. At the same
average speed, how many minutes will it take the boat to race from buoy 10 to buoy 15?
A. 32.5
B. 0
C. 1
D. 2
E. 22.5
Answer: C
Explanation:
In 45 minutes, the boat traveled 9 (not 10) segments (from one buoy to the next). That
means it takes the boat 5 minutes to travel the length of 1 segment. At buoy 10, 5 additional segments
remain ahead. Since the boat's average speed is the same before and after buoy 10, it takes the boat
5 x5=25 minutes to travel from buoy 10 to buoy 15.

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