Miscellaneous
Last Modified:
7/12/2016 9:30:50 AM
4: Network Layer
4a-1
Grades
 Resubmit graded HW2 so that we can note
who graded each assignment
Will compute means for each grader and adjust
accordingly
 If you are not satisfied with that can submit
written request for regrade

 Midterm – one person graded all instances
of a given question so should be fair

If you have a question, may submit written
request for regrade
4: Network Layer
4a-2
Programming Assignment 2
Resources
 Standard Assignment
 Linux home directories (coming today)
 All: keiko, splash, stella, namu, nova
(.csuglab.cornell.edu)
 Meng Additional: cfs01 - cfs20 (.cs.cornell.edu)
 20 GB for everyone (~40 MB per team; if run out of disk
space let Dora Abdullah know)
 Optional Assignment
 7 machines in 330 Upson Linux Installation (marked with
red dots)
 7 machines in 315 Upson Linux Installation (marked with
red dots)
 Get more information (including passwords) from Ranveer
 You must officially sign up!! ASAP please
4: Network Layer
4a-3
Note on homeworks
 Details in homework; High level information
in class – why?
Any particular gory detail is not essential; a
feel for reality is
 Homeworks are the place for gory details in
this class (too many details to cover in lectures)
 Confront “inconsistencies” of reality when have
all references available to you
 Test what you are learning against the real
world

 Would people like to see us cover
homework solutions in class?
4: Network Layer
4a-4
Clarifications
 Path MTU Discovery
 Count to Infinity
4: Network Layer
4a-5
Path MTU Discovery:TCP
 If doing Path MTU Discovery, start with
minimum of receiver’s specified MSS or
local MTU and set the Don’t Fragment Bit
 If ICMP message received indicating that
fragmentation was required, then segment
size will be reduced
 Periodically (every 10 min or so), TCP will
try a higher segment size up to the
receiver’s MSS to see if new route is being
used that would allow larger segments
 Not all implementations support this
4: Network Layer
4a-6
Path MTU Discovery: UDP
 Not like TCP where sender sends stream in chunks




as they see fit and receiver reads in chunks as
they see fit
With UDP, the size of the UDP packet is much
more visible to the application
May send with DF bit off
May send with DF bit on and if get ICMP messages
then IP on host may fragment before sent but not
exposed to application layer to encourage smaller
amounts of data sent
Again not all implementations support
4: Network Layer
4a-7
Path MTU Discovery
 Look in Ethereal at TCP segments, will see
Do Not Fragment Bit is set
If on Ethernet, don’t usually see adjustment
 Ethernet has one of the smaller MTUs so never
get an ICMP error saying needs to be smaller
 If sent from local network with larger MTU
might see this activity

4: Network Layer
4a-8
Distance Vector: link cost changes
Link cost changes:
 node detects local link cost change
 updates distance table (line 15)
 if cost change in least cost path,
notify neighbors (lines 23,24)
“good
news
travels
fast”
1
X
4
Y
50
1
Z
algorithm
terminates
4: Network Layer
4a-9
Distance Vector: link cost changes
Link cost changes:
 good news travels fast
 bad news travels slow -
“count to infinity” problem!
60
X
4
Y
50
1
Z
algorithm
continues
on!
4: Network Layer 4a-10
Distance Vector: poisoned reverse
If Z routes through Y to get to X :
 Z tells Y its (Z’s) distance to X is
infinite (so Y won’t route to X via Z)
 will this completely solve count to
infinity problem?
60
X
4
Y
50
1
Z
algorithm
terminates
4: Network Layer 4a-11
Bigger Loops and Poison Reverse
E
D
D (A,D) = c(E,D) + min {D (A,w)}
= 2+3 = 5
w
Loop back through E! Poison reverse will fix this
7
A
1
B
C
2
8
1
E
D
2
E
D (A,B) = c(E,B) + min {D B(A,w)}
= 8+6 = 14
w
Loop back through E! Poison reverse will not fix this
E will try to send through B
B’s route is through C so no poison reverse
A
50
50
B
1
2
8
E
C
2
D
4: Network Layer 4a-12
Count to Infinity Example with
Bigger Loop
B will learn bad news
C will have told B infinity because its route is
through B, so B won’t reroute through C
E however will have told B about a good route
through D (cost 6)
B will choose that route instead and advertise it as
the new best to C (cost 6+8 = 14); it will be worse
than the old one it advertised to C (old cost = 1)
C will propagate this updated “best” route to D
(cost 15)
D will propagate this new “best” route to E (cost
17)
E will update the “best” route to B (cost 19)
Last time it advertised cost 6 to B
It will loop around adding 13 each time (cost of
loop)
Will continue until cost E advertises to B is bigger
than 500
A
1
1
B
C
2
8
E
A
500
D
2
B
1
2
8
E
C
2
D
4: Network Layer 4a-13