Universität Stuttgart Communication Networks II Sample Solution

INSTITUT FÜR
KOMMUNIKATIONSNETZE
UND RECHNERSYSTEME
Universität Stuttgart
Prof. Dr.-Ing. Andreas Kirstädter
Sample Solution
Communication Networks II
Examiner:
Professor Dr.-Ing. Andreas Kirstädter
Date:
September 29th, 2010
Duration:
90 Minutes
Required:
All Problems
Permitted Resources:
All
[56 P.] Problem 1
Traffic Engineering
Part 1
Shortest-Path Routing
Question 1
a)
eAB, eGB, eGE, eED
b)
Shortest path tree:
15 Points
C
B
G
E
D
A
F
c)
H
Forwarding table of node A:
Destination
Node
Outgoing Interface (edge)
B
eAB
C
eAB
D
eAB
E
eAB
F
eAF
G
eAB
H
eAF
d)
Question 2
10 Points
Forwarding table of node B (attention: new shortest path tree has to be
derived!):
Destination
Node
Outgoing Interface (edge)
A
eBA
C
eBC
D
eBG
E
eBG
F
eBG
G
eBG
H
eBH
a)
24 Mbit/s
b)
5 Mbit/s
c)
In case of IP routing, the edge eED along the shortest path is limiting the
capacity to 5 Mbit/s.
No, IP routing provides only a single path from source to destination.
Therefore, only up to 10 Mbit/s can be reached by modifying the link cost
values.
d)
Network operators should select link costs inversly proportional to the
transmission capacity of the link. In Figure 1 the ratio of cost to transmission capacity varies between 0.06 and 1.5. The high capacity links, i.e. the
50 MBit/s links, have low cost. For the 10 MBit/s links the cost values vary
between 7 and 15. The 5 MBit/s link is comparable cheap and must have
other advantages.
e)
Other criteria
•
Delay: Low delay links are required by some applications
•
Monetary Cost: The usage of some links might be charged by the
owner of the link
•
MTU: Usage of links with a small MTU might lead to packet fragmentation
•
Load: In dynamic scenarios the link load might be of interest. That
means low loaded links should be preferred to high loaded links.
Part 2
MPLS Switching
Question 3
a)
The LSR reads the label of the arriving MPLS packet and switches the
packet according to the information for this label and input link found in the
forwarding table. Also, the label is changed according to the entry in forwarding.
b)
The table entries can be set by
5 Points
Problem 1
Page 2
Question 4
•
Label distribution protocol: In case of dynamic MPLS path establishment a so called label distribution protocol is used to inform the LSR
about the used labels.
LSP A-B-C-D involves the edges eAB, eBC, and eCD.
The other paths provide less transmission capacity
b)
Node B: I-label = 1, O-interface = eBE, O-label = 2, swapping
Node C: I-label = 2, O-interface = eEH, O-label = 3, swapping
c)
Node A operates as a Label Edge Router: It reads the IP address of the
arriving packets and pushes a label to them according to the Forward Equivalence Class (FEC) they belong to and according to the LSP that they have
to follow in the MPLS network.
Node D also operates as a LER: It pops the label from the arrving packet.
a)
A TE link from the IP perspective are recognized as ordinary links to transport packets. They are realized in so called Multi-layer networks, e.g. by
MPLS LSP. That means a path on the lower layer (e.g. MPLS) appears as a
single link on the higher layer (e.g. IP).
(In many cases TE links use reserved resources, e.g. a certain amount of
bandwith.)
b)
The routing of OSPF can be influenced, that means specific traffic can be
directed via these links without affecting other traffic in the network.
a)
The MPLS-based TE links are considered during the IP routing like physical links. Virtual interfaces are created and added as outgoing interfaces.
That means the forwarding engine uses physical as well as virtual interfaces
for packet transport.
b)
The Ethernet link-layer frames contain a type field that indicates the type of
the packet contained in the arriving frame.
Ethernet frame type for IP packets: 0x0800
Ethernet frame type for MPLS: 0x8847
c)
The routers need to implement a priority scheduling at their output ports.
4 Points
Question 6
Network management: Via a network management interface the network operator is able to set MPLS paths manually. Such MPLS paths
are static.
a)
6 Points
Question 5
•
6 Points
Part 3
ECMP Routing
Question 7
a)
If several paths with the same minimum cost exist between the source and
destination node, the traffic will be evenly split among them at the current
node (source).
b)
By default packets belonging to the same flow (IP src/dst. address, transport
protocol, src/dst. port) are routed on the same path to avoid reordering of
packets. If ECMP wouldn’t be aware of flows reordering might take place,
because packets on different paths experience different delays. In case of
TCP, reordering of packets would lead to a significant goodput/throughput
decrease.
