E E 681 - Module 7
Introduction to rings: ring types, ring
sizing and ring loading
W. D. Grover
TRLabs & University of Alberta
© Wayne D. Grover 2002, 2003
Two main types of “survivable ring”....(1) UPSR
Unidirectional Path-switched Ring...Principle of operation
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
2
Two main types of “survivable ring”....(1) UPSR
Unidirectional Path-switched Ring ...
Unidirectional - because in normal
operation all working demand flows in one direction only.
i.e.,
A sends to B clockwise,
B also sends to A clockwise
Path-switched - because in restoration each receiver
selects an alternate end-to-end path
through ring, regardless of where
actual break occurred.
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
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UPSR Animation...
Working fibre
1
Tail-end Switch
5
2
Protection fibre
3
4
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
l1
4
UPSR ...line capacity requirement
A
• Consider a bi-directional demand
quantity between nodes A, B: dA,B.
- A to B may go on the short route
- then B to A must go around the longer route
A -> B
E
B
• Thus, every (bi-directional) demand pair
circumnavigates the entire ring.
B -> A
D
• Hence in any cross section of the ring,
we would find one unidirectional instance
of every demand flow between nodes
of the ring.
C
“ The UPSR must have a line rate
(capacity) greater (or equal to)
the sum of all the (bi-directional)
demand quantities between nodes of
the ring. “
E E 681 - Module 7
• Therefore, the line capacity of the UPSR
must be:
cUPSR   d ij
© Wayne D. Grover 2002, 2003
i j
5
Notes on the UPSR
• Can be thought of as a number of virtual 1+1 APS
set-ups sharing a single set of high-speed
transmission systems to obtain “economy of scale”.
• Economy of scale arises since one OC-96 (say)
optical Tx / Rx pair, is a lot less expensive
than 96 OC-1 Tx / Rx !
• UPSRs are inherently 2-fibre structures.
• Primary use is in “access” applications.
- distances are not great
- under pure “hubbed” demand pattern
UPSR is as efficient as BLSR.
• UPSR need not “revert” after protection switching.
The “access” demand pattern
• UPSR switching decisions are independent on
a tributary-by-tributary basis:
- switching on one channel has no effect on other channels.
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
0 d 21 d 31 d 41 d 51
0
0
0
0
0
0
0
0
0
0
6
Two main types of “survivable ring”....(2) BLSR
Bi-directional Line-switched Ring...Principle of operation (“4-fibre” BLSR illustrated)
Loop Back
Cable cut
(a) Normal Operation (before failure)
E E 681 - Module 7
(b) Protection Operation (after failure)
© Wayne D. Grover 2002, 2003
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Two main types of “survivable ring”....(2) BLSR
Bi-directional Line-switched Ring...Principle of operation (“4-fibre” BLSR illustrated)
Bi-directional - because in normal
operation working demand flows travel in
opposite directions over the same
route through the ring
Loop Back
Cable cut
(a) Normal Operation (before failure)
(b) Protection Operation (after failure)
“ The BLSR must have a line rate
(capacity) greater (or equal to)
the largest sum of demands routed
over any one span of
the ring. “
E E 681 - Module 7
Line-switched - because in restoration the composite
optical line transmission signal is switched to
the other direction around the ring (on the other fibre pair)
specifically around the failed section.
Note implication: Protection fibre capacity must equal
the largest-working capacity
cross-section of any span on the ring.
© Wayne D. Grover 2002, 2003
8
(4 fibre) BLSR Animation...
Working fibres
1
Loop-back
5
2
Protection fibres
3
4
Loop-back
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l1
© Wayne D. Grover 2002, 2003
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BLSR can also be in a “2-fibre” variant
fibre 1
Loop Back
Cable cut
Working channel group
Working channel group (in use)
Protection channel group
Protection channel group (in use)
(a) Normal Operation (before failure)
(b) Protection Operation (after failure)
Ex: OC-48 2 BLSR: fibre 1 (cw)
E E 681 - Module 7
• Each fibre has its
tributary channels
arranged in two groups
- Working
- Protection
fibre 2 • The set of 4 channel
groups on two fibres
then acts logically
just like a 4-fibre BLSR
• For the same demand
pattern the required
line rate is doubled
c2 BLSR  2  c4 BLSR
fibre 2 (ccw)
“loopback”
- Channels 1-24 Working
- Channels 1-24 Working
- Channels 25-28 Protection
- Channels 25-48 Protection
© Wayne D. Grover 2002, 2003
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BLSR ...line capacity requirement to serve its demands
• Start by considering how BLSR demand routing differs from UPSR....
1
Working fibre
Working fibres
1
5
2
5
2
Protection fibres
Protection fibre
3
4
4
l1
UPSR: every demand pair
circumnavigates ring
3
l1
BLSR: demand pair can be routed over
shortest path. Not all spans “see” any given
demand pair
 opportunity for “bandwidth reuse”
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
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BLSR ...Bandwidth re-use improves BLSR efficiency
• Concept of “bandwidth re-use” in a BLSR....
Demand 1-4
Demand 1-3
Q. what demand pattern lends itself
to perfect bandwidth re-use ?
1
Demand 1-4
Demand 3-4
Time Slot #1
4
2
• Now note:
the blue demand (1-3) could
equally well have gone on
route 3-4-1 as 3-2-1 (since
same distance used).
• If so, what would effect be on required
line-rate capacity ?
3
Demand 3-4
• The example shows one timeslot
(or “channel”) being reused on
4 spans to serve three different
demand pairs.
Demand 1-3
• Implication: BLSR line-rate requirement
depends on how the set of demands it
is to serve are loaded into it !
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
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Introduction to the ring sizing and
loading problems...
W. D. Grover
TRLabs & University of Alberta
© Wayne D. Grover 2002, 2003
BLSR … Line capacity dependence on internal demand routing
• A heuristic algorithm for BLSR ring loading (Wu ‘92)....
1. Rank all demands in descending order
2. Map any adjacent-node demands into the ring (and remove from list)
3. Repeat In descending order:
- map next largest demand into ring over its shortest route
- map the same demand into the ring over the complementary route
- choose the route that produces:



min max   wi  
 i


where wi is the accumulation of demands crossing span i.
- if each route produces the same {max wi } choose the shorter route
- if both routes are equal, alternate this route choice with that at the next
similar “tie”.
Until all demands are served
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
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BLSR ...Example of Wu’s heuristic loading algorithm
• Example of the heuristic BLSR loading algorithm....
A
B
Demands
(sorted in decreasing order):
AC
10
EB
8
EA
6*
ED
6*
DB
5
DC
4*
EC
4
BC
3*
AB
2*
C
E
step 1 : Place adjacentnode demands:
D
A wi = 2
wi = 6
B
wi = 3
C
E
wi = 6
D
wi = 4
* denotes demand between adjacent nodes
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
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BLSR ... Example of the heuristic BLSR loading algorithm
Remaining demands
step 2 : Consider routing of the AC demand:
(sorted):
AC
10
EB
8
wi = 6
DB
5
wi = 13
C
E
wi = 6
EC



min max   wi  
 i


Awi = 2+ 10 = 12
B
D
wi = 4
A wi = 2
4
wi = 16
B
wi = 3
C
E
 shorter route is preferred:
map AC via route A-B-C
(max wi = 13)
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
wi = 16
D
wi = 14
16
BLSR ... Example of the heuristic BLSR loading algorithm
Remaining demands
step 3 : Consider routing of the EB demand:
(sorted):
A
EB
wi = 20
8
wi = 14
DB
wi = 13
C
E
5
wi = 6
EC
B
D
wi = 4
A wi = 12
4
wi = 6
B
wi = 21
C
E
 shorter route is again preferred:
map EB via route E-A-B
wi = 14
D
wi = 12
(max wi = 20)
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
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BLSR ... Example of the heuristic BLSR loading algorithm
Remaining demands
step 4 : Consider routing of the DB demand:
(sorted):
A wi = 20
wi = 14
DB
wi = 18
C
E
5
wi = 6
EC
B
D
wi = 9
A wi = 25
4
wi = 19
B
wi = 13
C
E
 shorter route is again preferred:
map DB via route D-C-B
wi = 11
D
wi = 4
(max wi = 20)
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
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BLSR ... Example of the heuristic BLSR loading algorithm
Remaining demands
step 5 : Consider routing of the EC demand:
(sorted):
A wi = 24
wi = 18
B
wi = 22
C
E
wi = 6
EC
D
wi = 9
A wi = 20
4
wi = 14
B
wi = 18
C
E
 shorter route is again preferred:
map EC via route E-D-C
wi = 10
D
wi = 13
(max wi = 20)
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
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BLSR ... Example of the heuristic BLSR loading algorithm
Resultant ring loading and sizing plan:
Demand pair
AC
10
route
ABC
EB
8
EAB
EA
6*
direct
ED
6*
direct
DB
5
DCB
DC
4*
direct
EC
4
EDC
BC
3*
direct
AB
2*
direct
Resulting in these net span
loadings:
A
wi = 20
wi = 14
B
OC-24 4-fiber BLSR
wi = 18
C
E
wi = 10
and thus requiring
(in practise) an
D
wi = 13
or
OC-48 2-fiber BLSR
or
an “ideal” 4-fiber
OC-20 BLSR
possible project idea: implement Wu’s algorithm followed by a meta-heuristic search for improvement towards optimal
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
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BLSR ... Capacity efficiency / redundancy assessment
• consider the ring just designed... One measure of BLSR efficiency is:
 wi
~ capacity usefully
serving demands
i
A wi = 20
wi = 14
N  max{wi }
B
wi = 18
C
E
wi = 10
here...
w
i
i
D
wi = 13
~ redundant protection
capacity required
N  max{wi }

