Optical Networks
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OPTICAL NETWORKS
(Exercises and Solutions)
© POOMPAT SAENGUDOMLERT
ASIAN INSTITUTE OF TECHNOLOGY
2009
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© P. Saengudomlert, Asian Institute of Technology
Chapter 1: Introduction
Problem 1.1 (based on problem 7.7 of [RS02]): Consider the wavelength plane
switch architecture shown below. Consider the situation where we have a total of 4 input fibers, 4 output fibers, and 32 wavelengths on each fiber. We must design the node
such that any 4 signals can be dropped or added. (Note that this implies we could potentially drop or add all the 4 wavelengths on a particular fiber while passing through
all the wavelengths on the other fibers.) The wavelengths are added and dropped
through tunable transponders that have tunable lasers. The 32 wavelengths are split
into 4 bands of 8 wavelengths each, and a tunable laser can tune over a single band.
λ32
…
…
…
…
optical
switch
…
…
…
…
…
λ1
…
…
optical
switch
…
λ1, …, λ32
MUX
…
fiber
…
DMUX
optical switch
R T
T R
…
R T
T R
tunable
transponders
from/to end users
(a) Draw a block diagram of this node, indicate the sizes of optical switches used, and
indicate the minimum number of transponders needed. (Instead of drawing everything, e.g., drawing 32 switches, simply identify the numbers of components, as
done in the above diagram.)
(b) Now suppose that we have tunable lasers that can tune over two bands instead of
one. Modify the block diagram of part (a) and indicate the number of transponders
needed.
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Problem 1.2 (Optical bypassing of electronic switching): Consider the switching
node B in the network shown below. Each directed link is a single fiber. Assume that
there are 3 s-d pairs: A-C, A-D, and E-C. Each s-d pair sends and receives traffic at 4
Gbps. In addition, assume that one wavelength channel can carry up to 10 Gbps.
D
A
B
C
E
(a) Assume the use of electronic switching architecture at node B. Identify the amount
of traffic (in Gbps) that must be processed electronically at node B.
(b) Assume the use of transparent optical switching architecture at node B, and that we
bypass electronic processing of traffic at node B. Identify the number of wavelength channels used on the fiber from A to B.
(c) Repeat part (a) but with 8 Gbps for each s-d pair.
(d) Repeat part (b) but with 8 Gbps for each s-d pair.
Problem 1.3 (Constructing a large switch from smaller switches): Below is an architecture of an optical add-drop multiplexer (OADM). Assume that there are 4 wavelength channels in a fiber. The OADM shown consists of an optical multiplexer
(MUX), an optical demultiplexer (DMUX), a 6×6 optical switch, and 2 tunable transponders. (Each transponder contains a tunable transmitter and a tunable receiver.)
Note that this OADM allows the node to drop and add any set of 2 wavelengths.
DMUX
input fiber
λ1
λ1
λ2
λ3
λ4
6×6
optical
switch
tunable
transponders 1
λ2
λ3
λ4
MUX
output fiber
2
Suppose that, instead of a 6×6 optical switch, you only have available 4×4 switches.
Indicate how you can construct an OADM with the same functionality by drawing a
diagram of your OADM architecture.
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© P. Saengudomlert, Asian Institute of Technology
Problem 1.4 (Constructing a large switch from smaller switches): Consider a construction of a transparent optical switching node with 3 input fibers and 3 output fibers
shown below. Each fiber has 2 wavelengths. In addition to performing the switching
function, this switching node can also drop and add any single wavelength from any
fiber.
DMUX
input fiber
λ1
λ1
λ2
λ2
λ1
λ2
optical
switch
MUX
output fiber
λ1
λ2
λ1
λ1
λ2
λ2
tunable
transponder
Suppose that you only have available 4×4 switches. Indicate how you can construct a
switch with the same functionality by drawing a diagram of your switching node architecture. More specifically, draw the structure inside the area with the dashed boundary
in the above diagram.
Problem 1.5 (Using splitters and filters instead of MUXs and DMUXs): Consider a
construction of a transparent optical switching node with 2 input fibers and 2 output
fibers as shown below. Each fiber has 2 wavelengths. In addition to performing the
switching function, this switching node can also drop and add any single wavelength
from any fiber.
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DMUX
input fiber
λ1
λ1
λ2
λ2
λ1
λ2
optical
switch
MUX
output fiber
λ1
λ2
fixed transponder λ1
fixed transponder λ2
(a) Suppose you only have available the following components: fixed optical filters for
λ1 and λ2, passive splitters and combiners (made from optical couplers), 2×2 optical
switches, and fixed transponders (transmitter and receiver units) for λ1 and λ2.
Note that you do not have MUXs and DMUXs.
Indicate how you can construct a switch with the same functionality by drawing a
diagram of your switching node architecture. More specifically, draw the structure
inside the area with the dashed boundary in the above diagram. HINT: Demultiplexing can be done using a splitter followed by optical filters.
(b) What is the drawback of this alternative implementation?
Chapter 2: Building Blocks of WDM Networks
Problem 2.1: Specify each of the following in one or two sentences.
(a) Advantage of a single-mode fiber compared to a multi-mode fiber.
(b) Advantage of a multi-mode fiber compared to a single-mode fiber.
(c) Advantage of a laser compared to a light emitting diode (LED).
(d) Advantage of an LED compared to a laser.
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© P. Saengudomlert, Asian Institute of Technology
Chapter 3: Wavelength-Routed WANs
Problem 3.1 (RWA for Opaque vs. Transparent Switching Nodes): Consider the
integer linear programming (ILP) problem for routing and wavelength assignment
(RWA) shown below. Note that this is the same as in your class notes.
Static RWA (minimizing the maximum link load)
minimize fmax
subject to
∀l ∈ L, f max ≥
∑ fl ,sw
s∈S , w∈{1,...,W }
∀l ∈ L, w ∈ {1,...,W },
∑f
s∈S
s
l ,w
∀j ∈ N , s ∈ S , w ∈ {1,...,W },
≤1
∑
l∈L( ⋅ , j )
∀s ∈ S ,
∑
w∈{1,...,W }
fl ,sw −
∑
l∈L( j ,⋅ )
f l ,sw
− f ws , j = source of s

=  f ws , j = destination of s
 0,
otherwise

f ws = t s
∀l ∈ L, s ∈ S , w ∈ {1,..., W },
f l ,sw ∈ {0,1}
NOTE: Recall the following definitions.
Given parameters
• W : number of wavelengths in each fiber/link
• N : set of nodes
• L : set of links
• L( j ,⋅) : set of links that leave from node j
•
L(⋅, j ) : set of links that go to node j
•
•
S : set of source-destination (s-d) pairs with nonzero traffic
t s : traffic demand (in wavelengths) for s-d pair s
Variables
• fl ,sw ∈ {0, 1} : traffic flow (in wavelength) for s-d pair s on link e on wavelength w
•
f ws ∈ {0, 1, 2, …} : traffic flow for s-d pair s on wavelength w
•
f max : maximum link load
(a) Based on the above ILP formulation, identify the number of variables and the number of constraints (not including the integer constraints) for the ILP problem given
the network topology and traffic shown below. Assume that there are 4 wavelengths in each fiber, i.e., W = 4.
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j= 1
2
3
node 1
4
5
2
3
4
5 6
i = 1 0 0 2 0 0 1
0 0 0 0 2 0
2


0 0 0 1 0 0
3
i -j
t  =


4
2 0 0 0 0 0
0 0 3 0 0 1
5


6
6
0 1 0 0 0 0
traffic matrix: entry (i, j) is
traffic from node i to node j.
(b) Assume that we use opaque switching nodes instead of transparent switching nodes.
Modify the above ILP formulation to correspond to the opaque optical switching
architecture, i.e., write down a new ILP formulation.
Based on this new formulation, identify the number of variables and the number
of constraints (not including the integer constraints) for the ILP problem given the
network topology and traffic shown in part (a).
HINT: For opaque switching nodes, there is no wavelength assignment problem.
Instead of using f l ,sw and f ws , use fl s ∈ {0,1, 2,...} defined as the amount of traffic
flow for s-d pair s on link l as the decision variables.
Problem 3.2 (WA in line networks, based on problem 8.14 in [RS02]): Consider a
line network with N nodes shown below. Assume that adjacent nodes are connected by
two fibers, one for the transmission in each direction. Suppose that, after setting up all
the lightpaths to support a given traffic matrix, the maximum fiber/link load in the line
network is equal to L wavelengths. In this problem, you will show that L wavelengths
are sufficient to support the traffic with maximum link load equal to L.
node 1
2
3
…
N
Consider the following simple wavelength assignment (WA) algorithm.
1. Number the wavelengths from 1 to L. Start with the first lightpath from the left and
assign to it wavelength 1. If there are several choices, we can arbitrarily select one.
2. Go to the next lightpath starting from the left and assign to it the available wavelength with the least index. Repeat until all the lightpaths are assigned wavelengths.
(a) Assume that N = 4. Use the above WA algorithm to assign wavelengths to the lighpaths for the following source-destination (s-d) pairs: (1,2), (1,3), (1,4), (2,4), (2,4),
(3,4). Note that there are two lightpaths for s-d pair (2,4).
(b) Argue that the above WA algorithm never uses more than L wavelengths when the
maximum link load is equal to L wavelengths. (HINT: Use contradiction. Consider
a node at which the (L + 1)th wavelength must be used. How many lightpaths must
pass through this node?)
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© P. Saengudomlert, Asian Institute of Technology
Problem 3.3 (Supporting uniform all-to-all traffic in line networks): Consider
again an N-node line network in problem 3.2. Suppose the traffic to be supported is the
uniform all-to-all traffic in which each node wants to send one wavelength of traffic to
each of the other nodes. What is the minimum number of wavelengths required to support the traffic, denoted by Wmin, in this case? Justify your answer.
Problem 3.4 (Node coloring for WA): Consider the six lightpaths and their routes as
shown below. Assume that the network uses transparent optical switching, and the
wavelength continuity constraint must therefore be satisfied.
2
3
node 1
4
6
5
(a) Draw the path graph associated with the given lightpaths and their routes.
(b) Consider node coloring of the path graph. Identify the node order (from first to last)
for largest-first (LF) and smallest-last (SL) sequential coloring heuristics.
(c) Color the nodes of the path graph according to the SL heuristic, and identify the
number of colors used.
(d) What is the chromatic number of the path graph? Justify your answer. (HINT: You
may use the fact that an N-node ring has chromatic number equal to 3 when N is
odd.)
(e) What is the minimum number of wavelengths needed to support the same routing of
lightpaths if we use opaque optical switching instead of transparent optical switching.
Problem 3.5 (Node coloring for WA):
(a) Consider the network topology shown below. Each undirected link represents two
fibers, one for the transmission in each direction. There are 4 end nodes and 2 hub
nodes.
end node 1
2
3
hub
A
hub
B
4
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Consider the following s-d pairs each of which has 1 wavelength unit of traffic: 1-3,
1-4, 2-3, 2-4, 3-1, 3-2, 4-1, 4-2, 4-3. Specify the wavelength assignment (WA) that
uses the minimum number of wavelengths. Justify your answer.
(b) Consider the unidirectional ring and the traffic (in wavelength unit) shown below.
Note that each directed link represents one fiber in the clockwise direction. Specify
the WA according to the smallest last (SL) sequential coloring heuristics. How
many colors are used?
j= 1 2 3 4 5 6
node 1
2
3
6
5
4
i = 1 0
0
2

