Data Broadcast in Asymmetric Wireless Environments Nitin H. Vaidya Sohail Hameed
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Data Broadcast in Asymmetric Wireless Environments
Nitin H. Vaidya Sohail Hameed
SUBJECT OF THE PAPER
– Mechanisms that efficiently decide what and when to transmit
CHARACTERISTICS OF THE SYSTEM– Wireless communications (server - clients)
– Asymmetric environment
– Not explicit requests from the clients to server
– Minimization of the wait time of clients
ALSO COVERED
– Environments with errors
– Multiple number of Broadcast channels
METRICS USED TO EVALUATE THE PERFORMANCE
– Access time
– Tuning time
CONTRIBUTIONS OF THE PAPER
– Square root rule
– Lower bound on the achievable access time
– “on - line” Broadcast Scheduling algorithm
– Modified “on - line” algorithm
PRELEMINARIES– Database with information Items– Time unit– M = Total number of items– li = Length of item i – Broadcast cycle with N time units– Instance of an item– Schedule of Broadcast– Frequency of an item– Spacing
I t e m 1 I t e m 3I t e m 2 I t e m 1 I t e
10 8 4 10
S p a cin g b e tw een 1 st a n d 2 n d
in stan ces of Item 1
1st in stan ce ofItem 1
2n d in stan ce ofItem 1
PRELEMINARIES Continued
– Item Mean Access Time
– Demand Probability
– Overall Mean Access time
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ijij
i
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Ns
Nss
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1
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1
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2
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i s1 1
2
1 2
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1
CONSTRUCTION OF BROADCAST SCHEDULING ALGORITHMS
Transform ing probabilities intofrequencies
M apping frequencies to aschedule
demandprobabilites
frequenciesof pages
broadcastschedule
MAPPING DEMAND PROBABILITIES TO ITEM FREQUENCIES
Lemma 1: The Broadcast Schedule with minimum Overall Mean Access Time results when the instances of each item are equally spaced
Theorem 1 (Square Root Rule): Given the Demand Probability of each item i, the minimum Overall Mean Access Time, t, is achieved when frequency of each item i is proportional to
and inversely proportional to , assuming that instances of each item are equally spaced.
That is
pi
f i
l
pf
i
ii
pi
l i
BROADCAST SCHEDULING ALGORITHMS
Algorithm A: ON - LINE algorithm:
Define
Step 1: Determine maximum F(i) over all items i, .
Let denote the maximum value of F(i).
Step 2: Choose item i such that F(i) = . If this equality holds for more than one item, choose any one of them arbitrarily.
Step 3: Broadcast item i at time Q.
Step 4: = Q
Mi 1F max
F max
iR
l
piF
i
iiRQ2
EXAMPLE A:
p1 p2 p3 l1 l2 l3= 1/2 = 3/8 = 1/8 = 1 = 2 = 4
12 2 1 3 ?
92 93 95 96 100
1 2 1 4
time
2F 3F 1F = 12.5 = 9.18 = 0.5
EXAMPLE B:
l1 = l2 = 1 p1 = 0.2 + ε p2 = 1 - p1
On - line Algorithm A: Schedule (1,2), t = 1
Schedule (1,2,2), t = 2.9/3 +2ε/3 < 1
ON- LINE ALGORITHM B WITH BUCKETING
Complexity of algorithm A O(M)
Complexity of algorithm B O(k)
Divide the database into k buckets
Bucket i contains items
Average Demand Probability of items in bucket i
Average Length of items in bucket i
mi
k
jj
Mm1
mpq ij
ii
mi
/1
mld ij
ii
mi
/1
ALGORITHM B
Define
Step 1: Determine maximum G(i) over all buckets i, .
Let denote the maximum value of G(i).
Step 2: Choose a bucket i such that G(i) = . If this equality
holds for more than one bucket, choose any one of them arbitrarily.
Step 3: Broadcast item Ij from the front of the bucket Bi at time Q.
Step 4: Dequeue item Ij at the front of the bucket Bi and enqueue it
at the rear of Bi .
