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

f

j

ijf

j

ijij

i

ii

Ns

Nss

t1

2

1 2

1

2ss

Ns

fNs

Ns

t iiij

i

f

j

if

j

iji

ii

2

1

2

1

2

1

2

1

2

1 2

1

2

1

2

M

iii lfN

1fNs ii /

pi

M

ii

j

ijM

iii p

f

Nptt

i s1 1

2

1 2

1 fjss iiij 1

M

iii pst

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

iiierropt

)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