› ~bohacek › SensorAndData... · Bit-rate Selection over a Single Hop I Recall that Shanon capacity says bit-rate=log 2 (1+SNR)=log 2 1+ H P t N whereP t is the transmit power,H

Bit-rate Selection over a Single HopI Recall that Shanon capacity says

bit-rate = log2 (1+ SNR) = log2

�1+

H � PtN

�where Pt is the transmit power, H is the channel, and N is thenoise.

I The energy required to transmit a packet is

I

Energy =packet size

log2�1+ H�Pt

N

� � �PtxSys + α+ βPt�=

z (B + CP)log2 (1+ AP)

I for AP � 0, log2 (1+ AP) � 0+ 1log(2)AP

Energy=z log (2) (B + CP)

AP

I If A > z log (2)C , then P� = ∞, in which case AP � 0I Take the derivative and set to zero

ddPz log (2) (B + CP)

AP=

z log (2)CAP � z log (2) (B + CP)A(AP)2

=�z log (2)BA(AP)2

I which is zero only when P = ∞, and again, AP � 0



�1+

H � PtN


I The energy required to transmit a packet is

I

Energy =packet size

log2�1+ H�Pt

N





AP



AP=






�1+

H � PtN


I The energy required to transmit a packet isI

Energy =packet size

log2�1+ H�Pt

N





AP



AP=






�1+

H � PtN



Energy =packet size

log2�1+ H�Pt

N





AP



AP=






�1+

H � PtN



Energy =packet size

log2�1+ H�Pt

N





AP

I If A > z log (2)C , then P� = ∞, in which case AP � 0

I Take the derivative and set to zero


AP=






�1+

H � PtN



Energy =packet size

log2�1+ H�Pt

N





AP



AP=






�1+

H � PtN



Energy =packet size

log2�1+ H�Pt

N





AP



AP=




Bit-rate Selection over a Single Hop (2)I Take the derivative and set to zero

ddP


=zC log2 (1+ AP)�

�z (B+CP )1+AP

�log2 (1+ AP)

2 = 0

C log2 (1+ AP) =

�(B + CP)1+ AP

�

I if AP>>1,

C log2 (1+ AP) =

�(B + CP)AP

�log2 (1+ AP) =

BCP

+ 1

I Consider the two curves, BCP + 1 and log2 (1+ AP) as functions

of P. The optimal P is where these intersect.

I Observed that as B (other energy for transmit) increases, the P�

increasesI As C (e¢ ciency) increase, P� decreasesI as A increases, P� decreases (better channel results in lowertransmit power)


ddP


=zC log2 (1+ AP)�

�z (B+CP )1+AP

�log2 (1+ AP)

2 = 0

C log2 (1+ AP) =

�(B + CP)1+ AP

�I if AP>>1,

C log2 (1+ AP) =

�(B + CP)AP

�log2 (1+ AP) =

BCP

+ 1






ddP


=zC log2 (1+ AP)�

�z (B+CP )1+AP

�log2 (1+ AP)

2 = 0

C log2 (1+ AP) =

�(B + CP)1+ AP

�I if AP>>1,

C log2 (1+ AP) =

�(B + CP)AP

�log2 (1+ AP) =

BCP

+ 1






ddP


=zC log2 (1+ AP)�

�z (B+CP )1+AP

�log2 (1+ AP)

2 = 0

C log2 (1+ AP) =

�(B + CP)1+ AP

�I if AP>>1,

C log2 (1+ AP) =

�(B + CP)AP

�log2 (1+ AP) =

BCP

+ 1


of P. The optimal P is where these intersect.I Observed that as B (other energy for transmit) increases, the P�

increases

I As C (e¢ ciency) increase, P� decreasesI as A increases, P� decreases (better channel results in lowertransmit power)


ddP


=zC log2 (1+ AP)�

�z (B+CP )1+AP

�log2 (1+ AP)

2 = 0

C log2 (1+ AP) =

�(B + CP)1+ AP

�I if AP>>1,

C log2 (1+ AP) =

�(B + CP)AP

�log2 (1+ AP) =

BCP

+ 1



increasesI As C (e¢ ciency) increase, P� decreases

I as A increases, P� decreases (better channel results in lowertransmit power)


ddP


=zC log2 (1+ AP)�

�z (B+CP )1+AP

�log2 (1+ AP)

2 = 0

C log2 (1+ AP) =

�(B + CP)1+ AP

�I if AP>>1,

C log2 (1+ AP) =

�(B + CP)AP

�log2 (1+ AP) =

BCP

+ 1




Many short hops or a few long hops?I If the maximum transmit power does not allow transmission tothe destination, then multi-hop must be used.