Step 5: = Q
ki 1
d
qiG
i
iiRQ2
Gmax
Gmax
iR
Optimal Mean Access time
Heuristic that determines membership of each item into buckets
– Calculate Rmin = , Rmax =
– Divide δ = Rmin - Rmax into k equally sized sub - intervals– Calculate for all items. Item i is into bucket j Bj if
R m in R m ax
d /k
B3B1 B2 B4 B5
k =5
k
jjjj dqmt buckopt
12
12
lp iiimin lp iiimax
lp ii
kjlipikj R //1min
Effect of Transmission Errors on Scheduling Strategy
– E(l)
– Overall Mean Access Time
Theorem 2: Given that the probability of occurrence of uncorrectable errors in an item of length l is E(l), the overall mean access time is minimized when
)E(-1)E(1
2/1
ll
pls
i
i
i
ii
)E(-1)E(1
2/1
ll
lp
fi
i
i
i
i
)E(-1)E(1
2/1
12
1
2
ll
lpti
iM
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)E(-1
)E(1
2
11 l
lsi
iM
iii
pt
Multiple Broadcast Channels
Divide the available bandwidth into c channels
Define properly iR
On - line algorithm for channel h, 1 h c
Step 1: = , 1 i M
Step 2: Determine maximum F(j) over all items j. Let Fmax denote the maximum value of F(j).
Step 3: Choose i such that F(i)= Fmax. If this equality holds for than one item, choose any one of them arbitrarily.
Step 4: Broadcast item i on channel h at time Q.
Step 5: Set = Q
iR iR jhjmax1
iR
A heuristic for initializing values iR j
M
jii sp
1
/1
M
jiijji pllps
1
Step 1: Set time=1Step 2: For every item in database {Step 3: For every item {Step 4: if {Step 5:Step 6: time=time+ } }Step 7:Step 8: = timeStep 9: time=time+Step 10: For
}Step 11: Find , , , by rotating the values of by an amount of
Mi 1
Mj 1
0j
s jj
l i
sii
iR1
l i
ij 1
l ijj
iR j cj 1 Miq iR j 1
Performance Evaluation
Demand Probability Distribution
M
iic
1/1/1
1cp
i
1e - 05900200 8007006005004003001000 1000
1
0.1
0.01
0.001
0.0001
Item Num ber
Access P
ropabili
ty
THETA=0.25
THETA=1
THETA=1.25
THETA=0.75
THETA=0.5
THETA=1.5
Zipf distribution for various values of θ
Length Distribution
LLLl o
oi
iM
round 11
1 Mi 1
Uniform Length Distribution
Increasing Length Distribution
Decreasing Length Distribution
10,1 1 LLo
10,1 1 LLo
1,10 1 LLo
900200 8007006005004003001000 1000Item Num ber
Leng
th
0
2
4
6
8
10
increasing
decreasing
uniform
Length Distribution
Performance evaluation in the Absence of Uncorrectable Errors
THETA
Ove
ral M
ean
Acc
ess
Tim
e
0.25 1.51.2510.750.5
0
2500
2000
1500
1000
500
3000
optimal
on - line
1 - bucket
5 - bucket
10 - bucket
Increasing Length Distribution
Ove
ral M
ean
Acc
ess
Tim
e
0.25 1.51.2510.750.5
2500
2000
1500
1000
500
3000
on - line
1 - bucket
10 - bucket
THETA
5 - bucket
optimal
Decreasing Length Distribution
0.25 1.51.2510.750.5THETA
Ove
ral M
ean
Acc
ess
Tim
e
0
2500
2000
1500
1000
3000
on - line
10 - bucket
5 - bucket
optimal
1 - bucket
500
Random Length Distribution
Performance evaluation in the Presence of Uncorrectable Errors
0 0.05 0.1 0.15 0.2
0
2000
4000
6000
8000
10000
12000
14000
16000
18000 THETA=0
THETA=1.5
THETA=1
LAMBDA
Ove
rall
Mea
n A
cces
s T
ime
Increasing Length
Distribution
ell
iE 1 12
2 eiRQ l
i
i i
l
piF
0 0.05 0.1 0.15 0.2
0
2000
4000
6000
8000
10000
12000
14000
16000
18000
THETA=0
THETA=1.5
THETA=1
LAM BDA
Ove
rall
Mean A
ccess
Tim
e
Decreasing Length Distribution
Performance with Multiple broadcast Channels
Num ber of Channels
1 2 3 4
Ove
rall
Mea
n A
cces
s T
imes
100
150
200
250
300
350
400
450
500
THETA=0
THETA=1
THETA=0.5
Uniform Length
Distribution
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