0
3
i -j
t  =

4
0
0
5

6
0
0 1 0 0 0
0 0 1 0 0 
0 0 0 2 0

0 0 0 0 1
2 0 0 0 0

0 0 0 0 0
traffic matrix: entry (i, j) is
traffic from node i to node j.
Problem 3.6 (RWA for two connected rings with uniform all-to-all traffic): In the
topology shown below, consider supporting uniform all-to-all traffic in which each
node wants to send 1 wavelength of traffic to each of the other nodes. Each undirected
link represents two fibers, one for the transmission in each direction.
Specify the minimum number of wavelengths needed to support the traffic. Justify
your answer. HINT: You may use the facts that the above topology contains a ring, and
the uniform all-to-all traffic can be supported on an N-node ring using ( N 2 − 1) 8
wavelengths for odd N.
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© P. Saengudomlert, Asian Institute of Technology
Problem 3.7 (RWA for 1+1 protection): Consider the ILP problem for RWA with
1+1 dedicated protection shown below. Note that this is the same as in your class
notes.
Path-based static RWA with dedicated (1+1) path protection:
minimize fmax
subject to
∀l ∈ L, f max ≥
g wp ,b
∑ f wp +
∑
p∈Pl , w∈{1,...,W }
∑f
∀l ∈ L, w ∈ {1,..., W },
∀s ∈ S ,
∑
p∈Pl
f =t
p∈P s , w∈{1,...,W }
∀p ∈ P , w ∈ {1,...,W },
p
w
p∈P ,b∈B p ∩ Bl , w∈{1,...,W }
p
w
p∈P ,b∈B p ∩ Bl
g wp ,b ≤ 1
s
∑g
b∈B
∑
+
p
∀p ∈ P , b ∈ B , w ∈ {1,..., W },
p
p ,b
w
= f wp
f wp , g wp ,b ∈ {0,1}
NOTE: Recall the following definitions.
Given information
Network parameters
• W : number of wavelengths in each fiber (Index them from 1 to W.)
• N : set of switching nodes in the network
• L : set of directional links/fibers
• S : set of source-destination (s-d) pairs with nonzero traffic
Traffic parameters
• t s : traffic demand (in wavelength) for s-d pair s
Path parameters
• P : set of all working paths
• P s : set of working paths for s-d pair s
• Pl : set of working paths that use link l
•
•
B p : set of backup paths for working path p
Bl : set of backup paths that use link l
Variables
• f wp ∈ {0, 1}: working traffic flow (in wavelength) on path p on wavelength w
•
g wp ,b ∈ {0, 1}: backup traffic flow (in wavelength) on path b for the working path p
on wavelength w
(a) Based on the above ILP formulation, identify the number of variables and the number of constraints (not including the integer constraints) for the ILP problem given
the network topology, the traffic matrix, and the candidate paths shown below. Assume that adjacent nodes are connected by two fibers, one for the transmission in
each direction. In addition, assume that there are 8 wavelengths in each fiber.
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2
j= 1 2 3 4 5 6
3
node 1
4
6
0 2 0 0 1
0 0 0 2 0 
0 0 1 0 0

0 0 0 0 0
0 3 0 0 1

1 0 0 0 0
traffic matrix: entry (i, j) is
traffic from node i to node j.
5
s-d pair
1-3
1-6
2-5
3-4
4-1
5-3
5-6
6-2
i = 1 0
0
2

0
3
t i -j  =

4
2
0
5

6
0
Primary path
1→2→3
1→5→6
2→1→5
3→4
4→5→1
5→4→3
5→6
6→3→2
Backup paths for each primary path
1→5→4→3, 1→5→6→3
1→2→3→6, 1→2→3→4→6
2→3→4→5, 2→3→6→5
3→6→4, 3→2→1→5→4
4→3→2→1, 4→6→3→2→1
5→1→2→3, 5→6→3
5→4→6, 5→4→3→6
6→5→1→2, 6→4→5→1→2
(b) Repeat part (a) but with the change of the paths for s-d pair 6-2 as follows. (All the
other paths are kept the same.)
s-d pair
1-3
1-6
2-5
3-4
4-1
5-3
5-6
6-2
Primary path
1→2→3
1→5→6
2→1→5
3→4
4→5→1
5→4→3
5→6
6→3→2
Backup paths for each primary path
1→5→4→3, 1→5→6→3
1→2→3→6, 1→2→3→4→6
2→3→4→5, 2→3→6→5
3→6→4, 3→2→1→5→4
4→3→2→1, 4→6→3→2→1
5→1→2→3, 5→6→3
5→4→6, 5→4→3→6
6→5→1→2
Problem 3.8 (WA for 1+1 protection): Consider the network topology and the traffic
matrix shown below. Assume that adjacent nodes are connected by two fibers, one for
the transmission in each direction.
2
j= 1 2 3 4 5 6
3
node 1
4
6
5
i = 1 0
0
2