I But if both options are posisble, which results in lower energyI First approach, neglect start-upI Multihop case

I Node 1 transmits, and then sleeps. Node 2 listens, then transmits.Node 3 sleeps, then recieves.

I Node 1 and 3 use the same energy. node 1 and 2 use the sametransmit power, etc

I Using the energy models from beforeI

transmit energy + recieve energy + transmit energy + receive energy +...

Iktransmit energy + kreceive energy

I

ksizerate

((Psys + α+ βPtx )) + ksizerate

Prx


I But if both options are posisble, which results in lower energy

I First approach, neglect start-upI Multihop case






I

ksizerate


Prx


I But if both options are posisble, which results in lower energyI First approach, neglect start-up

I Multihop case






I

ksizerate


Prx








I

ksizerate


Prx








I

ksizerate


Prx








I

ksizerate


Prx





I Using the energy models from before

I



I

ksizerate


Prx








I

ksizerate


Prx








I

ksizerate


Prx








I

ksizerate


Prx

I Distance between hops = d/k

I Received signal strength

1(d/k)γP

I bit-rate

log2

�1+

1N (d/k)γP

�I

k�

0@ 1

log�1+ Ptx

N (d/k )γ

� (Psys + α+ βPtx ) +1

log�1+ Ptx

N (d/k )γ

� (Prx )1A

I

k �

0@ 1

log�1+ Ptx

N (d/k )γ

� (Psys + α+ Prx + βPtx )

1AI

k �

0@ 1

log�1+ kγ Ptx

Ndγ


1AI or

energy = k � 1log (1+ kγA)

