cisco 3 (1)

tionwhat is the difference b/w pap and chap,and give comand to configure pap cahp on two router r1 and r2,

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Answer Posted By

Question Submitted By :: Jitenderakumar Sinha

I also faced this Question!! © ALL Interview .com

Answer Pap is a two way hand shaking. the password is sent clear text, where as the password is hide chapconfiguration on r1G#enable secrete passwordnameG#username r2 password passwordnameG#ints0G#encap pppG#ppp anthentication CHAP

configuration on r2G#enable secrete passwordnameG#username r1 password passwordnameG#int s0G#encap pppG#ppp anthentication CHAP

0 Chandrakanth

Question

WHAT IS SERIAL TRANSMISSIONS,PARALLELL TRANSMISSION

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Answer Posted By

Question Submitted By :: Guest

This Interview Question Asked @ ASD-LabI also faced this Question!! © ALL Interview .com

Answer In serial Transmission date goes one by one but in paralleltransmission we can send date simin.....

0 Vijay

Answer The difference is that serial transmissions

only use 2 wiresfor transmitting and receiving data, and only 1 for sendingor receiving, and can only send one bit at a time.

Parallel cables use several wires for data transmission andis much faster than serial thus 8 or more wires are carrying

0 Mr.shahin07

data *parallel* to each other to it's destination.

In Serial Transmission, Data goes out one by one bit basisat a time. Ex - 1 0 1 0 1 1 0 1 0 1 0 > out

In Parallal Transmission, Data goes out group of bits(Basically 1 byte) basis at a time.Ex - 1 > out 0 > out 0 > out 1 > out 0 > out 1 > out 0 > out 1 > out

Hope u can understand little from this topic.

Mark it & let me knw the feedback.

Thnx & RegardsShahin

Questio

nwhat is LAPB Ran

kAnswer

Posted By Question Submitted By :: GuestThis Interview Question Asked @ ASD-Lab


Answer Link Access Procedure, Balanced (LAPB) is a data link layer protocol used to manage communication and packet framing between data terminal equipment (DTE) and the data circuit-terminating equipment (DCE) devices in the X.25 protocol stack. LAPB, a bit-oriented protocol derived from HDLC, LAPB makes sure that frames are error free and properly sequenced. LAPB cause trmoundes amount of overhead becoz of its strict timeout and windowing techniques. it is a connection-oriented protocol.

0 Mahesh

Questio

nwhat is burst rate Ran

kAnswer

Posted By Question Submitted By :: Jitendera Kuamr SinhaThis Interview Question Asked @ ASD-Lab


Answer when aceess rate is grtaer then commted rate then this is called brust ratejitenderakuar sinha

0 Jitendera Kuamr Sinha

Questio

nwhat is the difference between hdlc and ppp protocol

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Answer Posted By

Question Submitted By :: Jitendera Kumar SinhaThis Interview Question Asked @ ASD-Lab , HCL


Answer HDLC and PPP are the protocols used in Leased lines and also in ISDN.HDLC-High Level Data Control,PPP-Point To Point Protocol.mainly HDLC is cisco proprietary protocol and PPP is OSI protocol.PPP checks for authorization.It inter=grates with 2 protocols PAP and CHAP.PPP supports compression and HDLC doesnt support compression.

0 Pavan Kumar

Answer sejirej sern

0 R

Answer HDLC and PPP are technologies that works with

leased line means dedicated line and also in ISDN.HDLC-High Level Data link Control,PPP-Point To Point Protocol.

3 Umesh

mainly HDLC is cisco proprietary protocol and PPP is OSI protocol.PPP It inter=grates with 2 protocols PAP and CHAP.PPP supports compression and HDLC doesnt support compression.

Answer ppp point to point is used in frame relay tech.

HDLC is a cisco made protocol used in leased line (WAN connection)i.ein local loop.

5 Durgesh Singh

Rajput

Question

How many subnets can be gained by subnetting 172.17.32.0/23 into a /27 mask, and how many usablehost addresses will there be per subnet?

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Answer Posted By

Question Submitted By :: Zhandan8207I also faced this Question!! © ALL Interview .com

Answer There is 16 subnet being made when we extend /23 to /27.Because we are using 4 bits from host bits.

