Thanks Avinash for doing the proper calculation and showing.

I think there is a slight mistake in (B). It should be:
(B) R->R5
=(100 Mbps *1024) / 64Kbps
= 102400 Kbps / 64 Kbps (remember you have to use the same metric when dividing!)
= 1600

..so you just have to convert the 100Mbps above to Kbps and then divide. So the 64Kpbs link is the highest cost link when ospf calculates. thus it’s the slowest link

I mean take it logically. The two T1 links(1.544Mbps) should be faster than the 64kbps(dial-up speed) link anyway.. .

The the two T1 links path is the second highest cost and the ospf cost is around 128 as you have correctly calculated.

And of course the fastest path is R1->R3->R4->R5 where it uses the two 100Mbps links and one 10Mpbs link comindg to a cost of 12 as you have again correctly calculated.

Cheers

can anybody explain the concept of TTL.

That’s my little contribution and i hope it makes sense to you

RIP is easy , no calculations just hops….min hop count irrespective of the link speed.

Thanks indeed 9tut.. tut

You are soooo right, BUT, read the question carefully, it is asking just for london to forward traffic to manchester.

we can understand easily…static touting..

OK, Answer E is fine, but to establish two-way communication we also need static route on Manchester router to direct all traffic destined for 192.168.10/23 to 10.1.1.1. One-way communication is rather useless except multicast and DOS attacks ;-)

ip route 172.16.0.0 255.255.252.0 (exit interface of london)

but we don’t have any information about interfaces on the exhibit.

ip route 172.16.0.0 255.255.252.0

but we don’t have any information about interfaces on the exhibit.

The simplest solution this issue is to set static route. The static route is:

ip route destination-network-address subnet-mask {next-hop-IP-address | exit-interface}

so we need to do:

ip route 172.16.0.0 255.255.252.0 10.1.1.2

-> 10.1.1.2 is the address of next hop (interface of next router)

On this exhibit is very simply connection so we don’t need to use dynamic routing.

Refer to the exhibit. The network administrator must establish a route by which London workstations can forward traffic to the Manchester workstations. What is the simplest way to accomplish this?

A. Configure a dynamic routing protocol on London to advertise all routes to Manchester.
B. Configure a dynamic routing protocol on London to advertise summarized routes to Manchester.
C. Configure a dynamic routing protocol on Manchester to advertise a default route to the London router.
D. Configure a static default route on London with a next hop of 10.1.1.1.
E. Configure a static route on London to direct all traffic destined for 172.16.0.0/22 to 10.1.1.2.
F. Configure Manchester to advertise a static default route to London.

Plz help why the answer is E

T1:64
T3:2
FastEhernet:F0/0 : 1
Ethernet e0/0 :10
56Kbps:1785
E1:48

At which layer of the OSI model does IPsec operate?

A. network
B. transport
C. session
D. application

Q:4
Q:7

In “ip default-network 0.0.0.0″ what network is represented by 0.0.0.0? I understand the syntax and just need to know the network represented by “0.0.0.0″
For example, the command “ip default-network 192.168.10.0″ will use the network address “192.168.10.0″ as gateway of last resort. In a topology with 20 networks, which one will be represented by “0.0.0.0″? I tend to believe that the nework address 0.0.0.0 is the same as “any (classful) network”, in which case the command “ip default-network 0.0.0.0″ would mean “send packets detsined to any (unknown) network” to “any network”.

My question is this: What (classful) network is represented by 0.0.0.0? Is it “any network”? If it is, how doess the router decide which network to send the packets to?

In other words, which network is the “any” network? “Any network” as “source” is easy to identify because it means I don’t care where you come from. However, “any network” as “destination” is complicated for obvious reason and that’s why it’s hard for me to believe that the “0.0.0.0″ network can mean “any destination network”.

In “ip default-network 0.0.0.0″ what network is represented by 0.0.0.0? I understand the syntax and just need to know the network represented by “0.0.0.0″
For example, the command “ip default-network 192.168.10.0″ will use the network address “192.168.10.0″ as gateway of last resort. In a topology with 20 networks, which one will be represented by “0.0.0.0″? I tend to believe that the nework address 0.0.0.0 is the same as “any (classful) network”, in which case the command “ip default-network 0.0.0.0″ would mean “send packets detsined to any (unknown) network” to “any network”.

My question is this: What (classful) network is represented by 0.0.0.0? Is it “any network”? If it is, how doess the router decide which network to send the packets to?

In other words, which network is the “any” network? “Any network” as “source” is easy to identify because it means I don’t care where you come from. However, “any network” as “destination” is complicated for obvious reason and that’s why it’s hard for me to believe that the “0.0.0.0″ network can mean “any destination network”.

If you comment on a question, can you make the first line of your comment the question #? That will help everyone. For example, (if) Ihave a comment or question about question #5 above, I would start like this:

Question #5:
Can someone explain the line “ip default-network 0.0.0.0″ and why C is the correct answet?

If you want to answer me or leran more about my question, all you have to do is go to question #5 for more information and background about my question.

Today I have passed the CCNA. (860/825)
50 questions 3 labs (VTP, EIGRP, ACL). 35 from 9tut.

Also thanks a lot Brar and Sekhar (still valid from examcollection)

Ty again 9tut

By the way thanks to 9tut for the help. Hey guys let’s donate to 9tut, how ever much you can afford. he’s doing a great job!

Network Interface Next Hop
10.1.1.0/24 e0 DC

10.1.2.0 /24 e1 DC

10.1.3.0 /25 s0 DC

10.1.4.0 /24 s1 DC

10.1.5.0 /24 e0 10.1.1.2

10.1.5.64 /28 e1 10.1.2.2.

10.1.5.64 /29 s0 10.1.3.3

10.1.5.64 /27 s1 10.1.4.4

10.1.3.0

helpppppp meeeeee :)

here in your question regarding dhcp server confusion can be between two choices B and C.
but you can differentiate by taking simple example. when dhcp is configured there is option to set time limit for a ip assigned to a host. suppose that timing is small amount and after that time server will automatically deassign ip allocated to your host without asking for renewal it will be very boring task to ask for ip to dhcp again and again while you are doing some important work. so according to option C either server will ask for renewal or host will request for renewal of currently using ip.

i hope i am right. if wrong let me know

EIGRP will use the link with the highest bandwidth and lowest delay. The T1 link right through the middle is worse than traversing the T3 links and extra router even though it may seem that T1 is more “direct”. And think about it… the T3 links can handle many times the traffic than T1 links, so you’ll want your traffic going through it as to avoid any congestion issues.

RIP will prefer the route with leasts hops, in RIPs opinion JAX-ORL is the best route.

OSPF:

Metric is cost, based on interface bandwidth by default
Link-state routing protocol.

they uses the interface with the highest bandwidth .

+ T1: 1.544 Mbps
+ T3: 45 Mbps (each T3 line consists of 28 T1 lines)
+ 10BaseT: 10 Mbps
+ 100BaseT (often referred to as FastEthernet): 100Mbps

Can u explain how best path is chosen by ospf..? Plz