B

I Distance between hops = d/kI Received signal strength

1(d/k)γP

I bit-rate

log2

�1+

1N (d/k)γP

�I

k�

0@ 1

log�1+ Ptx

N (d/k )γ


log�1+ Ptx

N (d/k )γ

� (Prx )1A

I

k �

0@ 1

log�1+ Ptx

N (d/k )γ


1AI

k �

0@ 1

log�1+ kγ Ptx

Ndγ


1AI or


B


1(d/k)γP

I bit-rate

log2

�1+

1N (d/k)γP

�

I

k�

0@ 1

log�1+ Ptx

N (d/k )γ


log�1+ Ptx

N (d/k )γ

� (Prx )1A

I

k �

0@ 1

log�1+ Ptx

N (d/k )γ


1AI

k �

0@ 1

log�1+ kγ Ptx

Ndγ


1AI or


B


1(d/k)γP

I bit-rate

log2

�1+

1N (d/k)γP

�I

k�

0@ 1

log�1+ Ptx

N (d/k )γ


log�1+ Ptx

N (d/k )γ

� (Prx )1A

I

k �

0@ 1

log�1+ Ptx

N (d/k )γ


1AI

k �

0@ 1

log�1+ kγ Ptx

Ndγ


1AI or


B


1(d/k)γP

I bit-rate

log2

�1+

1N (d/k)γP

�I

k�

0@ 1

log�1+ Ptx

N (d/k )γ


log�1+ Ptx

N (d/k )γ

� (Prx )1A

I

k �

0@ 1

log�1+ Ptx

N (d/k )γ


1A

I

k �

0@ 1

log�1+ kγ Ptx

Ndγ


1AI or


B


1(d/k)γP

I bit-rate

log2

�1+

1N (d/k)γP

�I

k�

0@ 1

log�1+ Ptx

N (d/k )γ


log�1+ Ptx

N (d/k )γ

� (Prx )1A

I

k �

0@ 1

log�1+ Ptx

N (d/k )γ


1AI

k �

0@ 1

log�1+ kγ Ptx

Ndγ


1A

I orenergy = k � 1

log (1+ kγA)B


1(d/k)γP

I bit-rate

log2

�1+

1N (d/k)γP

�I

k�

0@ 1

log�1+ Ptx

N (d/k )γ


log�1+ Ptx

N (d/k )γ

� (Prx )1A

I

k �

0@ 1

log�1+ Ptx

N (d/k )γ


1AI

k �

0@ 1

log�1+ kγ Ptx

Ndγ


1AI or


B

I As k ! ∞, energy! ∞

I DerivativeI

ddkk� 1

log (1+ kγA)B

=B log (1+ kγA)� kγBγAk

(1+kγA)

log (1+ kγA)2

I Derivaitve at k = 1I

B log (1+ A)� BγA(1+A)

log (1+ A)2?< 0 then energy decreases with k

I

B log (1+ A)?<

BγA(1+ A)

log (1+ A)?< γ

A(1+ A)

I As k ! ∞, energy! ∞I Derivative

I

ddkk� 1

log (1+ kγA)B


(1+kγA)

log (1+ kγA)2




I

B log (1+ A)?<

BγA(1+ A)

log (1+ A)?< γ

A(1+ A)

I As k ! ∞, energy! ∞I DerivativeI

ddkk� 1

log (1+ kγA)B


(1+kγA)

log (1+ kγA)2




I

B log (1+ A)?<

BγA(1+ A)

log (1+ A)?< γ

A(1+ A)


ddkk� 1

log (1+ kγA)B


(1+kγA)

log (1+ kγA)2

I Derivaitve at k = 1

I



I

B log (1+ A)?<

BγA(1+ A)

log (1+ A)?< γ

A(1+ A)


ddkk� 1

log (1+ kγA)B


(1+kγA)

log (1+ kγA)2




I

B log (1+ A)?<

BγA(1+ A)

log (1+ A)?< γ

A(1+ A)


ddkk� 1

log (1+ kγA)B


(1+kγA)

log (1+ kγA)2




I

B log (1+ A)?<

BγA(1+ A)

log (1+ A)?< γ

A(1+ A)

log (1+ A)?< γ

A(1+ A)

then energy decreases with k

I Conclusions

I For large γ, energy decreases with k.I Indpendent of BI Note

I 0 � A(1+A) � 1,

I log (1+ A) is the single hop bit-rate.I If the single hop bit-rate is low (i.e., less than one), then energydecreasing with k . (As our intuition suggests)

I The actual optimal value of k depends on the parameters(including B). Note that even if the energy is decreasing at k = 1(i.e., the deriviative is negaitve at k = 1), the optimal value couldstill be k = 1

log (1+ A)?< γ

A(1+ A)


I ConclusionsI For large γ, energy decreases with k.

I Indpendent of BI Note

I 0 � A(1+A) � 1,



log (1+ A)?< γ

A(1+ A)


I ConclusionsI For large γ, energy decreases with k.I Indpendent of B

I Note

I 0 � A(1+A) � 1,



log (1+ A)?< γ

A(1+ A)


I ConclusionsI For large γ, energy decreases with k.I Indpendent of BI Note

I 0 � A(1+A) � 1,



log (1+ A)?< γ

A(1+ A)



I 0 � A(1+A) � 1,



log (1+ A)?< γ

A(1+ A)



I 0 � A(1+A) � 1,

I log (1+ A) is the single hop bit-rate.

I If the single hop bit-rate is low (i.e., less than one), then energydecreasing with k . (As our intuition suggests)


log (1+ A)?< γ

A(1+ A)



I 0 � A(1+A) � 1,



log (1+ A)?< γ

A(1+ A)



I 0 � A(1+A) � 1,



Energy Optimization - �xed data sizeI suppose that the data to be sent over link (i , j) is Zi ,j

I e.g., sensors with node 5 data sink

13

24

5

1

1

2

2

I Z1,3 = 1, Z2,4 = 1, Z3,5 = 1 and Z4,5 = 1I Assume that each link can transmit at the same time.

