Hotspot

Hi guys,

I think that an answer to question 2 doesn’t have to be A-E-C-D (however, it could be). I don’t agree with 9tut that all traffic must go through root switch. In my understanding, a root switch is only a point where the STA algorithm starts to determine that the network is loop free. I believe that there is not enough information to choose only one answer in question 2. Either it could be A-E-C-D or just A-D, depending on Switch-D port priority and port numbers. Could anybody share some comments on this topic?

@azizz – I’m not 100% on 9tut’s answer as well actually. Can’t recalled that all traffic goes through the root.

Q2 Came on p4sure dumps and the answer there is A E A D. pls advise.

I think Sw-A, Sw-E, Sw-C, Sw-D –

for Question 2 the answer is A .. Sw-A, Sw-D

because stp block redundant connection at the
layer 2, and the stp(pvst) search the best path
to the root switch it’s correct..
but the switch will fowerd packets destained
in is CAM Table.. Otherwise why we have
CAM Table on Switchs?!
if the switch learned about its ports and their
MAC Address, it go stright to it..

kobi.j
ccent ccna ccnp

There has been a lot of confustion regarding this question. I myself have seen this answered A/E/C/D, and A/D. Because there were inconsistencies, I decided to test the theory myself.

I set up a simulation test using Cisco’s Packet Tracer (5.2). While in simulation mode, I ran a Ping PDU from Host A to Server 1. Sure enough, an ARP broadcast went out to switches D/E & B, but only switch E forwarded the ARP broadcast (the ARP broadcast packets time-out and died on D & B). It forwarded to Switch C, then switch C forwarded forwarded the ARP broad cast to switch B & switch D. (Again broadcast timed out and died on SwB)
Switch D then forwarded it on to the port Server 1 was connected to.

Once then the ARP response followed the same path in reverse. Once the ARP request had completed back to Host A, Ping packets were sent.

Guess which path the ping packets went? SAME PATH as the ARP broadcast. Switch A, E, C, then D.

No disrespect to Kobi.j, but while CAM tables are used in switches, STP governs the paths of that CAM tables use. The CAM table wasn’t manually created, it was learned through a series of broadcasts. It’s STP’s role to also prevent broadcast storms, not just select the best route in a redundant switched environment.

That is why you have to take care on where the root switch is located in your network.

Cisco Press, ICND2, Wendell Odom, Page 63:
“All the ports in Forwarding State are considered to be the be in the current spanning tree. The collective set of forwarding ports create a SINGLE path over which frames are sent between Ethernet segments. ”

If anyone is interested, I can upload the Cisco Packet Tracer .pkt simulation and you can watch it for yourself.

Tim
ccent

I am trying to figure out questions 1 and 2. Pass4Sure says that question 1 is sw-B, not sw-E as you have. They specify Vlan20 also as you do. Looks like an identical question, picture, options. The only difference is the answer they get.

Having said that I like your Sw-E Answer better, It would give some shorter routes than if B were the root

Next question, Question 2. You say that the route cannot go back thru sw-A, I’ll buy that. Why can it not go back thru sw-A though? What property of switches disallows that? If it were a router with Rip configured, i’d say somethng about split horizon. If we had more information we could see which switches are blocking via STP.

Pass4sure says that the correct route is A – E – A – D.

So who is right here?

One more question for me, sort of….

Your answer for question 4 is Router, and i agree, because they are on different VLANs. Pass4sure has “a trunk from SwitchB to SwitchE,” that makes no f’ing sense, does it? Why would they put that? Is there any way that pass4sure is right?

Thank you!

The question said “If most of the communication is between hosts and the servers” so Sw-E is better than Sw-B, which is far away from servers.

Sw-E will not forward the traffic back to Sw-A because it can easily create a loop. So the best option is A-E-C-D.

For question 4, a router is better choice than a trunk although both are correct.

a trunk routes the packet from one VLAN to another? I thought you needed a layer 3 device to do that? how could the answer to question 4 be a trunk?

To forward a packet from one VLAN to another, we need:

1. A router (called “router on a stick”)
2. A trunk link between the router and the switch

The answer to question 4 should be “a router” than “a trunk link” although we need both.