4 Points
Problem 1
Page 3
Question 8
a)
The traffic has to use all three links leading to the destination node:
eCD, eHD, and eED.
As B shall be the only splitting node, the following three branches have to
be used: A-B-C-D, A-B-H-D, and A-B-G-E-D.
Therefore we have to set: cBC = 5, cBH = 3.
b)
Modified link costs potentially change the routing of other traffic in the network. In case of traffic from H to C, the route is changed. Originally, the
traffic has been routed by node D. After the change, traffic is routed via
node B.
6 Points
Problem 1
Page 4
[49 P.] Problem 2
Performance Evaluation of Communication Networks
Part 1
Modelling
Question 1
a)
Max. frame size 1518 bytes = 12144 bits
10e9 bits/s / 12144 bits/frame ≈ 823452 frames/s
b)
x/100 * 823452 frames/s
c)
1/λ = x/100 * 823452
6 Points
fX ( y ) = λ ⋅ e
– λy
d)
Question 2
a)
16 Points
The link delays the packets by the transmission time and by the propagation
delay
1518bytes ⁄ packet ⋅ 8bits ⁄ byte
Transmission Time: -------------------------------------------------------------------------------- ≈ 1.2μs
10e9 bits/s
200m
Propagation Delay: ------------------------ = 1μs
8
2 ⋅ 10 m/s
Total delay: t del = 1.2μs + 1μs = 2.2μs
Problem 2
b)
Because the link delay is shorter than the transmission time of a packet, a
packet does not fit completely on the link.This means that 2 packets can be
simultaneously on the link at maximum.
c)
3 states are needed
•
Wire idle
•
1 frames on the wire
•
2 frames on the wire
Page 5
d)
e)
Time indicating the complete reception of a frame
•
during the tranmission of the previous frame:
t2=T1+ttrans
•
after the previous frame has left the sender completly:
t2=NOW+ttrans+tprop
Question 3
4 Points
Problem 2
"
"
!
Page 6
Part 2
Simulation Methodology
Question 4
Transient/Warm-Up Phase:
2 Points
•
System/buffers get loaded with events
•
Reach the stationary system state
Evaluation Phase:
•
Collect measured values
The differentiation is necessary to avoid bias w.r.t. the stationary state of the system.
Question 5
5 Points
Mean:
( 10 + 7 + 8 + 15 + 5 + 10 + 9 + 6 + 10 + 5 )ms
x = ---------------------------------------------------------------------------------------------------------------- = 8.5ms
10
Sample variance: S =
1
2
2
--- ( ( 10 – 8.5 ) + ( 7 – 8.5 ) + … )ms ≈ 3.028ms
9
1 + 0.99 Var
3.028ms
Half length: l = t 9 ⎛ ⎛ -------------------⎞ ⋅ ----------⎞ ≈ 3.25 ⋅ --------------------- ≈ 3.11ms
⎝⎝
2 ⎠
10⎠
10
Confidence interval: [5.39 ms , 11.61 ms]
Question 6
a)
Those parts, with small confidence intervals, i.e. the first 3 values are significant. The last 3 values show trend with a more or less acceptable conficende interval.
b)
The forth simulation value has a low significance, since the confidence
interval is very high. That means the mean waiting time can be in area of
the confidence interval, which might lead to very different results (compare
waiting time at the upper bound vs. lower bound). Moreover there are not
much simulation values around 4th value.
In order to improve the significane it is required to add additional simulation values below and above the 4th simulation values. In addition it is
required to simulated longer in order improve the significance of the 4th
value itself.
c)
The number of samples has a quadratic influence on the length of the confidence interval. That means that we have to simulate four times as much to
halve the confidence intervals.
7 Points
Problem 2
Page 7
Part 3
Capacity Extension
Question 7
It is required that the switch deals with the two physical ports as one virtual port
in order to jointly learn MAC addresses. Manual configuration is needed out of
two reasons
3 Points
•
Avoid infinite cyling of IP packets: If the port aggregation feature is not
configured on both ports, frames might cycle for infinity. Basically it would
be possible to use the spanning tree protocol with the drawback of having
only 1 GBit/s between the switches.
•
Port Aggregation provides a couple of different configuration options that
have to be the same on both switches. This has to be configured manually.
(Not expected as an answer in the exam)
Question 8
6 Points
Problem 2
Round Robin
Random
Hash Function
Implementation
Effort
Low
Low
Depends on hash
function
State
Yes
No
No
Reordering of Pak- Possible
kets within Flow
Possible
Not possible
Link Overload Pos- No
sible
Yes
Yes
Equal Link Utilization
Not Guaranteed
Depends on Hash
Function
Yes
Page 8