I link (1, 3) and (2, 4) might be able to transmit at the same timeas long as the interference is not too high. Or, they could usedi¤erent channels

I link (3, 5) and (4, 5) can transmit at the same time if they useCDMA

I Suppose we want to minimze energy

Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+ pj ,iZi ,jrj ,i

�I

ri ,j = W log (1+Hi ,jpi ,j )

pi ,j =1Hi ,j

exp (ri ,j/W )�1Hi ,j

I

Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�I

minE

subject to : ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�< E for all j

Energy Optimization - �xed data sizeI suppose that the data to be sent over link (i , j) is Zi ,jI e.g., sensors with node 5 data sink

13

24

5

1

1

2

2





Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+ pj ,iZi ,jrj ,i

�I


pi ,j =1Hi ,j


I

Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�I

minE

subject to : ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�< E for all j


13

24

5

1

1

2

2

I Z1,3 = 1, Z2,4 = 1, Z3,5 = 1 and Z4,5 = 1

I Assume that each link can transmit at the same time.




Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+ pj ,iZi ,jrj ,i

�I


pi ,j =1Hi ,j


I

Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�I

minE

subject to : ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�< E for all j


13

24

5

1

1

2

2





Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+ pj ,iZi ,jrj ,i

�I


pi ,j =1Hi ,j


I

Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�I

minE

subject to : ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�< E for all j


13

24

5

1

1

2

2





Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+ pj ,iZi ,jrj ,i

�I


pi ,j =1Hi ,j


I

Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�I

minE

subject to : ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�< E for all j


13

24

5

1

1

2

2





Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+ pj ,iZi ,jrj ,i

�I


pi ,j =1Hi ,j


I

Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�I

minE

subject to : ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�< E for all j


13

24

5

1

1

2

2





Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+ pj ,iZi ,jrj ,i

�

I


pi ,j =1Hi ,j


I

Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�I

minE

subject to : ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�< E for all j


13

24

5

1

1

2

2





Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+ pj ,iZi ,jrj ,i

�I


pi ,j =1Hi ,j


I

Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�I

minE

subject to : ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�< E for all j


13

24

5

1

1

2

2





Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+ pj ,iZi ,jrj ,i

�I


pi ,j =1Hi ,j


I

Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�

I

minE

subject to : ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�< E for all j


13

24

5

1

1

2

2





Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+ pj ,iZi ,jrj ,i

�I


pi ,j =1Hi ,j


I

Ej = ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�I

minE

subject to : ∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�< E for all j

ConvexityI is

∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i

�convex? Why convex?

I Sum of convex functions is convex.I Check second derivativeI 1/r� > �r�2� > +r�3 > 0 so convexI

exp (r/W )r

ddr� > 1

rWerW � 1

r2erW

Id2

dr2� >

�2r3� 2r2W

+1rW 2

�erW�

2W 2 � 2rW + r2

r3W 2

�erW

(r �W )2 +Wr3W 2

!erW

and(r �W )2 +W > 0

hence convex!I HW: solve such a problem with Matlab

ConvexityI is

∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i


I Sum of convex functions is convex.

I Check second derivativeI 1/r� > �r�2� > +r�3 > 0 so convexI

exp (r/W )r

ddr� > 1

rWerW � 1

r2erW

Id2

dr2� >

�2r3� 2r2W

+1rW 2

�erW�

2W 2 � 2rW + r2

r3W 2

�erW

(r �W )2 +Wr3W 2

!erW

and(r �W )2 +W > 0


ConvexityI is

∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i


I Sum of convex functions is convex.I Check second derivative

I 1/r� > �r�2� > +r�3 > 0 so convexI

exp (r/W )r

ddr� > 1

rWerW � 1

r2erW

Id2

dr2� >

�2r3� 2r2W

+1rW 2

�erW�

2W 2 � 2rW + r2

r3W 2

�erW

(r �W )2 +Wr3W 2

!erW

and(r �W )2 +W > 0


ConvexityI is

∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i


I Sum of convex functions is convex.I Check second derivativeI 1/r� > �r�2� > +r�3 > 0 so convex