Thanks for the thorough explanation, Tim Mann. Good research! Another example of why Packet Tracer and Cisco Press will always “trump the dumps”

Question 2 – answer is def A-E-C-D.

reason is that the link between A and D will be blocked as will the link between A and B or B and C (Unable to tell from diagram without knowing priorities and mac address’s of all switches)
Reason why these two links are blocked is to prevent looping.
Switch A has 3 paths to reach the root bridge-
A to E (the quickest and therefore the Root Port)
A to D to C to E
and A to B to C to E
therefore both the last two paths must be blocked to prevent a loop.
Meaning link A -D is blocked and either A to B or B to C is blocked.
You could also see the same links needing blocked if checking from switch C which also has 3 potential paths to root switch.

Actually I take back what I said, after also setting up the scenario in packet tracer I got different results depending on the mac address of the switches.
If switch C has a lower mac address than switch A then the packets do indeed go via A-E-C-D and the D switch port connecting to A is blocked, as is the B port connecting to A.
However!- if the mac addrsss for switch A is lower than the mac address for Switch C then different ports are blocked as a result.
Switch D port connecting to C is blocked and Switch B port connecting to C is blocked…meaning in this case the packets would travel from A-D!!!.
Therefore it is not possible to answer this question with the info given unless the mac address;s are given for the switches.

You my friend Have answered your own question, and has well have shown your inability to pass this test…The reason why the mac address thing works, is the fact that you are setting the Root Switch…Which is determined by the lowest mac address, So you just set the server and proved the senior right, because it does in fact tell you that E is the root switch.

and you my friend are talking nonsense…I did not say I made switch C or A the root switch,in my tests switch E was always the root,however depending on whether A or C has the lower of the two’s mac address the packets took different routes.
With Switch E as root and switch A having a lower mac than switch C the packets go via A-D.
With Switch E the root and switch C having a lower mac than A the packets go via A-E-C-D.
therefore without knowing what the mac address’s are of A and C you cannot answer this question.
Looks like you may wish to study up further before you sit the test.

Do you understand how a switch determines the root switch? It uses priority, which you didn’t change, and then the mac address…….SOOOOOOOOOOOOOOOOOOOOOOOOOO by making changing the mac address, you are changing the root switch. And thereby proving my point.

please try and set it up yourself and you will see I am right.
Yes I understand how it determines the root…and I did change the priority….I set the priority of switch E low to make it the root always.
Then it doesnt matter if you change mac address’s on the others as the low priority on E always overides any lower mac address’s..however changing macs on A and C does then have an affect on which ports the switches A and C decide to block based on which one has the lower mac.
I could also have achieved the same outcome changing their priority as well as long as they werent set lower than switch E.
I didnt think I needed to expalin every detail I did to test it, but apparantly your not happy accepting my word.
So as I said try it yourself before you start making comments..i’m trying to help people to realise this question is missing important info to answer it correctly and I dont believe it would appear on the exam as it is written here.
Either the mac address’s or priorities need to be given.

@ mark
i totally agree with you , without even trying the scenrio in packet tracer its obvious that in some cases like the ones you just said that path A-D would be a possible solution but apparently for the exam i believe we must assume that all packets needs to pass through the root switch.
But again without further details we cannot be sure .

I agree Mark’s opinions. Yes, the important factor is which switch is the second “root”. Switch E is the root, if switch A has a lower priority number than switch C, or has a lower mac address then switch C when they have the same priority number, switch A became the second “root” , B-C and C-A will be blocked, the answer should be A-D

i agree wit mark, its not possible to trace a path with the given info.

None of these on test today 9/6

@mark,
Good Explanation.
I totally agree with your theory and it makes sense. All will depend on who has the 2nd lowest Mac Address. Its a bit unfortunate if CISCO do not post enough details in the exam. If I get exactly the same question, my answer to Q2 will be A-E-C-D

@Binary Fairy,
Please open “Wendell Odoms book” and understand how Rot Priority etc. works in determining Root Bridge.
This may still not be enough, so I advise you to practice the same by testing on some real CISCO switches.

I came across this question:

Identify the four valid IPv6 addresses. (Choose four.)

A. 2000::
B. 2002:c0ad:01
C. 2003:dead:beef:4dad:23:46:bb:101
D. ::192.168:0:1
E. 2001:3452:4952:2837::
F. ::

The answers are B,C,D and F. Can someone explain why A and E are not valid addresses? A looks to me like it is short notation for 2000:0:0:0:0:0:0 and E looks like short notation for 2001:3452:4952:2837:0:0:0:0. I don’t understand why they are not valid. Thanks.

same question with Paul. I posted same Q in CLN and said that A,E are valid ipv6 addresses. Ty!

try this link Paul http://www.intermapper.com/ipv6validator, very useful.

