bohacek/sensoranddatawirelessnets_2010/en… · bit-rate selection over a single hop i recall that...
TRANSCRIPT
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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�PtN
� � �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
-
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�PtN
� � �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
-
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 isI
Energy =packet size
log2�1+ H�PtN
� � �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
-
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 isI
Energy =packet size
log2�1+ H�PtN
� � �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
-
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 isI
Energy =packet size
log2�1+ H�PtN
� � �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 � 0
I 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
-
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 isI
Energy =packet size
log2�1+ H�PtN
� � �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
-
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 isI
Energy =packet size
log2�1+ H�PtN
� � �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
-
Bit-rate Selection over a Single Hop (2)I Take the derivative and set to zero
ddP
z (B + CP)log2 (1+ AP)
=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 functionsof 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)
-
Bit-rate Selection over a Single Hop (2)I Take the derivative and set to zero
ddP
z (B + CP)log2 (1+ AP)
=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 functionsof 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)
-
Bit-rate Selection over a Single Hop (2)I Take the derivative and set to zero
ddP
z (B + CP)log2 (1+ AP)
=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 functionsof 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)
-
Bit-rate Selection over a Single Hop (2)I Take the derivative and set to zero
ddP
z (B + CP)log2 (1+ AP)
=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 functionsof 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)
-
Bit-rate Selection over a Single Hop (2)I Take the derivative and set to zero
ddP
z (B + CP)log2 (1+ AP)
=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 functionsof 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� decreases
I as A increases, P� decreases (better channel results in lowertransmit power)
-
Bit-rate Selection over a Single Hop (2)I Take the derivative and set to zero
ddP
z (B + CP)log2 (1+ AP)
=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 functionsof 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)
-
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
-
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 energy
I 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
-
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-up
I 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
-
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
-
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
-
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
-
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 before
I
transmit energy + recieve energy + transmit energy + receive energy +...
Iktransmit energy + kreceive energy
I
ksizerate
((Psys + α+ βPtx )) + ksizerate
Prx
-
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
-
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
-
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 Distance between hops = d/k
I Received signal strength
1(d/k)γ
P
I bit-rate
log2
�1+
1N (d/k)γ
P�
I
k�
0@ 1log�1+ Ptx
N (d/k )γ
� (Psys + α+ βPtx ) + 1log�1+ Ptx
N (d/k )γ
� (Prx )1A
I
k �
0@ 1log�1+ Ptx
N (d/k )γ
� (Psys + α+ Prx + βPtx )1A
I
k �
0@ 1log�1+ kγ PtxNdγ
� (Psys + α+ Prx + βPtx )1A
I orenergy = k � 1
log (1+ kγA)B
-
I Distance between hops = d/kI Received signal strength
1(d/k)γ
P
I bit-rate
log2
�1+
1N (d/k)γ
P�
I
k�
0@ 1log�1+ Ptx
N (d/k )γ
� (Psys + α+ βPtx ) + 1log�1+ Ptx
N (d/k )γ
� (Prx )1A
I
k �
0@ 1log�1+ Ptx
N (d/k )γ
