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NEW QUESTION: 1



A. SwitchB, Gi0/2, root
B. SwitchC, Fa0/2, root
C. SwitchD, Gi0/2, root
D. SwitchB, Gi0/1, designated
E. SwitchA, Fa0/1, root
F. SwitchA, Fa0/2, designated
Answer: C,E,F
Explanation:
The question says "no other configuration changes have been made" so we can understand these switches have the same bridge priority. Switch C has lowest MAC address so it will become root bridge and 2 of its ports (Fa0/1 & Fa0/2) will be designated ports. Because SwitchC is the root bridge so the 2 ports nearest SwitchC on SwitchA (Fa0/1) and SwitchD (Gi0/2) will be root ports. Now we come to the most difficult part of this question: SwitchB must have a root port so which port will it choose? To answer this question we need to know about STP cost and port cost. In general, "cost" is calculated based on bandwidth of the link. The higher the bandwidth on a link, the lower the value of its cost. Below are the cost values you should memorize:

SwitchB will choose the interface with lower cost to the root bridge as the root port so we must calculate the cost on interface Gi0/1 & Gi0/2 of SwitchB to the root bridge. This can be calculated from the "cost to the root bridge" of each switch because a switch always advertises its cost to the root bridge in its BPDU. The receiving switch will add its local port cost value to the cost in the BPDU. One more thing to notice is that a root bridge always advertises the cost to the root bridge (itself) with an initial value of 0. Now let's have a look at the topology again SwitchC advertises its cost to the root bridge with a value of 0. Switch D adds 4 (the cost value of 1Gbps link) and advertises this value (4) to SwitchB. SwitchB adds another 4 and learns that it can reach SwitchC via Gi0/1 port with a total cost of 8. The same process happens for SwitchA and SwitchB learns that it can reach SwitchC via Gi0/2 with a total cost of 23 -> Switch B chooses Gi0/1 as its root port ->. Now our last task is to identify the port roles of the ports between SwitchA & SwitchB. It is rather easy as the MAC address of SwitchA is lower than that of SwitchB so Fa0/2 of SwitchA will be designated port while Gi0/2 of SwitchB will be alternative port.

Below summaries all the port roles of these switches:

+
DP: Designated Port (forwarding state)
+
RP: Root Port (forwarding state)
+
AP: Alternative Port (blocking state)

NEW QUESTION: 2
Which HP commercial desktop portfolio satisfies the requirement of an SMB client that is looking for commercial desktops that offers performance, security, durability, expandability, and manageability?
A. HP Desktop Mini (DM)
B. HP ProDesk/ProOne
C. HP Business Desktop
D. HP EliteDesk/EliteOne
Answer: B
Explanation:
Explanation/Reference:
Explanation:
Reference: https://www.webantics.com/hp-prodesk-600-g2-i5-6500-microtower-desktop-pc

NEW QUESTION: 3
A customer Synergy solution configuration needs redundant paths to the drives used in the HPE Synergy D3940 storage module.
Which component must be redundant to meet the customer's requirement?
A. HPE Synergy 12GB SAS Connection Module
B. HPE Synergy D3940 storage module
C. HPE Compute Module Smart P542D SAS Cable
D. HPE Smart Storage Battery with 260mm Cable Kit
Answer: A

NEW QUESTION: 4
あなたは、あなたは、学生の年半ば成績を記録するためにStudentCode、SubjectCodeおよび成績カラムを持っているテーブルを作成します。 テーブルには様々な主題のために50の学生によって得られたマークがあります。あなたは、次の要件が満たされていることを確認する必要があります。
* 学生は彼らの平均点に基づいて評価されていなければなりません。
* 一人以上の学生が同じ平均点を持っている場合、同じランクはこれらの学生に与えられなければなりません。
* 同じランクが割り当てられる場合、連続するランクがスキップされなければなりません。
どのTransact-SQLクエリを使うべきでしょうか。
A. SELECT StudentCode AS Code,Marks AS Value FROM (
SELECT StudentCode, Marks AS Marks,
RANK() OVER(PARTITION BY StudentCode ORDER BY Marks ASC) AS Rank
FROM StudentMarks) tmp
WHERE Rank = 1
B. SELECT StudentCode as Code,
RANK() OVER(ORDER BY AVG (Marks) DESC) AS Value
FROM StudentMarks
GROUP BY StudentCode
C. SELECT StudentCode AS Code,Marks AS Value FROM (
SELECT StudentCode, Marks AS Marks,
RANK() OVER(PARTITION BY SubjectCode ORDER BY Marks DESC) AS Rank
FROM StudentMarks) tmp
WHERE Rank = 1
D. SELECT Id, Name, Marks,
DENSE_RANK() OVER(ORDER BY Marks DESC) AS Rank
FROM StudentMarks
E. SELECT StudentCode as Code,
NTILE(2) OVER(ORDER BY AVG (Marks) DESC) AS Value
FROM StudentMarks
GROUP BY StudentCode
F. SELECT StudentCode AS Code,Marks AS Value FROM (
SELECT StudentCode, Marks AS Marks,
RANXO OVER(PARTITION BY StudentCode ORDER BY Marks DESC) AS Rank
FROM StudentMarks) tmp
WHERE Rank = 1
G. SELECT StudentCode as Code,
DENSE_RANK() OVER(ORDER BY AVG (Marks) DESC) AS Value
FROM StudentMarks
GROUP BY StudentCode
H. SELECT StudentCode AS Code,Marks AS Value FROM (
SELECT StudentCode, Marks AS Marks,
RANK() OVER(PARTITION BY SubjectCode ORDER BY Marks ASC) AS Rank
FROM StudentMarks) tmp
WHERE Rank = 1
Answer: B
Explanation:
Reference:
http://msdn.microsoft.com/en-us/library/ms189798.aspx