Help me understand this question from Odom's Book

I'm on Chp 5 Fundamentals of IP Addressing and Routing and there is a Quiz Question that I don't understand the answer too.

Q: Imagine that PC1 needs to send some data to PC2, and PC1 and PC2 are separate by several routers. What are the largest entities that make it from PC1 to PC2?

A: Packet and L3PDU

The other answers to choose from are:

-Frame
-Segment
-L5 PDU
-L1 PDU

There isn't anything in the book that talks about size of each so I'm not sure how they got the answer.

thanks

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Comments

Ltat42a

Probably referring to the size of the TCP packet. It has more fields in it than the others that are listed.

JockVSJock

Ltat42a wrote: »

Probably referring to the size of the TCP packet. It has more fields in it than the others that are listed.

True, wasn't thinking about it in those terms. In Odom's book, TCP is way bigger then say HDLC and/or PPP.

lrb

The key part of the question is "separated by routers". The IP header and all the data (i.e. the L3 PDU) are retained from the source (PC1) and destination (PC2); the L2 headers would be discarded and rewritten each time the packet is routed from one subnet to another, making L1/L2 PDUs an incorrect answer here. L4 and L5 PDUs are encapsulated into an IP header which at least makes the L3 PDU bigger.

IP is end-to-end so in this case it is the largest entity (in size) that makes it all the way from PC1 to PC2.

Kind of a crap way of asking the question I admit. HTH

MrXpert

lrb wrote: »

Kind of a crap way of asking the question I admit. HTH

Something Mr Odom is a master of

JockVSJock

MrXpert wrote: »

Something Mr Odom is a master of

Agreed, Todd Lammle is guilty too.

sizeon

I disagree. To answer your question, a Protocol data unit(PSU) contains the control information at each layer of the osi model.

Forsaken_GA

lrb wrote: »

Kind of a crap way of asking the question I admit. HTH

No, it's not, it makes you think about it. Odom asks questions that, if you can answer them, demonstrate that you understand the core concept and can reason out any problem, not that you're merely memorizing and regurgitating. That's a *good* way of asking questions, it requires you to employ critical thinking skills.

In this case, the question makes sure you understand how the encapsulation/de-encapsulation process works. ie, a frame is a layer 2 construct, and since routers break up layer 2 domains, that data construct can't possibly survive end to end.

Once you run through your process of elimination on what it can't be, you're left with what it can be. If you knew the answer to begin with, then obviously this is unnecessary, but this is how you answer a question you don't know the answer to, by applying the knowledge you do have to the situation.

JockVSJock

I'm now into Chp 6 of Odom's book, which deals with TCP.

According to diagrams in his book, both TCP and IP have the same number of fields in their headers, 12.

This question doesn't make sense. He is asking about the bit size of the headers of the # of headers?

Forsaken_GA

JockVSJock wrote: »

I'm now into Chp 6 of Odom's book, which deals with TCP.

According to diagrams in his book, both TCP and IP have the same number of fields in their headers, 12.

This question doesn't make sense. He is asking about the bit size of the headers of the # of headers?

Not all of us have the book, so what's the question? And TCP and IP headers do not have the same number of fields. IP has 14, of which 13 are mandatory. A TCP header can have up to 19 (only 18 are usable though, as one of them is reserved), depending on what type of TCP packet it is (ie, if the ACK flag isn't set, then the ACK field isn't going to be populated, etc)

They do both contain a minimum of 20 bytes (they can contain more, if Options are used), so for any layer IP flow utilizing TCP as the transport protocol, you have to account for 40 bytes of overhead minimum, as opposed to only 28 bytes for a UDP flow (20 for the IP header, 8 for the UDP header)

JockVSJock

Forsaken_GA wrote: »

Not all of us have the book, so what's the question?

The question is #1 of this post:

Q: Imagine that PC1 needs to send some data to PC2, and PC1 and PC2 are separate by several routers. What are the largest entities that make it from PC1 to PC2?

A: Packet and L3PDU

In the book, Odom has header diagrams for both IP and TCP and they both have the same amount of fields, 12.

lrb

OK just so we're all on the same page, is this the two diagrams you are referring to in the book (I just grabbed the latest ICND1 book from safari so it might not be the same version of the book as you have)?

Cause yes when the header is shown like this they have the same amount of fields (even though he omitted the Options field from the IP header). However, most people tend to think of the DS Field (in the above diagram) as two fields: DSCP and ECN. Similarly, I would say that the Code Bits (in the diagram) field should actually be split up into one field per flag, e.g. URG, SYN, ACK, FIN, etc. Also if you add RFCs 3168 and 3540 to the mix, you also snatch 3 bits of the Reserved field for congestion notification purposes.

IPhdr.jpg

TCPhdr.jpg

JockVSJock

lrb wrote: »

OK just so we're all on the same page, is this the two diagrams you are referring to in the book (I just grabbed the latest ICND1 book from safari so it might not be the same version of the book as you have)?

Cause yes when the header is shown like this they have the same amount of fields (even though he omitted the Options field from the IP header). However, most people tend to think of the DS Field (in the above diagram) as two fields: DSCP and ECN. Similarly, I would say that the Code Bits (in the diagram) field should actually be split up into one field per flag, e.g. URG, SYN, ACK, FIN, etc. Also if you add RFCs 3168 and 3540 to the mix, you also snatch 3 bits of the Reserved field for congestion notification purposes.

I have the 2nd Edition of CCENT/CCNA ICND1 Official Certification Guide.