0
3
t i -j  =

4
1
0
5

6
0
0 1 0 0 0
0 0 0 1 0 
0 0 1 0 0

0 0 0 0 0
0 1 0 0 0

1 0 0 0 0
traffic matrix: entry (i, j) is
traffic from node i to node j.
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© P. Saengudomlert, Asian Institute of Technology
Assume that we want to provide 1+1 protection for failure recovery. In addition, assume that, after the routing problem, we find the following routes for all the lightpaths
as shown below. Notice that there are two routes for each source-destination pair, one
for the working path and the other for the protection path.
s-d pair
1-3
2-5
3-4
4-1
5-3
6-2
Routes
1→2→3, 1→5→4→3
2→1→5, 2→3→4→5
3→4, 3→6→4
4→5→1, 4→3→2→1
5→4→3, 5→6→3
6→3→2, 6→5→1→2
(a) Keeping in mind that we want to assign the same wavelength for both working and
backup paths, construct a path graph so that node coloring of the path graph can
give us the desired wavelength assignment for all the lightpaths.
(b) Use the SL heuristic to color the path graph in part (a). What is the chromatic number of the path graph? Justify your answer.
Problem 3.9 (Dynamic WA): Consider a 4-node transparent optical network shown
below. Assume that adjacent nodes are connected by two fibers, one for the transmission in each direction. In addition, assume that there are 2 wavelengths in each fiber.
node 1
2
4
3
Assume that calls (i.e., lightpath demands) arrive in the following sequence
2-1, 2-4, 4-3, 1-3, 2-4, …
where each value pair is the s-d pair for the call. Suppose that we use fixed routing
with the paths 1→4→3, 2→1, 2→1→4, and 4→3 for s-d pairs 1-3, 2-1, 2-4, and 4-3
respectively. Identify what happens to each call (i.e., put on λ1, put on λ2, or blocked)
for the following on-line WA schemes.
(a) First-fit WA: Assign the first possible wavelength starting from the smallest wavelength index.
(b) Most-used WA: Assign the wavelength with the highest utilization (before the new
call). The utilization of wavelength λi is the number of fibers on which wavelength
λi is used. If there is a tie, then first-fit WA is used.
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(c) Random WA: For the purpose of this exercise, assume that the random assignments
follow the following sequence of random values: λ2, λ1, λ1, λ2, λ2, … (You may not
need all these random values.)
NOTE: For simplicity, this problem does not take into account departures of calls. In
practice, calls arrive, get served, and depart.
Problem 3.10 (Computing blocking probability for opaque networks): Consider a
4-node opaque optical network shown below. Assume that adjacent nodes are connected by two fibers, one for the transmission in each direction. In addition, assume
that there are 4 wavelengths in each fiber.
node 1
2
4
3
Assume that 1-3, 1-4, and 2-4 are the only 3 s-d pairs with nonzero traffic. Assume that
the call arrival rates (in calls per time unit) for the s-d pairs are a1-3 = 0.6 , a1-4 = 0.1 ,
and a 2-4 = 0.7 . In addition, call arrivals for each s-d pair form a Poisson process; each
call holding time is exponentially distributed with unit mean and is independent of
other call holding times.
Suppose that we use fixed routing with the paths 1→4→3, 1→4, and 2→1→4. Under
the link decomposition method for approximating blocking probabilities, let Bl denote
the link blocking probability, where l is a link/fiber in the network.
(a) In terms of B(2,1) , B(1,4) , and B(4,3) , write down the call blocking probabilities for the
3 s-d pairs.
(b) Write down the set of Erlang fixed-point equations that implicitly give the values of
B(2,1) , B(1,4) , and B(4,3) . You can use the notation Erl(⋅,⋅) for the Erlang B formula.
(c) Using a calculator (MATLAB, Excel, or others), find the numerical values of the
call blocking probabilities in part (a).
Problem 3.11 (Computing blocking probability with 1+1 protection): Consider
again the network and traffic in problem 3.10. However, assume that we want to provide 1+1 path protection for each call. In particular, assume that we use fixed routing
with paths 1→4→3 and 1→2→3 for s-d pair 1-3, paths 1→4 and 1→2→4 for s-d pair
1-4, and paths 2→1→4 and 2→4 for s-d pair 2-4.
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© P. Saengudomlert, Asian Institute of Technology
(a) In terms of B(1,2) , B(1,4) , B(2,1) , B(2,3) , B(2,4) , B(4,3) , write down the call blocking
probabilities for the 3 s-d pairs. (HINT: What links must have a wavelength available for a call of s-d pair 1-3 to be accepted with 1+1 protection?)
(b) Write down the set of Erlang fixed-point equations that implicitly give the values of
B(1,2) , B(1,4) , B(2,1) , B(2,3) , B(2,4) , B(4,3) .
Chapter 4: Optical MANs
Problem 4.1 (UPSR and BLSR): Consider an optical metro ring network with 5
nodes. Adjacent nodes are connected by two fibers, one for the transmission in each
direction. Suppose that we have to support 3 lightpaths for the following sourcedestination (s-d) pairs: 1-3, 2-4, and 4-1. (NOTE: The lightpaths are from 1 to 3, from
2 to 4, and from 4 to 1. There are no other lightpaths in the opposite directions.)
1
5
2
4
3
(a) Assume that the ring is a unidirectional path switching ring (UPSR) in which the
working direction is clockwise (CW). Draw the routing and wavelength assignment
(RWA) for the working lightpaths such that the minimum number of wavelengths is
used. (NOTE: While these working lightpaths have backup capacities, you need
only draw the working lightpaths.)
(b) Assume that the ring is a bidirectional line switching ring (BLSR) in which the
working direction is CW for wavelengths λ1, λ2, λ3 and is counterclockwise (CCW)
for wavelengths λ4, λ5, λ6. Draw the RWA for the working lightpaths such that the
minimum number of wavelengths is used. Assume that shortest path routing is used
for working lightpaths where all links have the same distance.
(c) Assume that the ring is a UPSR, and that each of the same s-d pairs (1-3, 2-4, and 41) has 3 connections to set up (for a total of 9 connections). Each connection has
the rate of 5 Gbps, while the wavelength capacity is 10 Gbps. Use the first-fit
wavelength assignment to support the 9 connections in the following order:
1-3, 1-3, 1-3, 2-4, 2-4, 2-4, 4-1, 4-1, and 4-1.
Indicate the number of wavelengths used as well as the number of ADMs used. Assume that ADMs are only installed at the nodes that need them.
(d) Repeat part (c) but with a BLSR in which the working direction is CW for wavelengths λ1, λ2, λ3 and is CCW for wavelengths λ4, λ5, λ6.
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Problem 4.2 (Traffic grooming in a UPSR): Consider a feeder ring network with 1
EN and 7 ANs, as shown below. Assume that adjacent nodes are connected by two fibers, one for the transmission in each direction. Suppose that each wavelength can
carry traffic of rate g. In addition, suppose that AN i, i ∈ {1, 2, ..., 7}, transmits to the
EN at rate ri, and receives from the EN at rate ri.
Assume that the feeder ring operates as a UPSR, where working traffic is supported
in the CW direction. The CCW direction is for protection.
EN
AN 1
AN 7
AN 2
AN 6
AN 3
AN 5
AN 4
(a) Let g = 16 (traffic units) and r1 = r2 = ... = r7 = 6. Identify an ADM allocation that
yields the minimum total number of ADMs, i.e., specify the number and the
wavelengths of ADMs at each node. What is the number of wavelengths used in
your allocation?
(b) Suppose we restrict ourselves to use the minimum number of wavelengths. Identify
the ADM allocation that yields the minimum total number of ADMs under this
restriction. What is the number of wavelengths used in your allocation?
(c) Repeat part (a) but with (r1, r2, r3, r4, r5, r6, r7) = (5, 7, 3, 11, 8, 4, 8).
Problem 4.3 (Static traffic grooming with heavy in-bound traffic): Consider the
feeder ring network that is a UPSR with one EN and N ANs. Suppose that each wavelength can carry traffic rate g. In addition, suppose that each AN transmits to the EN at
rate rOUT, and receives from the EN at rate rIN, where rIN > rOUT. (This is a typical scenario when users tend to download more often than upload information.)
Let N = 8, g = 8, rIN = 3, rOUT = 1. What is the minimum number of ADMs, denoted by Amin, required to support the traffic? Is your answer different from the case
with rIN = rOUT = 3? Provide an argument to justify your answer.
Problem 4.4 (Static traffic grooming with two ENs): Consider the feeder ring network that is a UPSR with 2 ENs and N access nodes (ANs) as shown below. Suppose
that each wavelength can carry traffic rate g. In addition, suppose that each AN transmits to an EN (one of the two) at rate r, and receives from an EN at rate r.
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© P. Saengudomlert, Asian Institute of Technology
EN 1
EN 2
AN N
AN 1
AN N–1
…
AN 2
AN 3
(a) Let N = 8, g = 8, r = 3. What is the minimum number of add-drop multiplexers
(ADMs), denoted by Amin, required to support the traffic? What is the number of
wavelengths used in this case?
(b) Draw your RWA in part (a). In addition, draw your ADM allocations, i.e., specify
the number and the wavelengths of ADMs at each node.
‡
Problem 4.5 (Static traffic grooming in a UPSR with general traffic): Below is
the ILP problem for static traffic grooming in a BLSR with general traffic, as shown in
class notes.
Static traffic grooming for a BLSR:
minimize
∑ awi
i∈V , w∈{1,...,W }
subject to
∀e ∈ E , w ∈ {1,...,W }, t ∈ {1,..., g},
∀s ∈ S ,
∑
s
s: e∈ pWD
w
∑
f ws,t ≤ 1
f ws,t = t s
w∈{1,...,W },t∈{1,..., g }
∀i ∈ V , w ∈ {1,...,W },
∑
f ws,t ≤ gawi
∑
f ws,t ≤ gawj
s∈S( i ,⋅ ) ,t∈{1,..., g }
∀j ∈ V , w ∈ {1,...,W },
s∈S( ⋅ , j ) ,t∈{1,..., g }
∀s ∈ S , w ∈ {1,..., W }, t ∈ {1,..., g},
f ws,t ∈ {0,1}
∀i ∈ V , w ∈ {1,...,W }, awi ∈ {0,1}
Recall that the parameters and variables are defined as follows.
• W: number of wavelengths in each fiber (Index them from 1 to W.)
• g: capacity of a wavelength in time slots (Index time slots from 1 to g.)
• V: set of switching nodes in the network
•
E: set of directional links/fibers
•
S: set of s-d pairs with nonzero traffic
•
S(i,⋅): subset of S with all s-d pairs whose sources are node i
‡
Problems that are marked with this symbol are optional.
Optical Networks
E17
•
S(⋅,j): subset of S with all s-d pairs whose destinations are node j
•
•
WDw ∈ {CW, CCW}: working direction of wavelength w
(w, t): time slot t on wavelength w in the direction WDw. (We shall use the term
circle to refer to such a pair (w, t).)
t s : traffic demand (in time slots) for s-d pair s
pcs : path for s-d pair s in ring direction c (CW or CCW)
•
•
•
•
f ws,t ∈ {0, 1}: working traffic flow for s-d pair s on circle (w, t)
awi ∈ {0, 1}: awi is 1 if an ADM is used at node i for wavelength w, and is 0 otherwise.
Suppose we have a UPSR instead of a BLSR. Modify the above ILP problem formulation so that it can be applied to a UPSR.
Chapter 5: Optical LANs
Problem 5.1 (Bus broadcast PONs): Consider the single-bus architecture for a
broadcast PON that supports N users as shown below. A node transmits through a coupler in the top section of the fiber, and receives through a coupler in the bottom section
of the fiber.
input 1
coupler output 1
input 2
output 2
transmitter
node 1 receiver
fiber
...
transmitter
node N receiver
output 2
input 2
output 1
input 1
For each coupler, assume that the relationship between input and output powers is
given by
 Pout,1 
P
 =γ
 out,2 
α   Pin,1 
1 − α

,
 α
1 − α   Pin,2 

where α is the splitting loss and γ is the excess loss. Let PT be the transmit power from
each node, and Pmin be the minimum required receive power at each node.
P
1
and T = 16384 (= 214). Find the maximum number of
2
Pmin
users Nmax that can be supported in this PON architecture.
(a) Assume that α = γ =
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© P. Saengudomlert, Asian Institute of Technology
(b) Repeat part (a) but with the double-bus architecture shown below.
coupler
input 1
output 1
input 2
fiber
output 2
transmitter receiver
transmitter receiver
...
receiver transmitter
receiver transmitter
node 1
node N
(c) Briefly describe an advantage and a disadvantage of the architecture in part (b)
compared to the one in part (a).
Problem 5.2 (Bus access PONs): Consider the bus access PON that supports N users
as shown below. A user node transmits on the top fiber, and receives on the bottom
fiber.
upstream
fiber
coupler
output 1
coupler
output 2
CO
user 1
transmitter
receiver
...
user N – 1
input 2
downstream
fiber
input 1
input 1
input 2
transmitter
receiver
user N
transmitter
receiver
output 2
output 1
For each coupler, assume that the relationship between input and output powers is
given by
α   P0IN 
 P0OUT 
1 − α
=
γ
 OUT 
 α
  IN  ,
−
1
α
P

  P1 
 1 
where α is the coupling ratio and γ is the insertion loss parameter. Let PT be the transmit power from each node, and Pmin be the minimum required receive power at each
node. Assume that PT Pmin = 1024 (= 210).
(a) Assume that α = 1 2 and γ = 1. Find the maximum number of users, denoted by
Nmax, that can be supported in the given PON architecture.
(b) Repeat part (a) for α = 1 2 and γ = 1 2 .
Optical Networks
E19
Now consider the case in which we can choose the values of the coupling ratio α differently for different couplers. More specifically, for downstream traffic, we can set
the coupling ratios α1 ,..., α N −1 as follows.
α1 is chosen so that the receive power at user 1 is Pmin, yielding γα1 PT = Pmin or
equivalently α1 = Pmin (γ PT ) .
• For k ∈ {2, …, N – 1}, α k is chosen so that the receive power at user k is Pmin.
•
In what follows, we still assume that PT Pmin = 1024.
(c) Assume that γ = 1. Find the maximum number of users Nmax that can be supported
in the given PON architecture with the above choices of coupling ratios.
(d) Repeat part (c) for γ = 1 2 . (HINT: You may find the following identity useful:
1 + x + ... + x
k −1
1 − xk
for x > 0.)
=
1− x
Problem 5.3 (Tree broadcast PONs): Consider a broadcast PON with the tree topology shown below.
50-50 coupler (α = 1/2)
Tx: transmitter
user 1
user 2
Tx Rx Tx Rx
…
Tx Rx Tx Rx
Rx: receiver
user N
For each coupler, assume that the relationship between input and output powers is
given by
 Pout,1 
α   Pin,1 
1 − α
P
 =γ 