Iexp (r/W )

rddr� > 1

rWerW � 1

r2erW

Id2

dr2� >

�2r3� 2r2W

+1rW 2

�erW�

2W 2 � 2rW + r2

r3W 2

�erW

(r �W )2 +Wr3W 2

!erW

and(r �W )2 +W > 0


ConvexityI is

∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i



exp (r/W )r

ddr� > 1

rWerW � 1

r2erW

Id2

dr2� >

�2r3� 2r2W

+1rW 2

�erW�

2W 2 � 2rW + r2

r3W 2

�erW

(r �W )2 +Wr3W 2

!erW

and(r �W )2 +W > 0


ConvexityI is

∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i



exp (r/W )r

ddr� > 1

rWerW � 1

r2erW

Id2

dr2� >

�2r3� 2r2W

+1rW 2

�erW�

2W 2 � 2rW + r2

r3W 2

�erW

(r �W )2 +Wr3W 2

!erW

and(r �W )2 +W > 0

hence convex!

I HW: solve such a problem with Matlab

ConvexityI is

∑i

�βZi ,jri ,j

�+∑

i

�αZi ,jrj ,i

+

�1Hi ,j


�Zi ,jrj ,i



exp (r/W )r

ddr� > 1

rWerW � 1

r2erW

Id2

dr2� >

�2r3� 2r2W

+1rW 2

�erW�

2W 2 � 2rW + r2

r3W 2

�erW

(r �W )2 +Wr3W 2

!erW

and(r �W )2 +W > 0


Energy Optimization - with routing and single sinkI We can adjust the routing and the power

I First consider the routing problem (ignore energy)I Simple networking problem - max �ow

I route data from sources to a sinkI maximize the data, where each source sends the same amount ofdata

I Let Zi be the amount of data generated by source iI Let fi ,j be the data �ow from node i to node jI conservation of data: data coming in must be the same as datagoing out

∑j is neighbor of i

fi ,j = ∑j is neighbor of i

fj ,i + Zi

I except for the sink

Energy Optimization - with routing and single sinkI We can adjust the routing and the powerI First consider the routing problem (ignore energy)

I Simple networking problem - max �ow





fj ,i + Zi


Energy Optimization - with routing and single sinkI We can adjust the routing and the powerI First consider the routing problem (ignore energy)I Simple networking problem - max �ow





fj ,i + Zi



I route data from sources to a sink

I maximize the data, where each source sends the same amount ofdata




fj ,i + Zi







fj ,i + Zi




I Let Zi be the amount of data generated by source i

I Let fi ,j be the data �ow from node i to node jI conservation of data: data coming in must be the same as datagoing out



fj ,i + Zi




I Let Zi be the amount of data generated by source iI Let fi ,j be the data �ow from node i to node j

I conservation of data: data coming in must be the same as datagoing out



fj ,i + Zi







fj ,i + Zi







fj ,i + Zi


Put constriant into matrix formI Put constriant into matrix form

I e.g.,

13

24

5

I Make a vector of link �ows

f1,3 f2,4 f3,1 f3,4 f3,5 f4,2 f4,3 f4,5 f5,3 f5,4

I Let A 2 f0, 1,�1gnumber nodes�number links, An,J = 1 if fI goesfrom node n and An,I = �1 if fI goes into node n

II

A =

f1,3 f2,4 f3,1 f3,4 f3,5 f4,2 f4,3 f4,5 f5,3 f5,41 1 �12 1 �13 �1 1 1 1 �1 �14 �1 1 1 1 �15 �1 �1 1 1

Note that the columns sum to 0. Rows sum to zero if all links arebidirectional

I ∑j is neighbor of i fi ,j �∑j is neighbor of i fj ,i = ∑J Ai ,J fJI


fi ,j � ∑j is neighbor of i

fj ,i = Zi

Af = Z

where Z5 = Z1 + Z2 + Z3 + Z4

I

max z

Zi � z for all i 6= 5Z5 � 4z

Af = Z


I e.g.,

13

24

5


f1,3 f2,4 f3,1 f3,4 f3,5 f4,2 f4,3 f4,5 f5,3 f5,4


II

A =

f1,3 f2,4 f3,1 f3,4 f3,5 f4,2 f4,3 f4,5 f5,3 f5,41 1 �12 1 �13 �1 1 1 1 �1 �14 �1 1 1 1 �15 �1 �1 1 1