Ok guys, there is a lot of confusion about A & E and being legal or not.
Answer is they are not legal!!!

Let me explain –
2000::/3 is reserved for global unicast addressing. By Subnet 3, we mean to say that it must have 2 or 001 to start with. This means anything that starts with 2(001) is a global unicast address. Think of it as a PUBLIC IP the ISP assigns.
Now the IANA authority(I think this is what its called), assigns the whole IPv6 Subnet to organizations with large requirement. They can assign them an IP Subnet say – 2000:1111:AAAA:FFFF::/64 or even 2001:3452:4952:2837::/64(as above), but remember 2001:3452:4952:2837:: is a SUBNET and not an IP ADDRESS and also without the /64(or whatever) at the end is invalid. This exception only applies to anything starting with 2..Please read Wendell Odom Pgs 588-592. Hope this helps..

Re: That validator website – Its wrong and I would not trust it.

The answer is A,B,E,F

dude I do not think you are right!!

thanx 4 the input guys i’m motivated to go and write my exam b4 end of this month I hope everything hasn’t changed much.

Question 2:

With the information given, the solution of this question is not possible!

Starting from SW-D, you won’t be able to find out which port is its root port (Link SWD-SWA, or is it SWD-SWC), and therefore you are not able to answer this question. There is information missing.

Otto is correct, There is no way to tell because without extra infor wo dont know which port D would use as its route. this is proboly why this question hasnt appeard on an exam in the last 3 years.

Hi 9tut!!
Please post some new hotspot with some drag and drop questions as well for icnd2.
Went through the exams last week got 30 questions out of 40 which didn’t came from the website nor from pass4sure. Could score only 750marks!!
Thanks.

@CCNA-icnd2

Study concepts from the book/videos. Don’t rely on dumps only to pass the test.

Good Luck

The correct answer is Sw-A/Sw-E/Sw-C/Sw-D. Let me try to explain: at STP learning process there will be two blocked ports (Sw-D port to Sw-A AND Sw-B port to Sw-A), now Host A is going to send arp req to server 1 so after Sw-A receives packet (it does not know where to forward it yet) it floods it to all ports except the receiving one (Sw-D, Sw-E and Sw-B) because Sw-D and Sw-B ports are in blocking state the packet is received only by Sw-E, now Sw-E sends packet to .. Sw-C (because Sw-E will never forward packet out of interface from which it received that packet). Then Sw-C will forward the packet to Sw-D and Sw-B (the packet will be dropped at Sw-B (because as you remember port Sw-B to Sw-A is blocked). So the only path is Sw-A/Sw-E/Sw-C/Sw-D. Sorry for bad english.

I agree with “me” & 9tut with regards to the regards to Question 2.

If you don’t then I would (as i did) re-read all the training materials available on the fundamentals of a Ethernet Switch and the STP. I have also discussed this with CCNPs.

It can be argued there is not enough information. However, I don’t think that will do you much good AFTER the exam.

I personally do not want to see this question pop up because I think I am more likely to be caught out by the wording of the question than understanding the actual theory behind STP.

I did not get the switch Q today, with the controversial Q2. However, I set ip up in packet-tracer, then removed the trunks for SW-A and SW-C, swapped their locations, and re-established new trunks, effectively swapping the MAC/port role of SW-C and SW-A. Sure enough, A-D is a valid path one way, and A-E-C-D the other way. I was planning to answer A-E-C-D and post a comment if I got this one (saying that A-D is also possible).

I have been trying to confirm an answer to a question but I have had no luck. I recently took the part 2 test and had a question about limiting telnet to only one connection.

I chose

Line Vty 0
Password (password)
Login

Does that sound correct? Does only enabling the password on the first line only allow one connection to telnet? Would setting a password on teh first two lines only allow two connections?

For instance would:

Line VTY 0-1
Password (password)
Login

Allow two connections?

Also if its correct does it have to be the first line IE 0 or could it be any line?

Would

Line VTY 2
Password (password)
Login

Do the same and allow only one connection?

Thanks in advance!

Answered my own question with some help. If you are interested the answer is under share your ICDN2 experience, and this question was on a recent test.

Come on guys question 2 is not too hard.
You can only go with the options given and not ask why other options aren’t there
You dont have to know your STP to work this one out.