2*2*2*2 = 16

so the subnets are :-

172.17.32.0 to 172.17.47.0

Thanks

0 Devender Chauhan

Answer There is 16 subnet being made when we extend

/23 to /27.Because we are using 4 bits from host bits.

2*2*2*2 = 16


172.17.32.0 to 172.17.47.0

and the total number of host is 16x255 =4080

0 Devender Chauhan

host

Answer There is 16 subnet being made when we extend

/23 to /27.Because we are using 4 bits from host bits.

2*2*2*2 = 16


172.17.32.0 to 172.17.32.31

172.17.32.32 to 172.17.32.63

and continue...with 32 IPs per subnet...

0 Devender Chauhan

Answer 172.17.32.0/23

mask of that ip is sunated in the 172.17.32.0/27sunating of 4 bittotal number of n/w2^4-2=14(according to cisco)now taol nomber of valabe bit for host is 16-11=5so taotal nomber of avilable host is 2^5-2=30note**(why eleven is redused from 16.ans becoz this is a xlass b adress and it have 16 bit for n/w and 16 bit for host but we have the mask of 27 then we have taken 11 bit from host for n/w.so we redused that eleven bit from the 16)nownow hat is the mask of that ip 255.255.128+64+32+16+8+4+2+1.128+64+32i.emask is 255.255.255.224/27so the block size is 256-224=32so the n/w is 172.17.32.0---------------------.31.32------------------------------.63 and so on

5 Jitendera Kumar Sinha

Answer 172.17.32.0/23 into a /27 how mant subnet?

n=23-27=4

0 Bhisham

total no subnet mask =2^n - 2=2^4-2=14where n is the no host host =2^n-2=14subnet =2^n=2^4=16

Answer Dealing with a 172 IP places you into the Class

B Network. The default subnet mask is 255.255.0.0. However, with theclassless identifiers, or 'prefix length', of /23 and /27your subnet masks will be as follows:

/23 255.255.254.0/27 255.255.255.224

Which is broken down into binary as follows:Default 11111111.11111111.00000000.00000000/23 11111111.11111111.11111110.00000000/27 11111111.11111111.11111111.11100000

If you'll notice there are 23 ones in the second subnet maskand 27 in the third. Hence the /23 and /27. Pay attentionto the 'overlap' (area where the default reads 0 and theprefix length reads a 1. In this case the last two octets) For the /23 there are 7 and the /27 there are 11. Rememberthese as they will come up later.

Note: There is a proper term for the 'overlap' but I don'tremember it

Beginning at the right of the binary version of the /23subnet mask. Start with the number 1 and double it for every'0' you have, stopping ON the first '1'. i.g.

512 256 128 64 32 16 8 4 2 1 /23 11111111.11111111.111111 1 0. 0 0 0 0 0 0 0 0/27 11111111.11111111.111111 1 1. 1 1 1 0 0 0 0 0

As you can see for the /23 subnet mask, 512 is the TOTALnumber of hosts you can have per subnet. You'll have tosubtract 2 hosts for the network and broadcast

0 Robert Davenport

addressesleaving you with 510 USABLE hosts per subnet. So, both looklike this:

Prefix TOTAL USABLELength Subnets Subnets/23 512 510/27 32 30

Remember the 'overlap' areas from above? if not I'll givethem to you again:

/23 7/27 11

This number will be used to find out the number of subnetsby using the following formula:

2^nWhere n is the the 'overlap'.

So... 2^7= 2x2x2x2x2x2x2=128 or2x2=4x2=8x2=16x2=32x2=64x2=1281 2 3 4 5 6 7

2^11= 2x2x2x2x2x2x2x2x2x2x2=2048 or2x2=4x2=8x2=16x2=32x2=64x2=128x2=256x2=512x2=1028x2=20481 2 3 4 5 6 7 8 9 10 11

Like with the TOTAL hosts, we have to remove 2 from theTOTAL subnets for network and broadcast ranges.