fj ,i = Zi

Af = Z

where Z5 = Z1 + Z2 + Z3 + Z4

I

max z


Af = Z


I e.g.,

13

24

5


f1,3 f2,4 f3,1 f3,4 f3,5 f4,2 f4,3 f4,5 f5,3 f5,4


II

A =

f1,3 f2,4 f3,1 f3,4 f3,5 f4,2 f4,3 f4,5 f5,3 f5,41 1 �12 1 �13 �1 1 1 1 �1 �14 �1 1 1 1 �15 �1 �1 1 1





fj ,i = Zi

Af = Z

where Z5 = Z1 + Z2 + Z3 + Z4

I

max z


Af = Z


I e.g.,

13

24

5


f1,3 f2,4 f3,1 f3,4 f3,5 f4,2 f4,3 f4,5 f5,3 f5,4


II

A =

f1,3 f2,4 f3,1 f3,4 f3,5 f4,2 f4,3 f4,5 f5,3 f5,41 1 �12 1 �13 �1 1 1 1 �1 �14 �1 1 1 1 �15 �1 �1 1 1





fj ,i = Zi

Af = Z

where Z5 = Z1 + Z2 + Z3 + Z4

I

max z


Af = Z


I e.g.,

13

24

5


f1,3 f2,4 f3,1 f3,4 f3,5 f4,2 f4,3 f4,5 f5,3 f5,4


II

A =

f1,3 f2,4 f3,1 f3,4 f3,5 f4,2 f4,3 f4,5 f5,3 f5,41 1 �12 1 �13 �1 1 1 1 �1 �14 �1 1 1 1 �15 �1 �1 1 1





fj ,i = Zi

Af = Z

where Z5 = Z1 + Z2 + Z3 + Z4

I

max z


Af = Z


I e.g.,

13

24

5


f1,3 f2,4 f3,1 f3,4 f3,5 f4,2 f4,3 f4,5 f5,3 f5,4


II

A =

f1,3 f2,4 f3,1 f3,4 f3,5 f4,2 f4,3 f4,5 f5,3 f5,41 1 �12 1 �13 �1 1 1 1 �1 �14 �1 1 1 1 �15 �1 �1 1 1





fj ,i = Zi

Af = Z

where Z5 = Z1 + Z2 + Z3 + Z4

I

max z


Af = Z


I e.g.,

13

24

5


f1,3 f2,4 f3,1 f3,4 f3,5 f4,2 f4,3 f4,5 f5,3 f5,4


II

A =

f1,3 f2,4 f3,1 f3,4 f3,5 f4,2 f4,3 f4,5 f5,3 f5,41 1 �12 1 �13 �1 1 1 1 �1 �14 �1 1 1 1 �15 �1 �1 1 1


I ∑j is neighbor of i fi ,j �∑j is neighbor of i fj ,i = ∑J Ai ,J fJ

I



fj ,i = Zi

Af = Z

where Z5 = Z1 + Z2 + Z3 + Z4

I

max z


Af = Z


I e.g.,

13

24

5


f1,3 f2,4 f3,1 f3,4 f3,5 f4,2 f4,3 f4,5 f5,3 f5,4


II

A =

f1,3 f2,4 f3,1 f3,4 f3,5 f4,2 f4,3 f4,5 f5,3 f5,41 1 �12 1 �13 �1 1 1 1 �1 �14 �1 1 1 1 �15 �1 �1 1 1





fj ,i = Zi

Af = Z

where Z5 = Z1 + Z2 + Z3 + Z4

I

max z


Af = Z


I e.g.,

13

24

5


f1,3 f2,4 f3,1 f3,4 f3,5 f4,2 f4,3 f4,5 f5,3 f5,4


II

A =

f1,3 f2,4 f3,1 f3,4 f3,5 f4,2 f4,3 f4,5 f5,3 f5,41 1 �12 1 �13 �1 1 1 1 �1 �14 �1 1 1 1 �15 �1 �1 1 1





fj ,i = Zi

Af = Z

where Z5 = Z1 + Z2 + Z3 + Z4

I

max z


Af = Z

Linear Programming can be solved E¢ ciently

I

x+y=C

x+2y=1

2x+y=1

Multi�ow versionI Now we consider multiple �ows, with each �ow passing fromsource k to sink S (k)