A – Sw-A, Sw-D
B – Sw-A, Sw-E, Sw-D
C – Sw-A, Sw-B, Sw-C, Sw-D
D – Sw-A, Sw-E, Sw-C, Sw-D
E – Sw-A, Sw-E, Sw-A, Svv-D

Option A can’t be true because SW-D is not conected to the Server directly.
Option B can’t be true because SW-D & SW-E are not directly conected.
Option C can’t be true because it would form a loop. SW-D would only have the option of sending it back to SW-C or SW-A
Option E can’t be true because it just makes no sense, firstly why would SW-A send the packet to SW-D when the end device is directly connected to SW-A?

Regarding question2, how do you know which port is Sw-D’s RP?

If Sw-D’s RP goes to Sw-A, would Sw-A interface to Sw-D have to remain in a Forwarding State? If so, wouldn’t traffic from Host A to Server 1 goes from Sw-A to Sw-D directly?

So I did a quick lab in Packet Tracer using the same topology. I specified Sw-E as the root and once STP converged, Sw-D blocked it’s link to Sw-C (meaning the path of A > E > C > D is no valid), and chose the interface to Sw-A as its RP.

Sw-A link to Sw-D is a DP and is in a Forwarding State.

My though process on this question is as follow:

Sw-E is the root and all its port are DP
Sw-A’s link to Sw-E is its RP and remains Forwarding
Sw-C’s link to Sw-E is its RP and remains Forwarding

From that point, there’s not enough information to answer the question.

When Sw-D determines its RP, it looks at the lowest cost route. If both link have the same cost, it looks at the BID of Sw-A and Sw-C. If that is equal as well, Sw-D uses the port number as the tie breaker. So if the connection to Sw-A is on Fa 0/1 and Sw-C is on Fa 0/2, it will choose Fa 0/1 as its RP.

As such, unless we know what the BID, cost and interface of all the switch, I don’t think you can accurately answer that question.

Jesus Christ. It’s no wonder I’m dominating the IT world. Has anyone actually opened a book and studied for their CCNA part2 exam? Because the majority of posters on this site seem border line mentally deficient. Not enough information? What is wrong with you morons.

The material covered on the ICND2 focuses on (among other things) looping and split horizon rules. Split horizon prevents the sending of data back along the path by which it was learned… therefore- when root switch E learns of switch A (from its direct connection from switch A), it will refuse to send information passed along from switch A back to switch A. As the root switch MUST be used, the correct answer to question 2 is:

A -> E -> C -> D.

This is fundamental networking concepts. Don’t make it more complicated than it is.

I saw some of the feeble minded debating Question #4 as well. The answer is indeed a router.

http://www.dell.com/downloads/global/products/pwcnt/en/app_note_8.pdf
“Any inter-VLAN traffic must first traverse a layer-3 device such as a router in order to communicate with another VLAN.” – 3rd paragraph, second page.

Russ, what are you talking about? Split horizon is a loop prevention mechanism for the RIP routing protocol, what does that have to do with STP?

STP is a layer 2 technology, it has nothing to do with split horizon or routes.

Answer this, without knowing what the BID of each switch in the topology and the cost of each link, how can you tell which port on Sw-D is the RP?

Umm I hope you joking. Split Horizon has nothing to do with LAN Switching. Testing out tomorrow I hope I make it.

Question 2: From what I can tell, a packet does not have to go through the root switch to reach its destination. The only way we can know if a packet has to go through the root switch to reach its destination is by knowing which switch is the root and which ports are in blocking or forwarding modes on all other switches. Without this info we cannot tell for sure which direction a packet will take. From what I can tell anyway…
As for Russ: Split horizon rule = Layer 3. STP = Layer 2. Question 2 has nothing to do with split horizon. Think you might have gotten the graphic for switches and routers mixed up.

Q2, I think A is the correct. A sends a frame -> SW-A floods it -> SW-D receives it, looks at the mac table -> Server1 gets it.

Edited: it COULD be correct. But we need to know which ports are in blocking state.

about ques 2,
by just looking at the diagram given it is impossible to ans which will be the correct path
i may be a-e-a-d or it may be a-e-c-d.
it totally depend on the bid values of switch a and switch.

about ques 2,
by just looking at the diagram given it is impossible to ans which will be the correct path
i may be a-e-a-d or it may be a-e-c-d.
it totally depend on the bid values of switch a and switch c.

question 1 and 2 are indeed correct…got 972 and used those answers….checked allto of sites and the reviews were mixed..i am telling you now…they r ineeed e and d respectively

Ok, so.. Let’s E be the root. All costs are equal. I think these links well be blocking by the STP: AD, AB. When you send a packet, A can send it out to E and B, but B is in blocking state. So E is the right way. E can’t send it back to A, so it must send it to C. C can send it to B and D, but B is not a right choice. So the way is A-E-C-D.