Now we know:

Prefix TOTAL USABLELength Subnets Subnets/23 128 126/27 2048 2046

Lets compile everything:

Prefix USABLE USABLE Total NetworkLength Subnets Hosts HOSTS/23 126 510 64,260/27 2046 30 61,380

So, even though we gain 1920 subnets. We lose 480 hosts per

subnet. Which means we will lose a total potential of 2880hosts.

Answer I have one bit of clarification on my answer.

As the IPaddress of, 172.17.32.0, was given that will reduce theamount of subnets. The third octet (.32.) has restricted usto one subnet on the /23 prefix. so that will change thetotal subnets for the /27, see the following:

Prefix Subnets in Hosts perLength Range Given Subnet/23 1 510/27 14 30

So, the Subnets will increase by 13 and the hosts will dropto 420 (14x30). Hope I got it this time.

0 Robert Davenport

Answer Hi all i read all answers and in every answer

you have some mistake.IT ask how many subnets and how many hosts have per subnet not how many host have totally!So here is the most logical and easy understanding of sub-netting with out need to convert to binary we all know to make binary no. from the regular number.23 subnet mask =255.255.254.027 subnet mask =255.255.255.224it ask for 27 how many subnets will have and how many hosts so we startwe all should know that totally in subnet mask have 32-bits here first 27 are taken for the subnet mask so we left 5 bits for host 32-27=5number of the host we get from formula 2^n-2=No of host per subnetso we get 2^5-2=32-2=30we take 2 address for the broadcast and subnet thats why is 2^n-2We finish with the hosts no lets get the no. of subnetsfirst given was /23 bits and the second was /27-bits

0 Metodija Pankovski

we just do this27-23=4no. of subnet is 2^n we don`t have here -2!!! some one make mistake in the answers so remember you don`t take minus 2 when you solve no. of subnets thats for hosts!!so27-23=42^4=16ANSWER OF THE QUESTION IS:Subnets 16HOST 30 per subnetaddress range like this172.17.32.0 first subnet172.17.32.32172.17.32.64172.17.32.96172.17.32.128172.17.32.160172.17.32.192172.17.32.224172.17.33.0172.17.33.32172.17.33.64172.17.33.96172.17.33.128172.17.33.160172.17.33.192172.17.33.224 last oneI hope i help to some of you:)THIS IS TE REAL ANSWER!

Question

What are the Timer of RIP, IGRP, EIGRP and OSPF Routing Protocol?

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Answer Posted By

Question Submitted By :: KapilasdhirThis Interview Question Asked @ IBM


Answer RIP-It sends complete routing table to all its active interface in evert 30 seconds.Hold down Timer:-it is a time during the destination hop is unreachable=240 seconds..

IGRP:-Timer 90 seconds

0 Ajay Kumar

Answer rip igrp eigrp ospf

30sec none 6osec 10se0 Dveashish

Kumar

Gupta

[Pioneer] Answer 180 se

0 Rajesh Tiwari

[Pioneer] Answer rip-30 sec

igrp-90 seceigrp-5 secospf- 10 sec

0 Vivek Singh [Pioneer]

Answer rip 120

igrp 90eigrp 90ospf 110

0 Chetan [Pioneer]

Answer RIP

30 Sec(Update) 180 Sec(Timeout) 120 Sec(Remove)

IGRP 90 Sec(Update) 3*Update(Invalid) 3*Update+10(Hold_down)7*Update(Flashtime)

EIGRP For LAN5 Sec(Hello) 3*Hello(hold-timer)

For WAN60 Sec(Hello) 3*Hello(hold-timer)

OSPF10 Sec(Hello) 4*Hello(Dead-interval)

4 Aruna [Pioneer]

Answer rip-30 sec


0 Upendr Kushwaha

[Pioneer]

Answer rip-30 sec


0 Raja Imran [Pioneer]

Answer 0

10011090

0 Shainu [Pioneer]

Answer RIP-30 Sec Periodically

IGRP-90 Sec PeriodicallyEIGRP-Only if there is a change in network structure.NoperiodicallyOSPF-Only Either if there is a change in network structureor 30 Min Whichever is earliar.

0 Navdeep Manchanda

[Pioneer]

Answer RIP,15 seconds.