,
1 − α   Pin,2 
 α
 out,2 
where α is the splitting loss and γ is the excess loss. Let PT be the transmit power from
each node, and Pmin be the minimum required receive power at each node.
1
P
and T = 4096 (= 212). Find the maximum number of usPmin
2
ers Nmax that can be supported in this PON architecture.
(a) Assume that α = γ =
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© P. Saengudomlert, Asian Institute of Technology
(b) Repeat part (a) but with γ = 1, i.e., no excess loss.
Problem 5.4 (Further improvement on modified SA/SA): Consider the modified
slotted Aloha/slotted Aloha (SA/SA) system for a broadcast PON with an additional
condition that transmissions must be synchronized on data channels, as illustrated below. In particular, time on data channels are divided into transaction slots each of
which contains L mini slots. Transaction sizes are the same and equal to 1 time unit or
L mini slots. Let W be the number of data wavelength channels.
Dashed lines indicate starting time for transmissions.
user 1
λ2
maximum round-trip time
λ1
user 2
user 3
control wavelength
0
time
(a) Let g be the total transmission rate of both new and retransmitted transactions (in
transaction/mini slot). Derive the throughput expression (in transactions/time
unit/wavelength) in terms of g, L, and W.
W 1
< (reasonable for large L and small W). Find the expression for
L e
the maximum throughput (in transaction/time unit/wavelength) by differentiating
the expression in part (a) with respect to g. Show that, in the limit when L → ∞, the
maximum throughput approaches 1/e, which is the maximum throughput of slotted
Aloha. (Note that we have improved the throughput of modified SA/SA by roughly
a factor of 2.)
(b) Assume that
‡
Problem 5.5 (Reservation with no central scheduler): Consider the reservation
scheme with no central scheduler for a broadcast PON with N nodes. Each node is
equipped with a fixed transmitter and a fixed receiver for the control channel, as well as
a tunable transmitter and a tunable receiver for data channels. Nodes transmit control
packets on the control channel using slotted Aloha. Assume that each control packet
lasts 1 mini slot. Assume that transaction sizes are the same and equal to 1 time unit or
L mini slots.
Let W be the number of data wavelength channels. Once a node successfully
transmitted a control packet, a reservation is made for that node’s transaction on a data
wavelength in a round-robin fashion, i.e., reservation j on wavelength (j mod W) + 1,
where j ∈ {0, 1, 2, …} and we start counting from 0. Each reservation is made starting
in the earliest possible mini slot on the selected data wavelength. Shown below are example operations of this reservation scheme. (This scheme requires every node to keep
Optical Networks
E21
track of all the latest reservations on all data channels. This requirement may not be
desirable in practice.)
λ2
user 1
user 2
maximum round-trip time
λ1
user 3
control wavelength
0
time
(a) Assume that slotted Aloha on the control channel operates at the maximum
throughput of 1/e. Assume that the network is always busy. Argue that the expected time between reservations on a given data wavelength channel is We mini
slots.
(b) Argue that the throughput T (in transaction/time unit/wavelength) of this reservation
system is given by
 L 
T = min 
,1 .
 We 
(Note that, for large L, the throughput of this system is essentially 1 when the network is always busy. For small L, the throughput is limited by the throughput of
slotted Aloha on the control channel.)
Problem 5.6 (Reservation with central scheduler): Consider a star broadcast PON
with central scheduler as shown below.
combiner
splitter
control
wavelengths
λC '
Rx: receiver
λC
Tx Rx
Tx Rx
Tx Rx Tx Rx
scheduler
user 1
user N
…
Each user transmits to the central scheduler on wavelength λC and receives from the
central scheduler on wavelength λC ' , where λC ' ≠ λC . For transmissions of control
packets to the central scheduler, assume that each user node follows the following protocol. Assume that time is slotted into intervals whose lengths are constant and equal
to the control packet length. Each such interval is called a mini slot.
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•
•
© P. Saengudomlert, Asian Institute of Technology
Each user transmits a control packet to the scheduler on λC in each mini slot with
probability p, independently of other mini slots.
A user does not retransmit if there is a collision on λC. It simply repeats the same
process of transmitting in each mini slot with probability p, independently of other
mini slots.
(a) With N users in the network, what is the probability that a transmitted control
packet is successful, i.e., no collision? What is the expected number of mini slots
taken, denoted by Λ, for a user’s successful control packet transmission? Express
your answers in terms of p and N.
(b) Suppose that we can choose the value of p. In term of N, what is the optimal value
of p that minimizes Λ? What is the corresponding minimum Λ?
Problem 5.7 (Reservation with central scheduler): Consider a star broadcast PON
with the central scheduler as shown below.
combiner
splitter
control
wavelengths
λC '
Rx: receiver
λC
Tx Rx
Tx Rx
Tx Rx Tx Rx
scheduler
user 1
user N
…
Each user transmits to the central scheduler on wavelength λC and receives from the
central scheduler on wavelength λC ' , where λC ' ≠ λC . For transmissions of control
packets to the central scheduler, assume that each user node follows the following protocol. Time is slotted into intervals whose lengths are constant and equal to the control
packet length. Each such interval is called a mini slot. In addition, F consecutive mini
slots are grouped intro a frame.
• At the beginning of each frame, each user generates a random number in the set {1,
…, F} with equal probabilities.
• Each user transmits a control packet exactly once in each frame, using the mini slot
whose number is equal to the randomly generated value in that frame.
Below is an illustration of the protocol for transmitting control packets from 3 users. In frame 1, user 1 chooses mini slot 2 (S = success), while users 2 and 3 choose
mini slot 3 (C = collision). In frame 2, user 1 chooses mini slot 1 (S), user 2 chooses
mini slot 2 (S), and user 3 chooses mini slot 4 (S).
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E23
frame 1
…
frame 2
F mini slots
1
2
3
1
2
3
S
C
S
S
S
time
(a) With the frame size F and N users in the network, what is the probability that a
given user has a successful control packet in each frame of F mini slots?
(b) Define the throughput to be equal to the number of successful control packets (from
all users) per mini slot. Suppose that we can choose the value of F. What is the
value of F (in term of N) that maximizes this throughput?
(c) Show that the associated maximum throughput in part (b) can be approximated as
x
1 e for large N. (HINT: You can use lim (1 + a x ) = e a .)
x →∞
Problem 5.8 (FCFS scheduler with look ahead): Consider a first-come-first-serve
(FCFS) scheduler with a look ahead factor k. This scheduler can be used for a broadcast PON with a central scheduling node. Assume that transaction sizes are the same
and equal to 1 time unit or 1 time slot. In addition, assume there are 3 data wavelength
channels. Consider the current queue states given below.
Labels are the destination nodes.
Queue j is stored at node j.
2 2 3 4
queue 1
λ1
queue 2
3 4 4 4
queue 3
2 2 1 4
queue 4
3 2 1 1
λ2
λ3
receiver 1
receiver 2
receiver 3
receiver 4
(a) Suppose that the node order to be served is 1, 2, 3, 4. Identify the transactions (by
specifying their sources and destinations) that will be scheduled in the next time
slot by the FCFS scheduler with k = 1.
(b) Repeat part (a) but with k = 2.
(c) Repeat part (a) but with k = 3.
Problem 5.9 (Operations of GPS and ACT): Consider sharing a downstream transmission channel between 2 users using a central scheduler in an access PON. Assume
that the channel rate is 1 transaction per time unit. Transaction arrival times are shown
below. Assume that transaction sizes are the same and equal to 1 time unit. Let
Ai (0, t ) denote the amount of traffic (in transaction) for user i that arrives in (0, t]. Let
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© P. Saengudomlert, Asian Institute of Technology
Si (0, t ) denote the amount of traffic for user i that is transmitted in (0, t]. The curves
for Ai (0, t ) are already shown in below.
arrivals
1 2 3
8
1 2 3
5
time
φ1 = φ2 = 1
4
4
A1 (0, t )
2
S1 (0, t ) = ?
1 2 3
A2 (0, t )
2
8
S 2 (0, t ) = ?
1 2 3
5
φ1 = 1,φ2 = 2
4
4
A1 (0, t )
2
S1 (0, t ) = ?
1 2 3
8
A2 (0, t )
2
S 2 (0, t ) = ?
1 2 3
5
(a) Draw the curves for Si (0, t ) when a central scheduler is a generalized processor
sharing (GPS) scheduler. Consider two cases: φ1 = φ2 = 1 and φ1 = 1, φ2 = 2 .
(b) Repeat part (a) for an adaptive cycle time (ACT) scheduler. Assume Wmax,i = φi .
Problem 5.10 (Operations of Multi-Wavelength GPS and ACT): Consider sharing
2 downstream transmission channels between 2 users using a central scheduler in an
access PON. Assume that each channel rate is 1 transaction per time unit. Transaction
arrival times are shown below. Assume that transaction sizes are the same and equal to
1 time unit. Let Ai (0, t ) denote the amount of traffic (in transaction) for user i that arrives in (0, t]. Let Si (0, t ) denote the amount of traffic for user i that is transmitted in
(0, t]. The curves for Ai (0, t ) are already shown below.
Optical Networks
2
E25
arrivals
number of arriving
transactions
2
1
1
1 2 3
6
8
φ1 = φ2 = 2
1 2 3
6
A1 (0, t )
4
5
time
A2 (0, t )
4
2
S1 (0, t ) = ?
1 2 3
6
number of arriving
transactions
2
8
φ1 = 2,φ2 = 4
1 2 3
6
A1 (0, t )
4
S 2 (0, t ) = ?
5
A2 (0, t )
4
2
S1 (0, t ) = ?
1 2 3
8
2
S 2 (0, t ) = ?
1 2 3
5
(a) Draw the curves for Si (0, t ) when a central scheduler is a GPS scheduler. For GPS,
assume that there is a single transmission channel with the rate of 2 transactions per
time unit. Consider two cases: φ1 = φ2 = 2 and φ1 = 2, φ2 = 4 .
(b) Repeat part (a) for a multi-wavelength ACT scheduler. Assume Wmax,i = φi .
Problem 5.11 (Operations of IPACT): Consider two cycles of operations of the interleaved polling with adaptive cycle time (IPACT) shown below. Assume that there are
only 2 users in the network, and each user originally has no transaction in the queue.
Assume also that the CO originally knows that each user has no transaction in the
queue. Notice that each request message to the CO reports the queue length at its
transmission time.
For simplicity, the control packet length, the transaction length, all one-way propagation delays, and the guard time are all equal to 1 time unit or 1 time slot. Assume the
packet processing time is negligible. The dynamic bandwidth allocation (DBA)
scheme used below is based on limited service. In each scheduling cycle, the allocated
transmission window size for user i, denoted by Wi, is given by
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© P. Saengudomlert, Asian Institute of Technology
Wi = min (Vi ,3) , i ∈ {1, 2},
where Vi denotes the number of remaining transactions from node i known by the CO.
guard time
for Rx
CO
0
Tx 0
Rx
1
1
2
0
1
2
node 2 Tx
Rx
3
2
1
Tx
node 1 Rx
2
2
3
0
2
transaction
arrivals
at node 1
transaction
arrivals
at node 2
0
2
4
6
8
time
Suppose now that we use constant credit service in which
Wi = min (Vi + 1,3) , i ∈ {1, 2}.
Redraw two cycles of operations for IPACT with the above constant credit service.
(Note that, in two cycles, there are 4 grant messages and 4 request messages.)
Optical Networks
E27
Solutions
Solution 1.1 (based on problem 7.7 of [RS02]):
(a) The block diagram of the switching node is shown below. Note that, instead of a
single 144×144 switch to connect to transponders, it is possible to use four 36×36
switches, one for each wavelength band. Since we need at least 4 transponders for
each wavelength band, the minimum number of transponders is 16.
λ1
128
output
ports
λ32
…
…
…
optical
switch
…
…
…
4 add/drop
ports
…
4 input
fibers
…
switch
…
optical
MUX
…
…
…
λ1, …, λ32
8 output
ports
…
fiber
8 input
ports
…
DMUX
128
input
ports
optical
switch
4 output
fibers
16 input/output ports
8×8 144×144
optical optical
switch switch
RT
TR
…
RT
TR
from/to end users
tunable
transponders
16 transponders
(4 for each band)
(b) The modified block diagram of the switching node is shown below. We can argue
that the minimum number of transponders is now 8 as follows. Note that each
transponder can tune over two adjacent bands. For band 1, we must have 4 transponders tunable to bands 1 and 2. For band 4, we must have 4 transponders tunable
to bands 3 and 4. In total, we must have at least 8 transponders.
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© P. Saengudomlert, Asian Institute of Technology
λ1
128
output
ports
λ32
…
…
…
optical
switch
…
…
…
4 add/drop
ports
…
4 input
fibers
…
switch
…
optical
MUX
…
…
…
λ1, …, λ32
8 output
ports
…
fiber
8 input
ports
…
DMUX
128
input
ports
optical
switch
4 output
fibers
8 input/output ports
8×8 136×136
optical optical
switch switch
tunable
transponders
8 transponders
(4 for bands 1-2,
from/to end users 4 for bands 3-4)
RT
TR
…
RT
TR
Solution 1.2 (Optical bypassing of electronic switching):
(a) All the traffic from the 3 s-d pairs is electronically processed at node B. Hence, the
total of traffic processed is 3×4 = 12 Gbps.
(b) The fiber from A to B must carry traffic for s-d pairs A-C and A-D. Since each traffic session bypasses electronic processing at node B, they cannot be on the same
wavelength (even though the wavelength capacity of 10 Gbps is greater than 2×4 =
8 Gbps). Hence, the number of wavelength channels used is 2.
(c) The total of traffic processed is 3×8 = 24 Gbps.
(d) The 8-Gbps traffic sessions for A-C and A-D can each be supported on one wavelength (with capacity 10 Gbps). Hence, the number of wavelength channels used is
2.
Solution 1.3 (Constructing a large switch from smaller switches): An alternative
switching architecture based on 4×4 switches is shown below. Note that each tunable
transponder can be connected to any wavelength. The architecture is based on the node
structure in problem 1.
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E29
The second diagram shows another switching architecture, which actually uses
fewer 4×4 switches than the first solution!
DMUX
λ1
λ2
4×4
optical
switch
λ3
λ4
input fiber
λ1
λ2
λ3
λ4
4×4
optical
switch
4×4
optical
switch
1st solution
2
MUX
DMUX
input fiber
output fiber
4×4
optical
switch
tunable
transponders 1
λ1
λ2
λ3
λ4
MUX
4×4
optical
switch
4×4
optical
switch
output fiber
4×4
optical
switch
tunable
transponders
1
2
2nd
solution
Solution 1.4 (Constructing a large switch from smaller switches): The node with
the same switching function can be constructed from 4×4 switches as follows .
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© P. Saengudomlert, Asian Institute of Technology
λ1
DMUX
MUX
4×4
optical
switch
input fiber
output fiber
λ2
4×4
optical
switch
4×4
optical
switch
tunable
transponder
Solution 1.5 (Using splitters and filters instead of MUXs and DMUXs):
(a) The node with the same switching function can be constructed from 4×4 switches as
follows.
splitter
input fiber
filter λ
1
2×2 switch
λ1
λ2
λ2
λ1
λ1
λ2
λ2
λ1
λ2
fixed transponder
combiner
output fiber
Optical Networks
E31
(b) One drawback of the architecture in part (a) is the high power loss resulting from
using splitters and combiners instead of DMUXs and MUXs. Each time the signal
travels through a splitter/combiner, there is at least a 3-dB power loss due to the
property of optical couplers.
Solution 2.1:
(a) A single-mode fiber does not possess the intermodal dispersion problem.
(b) We can couple more light from a transmitter into a multi-mode fiber. So it is preferred for short-distance transmission where dispersion is not a major problem.
(c) A laser has more transmit power and transmits light whose bandwidth is smaller.
Hence, the transmitted signal suffers less chromatic dispersion.
(d) An LED is cheaper and less sensitive to environmental conditions such as temperature. Without proper control, a laser output frequency can vary depending on the
temperature of its surrounding.
Solution 3.1 (RWA for Opaque vs. Transparent Switching Nodes):
(a) The number of s-d pairs with nonzero traffic is 8. The number of fibers is 16. The
number of nodes is 6. There are 4 wavelengths in each fiber. The number of variables and constraints are listed below.
Variables
fl ,sw
Count
8×16×4 = 512
8×4 = 32
f ws
1
f max
TOTAL
512+32+1 = 545
Constraints
∀l ∈ L, f max ≥
∑
f
s∈S , w∈{1,...,W }
∀l ∈ L, w ∈ {1,..., W },
∑f
s∈S
Count
16
s
l ,w
s
l ,w
≤1
∀j ∈ N , s ∈ S , w ∈ {1,...,W },
∑
l∈L( ⋅ , j )
f l ,sw −
∀s ∈ S ,
∑
l∈L( j ,⋅ )
∑
fl ,sw
w∈{1,...,W }
TOTAL
16×4 = 64
6×4×8 = 192
− f ws , j = source of s