Yes, those are the two diagrams that I'm referring to.

The reason I'm confused is that if you count the number of fields, as they are presented in this boo, for both IP and TCP, the total for each is 12.

So if you are reading into this like I am, its hard to answer that question.

Forsaken_GA

JockVSJock wrote: »

The question is #1 of this post:

Well, as I stated before, use your process of elimination.

You can throw out Layer 1 PDU's right away, as that would be the bit. The bit is the smallest possible entity to make it from point a to point b, so it's obviously not your answer.

Layer 5 PDU's don't exist. Layer 5 and above are raw data, they have not yet been encapsulated by any PDU's.

So this leaves you Frame, Packet, and Segment.

Layer 4 = Segment

Layer 3 = Packet

Layer 2 = Frame

Well, think about how encapsulation works. Encapsulation is basically adding a layer, or a coating to what's passed into it. So of the three left, the TCP segment is the initial PDU. Then it gets encapsulated into a packet, which makes it bigger. The packet then gets encapsulated into a frame, which makes it the frame bigger than the packet.

However, the frame doesn't survive end to end, the frame is link local. The frame that PC1 sends to it's default gateway will not be the same frame that PC2 receives from it's default gateway (indeed, depending on the MTU's involved, the frame that PC2 receives may not even be the same size as the one PC1 sends), PC1's default gateway will strip the frame it receives, and then put on a new frame when it forwards it out the appropriate link. The IP packet and the TCP segment are the only portions that makes it through end without any alterations. And since the TCP segment lives within the IP packet, by definition, it must be smaller than the IP packet, making the packet/layer 3 PDU (since it may not be IP) the largest entity that make it from PC1 to PC2

Now note, if the question had not specified that there were routers between PC1 and PC2, and you could reasonably make the assumption that the PC's shared the same link, then Frame *could* be the correct answer, but that depends entirely on the size of the IP packets. If they're large enough to exceed the MTU of the layer 2 segment, then they'll be fragmented, and when they're reassembled, the packet will end up being bigger than the frame.

In the book, Odom has header diagrams for both IP and TCP and they both have the same amount of fields, 12.

Actually, no. The TCP diagram has Data included in the header... that's wrong. Data is not part of the TCP header. But it looks like the book is just combining the flags field into one.

JamesFigueroa

Forsaken_GA wrote: »

Well, as I stated before, use your process of elimination.

You can throw out Layer 1 PDU's right away, as that would be the bit. The bit is the smallest possible entity to make it from point a to point b, so it's obviously not your answer.

Layer 5 PDU's don't exist. Layer 5 and above are raw data, they have not yet been encapsulated by any PDU's.

So this leaves you Frame, Packet, and Segment.

Layer 4 = Segment

Layer 3 = Packet

Layer 2 = Frame

Well, think about how encapsulation works. Encapsulation is basically adding a layer, or a coating to what's passed into it. So of the three left, the TCP segment is the initial PDU. Then it gets encapsulated into a packet, which makes it bigger. The packet then gets encapsulated into a frame, which makes it the frame bigger than the packet.

However, the frame doesn't survive end to end, the frame is link local. The frame that PC1 sends to it's default gateway will not be the same frame that PC2 receives from it's default gateway (indeed, depending on the MTU's involved, the frame that PC2 receives may not even be the same size as the one PC1 sends), PC1's default gateway will strip the frame it receives, and then put on a new frame when it forwards it out the appropriate link. The IP packet and the TCP segment are the only portions that makes it through end without any alterations. And since the TCP segment lives within the IP packet, by definition, it must be smaller than the IP packet, making the packet/layer 3 PDU (since it may not be IP) the largest entity that make it from PC1 to PC2

Now note, if the question had not specified that there were routers between PC1 and PC2, and you could reasonably make the assumption that the PC's shared the same link, then Frame *could* be the correct answer, but that depends entirely on the size of the IP packets. If they're large enough to exceed the MTU of the layer 2 segment, then they'll be fragmented, and when they're reassembled, the packet will end up being bigger than the frame.

Actually, no. The TCP diagram has Data included in the header... that's wrong. Data is not part of the TCP header. But it looks like the book is just combining the flags field into one.

Dude your a beast...thanks for the breakdown

carpadz@yahoo.com

hi, JockVSJock.
Chp 6 of Odom's Book is "Fundamentals of TCP/IP Transport, Applications, and Security" i can't find in the book which you've mentioned TCP and IP have the same number of fields in there headers, 12? according to diagram? it is mention in the book TABLE and FIGURE only. and also, there's no question asked as you said... I've check read again but sorry it seems not found. thanks,

JockVSJock

carpadz@yahoo.com wrote: »

hi, JockVSJock.
Chp 6 of Odom's Book is "Fundamentals of TCP/IP Transport, Applications, and Security" i can't find in the book which you've mentioned TCP and IP have the same number of fields in there headers, 12? according to diagram? it is mention in the book TABLE and FIGURE only. and also, there's no question asked as you said... I've check read again but sorry it seems not found. thanks,

Be sure you're referring to 2nd Edition, that is the one that I have.

carpadz@yahoo.com

JockVSJock wrote: »

Be sure you're referring to 2nd Edition,
that is the one that I have.

Reply: yes, 2nd Edition of Wendell Odom's book. Chp-6 is starting from page-128 upto page-163. correct me if I'm wrong... i really need to study hard and review before i get to enroll in obtaining a CCNA CERTIFICATION, and also, to pass
the exam is a way to go through first step in the networking world.. thanks,