(RIP Send routing table to all connected routers)IGRP,60 seconds. (IGRP Send routing table to all connected routers)EIGRP, (After it well converged than If there is a change in network structure it will send its routing table to all connected routers)

OSPF, It is same as EIGRP. (After it well converged than If there is a change in network structure it will send its routing table to all

0 Irfan [Pioneer]

connected routers)

Answer The correct RIP timers as per cisco doc

Hellow interval 30 seconddead interval 180hold down timers 180 secondn flush timers 240 seconds

the eigrp timershellow send every 5 second dead 15 second in point to pointin nbma is 60 hellow interval and dead is 180

in ospf ppp hellow 10 dead 40brodcast same

but in point to multipoing hellow is 30 secon n dead is 120 secondpoint multipoint brodcast samepoint to multipoint non brodcast

0 Rutuja [Pioneer]

Answer Timer given below:-

RIP (Routing Information Protocol):-Route update Timer 30Sec,2 Route Invalid n Holdown timer 180 Sec 3 Flush Out =240 Sec

IGRP (Interior gateway routing protocol):-Route Update 90 Sec.2Route Invalid =270 Sec3 holdown timer = 280 Sec 4 Flush out timer = 680 Sec

EIGRP (Enhance Interior gateway routing protocol):-For 1.updatation used HELLO Protocol2. Toplology used Dual All path entry 3. Routing Table Best path

OSPF (Open Shortest path First):-1. For updation we used HEllo Protocol2. LSATable Best path in N/w

0 Bhisham [Pioneer]

Answer rip:-

update 30 sec timeout 180 sec flush timer 120 sec

igrp:- update timer 90 sec

0 Ayesha Anwar

[Pioneer]

Answer RIP = 30 sec

OSPF = (When there is update it send the update part of routing)

IGRP = 90 sec

EIGRP = 30 sec

Proved during simulation.

0 Majid [Pioneer]

Questio

nWhat is the 5-4-3-2-1 rule of network design? Ran

kAnswer

Posted By Question Submitted By :: Kapilasdhir


Answer this is the ring topologies method.

0 Neeraj

Answer he 5-4-3-2-1 rule embodies a simple recipe for

network design. It may not be easy to find examples in practice, but this rule neatly ties together several important elements of design theory. To understand this rule, it's first necessary to understand the concepts of collision domains and propagation delay. Collision domains are portions of a network. When a network packet is transmitted over Ethernet, for example, it is possible for another packet from a different source to be transmitted close enough in time to the first packet to cause a collision on the wire. The total range over which a packet can travel and potentially collide with another is

0 Tushar [Pioneer]

its collision domain.Propagation delays are a property of the physical medium (e.g., Ethernet). Propagation delays help determine how much of a time difference between the sending of two packets on a collision domain is "close enough" to actually cause a collision. The greater the propagation delay, the increased likelihood of collisons.

The 5-4-3-2-1 rule limits the range of a collision domain by limiting the propagation delay to a "reasonable" amount of time. The rule breaks down as follows:

5 - the number of network segments4 - the number of repeaters needed to join the segments into one collision domain3 - the number of network segments that have active (transmitting) devices attached2 - the number of segments that do not have active devices attached1 - the number of collision domains Because the last two elements of the recipe follow naturally from the others, this rule is sometimes also known as the "5-4-3" rule for short.

Answer Ethernet and IEEE 802.3 implement a rule, known

as the 5-4-3 rule is also known as the IEEE way, for the number of repeaters and segments on shared access Ethernet backbones in a tree topology. The 5-4-3 rule divides the network into two types of physical segments: populated (user) segments, and unpopulated (link) segments. The 5-4-3 rule is also known as the IEEE way and is sometimes compared to the Ethernet way which is having 2 repeaters on the single network without having any hosts connected to a repeater.

5 Segments4 Repeaters3 Populated Segments2 Unpopulated Segments1 Collision Domain


[Pioneer]

Question

10. Which STP state populates the MAC address table but doesn't forward data frames? A. Blocking B. Listening C. Learning D. Forwarding

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Answer Posted By

Question Submitted By :: KapilasdhirThis Interview Question Asked @ Cisco


Answer C.