20  14  10  13  18
 75%
5  20
or conversely the redundancy is ... 133 %
si required to be 20 everywhere
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
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Compare to UPSR ...
• to serve the same set of demands, the UPSR would require the ring line rate to be :
cUPSR   d ij  10  8  6  6  5  4  4  3  2  48
i j
• but the amount of demand-serving capacity of the BLSR loading still applies as the
measure of useful service or utility:
 w  20  14  10  13  18  75
i
A wi = 20
wi = 14
i
B
wi = 18
C
E
wi = 10
D
E E 681 - Module 7
wi = 13
• Therefore, the redundancy measure (“spare
to working” ratio) for the UPSR can be
formed as:
total capacity - working capacity
working capacity
2  5  48  2  75
redundancy 
 220 %
2  75
© Wayne D. Grover 2002, 2003
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Effect of some “generic” demand patterns on BLSR
• From preceding it is evident that BLSR demand-serving ability depends in general on
the demand pattern.
• Some of the recognized tendencies in real demand patterns are:
Node-to-Adjacent Node
Uniform
or “mesh”
Single Hub
Double Hub
Demand
Hub
ideal case for BLSR
perfect bw re-use
BLSR much more
efficient than UPSR
no optimization
required
E E 681 - Module 7
this is the
general tendency in
inter-city backbone
network
optimization of ring
loading
same basic “access” demand
pattern but dual hubs employed
for access survivability
this is a fairly exact
model for access
ring applications
BLSR efficiency = UPSR
© Wayne D. Grover 2002, 2003
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Effectiveness of BLSR relative to UPSR depending on demand pattern
• systematic study of relative demand-serving ability of (2 fibre) BLSR to UPSR...
(Tom Flanagan, IEEE Communications Magazine, June 1990 - see web site “reading” for lecture 9)
with perfect bw re-use
BLSR gets proportionally
better as ring size increases
capability
serving
Total demand
relative
capacity
600%
500%
e
od
n
nt
ce
dja
400%
-a
- to
e
d
No
300%
ibuted
Typical distr
Uniform
200%
in this range optimized BLSR
loading (and ring selection)
can give significant benefits
over UPSR
rn
mesh patte
with perfect hubbing demand
patterns, BLSR never has
any advantage over UPSR
100%
Single and double hub
0%
2
3
4
5
6
7
8
9
10
no. of nodes
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
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BLSR relative to UPSR depending on demand pattern
A
A
B
C
E
B
C
E
D
D
• Under adjacent-node pattern we
see perfect bw re-use.
• The more nodes, the more
demands are served with the
same line rate of the ring.
• Under single-hub pattern. The ring is
“sized” by the cross-section of demands
accumulating in spans next to the hub.
• If no. nodes is odd, half the demands
appear in each such span.
• ~ UPSR like, but for a factor of 1/2
(If no. nodes is even, the best we can do is stay at same
sizing principle, by splitting the flow where needed.)
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
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Avoiding some confusions in working with rings
• Ring “capacity” - generally means the optical line rate capacity of the ring
but context matters:
- is it the line-rate of an actual given ring system ?
- or is someone speaking of the capacity required to serve some demands ?
also convention:
- usually the working capacity is referred to, with understanding for BLSR
that the protection capacity is identical.
e.g. “OC-48 4-BLSR” really represents two complete OC-48
bi-directional transmission systems
• Ring “size” should be avoided unless explicitly clarified ...
- does it mean the number of spans / nodes on the ring ? (“circumferential size”)
or
- does it mean the line capacity of the ring?
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
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Some other info about rings...
• SONET rings operate at OC-n line rates and the STS-1 tributaries are the “channels”
• The nodes of a ring are equipment called “Add-Drop Multiplexers” (ADMs)
• SONET rings may have a maximum of 16 active nodes, plus “glass-through” sites
• “Glass-throughs” are just nodes transited by the ring, but where no ADM is present
• “Glass-throughs” may be simply fibre splices or a regenerator point (“pass throughs”)
• Demand splitting refers to whether or not the total demand exchanged between
two nodes has to be kept together on the same route of a ring or can be ‘split’
• Time slot interchange (TSI) refers to whether the ADMs have the ability to crossconnect timeslot contents (assign a new time slot to a demand on the next span)
• More recent Optical rings have a DWDM optical line signal and add / drop single
wavelengths or wave-bands
- the logical “channel” is a wavelength (l) or waveband
- UPSR < - > OPPR (Optical Path Protection Ring)
- BLSR < - > OSPR (Optical Shared Protection Ring)
- ADM < - > OADM
- TSI (Time slot interchange) < - > l conversion
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
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BLSR related optimization problems (1)
1. Ring “Sizing”
- CONTEXT: A number of demand pairs are to be served by a BLSR
- QUESTION IS: What is the minimum line rate BLSR required?
Required BLSR
line capcity
• line rate = f (demands,
routing in ring)
Q. What is it that has to be optimally
decided to minimize the required
line rate ?
i.e. (What do we have control over here?)
demands that must
be served
E E 681 - Module 7
A. for each demand: cw, or ccw ?
© Wayne D. Grover 2002, 2003
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BLSR related optimization problems (2)
2. Ring “Loading”
- CONTEXT: A number of demand pairs are to be served, but not necessarily
all in same ring.
i.e., there is a “pool” of outstanding demands to consider for selection into a given ring.
- QUESTION IS: What is the maximum number of these demands
that a BLSR with given capacity can serve?
or... (alternate goal)
Which set of demands (and routings) achieves greatest
utilization of ring capacity?
pool of demands
needing to
be served
E E 681 - Module 7
fixed ring
capacity
? which demands
to pick ?
© Wayne D. Grover 2002, 2003
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( Aside: “A word to the wise” )
…. Many published papers on either ring sizing or loading problems are called
ring “loading” problems without distinction.
- One has to study each paper to see if it is really addressing a sizing or a
loading problem.
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
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BLSR related optimization problems
Ring “Sizing” : General Optimum design formulation
Inputs (“parameters”) D  set of demands to be served
k indexes set D
d k  a demand quantity in D
R  set of spans in the ring
i indexes set R
 k,i , k,i  l if routing demand k cw(+) [(ccw(-)] crosses span i
0 otherwise
Outputs (“variables”)
 k ,  k  l if demand k is routed cw(+) [ccw(-)]
0 otherwise
C  required line capacity of the ring
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
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BLSR related optimization problem formulations
Understanding how the 1 / 0 parameters or variables encode the
problem knowledge
A