Question number 4 is correct remember you have to read the question carefully. The question only ask what kind of link do we need to communicate? It does not ask what do we need to communicate it specifically states LINK….yes, you will also need a router or a layer 3 switch in order to completely get the VLANs to communicate. Always remember what are they asking if they ask what layer 3 device or what if anything…CISCO has ways of asking some really trick question watch every word when you test….good luck to all….

anyone have these Q’s on their exams

It’s definitely so that Q2 cannot be answered without knowing the MAC addresses of Switch A and C. I’ve also checked with Packet Tracer and had the same result as OneAndOne. Swapping the A and C Switch changes the paths because MAC addresses are considered. However, take the A-E-C-D solution, it’s the only one the counts in exam.

I’ve tried to make a lab in PT:

Host – SwitchA – SwitchB – SwitchC – SwitchD – Root Switch
|
Server

ICMP goes through A,B definetly to Server. So explanation “All the traffic must go through root switch ” is incorrect, isn’t it ?

@vadim
have you set the priorities accordingly?

@xallax
I’ve just made the last switch in chain “spanning-tree vlan 1 root primary ”

p.s. Sorry picture misformed, try again
…………………………………………………………………………………..
Host – SwitchA – SwitchB – SwitchC – SwitchD – Root Switch
……………………………|…………………………………………………….
………………………..Server………………………………………………..

@vadim
i am sorry to say this but… your design doesn’t need STP. it is a simple chain of switches.
STP applies when there are more ways to get from point A to point B.

in your design there is no need to block any ports and the normal and logical flow is from Host to Server via SwitchA and SwitchB.

Try connecting all 4 switches to each other
connect Host to SwA
connect Server to SwB
set SwD as root and watch what happens

you can also check on this small demo:
http://www.ciscovce.com/demo/activities/stp1.pkt

On this small demo delete line from C->D and add host to switch C

Now ping from host connected to A to host connected to C

And you will see scheme like here in question 2…
But Answer D will be wrong (Or will be just one of two right answers)
And explanation “All traffic must go through root” wil be wrong too

Question 2 in PKT. Sw E is Root primary. Answer A and D are correct.
http://depositfiles.com/files/aqy77c4ol

@vadim
i’ve tried it in packet tracer and notices this:

if switch D has a root port towards switch A then the path is A-D
having that root port towards A means that the port towards C will be discarding.
the path will be A-D because the frame can not go A-E-A-D.

if switch D has a root port towards switch C then the path is A-E-C-D…

the question doesnt give us the mac addresses of the switches so we must work with the least information: switch E is the root.

i’ve swapped the switches around in the activity i’ve uploaded at http://www.ciscovce.com/demo/activities/stp1.pkt so that the path is A-E-C-D
if you swap switches A and C you will get the A-D result

so… go with the A-E-C-D answer

@Q2

Guys! is this the result of people trying to pass these exams by practicing dumps rather than studying?

Think a little: if switch E is THE ROOT BRIDGE FOR ALL VLANS than it has to have all its ports as DP (this is something you will find out in all books so spend some time to read them). This will generate that switch A and C uplinks to E will be ROOT ports – so forwarding.
And now the most interesting part: A to D link will have one port in blocking and somewhere between A – B – C will be a blocking port – doesn’t matter where because the path from A to D can’t be in that direction.

Regards

I need to clarify my previous post related to Question 2 – because someone can understand it wrong.

When comes about STP there are some well defined steps in this process, as per below:

1) Electing ROOT bridge – this is the first requested step in order to see how the graph is going to be transformed in a tree. Here it comes about bridge ID and the way MAC addresses are used to break the tie (having 32768 as default ID value all switches will need a way to make the difference and as MACs are uniques, this is a good tie-breaker)
2) After the ROOT bridge was elected, now, the switches will need to determine which ports will have to be blocked in order to create the tree (meaning a loop free environment). In this step the MOST important value is the PATH COST. Path cost for a FA interface is 19, for a Gi is 4 and for 10 Gi is 2. IF THE PATH COST TO ROOT IS THE SAME THE BRIDGE ID WILL MAKE THE DIFFERENCE BETWEEN WHICH SWITCH WILL HAVE THE DP AND WHICH ONE WILL PUT ITS PORT INTO BLOCKING. IF THE BRIDGE IDs ARE THE SAME, THE PORT PRIORITY WILL BRAKE THE TIE.