I For the �ow kAf k = uBk

where Bkk = 1 and BkS (k ) = �1

I Suppose we seek to maximize

max u

Af k = uBk for each k





max u






max u


Energy Optimization - with routing and single sinkI We can adjust the routing and the power

I minimize energy and maximize data. How to do both? minimizeenergy put constraint on data

I

minE

st

Ej < E

Af = uB

where u is given.I where, as before

Ej = ∑i

�βfi ,jri ,j

�+∑

i

�αfj ,irj ,i+ pj ,i

fj ,irj ,i

�I or

Ej = ∑i

�βfi ,jri ,j

�+∑

i

�αfj ,irj ,i+

�1Hi ,j


�fj ,irj ,i

�I Is this convex? We added the fi ,j but the second derivative withrespoect to fi ,j is zero� 0. So it is still convex

Energy Optimization - with routing and single sinkI We can adjust the routing and the powerI minimize energy and maximize data. How to do both? minimizeenergy put constraint on data

I

minE

st

Ej < E

Af = uB


Ej = ∑i

�βfi ,jri ,j

�+∑

i


fj ,irj ,i

�I or

Ej = ∑i

�βfi ,jri ,j

�+∑

i

�αfj ,irj ,i+

�1Hi ,j


�fj ,irj ,i



I

minE

st

Ej < E

Af = uB

where u is given.

I where, as before

Ej = ∑i

�βfi ,jri ,j

�+∑

i


fj ,irj ,i

�I or

Ej = ∑i

�βfi ,jri ,j

�+∑

i

�αfj ,irj ,i+

�1Hi ,j


�fj ,irj ,i



I

minE

st

Ej < E

Af = uB


Ej = ∑i

�βfi ,jri ,j

�+∑

i


fj ,irj ,i

�

I or

Ej = ∑i

�βfi ,jri ,j

�+∑

i

�αfj ,irj ,i+

�1Hi ,j


�fj ,irj ,i



I

minE

st

Ej < E

Af = uB


Ej = ∑i

�βfi ,jri ,j

�+∑

i


fj ,irj ,i

�I or

Ej = ∑i

�βfi ,jri ,j

�+∑

i

�αfj ,irj ,i+

�1Hi ,j


�fj ,irj ,i

�

I Is this convex? We added the fi ,j but the second derivative withrespoect to fi ,j is zero� 0. So it is still convex


I

minE

st

Ej < E

Af = uB


Ej = ∑i

�βfi ,jri ,j

�+∑

i


fj ,irj ,i

�I or

Ej = ∑i

�βfi ,jri ,j

�+∑

i

�αfj ,irj ,i+

�1Hi ,j


�fj ,irj ,i


What about radio start-up?I Suppose that for each �ow that a node transmits (or receives) thereceiver radio must start and stop

I Let ESS be the energy to start and stop the radioI Then we add ni � ESS to the energy usage of node i , where ni isthe number of �ows that node i receives or transmits.

I How can we count the number of �ows. The earlier problem wejust computed the total data passing through a node. We did notdi¤erentiate between the source of the data

I To di¤erentiate �ows, we need multi-commodity �ow problem.Let f ki ,j be the �ow over link (i , j) that originated at node k

I

Ei = ∑i

β

∑k fki ,j

ri ,j

!+∑

i

α

∑k fkj ,i

rj ,i+ ESS1ff kj ,i>0g + pj ,i

∑k fkj ,i

rj ,i

!+∑

j∑k

ESS1ff ki ,j>0g +∑j

∑k

ESS1ff kj ,i>0g

I But this is not convex. Plot ESS1ff ki ,j>0g vs f . This is an integerprogramming problem

I

Ei = ∑i

β

∑k fki ,j

ri ,j

!+∑

i

α

∑k fkj ,i


∑k fkj ,i

rj ,i

!+∑

j∑k

Nki ,j +∑j

∑k

Nkj ,i

f ki ,j � Nki ,j � 100000Nki ,j 2 f0, 1g


I Let ESS be the energy to start and stop the radio

I Then we add ni � ESS to the energy usage of node i , where ni isthe number of �ows that node i receives or transmits.



