Broadcasting

Question

Consider the Network below : 

If flooding, is used to send packets from A to E, then total number of packets generated in the network when hop count from  A to E is 3 ................ ?

Comments

My answer is 8 Please verify ?

Because when a router floods the network it keeps an identification of the packet flooded and if that same packet comes again its discarded instead of flooding it again else there would be infinite number of packets in the network. It does this by keeping a counter that counter is basically the latest largest sequence number of the packet flodded by him if any packet with sequence number less then it comes its discarded because its already is being seen else flodded and counter is changed.

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i hope answer is 14 ( i didn't calculated now, i remember it.)

check this https://gateoverflow.in/226655/me-testseries

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Can you please explain the answer and approach i am getting 8 not 10 Please Watch my approach.

Ok please see my approach brother :-

1. A sent two packets to B and D. ------> 2 Packets

2. B sent two packets to C and D  -------> 2 Packets

3. C sent two packets to B and D ---------> 2 Packets

4. D sent two packets to C and E ---------->2 packets

Now after this all the routers and seen this packet and wont flood them again if they encouter one. So total of 8

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i apologies,

total 8 is correct ( i don't know why i didn't find that mistake in https://gateoverflow.in/226655/me-testseries )

Process is :- Starts with A ( hop count = 3 )

A can send B or C

i) A send packet to B, now packet at B ( hop count = 2 )

         B can send the packet to either C or D ( but can't A )

                  B send Packet to C ( hop count = 1 )

                             C can send Packet to A ( hop count = 0 )

                             C can send Packet to D ( hop count = 0 )

                  B send Packet to D ( hop count = 1 )

                             D can send Packet to E ( hop count = 0 )

                             D can send Packet to C ( hop count = 0 )

    Total = 4 ( ABCD, ABCA, ABDC, ABDE )

 

ii) A send packet to C, now packet at C ( hop count = 2 )

         C can send the packet to either B or D ( but can't A )

                  C send Packet to B ( hop count = 1 )

                             B can send Packet to A ( hop count = 0 )

                             B can send Packet to D ( hop count = 0 )

                  C send Packet to D ( hop count = 1 )

                             D can send Packet to E ( hop count = 0 )

                             D can send Packet to B ( hop count = 0 )

    Total = 4 ( ACBD, ACBA, ACDB, ACDE )

 

Finally = 4+4 = 8

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1. Brother is my approach correct ?

2. And according to me brother with the hop count statement werent given in question then also the answer would have been 8. Because of the reason i stated earlier.

Am i right brother :P

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And according to me brother with the hop count statement werent given in question then also the answer would have been 8. Because of the reason i stated earlier.

if they didn't give the hop count we can't say it is finite, it may lead to infinite.

 

 Because when a router floods the network it keeps an identification of the packet flooded and if that same packet comes again its discarded instead of flooding it again else there would be infinite number of packets in the network. It does this by keeping a counter that counter is basically the latest largest sequence number of the packet flodded by him if any packet with sequence number less then it comes its discarded because its already is being seen else flodded and counter is changed.

this is LSR algorithm, but not flodding.

Note that LSR uses Flodding, but those are not equal.

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Shaik Masthan 

B can send the packet to either C or D ( but can't A )

 

why ???

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Because Magma that packet was given to C by A itself then why would C flood the packet that was given by A itself to A again. In general, when a packet comes in to router from one interface it floods that packet to every other interface except the one on which it was arrived :) I hope its clear now

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Nice answer brother :)

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Is this is correct ???

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@Magma

yes, you are correct, thanks for correcting me.

 

@Na462

correct answer is 14 only.

we calculated the paths, but in the question they ask how many packets generated ?

 

A generates 2 packets 

 

           --(one packet) --> B generated 2 packets --(one packet) --> C generated 2 packets ( for D and A)

                                                               --(one more packet of B ) --> D generated 2 packets ( for C and E)

                       ===> packets of B + packets of C + Packets of D = 6

 

     --(one more packet from A) --> C generated 2 packets --(one packet) --> B generated 2 packets ( for D and A)

                                                               --(one more packet of C ) --> D generated 2 packets ( for B and E)

                          ===> packets of C + packets of B + Packets of D = 6

 

Total = packets of A + 6 + 6 = 2+6+6 = 14

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Brother i don't understand here when A generated two packets it went to B and C, then B generated two packets for C and D Now C generated two packets for A and D. Now u say when C gets packet of A it also floods that why ? it has already seen that packet from B why would he flood it again ?

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in flooding, only one rule... Except incoming line, you have to forward packets in all directions...

any node can't have the data, which is already received and which is already sent, it simply see at that time who sends the packet, except to that node broadcast in all directions.

 

in LSP, there is sequence number and timer(may be), so the node see the sequence number of a packet and note down the incoming line, then it will analyze whether this packet need to broadcast or not?

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Ok so thats why hop count is important here else there would be infinite number of packets floating around isn't it ?

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its ans is 14. I want to know how?

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Ok I got it.

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can anybody explan why A is not flooding to s. as A must flood to all its neighbours.

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@sandygate

due to A received packet from S

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where is it mentioned. is it considered that s was sending to r and a to e is subpart of it. how did you know it. As i cant find anywhere.

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A is receiving the packet from S. Then Why A will send to  S.

Read concept of flooding.

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where is it mentioned. is it considered that s was sending to r and a to e is subpart of it. how did you know it. As i cant find anywhere.

S to R means, S should send to A, because there is no other route ===> we can sense that packtet to A reached from S ===> By flooding we can't sent to the interface which is connected to S 

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Router A sending message to E.

So, we can assume that A got a message from S.Becuase there is no other way.

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total ans is 14 .

2+ 6+6=14

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yes correct answer is 14

nice answer @shaik masthan

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yes i was also getting 8 in first attempt and was doing the same mistake. now i understood. thankyou. superb explanation

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@Shaik Masthan  can u please explain this.

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should be read whole discussion https://gateoverflow.in/244634/broadcasting

Answer 1

Total packets sent from A to E are ...

Minimum hop count from A to E is 3

1: A sends 2 packets B and C.

2: B sends 2 packets (C & D), C sends 2 packets (B & D).

3: C sends 2 packets (A & D), B sends 2 packets (A & D), D sends 2 packets (C & E), D sends 2 packets (B & E).

Total amount of packets =  2 · 7 = 14.

where 

A	B	C	C	D	B	D	B	D	C	E	A	D	B	E
1	2	3	4	5	6	7	8	9	10	11	12	13	14	15

So total numbers of packets  = 14

Comments

Check the answer now, you have still doubt, let me know.

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Thanks for your effort .but i think no 8 should be A and 6 should be B.Because if C is receiving packet from B then C has two options either send packet to A or to D.Correct me if i am wrong ?. Or should I consider the hop limit also. If Hop limit of 3 is taken into consideration then your ans is correct bcz if I send packet to A the packet will not be able to reach the destination?

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Yes, you are right. I corrected it.