Considering this – THE ONLY WAY SEGMENT A –> D WILL REMAIN UN-BLOCKED WILL BE IF THE PATH COST BETWEEN A AND E IS LOWER USING OTHER SEGMENTS.
Now – my question for you is why on Earth will someone consider to get a lower path cost using A –> B –> C –> E OR A –> D –> C –> E WHEN YOU HAVE A DIRECT CONNECTION TO E. This is why I said A to E is a root port. If all links are using FA/Gi/10Gi – this will be the elected path.
Again, in the books, you will see design considerations and layered architectures making all other choices being not accepted ( I don’t want to say that some response options are impossible at all) by a person who is not insane.
About Packet Tracer – I am not sure it is bug free and is considering PATH COST – this is why the best way to test this is with real switches (not need to be Cisco ones if you don’t have)

Hope this helped.

Regards

You can’t answer Q2 with a proper explanation if you have been just studying dumps… the F*** answer is A-E-C-D

I set the topology in packet tracer and spent nearly two hrs examining every single thing

Hope this help

Which option is a valid IPv6 address?

A. 2001:0000:130F::099a::12a
B. 2002:7654:A1AD:61:81AF:CCC1
C. FEC0:ABCD:WXYZ:0067::2A4
D. 2004:1:25A4:886F::1

Answer: D
Explanation/Reference:
IPv6 address is 128bit hexadecimal where each octet is 16bit followed by colon : (16×8=128). The digits used to represent numbers using hexadecimal notation are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, and F. Consecutive zeros can be replaced Once only with either one zero or can be left blank, eg 0000:0000:0000:0000 – can be written in :: or :0:. Choice A is invalid because :: can be replaced with as many as zeros until ipv6 is 128bit but twice :: in an ipv6 addr is invalid. Choice B is invalid as 16×6=96 (16bit x 6 octet) which is not 128bit address. Choice C is invalid third octet has WXYZ which is invalid hexadecimal notation. Choice D is valid ipv6 address… 2004:0001:25A4:886F:0000:0000:0000:0001

Couple different answers for this one????????

Identify the valid IPv6 addresses. (Choose all apply)

A. ::
B. ::192:168:0:1
C. 2000::
D. 2001:3452:4952:2837::
E. 2002:c0a8:101::42
F. 2003:dead:beef:4dad:23:46:bb:101

Answer: ABCDEF
Explanation/Reference:
:: Unicast unspecified address, The IPv6 unicast unspecified address is equivalent to the IPv4 unspecified address of 0.0.0.0. The IPv6 unspecified address is 0:0:0:0:0:0:0:0:, or a double colon (::), all address are vaild ipv6 but address B,C,D,E,F can’t be configured to interfaces without prefix-length. As a CCNA Candidate you should be able to recognize ipv6 address but addresses with prefix-length is not represented here. ipv6 address 2000:0:0:0:0:0:0:0 is valid but can’t be configured without /3 true but none of this ipv6 addresses can be configured without valid prefix-length aka netmask.

Select the valid IPv6 addresses. (Choose all apply)

A – ::192:168:0:1
B – 2002:c0a8:101::42
C – 2003:dead:beef:4dad:23:46:bb:101
D – ::
E – 2000::
F – 2001:3452:4952:2837::

Answer: A B C D F

Explanation:

Answers A B C are correct because A and B are the short form of 0:0:0:0:192:168:0:1 and 2002:c0a8:0101:0:0:0:0:0042 while C are normal IPv6 address.

Answer D is correct because “::” is named the “unspecified” address and is typically used in the source field of a datagram that is sent by a device that seeks to have its IP address configured.

Answer E is not correct because a global-unicast IPv6 address is started with binary 001, denoted as 2000::/3 in IPv6 and it also known as an aggregatable global unicast address.The 2000:: (in particular, 2000::/3) is just a prefix and is not a valid IPv6 address.

The entire global-unicast IPv6 address range is from 2000::/128 to 3FFF:FFFF:FFFF:FFFF:FFFF:FFFF:FFFF/128, resulting in a total usable space of over 42,535,295,865,117,307,932,921,825,928,971,000,000 addresses, which is only 1/8th of the entire IPv6 address space!