0 Tzar

Answer c

it will learn the node who broadcast it

0 M.thangadurai

[Pioneer] Answer Answer: C

The purpose of the blocking state is to prevent the use of looped paths. A port in listening state prepares toforward data frames without populating the MAC address table. A port in learning state populates the MAC

address table but doesn't forward data frames. The forwarding port sends and receives all data frames on the bridged port. Lastly, A port in the disabled state is virtually nonoperational

0 Kapilasdhir [Pioneer]

Question

9. Which STP state prevents the use of looped paths? A. Blocking B. Listening C. Learning D. Forwarding

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Answer Posted By

Question Submitted By :: KapilasdhirI also faced this Question!! © ALL Interview .com

Answer a

0 Vivek

Answer Answer: A

The purpose of the blocking state is to prevent the use of looped paths. A port in listening state prepares to forward data frames without populating the MAC address table. A port in learning state populates the MACaddress table but doesn't forward data frames. The forwarding port sends and receives all data frames on the bridged port. Lastly, A port in the disabled state is virtually nonoperational.


Question

8. When is Frame Tagging used? A. When you install repeaters in your network B. When you install bridges in your network C. When you install routers in your network D. When you use Switches configured with multiple VLANS

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Answer Posted By


Answer answer is d

frame tagging methods are used in VLAN to commnunicate between different VLANs

0 Ram

Answer ans d

ram has write the wright answer but explanation is wrong.frmae taging is not used for communication b/w vlan.if frame tagging is used for intervaltion comunication then what the need of router thats called intervlan communicationsokthe wright expalnation is that when we trunk the two or more then two switch then the comunication bewwwen same vlan which are plased at diffrent vlan frme


[Pioneer]

taging is usedthis is the wright explanation of franetaging.

Answer if i am wrong that frame tagging is v-lan

identification method then what is frame tagingif farme tagging is used for comunication in diffrent vlan then plz tell me what is the need of implement router for intervlan routing.some one marke my answer wrong i just reqto him plz make the right naswer.plzwith humble requestjitendera kuamr sinha


[Pioneer]

Answer Yes Dear Jitendera Kumar Sinha is right

Answer: D

Frame tagging is used to uniquely identify a frame as it traverses the switch fabric on a VLAN


Question

7. What is an advantage to using switches in your network? A. Addressing of hosts by DHCP B. Ease of administration C. Stops broadcast storms D. Works like a repeater

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Answer Posted By

Question Submitted By :: KapilasdhirThis Interview Question Asked @ iGate


Answer d

0 Kushal

Answer Answer: B

Switches filter by MAC address and, by default, are typically easy to administrate. B is the "best" answer


Question

6. Which device can create smaller networks and stop broadcast storms by default? A. Repeater B. Bridge C. Router D. Gateway

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Answer Posted By


Answer bridge

0 Sudha

Answer Sorry Dear Sudha the answer is :--

Answer: C Routers can create internetworks and filter by logical address (IP) address, which can stop broadcaststorms


Answer yeah sudaha is write but i think explanation is

partially coreectthe explanation is scince rputer cretae dectase brodcast domain at each enterface so it create smaller network and sriop bradcast dpmain


[Pioneer]

Answer i have just suppose that sudah gives the answer

router.but she answersed bridges.router is the answer becoz router create the seprtae broadcast domain per inerface so it stop broadcast strom by default.this is the answer.jitendera kuar sinha


[Pioneer]

Question

5. What is an advantage of using routers in your network? (Choose all that apply) A. Stops broadcast storms B. Reduces downtime on your network C. Creates internetworks D. Can filter by logical address

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Answer Posted By


Answer B. & D.

0 Arijit

Mondal

Answer A.

0 Arijit Mondal [Pioneer]

Answer A, C, D - these are basic features of a router.

B. Downtime on of your network will not be reduced just because you have a router. Any properly configured device will reduce downtime.

0 Ray [Pioneer]

Answer a b c d

0 Yogeeshin

[Pioneer]

Answer Answer: A, C, D

Routers can stop broadcast storms, create smaller networks and filter by logical (IP) addresses.


Answer a b c d

cisco 3 (1)

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