=  f ws , j = destination of s
 0,
otherwise

f ws = t s
8
16+64+192+8 = 280
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© P. Saengudomlert, Asian Institute of Technology
(b) Using the newly defined variables f l s ’s as suggested by the hint, the new ILP formulation for opaque switching nodes is shown below.
Static routing (minimizing the maximum link load)
minimize fmax
subject to
∀l ∈ L, f max ≥ ∑ fl s
∀l ∈ L,
∑f
s∈S
s∈S
s
l
∀j ∈ N , s ∈ S ,
≤W
∑
l∈L( ⋅ , j )
∀l ∈ L, s ∈ S ,
fl s −
∑
l∈L( j ,⋅ )
fl s ∈ {0,1, 2,...}
−t s , j = source of s

f l s =  t s , j = destination of s
 0, otherwise

New variables
• fl s ∈ {0, 1, 2, …} : traffic flow for s-d pair s on link l
Based on the new ILP formulation, the number of variables and constraints are
listed below.
Variables
fl s
Count
8×16 = 128
f max
TOTAL
1
128+1 = 129
Constraints
∀l ∈ L, f max ≥ ∑ fl s
∀l ∈ L,
∑f
s∈S
s∈S
s
l
∀j ∈ N , s ∈ S ,
16
≤W
∑
l∈L( ⋅ , j )
TOTAL
Count
16
fl s −
∑
l∈L( j ,⋅ )
−t s , j = source of s