I

Ei = ∑i

β

∑k fki ,j

ri ,j

!+∑

i

α

∑k fkj ,i


∑k fkj ,i

rj ,i

!+∑

j∑k


∑k

ESS1ff kj ,i>0g


I

Ei = ∑i

β

∑k fki ,j

ri ,j

!+∑

i

α

∑k fkj ,i


∑k fkj ,i

rj ,i

!+∑

j∑k

Nki ,j +∑j

∑k

Nkj ,i

f ki ,j � Nki ,j � 100000Nki ,j 2 f0, 1g





I

Ei = ∑i

β

∑k fki ,j

ri ,j

!+∑

i

α

∑k fkj ,i


∑k fkj ,i

rj ,i

!+∑

j∑k


∑k

ESS1ff kj ,i>0g


I

Ei = ∑i

β

∑k fki ,j

ri ,j

!+∑

i

α

∑k fkj ,i


∑k fkj ,i

rj ,i

!+∑

j∑k

Nki ,j +∑j

∑k

Nkj ,i

f ki ,j � Nki ,j � 100000Nki ,j 2 f0, 1g





I

Ei = ∑i

β

∑k fki ,j

ri ,j

!+∑

i

α

∑k fkj ,i


∑k fkj ,i

rj ,i

!+∑

j∑k


∑k

ESS1ff kj ,i>0g


I

Ei = ∑i

β

∑k fki ,j

ri ,j

!+∑

i

α

∑k fkj ,i


∑k fkj ,i

rj ,i

!+∑

j∑k

Nki ,j +∑j

∑k

Nkj ,i

f ki ,j � Nki ,j � 100000Nki ,j 2 f0, 1g





I

Ei = ∑i

β

∑k fki ,j

ri ,j

!+∑

i

α

∑k fkj ,i


∑k fkj ,i

rj ,i

!+∑

j∑k


∑k

ESS1ff kj ,i>0g


I

Ei = ∑i

β

∑k fki ,j

ri ,j

!+∑

i

α

∑k fkj ,i


∑k fkj ,i

rj ,i

!+∑

j∑k

Nki ,j +∑j

∑k

Nkj ,i

f ki ,j � Nki ,j � 100000Nki ,j 2 f0, 1g





I

Ei = ∑i

β

∑k fki ,j

ri ,j

!+∑

i

α

∑k fkj ,i


∑k fkj ,i

rj ,i

!+∑

j∑k


∑k

ESS1ff kj ,i>0g


I

Ei = ∑i

β

∑k fki ,j

ri ,j

!+∑

i

α

∑k fkj ,i


∑k fkj ,i

rj ,i

!+∑

j∑k

Nki ,j +∑j

∑k

Nkj ,i

f ki ,j � Nki ,j � 100000Nki ,j 2 f0, 1g





I

Ei = ∑i

β

∑k fki ,j

ri ,j

!+∑

i

α

∑k fkj ,i


∑k fkj ,i

rj ,i

!+∑

j∑k


∑k

ESS1ff kj ,i>0g


I

Ei = ∑i

β

∑k fki ,j

ri ,j

!+∑

i

α

∑k fkj ,i


∑k fkj ,i

rj ,i

!+∑

j∑k

Nki ,j +∑j

∑k

Nkj ,i

f ki ,j � Nki ,j � 100000Nki ,j 2 f0, 1g





I

Ei = ∑i

β

∑k fki ,j

ri ,j

!+∑

i

α

∑k fkj ,i


∑k fkj ,i

rj ,i

!+∑

j∑k


∑k

ESS1ff kj ,i>0g


I

Ei = ∑i

β

∑k fki ,j

ri ,j

!+∑

i

α

∑k fkj ,i


∑k fkj ,i

rj ,i

!+∑

j∑k

Nki ,j +∑j

∑k

Nkj ,i

f ki ,j � Nki ,j � 100000Nki ,j 2 f0, 1g

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› ~bohacek › SensorAndData... · Bit-rate Selection over a Single Hop I Recall that Shanon capacity says bit-rate=log 2 (1+SNR)=log 2 1+ H P t N whereP t is the transmit power,H