Um Russ, you arent dominating anything associating split-horizon with layer 2 bridging, though I do agree there are some ppl here that are missing some basic STP understanding.

The root switch forwards on ALL ports, and all other switches use the least cost path to the root meaning you should know exactly which links will be blocked by STP in this particular diagram with those 2 pieces of info.

I should add that ports can be forwarding on one side and not on the other but that wouldn’t be the case here since the root has 2 links to 2 different switches.

Passed ICND2 today with 944!! I didn’t have any of these questions on my test today

Q. 2
It is working like Split Horizon rule ?

@big man, big mind
kind of… except that here the frames must be directed to and from the BIG ONE (the root bridge).

Woa. Q2 is pretty easy, I got it right without really thinking. If you read the material it makes sense. Then I read all the comments.. eeek!

I KEEP HEARING ABOUT ACL2, WHERE CAN I FIND THIS HERE ON 9TUT?

Q1 and 2 on my ICND2 exam today.

BIGF2 http://www.9tut.com/category/ccna-lab-sim

you need to use the study material that corresponds to the ICND2 from the CCNA area of this wbsite too. Good luck.

last question was on my test today…verbatim.

access port and trunk port in my exam

Hi NerroAzurro, you mentioned that frame relay found in hotspot in 9tut. But i couldn’t find it here.
Are you referring to other place?

passed with 944 mark. Got many questions switches and vtp, doesn’t seem to be listed here, but if you nailed down the concepts on switches, you should have no problem with passing. Questions were easy just don’t see them here.

Passed ICND2 today with 986/1000 🙂

Q3 was there

wrote exam 17/05/2012 – q3 was there

BTW, I was talking to everyone else cursing and having temper tantrums in the forum, not myself…

I just passed the ICND2 with 916……..There are new labs,Simlets and Drag & Drop……

Study Access List-WAN Commands-NATS……..The questions were not on the latest Test King dump.

Good Luck All!

I agree with people above that Q2 is actually cannot be answered properly without further info about switches’ mac addresses. I believe in the actual exam you will have a chance to log in and see for yourself. It may be the case as it’s actually part of a multiple-question scenario

I think I finally see the light of Q2. Actually there are two possible answers. The explanation is actually not correct.

STP only defines the forwarding path, blocking ports that cause loops. It’s not necessary to have all traffic going through the root switch. It is possible though. It depends on the final convergence.

But for this case, I assume the A-D path is disabled so that A-E-C-D path is working as intended. The convergence is made such that A-D is disable. However, once by manipulating the switches’ priority, A-D is possible, and no need to go through the root switch! Again the diagram has not enough info to deduct anything further.

I come across this link: https://learningnetwork.cisco.com/thread/40113 Beau explained it really well.

Need to clarify one thing, A-E is not necessary have to be enable. Although E is the root switch and the port towards A is designated port (forwarding state), SW-A port to E can be a blocking port, thereby disabling A-E. An alternate path A-B-C-E is possible. Again it all depends on the final convergence.

Oops, I mess up myself. SW-A port to E has to be a RP, as it’s closest to root. BP only happens at non-root switches. I need to think about it first before posting more. Sorry!

Ok I promise one last post as I think I find it out. As A-E and C-E needs to be enabled, the only possible way to avoid loop in the A-C-E is to block a port in the A-B-C path. The tricky part is SW-B as you can’t left it out, so either block A-B or B-C but not both!

Then the whole converged topology becomes D-A-E-C-B, (or D-A-E-C, with B connected to A, it’s ok as long as there’s no loop) And D-C needs to be disabled. You can’t have both A-D and A-E-C-D at same time to introduce loop. Then host A can send traffic via A-D to server 1. No need to go through the root switch E. Again this is not the absolute answer. A-E-C-D is still valid as long as A-D is disabled, but that is not showing all traffic must go through the root switch. The final topology makes it so. SW-E happens just in the way.

any one help me to take prepare my exam
thank

I really hope Q2 is not a real exam question, I have read all that has been said in previous posts and I have studied something silly for my ICND2 exam tomorrow. With all I have studied there is no way you can accurately answer the question. With STP you need the switch MAC’s and maybe if it gets down to ports port numbers. No where is it said that all traffic must go through the root, it’s recommended to be central.

Anyway for further peace of mind I set this exact layout in Packet Tracer and set SwE as the Root. As luck would have it traffic in Packet Tracer would flow directly from A to D. Things could have been different if my MAC’s were shuffled, but proves the point you can’t answer this question.