f l s =  t s , j = destination of s
 0, otherwise

6×8 = 48
16+16+48 = 80
Solution 3.2 (WA in line networks, based on problem 8.14 in [RS02]):
(a) The algorithm assigns wavelengths for lightpaths in the following order: (1,2), (1,3),
(1,4), (2,4), (2,4), (3,4). The resultant WA is illustrated below.
Optical Networks
node 1
E33
2
3
4
λ1
λ2
λ3
λ4
(b) We now argue that the given algorithm never uses more than L wavelengths when
the maximum link load is equal to L. We proceed by contradiction. Consider a
node j at which the (L + 1)th wavelength must be used. Since L + 1 wavelengths
are needed on the fiber leaving from j, it follows that there must be L lightpaths
passing through j. But these pass-through lightpaths together with the lightpath
leaving from j creates a load of L + 1, contradicting the assumption that L is the
maximum link load. We conclude that at most L wavelengths are needed.
Solution 3.3 (Supporting uniform all-to-all traffic in line networks): For an N-node
line network, each fiber is a cut. In particular, a cut that separates x nodes from reaching the other N – x nodes corresponds to the link load equal to x( N − x) .
For N even, the maximum link load L occurs on the cut that separates N 2 nodes
2
from the other N 2 nodes. More specifically, L = ( N 2 ) = N 2 4 . Since the fiber of
this cut must carry L wavelengths, it is clear that Wmin ≥ L. To show that Wmin ≤ L, we
can use the WA algorithm for a line network in problem 3.2 to argue that at most L
wavelengths are needed. Therefore, Wmin = L = N 2 4 .
For N odd, the maximum link load L occurs on the cut that separates ( N − 1) 2
nodes from the other ( N + 1) 2 nodes. More specifically, L = ( N − 1) 2 × ( N + 1) 2 =
( N 2 − 1) 4 . Repeating the same argument as for N even, we conclude that Wmin =
L = ( N 2 − 1) 4 .
Solution 3.4 (Node coloring for WA):
(a) The path graph is given below. Note that a node label indicates the nodes along
each path.
543
1546
6432
215
6321
563
(b) For the LF heuristic, one possible node order (from first to last) is
6321, 6432, 543, 1546, 215, 563.
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© P. Saengudomlert, Asian Institute of Technology
For the SL heuristic, one possible node order is
215, 1546, 543, 6432, 6321, 563.
(c) Using the SL heuristic, one possible node coloring is shown below. Note that 3 colors are used.
543
1546
6432
215
6321
563
(d) Let χ denote the chromatic number of the path graph. Since there is a 5-node ring
(consisting of 1546, 215, 6321, 6432, and 543) that cannot be colored with 2 colors,
we must have χ ≥ 3. The node coloring in part (c) shows that χ ≤ 3. It follows that
χ = 3.
(e) If we use opaque optical switching, then the number of colors used is the maximum
link load, which is equal to 2 in this case.
Solution 3.5 (Node coloring for WA):
(a) The path graph is given below. Note that a node label indicates the nodes along
each path.
1AB3
2AB3
3BA1
4BA1
3BA2
4BA2
4B3
1AB4
2AB4
Let χ denote the chromatic number of the path graph. Since 4 nodes 1AB3, 1AB4,
2AB3, 2AB4 are fully connected, χ ≥ 4.
Consider using the SL heuristics. One possible node order is
4BA2, 4BA1, 3BA2, 3BA1, 4B3, 2AB3, 1AB3, 2AB4, 1AB4.
The corresponding node coloring is shown below. Note that 4 colors are used.
Optical Networks
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1AB3
2AB3
3BA1
4BA1
3BA2
4BA2
4B3
1AB4
2AB4
The above node coloring shows that χ ≤ 4. It follows that χ = 4. Thus, the above
WA uses the minimum number of wavelengths.
(b) The path graph is given below.
123
234
5612a
345a
5612b
345b
456
One possible node order for SL heuristics is
345b, 345a, 234, 456, 5612b, 5612a, 123.
The corresponding node coloring is shown below. Note that 4 colors are used.
123
234
5612a
345a
5612b
345b
456
NOTE: Can you provide an argument to show that the chromatic number is 4?
Solution 3.6 (RWA for two connected rings with uniform all-to-all traffic): Let
Wmin be the minimum number of wavelengths needed to support the traffic. We first
use the cut set bound to obtain a lower bound on Wmin. Consider a cut that consists of 2
fibers as shown below.
E36
© P. Saengudomlert, Asian Institute of Technology
Since there are 5×4 wavelengths of traffic that has to travel across this cut, the cut set
bound is Wmin ≥  20 2  = 10 . To obtain an upper bound on Wmin, note that the given
topology contains a 9-node bidirectional ring, as indicated below.
From the hint, the traffic can be supported using ( 92 − 1) 8 = 10 wavelengths, yielding
Wmin ≤ 10 . In conclusion, Wmin = 10 .
Solution 3.7 (RWA for 1+1 protection):
(a) The number of s-d pairs with nonzero traffic is 8. Each s-d pair has 1 primary path
that has 2 backup paths. The number of fibers is 16. The number of nodes is 6.
There are 8 wavelengths in each fiber. The number of variables and constraints are
listed below.
Variables
f wp
Count
8×8 = 64
g wp ,b
8×2×8 = 128
f max
TOTAL
1
64+128+1 = 193
Constraints
∀l ∈ L, f max ≥
∑
p∈Pl , w∈{1,...,W }
∑f
∀l ∈ L, w ∈ {1,..., W },
∀s ∈ S ,
s
∑
∑
f +
p
w
p∈Pl
p
w
g
p∈P ,b∈B p ∩ Bl , w∈{1,...,W }
+
∑
p∈P ,b∈B p ∩ Bl
Count
16
p ,b
w
g wp ,b ≤ 1
16×8 = 128
8
f wp = t s
p∈P , w∈{1,...,W }
∀p ∈ P , w ∈ {1,...,W },
∑g
b∈B
p
p ,b
w
= f wp
8×8 = 64
TOTAL
16+128+8+64 = 216
(b) With the number of backup paths for s-d pair 6-2 reduced from 2 to 1, the number
of variables and constraints are listed below.
Variables
f wp
Count
8×8 = 64
g wp ,b
(7×2+1×1)×8 = 120
f max
TOTAL
1
64+120+1 = 185
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Constraints
∀l ∈ L, f max ≥
∑
p∈Pl , w∈{1,...,W }
∑f
∀l ∈ L, w ∈ {1,..., W },
∀s ∈ S ,
s
∑
∑
f +
p
w
p∈Pl
p
w
g
p∈P ,b∈B p ∩ Bl , w∈{1,...,W }
∑
+
p∈P ,b∈B p ∩ Bl
p ,b
w
g wp ,b ≤ 1
Count
16
16×8 = 128
8
f wp = t s
p∈P , w∈{1,...,W }
∀p ∈ P , w ∈ {1,...,W },
∑g
b∈B
p
p ,b
w
= f wp
8×8 = 64
TOTAL
16+128+8+64 = 216
Note that the numbers of constraints are the same as in part (a).
Solution 3.8 (WA for 1+1 protection):
(a) In the path graph for 1+1 protection, each node represents two paths: a working path
and its protection path. Two nodes are connected if there is at least one link in
common. The path graph is shown below.
123
1543
632
6512
215
2345
543
563
34
364
451
4321
(b) One possible node order in the SL heuristic is: 632-6512, 123-1543, 543-563, 4514321, 215-2345, 34-364. The corresponding coloring is shown below. Note that 4
colors are used.
123
1543
632
6512
215
2345
543
563
34
364
451
4321
E38
© P. Saengudomlert, Asian Institute of Technology
Let χ denote the chromatic number of the path graph. The SL heuristic tells us that
χ ≤ 4. Since nodes 123-1543, 543-563, 632-6512, and 451-4321 are fully connected, χ ≥ 4. In conclusion, χ = 4.
Solution 3.9 (Dynamic WA):
(a) First-fit WA yields the following results.
2-1 on λ1, 2-4 on λ2, 4-3 on λ1, 1-3 blocked, 2-4 blocked.
(b) Most-used WA yields the following results.
2-1 on λ1, 2-4 on λ2, 4-3 on λ2, 1-3 on λ1, 2-4 blocked.
(c) Random WA with the sequence of random values λ2, λ1, λ1, λ2, λ2, … yields the
following results.
2-1 on λ2 (random), 2-4 on λ1, 4-3 on λ1 (random), 1-3 on λ2, 2-4 blocked.
Solution 3.10 (Computing blocking probability for opaque networks):
(a) The call or path blocking probabilities are shown below. For clarity, we specify
each s-d pair by its path, e.g. writing 1-3 as 1→4→3.
B1→ 4→3 = 1 − (1 − B(1,4) )(1 − B(4,3) )
B1→ 4 = B(1,4)
B 2→1→4 = 1 − (1 − B(2,1) )(1 − B(1,4) )
(b) The set of Erlang fixed-point equations are shown below.
B(2,1) = Erl ( a 2→1→4 (1 − B(1,4) ),W )
= Erl ( 0.7 − 0.7 B(1,4) , 4 )
B(1,4) = Erl ( a1→ 4→3 (1 − B(4,3) ) + a1→4 + a 2→1→ 4 (1 − B(2,1) ), 4 )
= Erl (1.4 − 0.6 B(4,3) − 0.7 B(2,1) , 4 )
B(4,3) = Erl ( a1→ 4→3 (1 − B(1,4) ),W )
= Erl ( 0.6 − 0.6 B(1,4) , 4 )
(c) From repeated substitutions in numerical computation,
B(2,1) ≈ 0.00435 , B(1,4) ≈ 0.0397 , and B(4,3) ≈ 0.00258 .
Optical Networks
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It follows from part (a) that
B1→4→3 ≈ 0.0422 , B1→4 ≈ 0.0397 , B 2→1→4 ≈ 0.0439 .
Solution 3.11 (Computing blocking probability with 1+1 protection):
(a) Note that a call of s-d pair 1-3 is accepted only if there are free wavelengths on links
(1,4), (4,3), (1,2), and (2,3). Using the same approach, the call blocking probabilities are as follows.
B1→4→3 = 1 − (1 − B(1,4) )(1 − B(4,3) )(1 − B(1,2) )(1 − B(2,3) )
B1→4 = 1 − (1 − B(1,4) )(1 − B(1,2) )(1 − B(2,4) )
B 2→1→4 = 1 − (1 − B(2,1) )(1 − B(1,4) )(1 − B(2,4) )
Note that, under 1+1 protection, we view each path as consisting of all the links on
working as well as backup paths.
(b) The set of Erlang fixed-point equations are shown below. Note that the reduced
load approximations take into account all links on working as well as backup paths.
B(1,2) = Erl ( a1-3 (1 − B(2,3) )(1 − B(1,4) )(1 − B(4,3) ) + a1-4 (1 − B(2,4) )(1 − B(1,4) ), W )
= Erl ( 0.6(1 − B(2,3) )(1 − B(1,4) )(1 − B(4,3) ) + 0.1(1 − B(2,4) )(1 − B(1,4) ), 4 )
 a1-3 (1 − B(4,3) )(1 − B(1,2) )(1 − B(2,3) ) + a1-4 (1 − B(1,2) )(1 − B(2,4) ) 
,W 
B(1,4) = Erl 


+ a 2-4 (1 − B(2,1) )(1 − B(2,4) )


 0.6(1 − B(4,3) )(1 − B(1,2) )(1 − B(2,3) ) + 0.1(1 − B(1,2) )(1 − B(2,4) ) 
= Erl 
,4
+0.7(1 − B(2,1) )(1 − B(2,4) )


B(2,1) = Erl ( a 2-4 (1 − B(1,4) )(1 − B(2,4) ), W )
= Erl ( 0.7(1 − B(1,4) )(1 − B(2,4) ), 4 )
B(2,3) = Erl ( a1-3 (1 − B(1,2) )(1 − B(1,4) )(1 − B(4,3) ), W )
= Erl ( 0.6(1 − B(1,2) )(1 − B(1,4) )(1 − B(4,3) ), 4 )
B(2,4) = Erl ( a1-4 (1 − B(1,2) )(1 − B(1,4) ) + a 2-4 (1 − B(2,1) )(1 − B(1,4) ), W )
= Erl ( 0.1(1 − B(1,2) )(1 − B(1,4) ) + 0.7(1 − B(2,1) )(1 − B(1,4) ), 4 )
B(4,3) = Erl ( a1-3 (1 − B(1,4) )(1 − B(1,2) )(1 − B(2,3) ), W )
= Erl ( 0.6(1 − B(1,4) )(1 − B(1,2) )(1 − B(2,3) ), 4 )
E40
© P. Saengudomlert, Asian Institute of Technology
Solution 4.1 (UPSR and BLSR):
(a) For the given 3 lightpaths, the RWA using the minimum number of wavelengths in
the UPSR is shown below. Note that 3 wavelengths are used.
λ3
λ1
1
5
2
UPSR
4
3
λ2
(b) For the given 3 lightpaths, the RWA using the minimum number of wavelengths in
the BLSR is shown below. Note that 2 wavelengths are used.
λ1
λ1
1
5
2
BLSR
4
3
λ2
(c) For the given 9 connections, the RWA and ADM allocation using the first-fit WA in
the UPSR is shown below. Note that 5 wavelengths and 12 ADMs are used.
λ1, λ2, λ4, λ5
λ5
λ2
1
5 λ4
λ1
2
UPSR
λ2, λ3
λ3
4
3
λ2, λ3, λ4, λ5
λ1, λ2
(d) For the given 9 connections, the RWA and ADM allocation using the first-fit WA in
the BLSR is shown below. Note that 3 wavelengths and 9 ADMs are used.
λ1, λ2
λ2
5 λ1
4
λ1, λ2, λ3
λ2
1
λ1
BLSR
2
λ2, λ3
λ3
3
λ1, λ2
Optical Networks
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Solution 4.2 (Traffic grooming in a UPSR):
(a) For N = 7, g = 16, r = 6, the minimum number of ADMs used is
 N 
 7 
UPSR
Amin
= N +
 = 7+
 = 11 .
  g r  
 16 6  
The number of wavelengths used is the number of ADMs at the EN, which is equal
 N   7 
to 
=
 = 4 . One WA and ADM allocation is shown below.
  g r    16 6  
λ1
λ2
λ3
λ4
EN
Wavelengths
λ1
6
λ2
λ1
λ4
λ3
6
6
λ4
λ1
λ3
6
6
λ3
6
λ2
6
λ2
Only downstream transmissions
on primary paths are shown.
Upstream transmissions follow
circle-based routing.
Arrow labels are traffic units.
 Nr   7 × 6 
UPSR
= =
(b) The minimum number of wavelengths used is Wmin
 = 3 . Using
 g   16 
UPSR
wavelengths, one possible
the ADM allocation algorithm subject to using Wmin
RWA and ADM allocation is shown below. Note that 11 ADMs are used in total.
(In this problem, we are able to use the minimum number of wavelengths and the
minimum number of ADMs at the same time. This may not be possible in general.)
E42
© P. Saengudomlert, Asian Institute of Technology
λ1
λ2
λ3
Wavelengths
EN
λ1
λ1
6
λ2
λ1
λ2
4
λ3
6
2
λ1
λ3
6
6
λ3
6
λ2
6
λ2
Only downstream transmissions
on primary paths are shown.
Upstream transmissions follow
circle-based routing.
Arrow labels are traffic units.
(c) With (r1, r2, r3, r4, r5, r6, r7) = (5, 7, 3, 11, 8, 4, 8) and g = 16, the first-fit bin
packing heuristics will use 4 bins with the following contents: (5,7,3), (11,4), and
 5+7+3+11+8+4+8 
(8,8). Note that this is optimal since we need at least 
 = 3
16