EC ,1
l

 EC
,2  l
i =2
(+)
i =1
B


EC ,3
1
i =3
A


EC ,1
i =2
0

0
D
C
E
case (a) demand EC is considered
for clockwise routing

E E 681 - Module 7
i =5
i =4

 EC
,4  0

 EC
,3  0
i =3
i =5

EC ,5
B
i =1
C
E

 EC
,2  0

 EC
,5  1
D
(-)
i =4

 EC
,4  1
i.e.,  k  1
case (b) demand EC is considered
for counter-clockwise routing i.e.,    0
k
i.e.,  k  0
i.e.,  k  1
© Wayne D. Grover 2002, 2003
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BLSR related optimization problem formulations
Ring “Sizing” : General Optimum design formulation
min
C
minimize the required ring capacity
s.t.
d
k
 ( k   k,i   k   k,i )  C
kD
 k   k  1
 k ,  k  {0,1}
i  R
keep sum of all demands
crossing a span under the capacity
k  D
k  D
E E 681 - Module 7
every demand has to be routed
either cw or ccw, but not both
routing decisions are binary (cw or ccw)
© Wayne D. Grover 2002, 2003
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BLSR related optimization problem formulations
Ring “Loading” : General Optimum design formulation
Inputs (“parameters”)
C  given line capacity of the ring
and as before...
D  set of demands to be served
k indexes set D
d k  a demand quantity in D
R  set of spans in the ring
i indexes set R
 k,i , k,i  l if routing demand k cw(+) [(ccw(-)] crosses span i
0 otherwise
Outputs (“variables”)
 k  l if demand k is chosen and routed cw(+)
0 otherwise
 k  l if demand k is chosen and routed ccw(+)
0 otherwise
E E 681 - Module 7
© Wayne D. Grover 2002, 2003
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BLSR related optimization problem formulations
Ring “Loading” : General Optimum design formulation
max   k   k
or
kD
max  ( k   k )  d k
kD
maximize the number of demand
pairs wholly served or, maximize
total demand volume served
s.t.
d
kD





(







k
k
k ,i
k
k ,i )  C
  1

k

k
 k ,  k  {0,1}
i  R
you can refuse any demand, or
to select it and route it cw or ccw,
but not both
k  D
k  D
E E 681 - Module 7
keep the sum of all flows crossing
a span under the line capacity
decisions are binary
© Wayne D. Grover 2002, 2003
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