Without being given the priorities and sample MAC addresses question 2 would be a complete guess. It is possible that SW-E could have a higher priority value and therefore not become the root bridge. STP functions in conjunction to a priority value and if they are all the same it’s the lowest MAC address which this wins. By default the STP priority is 32768 so it’s down to the MAC. Not knowing the MAC makes it impossible to fathom.

i think Q6 which facilitates intervlan routing the access port also qualifies when you configure the router with two physical interfaces.

There is not enough information in question 2 to answer it correctly. And it’s really sad to see some people’s answers and thoughts on this question because many of them are fundamentally flawed and flat out misleading and incorrect.

As with all STP questions, you need to first determine the root bridge, then determine the root port of all non-root switches, then determine the designated port on each network segment between non-root switches in order to accurately determine which ports will be in a forwarding or blocking state.

We know that switch E is the root in this case since it was given to us. That means that all of E’s interface will be set as designated ports on their respective segments and will be forwarding. Next, we need to determine the root port of all of the other non-root switches.

That’s impossible because we don’t know the cost associated with any of these links. Switch A might not even use it’s direct link to Switch E as it’s root port. For instance, if the link between Switch A and Switch E was a 10mb connection, the STP cost would be 100. If the links from A to D to C to E were all using 100mb connections, the cost would be 10 + 10 + 10 = 30. Therefore, switch A would select it’s connection to switch D as it’s root port and would block on it’s direct connection to switch E.

Even if we were to assume that all of the links were the same speed and used default costs, we still can’t accurately determine the answer. If that were the case, we could deduce that Switch A would select it’s connection to Switch E as it’s root port and forward on that interface. We would also know that the designated port on the segment between Switch A and Switch D would be switch A’s since Switch A would have the lowest cost path to the root bridge, regardless of link speeds (A only has 1 link to cross to reach the root bridge whereas D has 2 links no matter which path it chooses). However, we still wouldn’t be able to determine which port Switch D would choose as it’s root port.

The cost of Switch D’s path to Switch E trough Switch A or through Switch C would be equal (2x whatever the default STP cost is for the link speeds used). Switch D would need to use tie-breaker rules to determine it’s root port, selecting the connection through the neighbor with the lowest advertised bridge ID. Bridge ID is made up of the bridge priority (which is based on the assigned priority + the System ID extension aka VLAN) and system ID, which is based on the MAC address of the switch. Since we don’t know if bridge priorities have been manually assigned or just left to defaults and we don’t know the MAC addresses used in the bridge ID, we have no way of knowing which port will be selected as the root port.

If Switch D selected it’s link to Switch C as it’s root port, then the listed answer for question 2 would be correct since Switch D’s interface to Switch A would be neither it’s root port nor the designated port for the LAN segment, which would put it in a blocking state. No traffic would pass directly between Switch A and Switch D, with all traffic between A and D passing through switches C and E.

However, if Switch D selected it’s link to switch A as it’s root port, traffic could pass directly between switch A and D and frames between devices connected to the respective switches would not have to travel through the root bridge.

Well..they are just laughing at us with a cup ‘f coffee while we’re struggling this….LOL

I think what many are not considering(I as did at first) in question 2 is to look at the topology from SW-D’s perspective.For SW-D there are only 2 paths to root switch SW-E, one through SW-C and the other which is obviously worse is through SW-A. So SW-D’s spanning tree will place port to SW-A in blocking state thus preventing SW-A learning mac address of Server-1 from SW-D. So the right answer is only D – Sw-A, Sw-E, Sw-C, Sw-D.

Sitting for ICND2 in 30 minutes..
Wish me luck!!

@Q2
Guys re-draw this picture like this 🙂 Now let’s assume all the links are Gi which means interface path cost is 4. So lets assume mac addresses addresses are like this: A:00 – B:01 – C:03 – D:02
Now pay attention!
1. switch A and C have interfaces to root switch and have cost to root 0
2. As we assumed mac address of switch A is lower then C’s switch B will pick it’s path to root witch through switch A and have root cost of 4. Link between switch B and C will be blocked by switch C because C has lowest mac address.
3. As the D will recieve BPDU’s from switch A and C with the same root cost of 4 the switch will look their mac address and pick A because again C has the lowest mac address.
So the link between C and D is blocked.
E (root)
/ \
A—–B—–C
\ /
switch D

So! If we don’t know the link speed’s and mac addresses the answer could be both correct.
Hope I didn’t miss anything 🙂