bins. The associated ADM allocation is shown below.
λ1
λ2
λ3
Wavelengths
EN
λ1
5
λ2
λ1
λ3
λ3
7
8
λ1
λ2
3
4
λ3
8
λ2
11
λ1
Only downstream transmissions
on primary paths are shown.
Upstream transmissions follow
circle-based routing.
Arrow labels are traffic units.
Note that a total of 10 ADMs are used.
Solution 4.3 (Static traffic grooming with heavy in-bound traffic): For N = 8, g = 8,
rIN = 3, rOUT = 1, if we consider supporting only downstream working traffic in a
UPSR, the minimum number of ADMs used is given by
 N 
 8 
UPSR
Amin
= N +
 = 8+ 
 = 12 .
g
r
8
3




IN








The argument is exactly the same as in the proof of theorem 4.1 in class notes. Note
UPSR
that, in the proof of theorem 4.1, the value of Amin
is derived based on downstream
Optical Networks
E43
working traffic only. Therefore, the answer is the same as for the problem with rIN =
rOUT = 3. Since rOUT < rIN, it is easy to see that we can use circle-based routing to support upstream traffic. The RWA and ADM allocation for backup traffic follows automatically.
Solution 4.4 (Static traffic grooming with two ENs):
(a) For N = 8, g = 8, r = 3, the minimum number of ADMs at ANs is obviously N = 8,
where one ADM is used at each AN. The minimum number of ADMs at the two
 N   8 
ENs combined is 
=
 = 4 , which is based on the same argument as
  g r    8 3 
in the proof of theorem 4.1. Thus, the minimum number of ADMs is 12. The
number of wavelengths used is the number of ADMs used at the ENs, i.e., 4.
(b) One possible RWA and ADM allocation with 12 ADMs is shown below.
λ3
λ4
EN
λ4
λ1
EN
λ2
Wavelengths
3
λ1
λ1
3
λ2
3
λ4
λ1
3
3
λ3
λ2
3
λ3
3
3
λ2
λ3
λ4
Only downstream transmissions
on primary paths are shown.
Upstream transmissions follow
circle-based routing.
Arrow labels are traffic units.
NOTE: If we require the protection of access ADMs in case one EN fails, then we
need ADMs at all wavelengths at each EN, for a total of 16 ADMs. One possible
RWA and ADM allocation with 16 ADMs is shown below.
E44
© P. Saengudomlert, Asian Institute of Technology
λ1
λ2
λ1
λ2
λ3
λ4
λ3
λ4
EN
λ4
EN
Wavelengths
3
λ1
λ1
3
λ2
3
λ4
λ3
λ1
3
λ4
3
λ3
Only downstream transmissions
on primary paths are shown.
Upstream transmissions follow
circle-based routing.
Arrow labels are traffic units.
λ2
3
λ3
3
3
λ2
‡
Solution 4.5 (Static traffic grooming in a UPSR with general traffic): Note that
there are two main differences between a UPSR and a BLSR.
1. In a UPSR, working traffic is supported only in one, say CW, direction.
2. Protection in a UPSR is 1+1 dedicated protection.
From statement 1, we can drop the variables WDw from the list of variables since,
in a UPSR, WDw = CW for all w. From statement 2, we modify the flow variables
from f ws,t to
f ws,t ,CW ∈ {0, 1}: working traffic flow for s-d pair s on circle (w, t)
f ws,t ,CCW ∈ {0, 1}: backup traffic flow for s-d pair s on circle (w, t)
Accordingly, we set up an additional constraint that the working traffic on a CW directed wavelength must be protected on a CCW directed wavelength, i.e.,
∀s ∈ S , w ∈ {1,...,W }, t ∈ {1,...g},
f ws,t ,CCW = f ws,t ,CW
In addition, we want no wavelength collision in the CCW direction, i.e.,
∀e ∈ E , w ∈ {1,..., W }, t ∈ {1,...g},
∑
s
s: e∈ pCCW
f ws,t ,CCW ≤ 1
Finally, note that the constraints on availabilities of ADMs are automatically satisfied for backup traffic as long as they are satisfied for working traffic. In summary, the
modified ILP is shown below.
•
•
•
W: number of wavelengths in each fiber (Index them from 1 to W.)
g: capacity of a wavelength in time slots (Index time slots from 1 to g.)
V: set of switching nodes in the network
•
E: set of directional links/fibers
Optical Networks
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•
S: set of s-d pairs with nonzero traffic
•
S(i,⋅): subset of S with all s-d pairs whose sources are node i
•
S(⋅,j): subset of S with all s-d pairs whose destinations are node j
•
•
•
(w, t): time slot t on wavelength w. (We use the term circle to refer to a pair (w, t).)
t s : traffic demand (in time slots) for s-d pair s
pcs : path for s-d pair s in ring direction c (CW or CCW)
•
f ws,t ,CW ∈ {0, 1}: working traffic flow for s-d pair s on circle (w, t)
•
f ws,t ,CCW ∈ {0, 1}: backup traffic flow for s-d pair s on circle (w, t)
•
awi ∈ {0, 1}: is 1 if an ADM is used at node i for wavelength w, and is 0 otherwise.
Static traffic grooming for a UPSR:
minimize
∑ awi
i∈V , w∈{1,...,W }
subject to
∀e ∈ E , w ∈ {1,..., W }, t ∈ {1,...g},
∑
s
s: e∈ pCW
∀s ∈ S , w ∈ {1,...,W }, t ∈ {1,...g},
∀s ∈ S ,
∑
w∈{1,...,W },t∈{1,... g }
f ws,t ,CW ≤ 1
f ws,t ,CCW = f ws,t ,CW
f ws,t ,CW = t s
∀i ∈ V , w ∈ {1,...,W },
∑
f ws,t ,CW ≤ gawi
∑
f ws,t ,CW ≤ gawj
s∈S( i ,⋅ ) ,t∈{1,..., g }
∀j ∈ V , w ∈ {1,..., W },
s∈S( ⋅ , j ) ,t∈{1,..., g }
∀s ∈ S , w ∈ {1,...,W }, t ∈ {1,..., g},
f ws,t ,CW , f ws,t ,CCW ∈ {0,1}
∀i ∈ V , w ∈ {1,...,W }, awi ∈ {0,1}
Solution 5.1 (Bus broadcast PONs):
(a) The maximum power loss Lmax occurs for the transmission from node 1 to itself.
Note that Lmax is γ 2 N α 2 (1 − α ) 2 N − 2 = 2−4 N .
(NOTE: With some MAC protocols, a node need not hear its own transmission,
e.g. reservation scheme in section 5.3. In this case, the maximum power loss Lmax
occurs for s-d pair 1-2 (or 2-1). It follows that Lmax = γ 2 N −1α 2 (1 − α ) 2 N −3 = 2− (4 N − 2) .)
By solving PTLmax = Pmin, we can write
24 N =
1
P
= T = 214
Lmax Pmin
which yields a realistic value Nmax = 3.
⇒ N=
14
,
4
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© P. Saengudomlert, Asian Institute of Technology
(b) With the double-bus architecture, the maximum power loss Lmax occurs for s-d pair
1-N (or N-1). Note that Lmax is γ N α 2 (1 − α ) N − 2 = 2−2 N .
By solving PTLmax = Pmin, we can write
22 N =
P
1
= T = 214
Lmax Pmin
⇒ N=
14
,
2
which yields Nmax = 7.
(c) The double-bus architecture in part (b) can support more users than the single-bus
architecture (advantage). However, it requires one more fiber as well as an extra
transmitter-receiver pair for each node (disadvantage).
Solution 5.2 (Bus access PONs):
(a) Without loss of generality, we can focus on downstream traffic. User N experiences
N −1
the maximum loss Lmax. For α = 1 2 and γ = 1, Lmax = γ N −1 (1 − α ) = 2− ( N −1) . We
solve for Nmax by setting Pmin = Lmax PT = 2− N +1 PT , yielding 2 N −1 = 210 . It follows
that Nmax = 11.
(b) For α = γ = 1 2 , Lmax = γ N −1 (1 − α )
N −1
= 2−2( N −1) . We solve for Nmax by setting
Pmin = Lmax PT = 2−2 N + 2 PT , yielding 22 N − 2 = 210 . It follows that Nmax = 6.
(c) Note that user k, k ∈ {1, …, N – 1}, has the receive power equal to Pmin. Since there
is no insertion loss, i.e., γ = 1, it follows that user N who experiences the maximum
loss receives with power PT − ( N − 1) Pmin . From the receive power requirement, we
solve for Nmax by setting PT − ( N − 1) Pmin = Pmin , yielding Nmax = 1024.
(d) Let us focus on the remaining power at different couplers along the way from the
CO to user N who experiences the maximum power loss.
• After coupler 1, the remaining power on the way to user N is
1
2
γ (1 − α1 ) PT = γ PT − γα1 PT = PT − Pmin ,
•
where the last equality follows from the choice of α1 .
After coupler 2, the remaining power is
Optical Networks
E47
1
2


