CSE 577 – Design and Analysis of Switching Systems
Review Questions
Buffered Networks with Static Routing
Jonathan Turner
1. Consider the fat tree topology F3,16. How many hosts can this network support?
16×16×32=8192 hosts.
If host addresses are assigned as outlined on page 3 of the lecture notes, what sequence of
routing decisions would be made to reach a host whose decimal address is 5237, from a top
level switch?
5237 can be expressed as 20×162+7×16 +5. So, at the top level switch, a packet would be sent out on
port 20, at the middle level switch, it would be sent to port 7 and at the bottom level, it would be sent
to port 5.
Suppose the host with address 260 sends a packet to the host with address 270, how many
switches would the packet pass through, assuming it takes a minimum length path? How
many different minimum length paths could the packet take?
Since ⎣260/16⎦=⎣270/16⎦=16 these two hosts are connected to the same switch in the first level, so the
packet would pass through one switch, and there is only one path.
Suppose the host with address 260 sends a packet to the host with address 500, how many
switches would the packet pass through, assuming it takes a minimum length path? How
many different minimum length paths could the packet take?
Since ⎣260/256⎦=⎣500/256⎦=1 these two hosts have the same set of “grandparents” in the fat-tree; or
viewed in another way, they are in the same “level-2 cluster”. So the packet would pass through three
switches, and there are 16 possible paths the packet could take.
Suppose the host with address 520 sends a packet to the host with address 2000, how many
switches would the packet pass through, assuming it takes a minimum length path? How
many different minimum length paths could the packet take?
Since ⎣520/256⎦≠⎣2000/256⎦ these two hosts have no common grandparents, so a packet would have
to go all the way to the top level to get from one to the other. This means it would pass through five
switches and there are 256 different paths that the packet could take (one for each top level switch).
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2. Draw the fat-tree F3,2. Number the switches in each level from left-to-right, starting with 0.
Suppose that all links are 1 Gb/s and that the host with address 1 has a 1 Gb/s flow to the
host with address 11 through the top level switch numbered 1 and that the host with
address 2 has a 1 Gb/s flow to the host with address 12 through top level switch 2. Show
these flows on your diagram. Now, suppose you want to establish a 1 Gb/s flow from host 0
to host 3. Is there a route of length 3 with sufficient capacity? Is there a route of length 5
with sufficient capacity?
0
1
2
3
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
0 1
2 3
4 5
6 7
8 9 10 11 12 13 14 15
There is no route of length 3 or 5 that has enough capacity from 0 to 3 with the capacity for this flow.
Can you reroute one of the existing flows to accommodate the new flow? Show how you
would do that.
The figure below shows an alternate routing of the 2-12 flow that leaves enough capacity for the 0-3
flow.
0
1
2
3
0
1
2
3
4
5
6
7
0
1
2
3
4
5
6
7
0 1
2 3
4 5
6 7
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8 9 10 11 12 13 14 15
3. The nonblocking condition on page 5 of the lecture notes gives a lower bound on the
speedup needed for a fat tree network. Create a chart showing how the speedup grows as a
function of the number of hosts in the fat tree network. That is, show the number of hosts on
the x-axis and the speedup on the y-axis. Plot three curves. One for d=8, B=0.1, one for d=24,
B=0.1 and one for d=24 and B=0.5. Let the number of hosts range from 10 to 50,000.
8
7
speedup (S)
6
d=8, B=0.1
5
d=24, B=0.5
4
d=24, B=0.1
3
2
1
0
50
500
5000
50000
number of hosts (n)
Assume that all the links in the network are 1 Gb/s. If we limit the hosts to send and receive
no more than Δ Mb/s, we can view the system as having a speedup of 1000/Δ. For a system
with 10,000 hosts, d=24 and a maximum per host flow bandwidth of 100 Mb/s, what is the
largest value of Δ, we can have if we want the network to be nonblocking?
We apply the nonblocking condition, substitution 1000/Δ for S and 100/ Δ for B. We also let d=24
and k=3. This gives
1000/∆
1
2/
100/∆
solving this for Δ yields Δ=232 Mb/s.
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2/
1
1
1
4. Consider a partial configuration of F3,24 in which there is just one top level switch and two
clusters that each support 24×24=576 hosts (see page 11 of the lecture notes). Draw a
diagram showing this configuration.
.. .
.. .
.. .
Suppose we wanted to add two more clusters. How many top level switches would we need
to add? Draw a diagram showing this configuration.
The original switch has 24 top level switches, so a system that supports four clusters would need
twice as many, or 48. So, 24 would have to be added. The new configuration is shown below.
.. .
.. .
.. .
How many connections to the original top level switch would we need to move in order to
go from the original configuration to the new one?
Half of the connections to the original switch would have to be moved to the new switches. So that’s
576 connections that would have to be moved and another 576 that would have to be added.
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