Upload
ganit
View
40
Download
0
Tags:
Embed Size (px)
DESCRIPTION
Capacity Allocation Paradox Isaac Keslassy. Joint Work with Asaf Baron and Ran Ginosar EE Department, Technion, Haifa, Israel. The Capacity Allocation Paradox. Node A. Router. C A. Node C. R A. C R. Node B. C B. R B. Finite (small) buffers. Unlimited queues. - PowerPoint PPT Presentation
Citation preview
Capacity Allocation ParadoxCapacity Allocation Paradox
Isaac KeslassyIsaac Keslassy
Joint Work with Asaf Baron and Ran Ginosar
EE Department, Technion, Haifa, Israel
2
The Capacity Allocation ParadoxThe Capacity Allocation Paradox
Node B
Node C
Node A
CA
CB
CR
Router
Unlimited queuesFinite (small) buffers
RA
RB
Capacity Allocation Paradox:Adding Capacity Can Destabilize the Network
3
Fast Security CheckSlow Security Check
StableUnStableMarakana Soccer StadiumMarakana Soccer Stadium
Brazillian Line
Argentinian Line
Fast Security Check
Fast Swipe Ticket Entrance
Safety Check
Enter the stadium
4
MotivationMotivation Small buffer networks are widely used
When QoS not met: add capacity [Guz et al., ’06]May destabilize the network
Network On-Chip Interconnection of Computers
SpaceWire
5
Previous Work: Selfish RoutingPrevious Work: Selfish Routing Braess’s Paradox (1968)
Difference: We assume fixed routing
6
Previous Work: Cyclic DependencyPrevious Work: Cyclic Dependency Kumar & Seidman (1990)
Instability even though capacity > data rate
Dai, Hasenbein & Vande Vate (1998) Adding capacity may destabilize a network
Differences: No cycles in dependency graph Single router
Each packet visits router only once Several simple arbitration policies Independent of initial conditions
New fundamental reason: Finite buffers
7
A General PhenomenonA General Phenomenon
Node B
Node C
Node A
CA
CB
CR
Router
Unlimited queues
Finite (small) buffers
RA
RB
When buffer is full:1. Blocking: Wormhole Routing2. Dropping (with retransmission): Store And Forward
Arrivals:Periodic, Poisson…
CA=2
CR=2CA=1
8
IntuitionIntuition
Node B
Node C
Node A
CB=1
Router
Buffer of 1 bit
1 [pkt/T]
1 [pkt/T]
Assume A has priority:
(a) CA=1A1
B1
A2
B2
A3
B3
2T
Share of CR
2
T
1
3T(a)
(b) CA=2 A1B1 (1)
2T
Share of CR
2
T
1
3T
(b)
A2B1 (2)
A3B2 (1)
T/2 3T/2 5T/2
9
What are the conditions for stability?What are the conditions for stability? Necessary conditions:
A AC R
R A BC R R
B BC R
Node A
Node B
Node C
Stability Regions, CR =0.273Mflits/sec
CA
[Mflit/sec]
CB [
Mfli
t/se
c]
0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
BC
AC
CR is constantRA = RB
10
Stability Regions, CR =0.273Mflits/sec
CA [Mflit/sec]
CB [
Mfli
t/se
c]
0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
Case #1: Case #1:
Buffers in the router hold no more than one data unit
A B RC C C
Node C
Queue A
Queue B
Buffer A
Buffer B
Necessary conditions are also sufficient.
Stability Regions, CR =0.273Mflits/sec
CA [Mflit/sec]
CB [
Mfli
t/se
c]
0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
?
CA
CB
CR
11
EPRR
CA [Kf/s]
CB [
Kf/
s]
50 100 150 200 250 300 350 400 450 500
50
100
150
200
250
300
350
400
450
500
Example 1: Analysis Stability PictureExample 1: Analysis Stability Picture
1
42
3 2
0
0
CA [Kf/s]
CB [
Kf/
s]
CR = 273[Kf/s] (Constant)
RA = RB = 100[Kf/s]
CB=110CB=150
CA=110CA=300CA=190
12
Analytic Stability regions, CRC
=0.273Mflits/sec
CAR
[Mflit/sec]
CB
R [
Mfli
t/se
c]
0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
Case #1Case #2Case #3
Example #1 – Capacity AllocationsExample #1 – Capacity Allocations
Node A
Node B
Node C
CR=273
StableUnStableStable1
42
3 2
RA = 100
RB = 100 1000 [flits/pckt]Buffer Size: 16 Flits
Exhaustive Round Robin, Wormhole
13
Analytic+Simulation Stability regions for CR =0.273Mf/s
CA [Kf/s]
CB [
Kf/
s]
125 150 175 200 225 250 275 300 325
125
150
175
200
225
250
275
300
325
Results – Simulation Stability Results – Simulation Stability RegionsRegions
1
42
3 2
CR = 273[Kf/s] (Constant)
RA = RB = 100[Kf/s]
14
EPRR
CA
[Kf/s]
CB [
Kf/
s]
550 600 650 700
150
200
250
300
350
400
450
500GPS
CA
[Kf/s]
CB [
Kf/
s]
550 600 650 700
150
200
250
300
350
400
450
500RRPF
CA
[Kf/s]
CB [
Kf/
s]
550 600 650 700
150
200
250
300
350
400
450
500
Example #2 – Wormhole RoutingExample #2 – Wormhole Routing
1000 [flits/pckt], Buffer Size: 16 Flits, RA = 500kf/s, RB = 100kf/s
Exhaustive Round Robin Round-Robin GPS
15
Exhastive, CRC=2.1Mbit/sec
CAR [Mbit/sec]
(b)
CB
R [
Mbi
t/se
c]
1.25 1.5 1.75 2 2.25 2.5
1.25
1.5
1.75
2
2.25
2.5
Priority, CRC=2.1Mbit/sec
CAR [Mbit/sec]
(a)
CB
R [
Mbi
t/se
c]
1.25 1.5 1.75 2 2.25 2.5
1.25
1.5
1.75
2
2.25
2.5
Example #3 – Store and forwardExample #3 – Store and forward
Poisson Arrivals with Parameters: A = 100, B = 100Packet Length 10^4 bitBuffer Size 3-4 packets
Strict Priority, CR = 2.1[Mbit/s] Exhaustive RR, CR = 2.1[Mbit/s]
16
Round-Robin, CRC=6.1Mbit/sec
CAR [Mbit/sec]
(c)
CB
R [
Mbi
t/se
c]
5.25 5.5 5.75 6 6.25 6.5
1.5
2
2.5
3
3.5
Exhastive, CRC
=6.1Mbit/sec
CAR
[Mbit/sec]
(d)
CB
R [
Mbi
t/se
c]
5.25 5.5 5.75 6 6.25 6.5
1.5
2
2.5
3
3.5
Example #3 – Store and forwardExample #3 – Store and forward
Poisson Arrivals: A = 500B = 100
Packet = 10^4 bitBuffer 3 packets
RR, CR = 6.1[Mbit/s]Exhaustive RR, CR = 6.1[Mbit/s]
All packets need to arrive
sometime
17
SummarySummary Adding capacity may destabilize even a
simple network
The scheduling algorithm affects the stability of the network (even if work-conserving)
GPS arbitration: always stable
18
Thank you.Thank you.