1

1
2

2
1
1
= PT − Pmin − Pmin ,
4
2


γ (1 − α 2 )  PT − Pmin  = γ  PT − Pmin  − γα 2  PT − Pmin 
•
where the last equality follows from the choice of α 2 .
More generally, after coupler k, the remaining power is
1
1
1
1
 1


P −  k −1 + k − 2 + ... + 1 Pmin = k PT − 2 1 − k
k T
2
2
2
2

 2

 Pmin ,

1 − xk
, x > 0, to establish the last
1− x
where we use the identity 1 + x + ... + x k −1 =
equality.
It follows that the remaining power for user N (after coupler N – 1) is 2− N +1 PT
−2 (1 − 2− N +1 ) Pmin . We solve for Nmax by setting
2− N +1 PT − 2 (1 − 2− N +1 ) Pmin = Pmin ,
yielding 2 N −1 = 342 . Since the solution is not an integer, we look for the maximum
N such that
2− N +1 PT − 2 (1 − 2− N +1 ) Pmin ≥ Pmin ,
yielding 2 N −1 ≤ 342 . Since 28 = 256 and 29 = 512, it follows that Nmax – 1 = 8 or
equivalently Nmax = 9.
Solution 5.3 (Tree broadcast PONs):
(a) With N nodes in the network, the power loss for each s-d pair is equal to (αγ )
The value of Nmax is obtained by solving for N from PT (αγ )
N max = 2
−
log 2 ( PT Pmin )
2log 2 (αγ )
=2
−
12
2⋅( −2)
= 8.
(b) With γ = 1, the value of Nmax becomes
N max = 2
−
log 2 ( PT Pmin )
2log 2 (αγ )
=2
−
12
2⋅( −1)
= 64.
2log 2 N
2log 2 N
= Pmin , yielding
.
E48
© P. Saengudomlert, Asian Institute of Technology
Solution 5.4 (Further improvement on modified SA/SA):
(a) For a given transmitted transaction on a data channel, note that there are L – 1 possible mini slots where a successful control packet can lead to a collision on the data
channel, as shown below. For convenience, call these mini slots contention slots.
Dashed lines indicate starting times for transmissions.
λ2
maximum round-trip time
λ1
control wavelength
0
time
Check marks indicate positions of control packets
that could lead to a collision on the data channel.
It follows that the probability of success for a transmitted transaction has the same
form as in modified SA/SA but with the following minor modification.
Pr{success}
≈ Pr{one control packet, no collision on data channel}
= Pr{one control packet} ⋅ Pr{no collision on data channel | one control packet}
= ge − g ⋅ Pr{no data on same wavelength in L – 1 contention slots}
= ge − g ⋅ (1 − ge − g W )
L −1
.
The above expression is throughput in transaction/mini slot. For throughput in
transaction/time unit/wavelength, we normalize to get the following expression.
L − g  ge − g 
ge ⋅ 1 −

W
W 

L −1
transaction/time unit/wavelength
(b) By differentiating the above throughput with respect to g and setting it equal to 0,
we see that, for local maxima/minima, g must satisfy
Optical Networks
E49
 ge − g 
e (1 − g )  1 −

W 

L −1
−g
 ge − g 
+ ge ( L − 1)  1 −

W 

L−2
−g
×
−1 − g
e (1 − g ) = 0
W
 ge − g
L −1 
⇒ e − g (1 − g )  1 −
− ge − g
=0
W
W 

L

⇒ e − g (1 − g )  1 − ge − g  = 0
W

W
⇒ g = 1 or ge − g =
L
W 1
< , it follows that, in addition to g = 1, we also have two more values
L e
of g with derivatives equal to 0. One of them is < 1 while the other is > 1. (Note
1
that ge − g ≤ , and consider the Aloha throughput curve in Fig. 5.10 of class notes.)
e
In the presence of 3 points with zero derivatives, the middle point, i.e., g = 1,
must be a local minimum since we must have the throughput approaching 0 as g
W
becomes large. Therefore, the maxima correspond to having ge − g =
and the
L
minimum corresponds to g = 1.
Since
Substituting ge − g =
W
in the throughput expression in part (a) yields the maxiL
 1
mum throughput of 1 − 
 L
‡
L −1
, which approaches
1
as L → ∞.
e
Solution 5.5 (Reservation with no central scheduler):
1
, there is
e
one successful control packet for every e mini slots on average. For round-robin
reservation, a wavelength is reserved once for every W successful control packets.
It follows that a wavelength is reserved once for every We mini slots on average.
(a) Since the probability of transmission success in each control packet is
(b) According to part (a), since time between successive reservations on each data
wavelength is We mini slots. If L < We, then there is one transaction transmission
L
for every We mini slots, yielding a normalized throughput of
transaction/time
We
unit/wavelength. If L > We, then on average each data wavelength is being used all
the time, yielding a throughput of 1 transaction/time unit/wavelength. In summary,
the throughput is
 L 
min 
,1 transaction/time unit/wavelength.
 We 
E50
© P. Saengudomlert, Asian Institute of Technology
Solution 5.6 (Reservation with central scheduler):
(a) With N users in the network, a transmitted control packet is not collided if N – 1
other users do not transmit in the same mini slot. Therefore,
Pr{transmitted control packet is successful} = (1 − p) N −1 .
It follows that the number of transmission attempts until success is a geometric random variable with parameter (1 − p) N −1 . Accordingly, the expected number of
transmission attempts until success is 1 (1 − p) N −1 = (1 − p) − N +1 . Since a control
packet is transmitted on average once every 1 p mini slots, the expected number of
mini slots taken for a user’s successful control packet transmission is
(1 − p ) − N +1
.
Λ=
p
(b) The optimal value of p, denoted by p*, that minimizes Λ can be found by solving
dΛ
= 0 . More specifically, we can write
dp *
dΛ
1
1
=
⋅ −(− N + 1)(1 − p*) − N −
(1 − p*) − N +1 = 0 ,
2
dp * p *
( p*)
yielding p* =
1
(1 − p ) − N +1
1

, Λ min = N  1 − 
. From Λ =
p
N
 N
− N +1
.
Solution 5.7 (Reservation with central scheduler):
(a) The probability that a given user has a successful control packet in a particular mini
slot is
Pr{user selects that mini slot, no other user selects that mini slot}
= Pr{user selects that mini slot}×Pr{no other user selects that mini slot}
1 
1
= × 1 − 
F  F
N −1
1
1
= 1 − 
F F
N −1
.
It follows that the probability that a given user has a successful control packet in a
1
1
frame is given by F ×  1 − 
F F
N −1
1

= 1 − 
 F
N −1
.
(b) From part (a), the successful probability for each user to have a successful control
1
1
packet in a mini slot is  1 − 
F F
N −1
. It follows that the successful probability for
Optical Networks
E51
any user to have a successful control packet, which is equal to the throughput, is
given by
1
1
T = N × 1 − 
F F
N −1
N
1
= 1 − 
F F
N −1
.
To find F that maximizes T, we solve dT dF = 0 , yielding F = N.
(c) Using F = N and considering the expression of T in part (b) for large N, we can approximate T as
T=
N
1
1 − 
N N
N −1
≈
1
.
e
Solution 5.8 (FCFS scheduler with look ahead):
(a) For convenience, let Q and R stand for “queue” and “receiver” respectively. For k =
1, two transactions are scheduled: from Q1 to R4, and from Q4 to R1.
(b) For k = 2, two transactions are scheduled: from Q1 to R4, and from Q3 to R1.
(c) For k = 3, three transactions are scheduled: from Q1 to R4, from Q3 to R1, and from
Q4 to R2.
E52
© P. Saengudomlert, Asian Institute of Technology
Solution 5.9 (Operations of GPS and ACT):
(a) For GPS, the service curves are shown below.
φ1 = φ2 = 1
4
4
A1 (0, t )
2
(7,3)
2
S1 (0, t )
1 2 3
A2 (0, t )
8
S 2 (0, t )
1 2 3
5
φ1 = 1,φ2 = 2
4
4
A1 (0, t )
2
(7,2)
S1 (0, t )
1 2 3
A2 (0, t )
2
8
S 2 (0, t )
1 2 3
5
(b) The service curves for ACT are shown below. Note that cycle boundaries are
marked with dotted lines. For comparison, the GPS service curves are shown in
dashed lines.
Wmax,1 = Wmax,2 = 1
4
4
A1 (0, t )
2
2
S1 (0, t )
1 2 3
A2 (0, t )
8
S 2 (0, t )
1 2 3
5
Wmax,1 = 1,Wmax,2 = 2
4
4
A1 (0, t )
2
A2 (0, t )
2
S 2 (0, t )
S1 (0, t )
1 2 3
8
1 2 3
5
Optical Networks
E53
Solution 5.10 (Operations of Multi-Wavelength GPS and ACT):
(a) For GPS assuming a single channel with rate 2 transactions per time unit, the service curves are shown below.
6
(8.5,6)
φ1 = φ2 = 2
(6.5,6)
6
A1 (0, t )
4
(6,5)
4
S1 (0, t )
2
S2 (0, t )
2
1 2 3
6
A2 (0, t )
8
φ1 = 2,φ2 = 4
1 2 3
(8.5,6)
(6.5,5)
4
(5.75,6)
A2 (0, t )
6
A1 (0, t )
5
4
(5.75,3.5)
S1 (0, t )
(2,1)
S 2 (0, t )
(1.75,1)
1 2 3
1 2 3
8
5
(b) For multi-wavelength ACT, the service curves are shown below.
6
Wmax,1 = Wmax,2 = 2
A1 (0, t )
6
4
4
S1 (0, t )
2
1 2 3
6
A2 (0, t )
S 2 (0, t )
2
8
Wmax,1 = 2,Wmax,2 = 4
A1 (0, t )
1 2 3
6
4
5
A2 (0, t )
4
S1 (0, t )
1 2 3
S 2 (0, t )
8
1 2 3
5
E54
© P. Saengudomlert, Asian Institute of Technology
Solution 5.11 (Operations of IPACT): Two cycles of operations for the given constant credit DBA scheme are shown below.
guard time
for Rx
CO
Tx 1
Rx
Tx
node 1 Rx
1
2
1
3
1
2
1
1
1
2
2
node 2 Tx
Rx
1
1
1
3
transaction
arrivals
at node 1
transaction
arrivals
at node 2
0
2
4
6
8
time