Your network contains one Active Directory forest named contoso.com. The forest contains a single domain. The domain contains the domain controllers is configured as shown in the following table.
Name Site
DC1 Site1
DC2 Site2
DC3 Site3
DC4 Site4
The replication topology is configured as shown in the following output.
Cost : 100
DistinguishedName : CN=SiteLink1, CN=IP, CN=Inter-Site Transports, CN=Sites, CN=Configuration, Dc=Adatum, DC=com
Name : SiteLink1
ObjectClass : SiteLink
ObjectGUID : e1c8c335-b75f-4612-8a9e-58a0edead21f
ReplInterval : 60
SiteList : {CN=Site4, CN=Sites, CN=Configuration, DC=Adatum, DC=Adatum, DC=com,
CN=Site2, CN=Sites, CN=Configuration, DC=Adatum, DC=Adatum, DC=com}
Cost : 100
DistinguishedName : CN=SiteLink1, CN=IP, CN=Inter-Site Transports, CN=Sites, CN=Configuration, Dc=Adatum, DC=com
Name : SiteLink2
ObjectClass :SiteLink
ObjectGUID : 9516948e-cd56-4a9b-b6ba-cdf3dd7fe0d1
ReplInterval : 60
SiteList : {CN=Site4, CN=Sites, CN=Configuration, DC=Adatum, DC=Adatum, DC=com,
CN=Site2, CN=Sites, CN=Configuration, DC=Adatum, DC=Adatum, DC=com}
Cost : 100
DistinguishedName : CN=SiteLink3, CN=IP, CN=Inter-Site Transports, CN=Sites, CN=Configuration, Dc=Adatum, DC=com
Name : SiteLink3
ObjectClass : SiteLink
ObjectGUID : 07a7a37e-a12c-40c4-8042-f5d2e737b8a9
ReplInterval : 60
SiteList : {CN=Site4, CN=Sites, CN=Configuration, DC=Adatum, DC=Adatum, DC=com,
CN=Site3, CN=Sites, CN=Configuration, DC=Adatum, DC=Adatum, DC=com}
Cost : 400
DistinguishedName : CN=SiteLink4, CN=IP, CN=Inter-Site Transports, CN=Sites, CN=Configuration, Dc=Adatum, DC=com
Name : SiteLink4
ObjectClass : SiteLink
ObjectGUID : 508810dc-30fd-4845-982a-d4552fba2e04
ReplInterval : 45
SiteList : {CN=Site4, CN=Sites, CN=Configuration, DC=Adatum, DC=Adatum, DC=com,
CN=Site2, CN=Sites, CN=Configuration, DC=Adatum, DC=Adatum, DC=com}
You discover that replication between Dc1 and DC3 takes a few hours.
You need to reduce the amount of time it takes to replicate Active Directory changes between DC1 and DC3.
What should you do?
A.
Create a site link that connects Site1 and Site3, has a cost of 350, and replicates every 15 minutes.
B.
Modify SiteLink4 to replicate every 15 minute.
C.
Disable Site Link bridging.
D.
Set the cost of SiteLink4 to 100.
D.
why D? or I am missing the site link on Site 1 and 3? I believe its A.
This question is wrong (think its not finished yet.) All the site links are 4 – 2. I had this question in my exam and there was an individual site link between all sites (eg. 1-2, 2-3 etc)
@All a new version of 412 (V3) is out on this platform.. go and check it out..
@Abdul the new version of 70-712(V3) they are equal to the other already present,i didn’t see no new questions.
You miss my point. The sites in the question are wrong. IT shouldnt be 4-2, 4-2, 4-3, 4-2 it should be something like 1-2, 2-3, 3-4, 1-4
D. Set the cost of SiteLink4 to 100.
If should be 1-2, 2-3, 3-4, 1-4, the correct answer could be D. because reducing the cost of synchronization between DC3 and DC4, DC3 can replicate faster with DC2 which in turn can replicate faster with DC1.
this question was on my exam, I chose D
I’m gonna post an explanation from a user named coodyscoops on another site, because i think he explained this pretty well. At first i wasn’t sure about the answer but after reading his comment i now am sure the correct answer is D.
Here is his post:
“…
Currently, Replication is being done through:
DC1=> DC2, DC4,
DC3=> DC4
DC4=>DC2
So replication between DC1 and DC3 is happening through DC4…. Its not like there is absolutely NO replication between DC1 and DC3
With that being said, route preference and optimization is calculated through cost. The cost of the route between DC1 and DC3 if you chose answer D would be 300… and since this falls within the 3 hop maximum, replication is performed:
DC1=>DC2: 100
DC2=>DC4: 100
(with modified cost) DC4=>DC3: 100
If you chose A. the cost between DC1=>DC3 is set to 350
Just because of cost alone here, answer D makes for a better answer because lower cost = better optimization(according to MS best practices) and the cost here is latency due to bandwidth usage.
If you add another site link with an adjusted replication time, more bandwidth will be used to accomodate the replication because by default, time of replication is 180 minutes and answer A is setting it to 15 mins which will use up more bandwidth than if you just performed answer D which is replicating every 45 mins.
Also, the question just wants to reduce time. Although it is not declared in which way to reduce time, its safer to assume that MS usually prefers in the most efficient way possible since this is what they preach. In this scenario, efficiency is not created by speed, but by cost because cost is a crucial factor in designing the replication topology.
Also, according to technet and the books the only instance a site link should be created, is when the existing one is down which inhibits replication completely between 2 DC’s.”
You can check his post here: http://www.certifychat.com/70-412-a/577-70-412-vce-wrong-answers-3.html
Thanks by the way coodyscoops. 🙂
That makes no sense.
Sitelink1 – site 4 and site 2
Sitelink2 – site 4 and site 2
Sitelink3 – site 4 and site 3
Sitelink4 – site 4 and site 2
There are no site links configured between DC1 and DC3.
If there are no site links, how will replication occur between DC1 and DC3? DC3 is getting replication data from DC4 through sitelink3, but it does not replicate data to or from DC1 from any of the site links whatsoever.
You should go back and review how site links work because that explanation shows poor knowledge of the basic fundamentals.
The ONLY way your explanation would work is if there is a common site being linked by both DC1 and DC3.
If A->B and B->C, then site link bridging will automatically link A->C.
In this case, DC1 is not linked to any of the sites.
http://serverfault.com/questions/757918/what-are-site-link-bridges
Surprisingly user pleasant website. Huge information offered on few gos to
buy madden 17 coins http://bbs.izhancms.com/home.php?mod=space&uid=320237&do=profile&from=space
I AGREE => D