� (Psys + α+ Prx + βPtx )1A
I
k �
0@ 1log�1+ kγ PtxNdγ
� (Psys + α+ Prx + βPtx )1A
I orenergy = k � 1
log (1+ kγA)B
-
I Distance between hops = d/kI Received signal strength
1(d/k)γ
P
I bit-rate
log2
�1+
1N (d/k)γ
P�
I
k�
0@ 1log�1+ Ptx
N (d/k )γ
� (Psys + α+ βPtx ) + 1log�1+ Ptx
N (d/k )γ
� (Prx )1A
I
k �
0@ 1log�1+ Ptx
N (d/k )γ
� (Psys + α+ Prx + βPtx )1A
I
k �
0@ 1log�1+ kγ PtxNdγ
� (Psys + α+ Prx + βPtx )1A
I orenergy = k � 1
log (1+ kγA)B
-
I Distance between hops = d/kI Received signal strength
1(d/k)γ
P
I bit-rate
log2
�1+
1N (d/k)γ
P�
I
k�
0@ 1log�1+ Ptx
N (d/k )γ
� (Psys + α+ βPtx ) + 1log�1+ Ptx
N (d/k )γ
� (Prx )1A
I
k �
0@ 1log�1+ Ptx
N (d/k )γ
� (Psys + α+ Prx + βPtx )1A
I
k �
0@ 1log�1+ kγ PtxNdγ
� (Psys + α+ Prx + βPtx )1A
I orenergy = k � 1
log (1+ kγA)B
-
I Distance between hops = d/kI Received signal strength
1(d/k)γ
P
I bit-rate
log2
�1+
1N (d/k)γ
P�
I
k�
0@ 1log�1+ Ptx
N (d/k )γ
� (Psys + α+ βPtx ) + 1log�1+ Ptx
N (d/k )γ
� (Prx )1A
I
k �
0@ 1log�1+ Ptx
N (d/k )γ
� (Psys + α+ Prx + βPtx )1A
I
k �
0@ 1log�1+ kγ PtxNdγ
� (Psys + α+ Prx + βPtx )1A
I orenergy = k � 1
log (1+ kγA)B
-
I Distance between hops = d/kI Received signal strength
1(d/k)γ
P
I bit-rate
log2
�1+
1N (d/k)γ
P�
I
k�
0@ 1log�1+ Ptx
N (d/k )γ
� (Psys + α+ βPtx ) + 1log�1+ Ptx
N (d/k )γ
� (Prx )1A
I
k �
0@ 1log�1+ Ptx
N (d/k )γ
� (Psys + α+ Prx + βPtx )1A
I
k �
0@ 1log�1+ kγ PtxNdγ
� (Psys + α+ Prx + βPtx )1A
I orenergy = k � 1
log (1+ kγA)B
-
I Distance between hops = d/kI Received signal strength
1(d/k)γ
P
I bit-rate
log2
�1+
1N (d/k)γ
P�
I
k�
0@ 1log�1+ Ptx
N (d/k )γ
� (Psys + α+ βPtx ) + 1log�1+ Ptx
N (d/k )γ
� (Prx )1A
I
k �
0@ 1log�1+ Ptx
N (d/k )γ
� (Psys + α+ Prx + βPtx )1A
I
k �
0@ 1log�1+ kγ PtxNdγ
� (Psys + α+ Prx + βPtx )1A
I orenergy = k � 1
log (1+ kγA)B
-
I As k ! ∞, energy! ∞
I DerivativeI
ddkk� 1
log (1+ kγA)B
=B log (1+ kγA)� k
γBγAk(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
=B log (1+ kγA)� k
γBγAk(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 DerivativeI
ddkk� 1
log (1+ kγA)B
=B log (1+ kγA)� k
γBγAk(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 DerivativeI
ddkk� 1
log (1+ kγA)B
=B log (1+ kγA)� k
γBγAk(1+kγA)
log (1+ kγA)2
I Derivaitve at k = 1
I
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 DerivativeI
ddkk� 1
log (1+ kγA)B
=B log (1+ kγA)� k
γBγAk(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 DerivativeI
ddkk� 1
log (1+ kγA)B
=B log (1+ kγA)� k
γBγAk(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)
-
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)
then energy decreases with k
I ConclusionsI 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)
then energy decreases with k
I ConclusionsI For large γ, energy decreases with k.I Indpendent of B
I 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)
then energy decreases with k
I ConclusionsI 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)
then energy decreases with k
I ConclusionsI 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)
then energy decreases with k
I ConclusionsI 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)
then energy decreases with k
I ConclusionsI 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)
then energy decreases with k
I ConclusionsI 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
-
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
exp (ri ,j/W )�1Hi ,j
�Zi ,jrj ,i
�I
minE
subject to : ∑i
�βZi ,jri ,j
�+∑
i
�αZi ,jrj ,i
+
�1Hi ,j
exp (ri ,j/W )�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
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
exp (ri ,j/W )�1Hi ,j
�Zi ,jrj ,i
�I
minE
subject to : ∑i
�βZi ,jri ,j
�+∑
i
�αZi ,jrj ,i
+
�1Hi ,j
exp (ri ,j/W )�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
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.
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
exp (ri ,j/W )�1Hi ,j
�Zi ,jrj ,i
�I
minE
subject to : ∑i
�βZi ,jri ,j
�+∑
i
�αZi ,jrj ,i
+
�1Hi ,j
exp (ri ,j/W )�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
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
exp (ri ,j/W )�1Hi ,j
�Zi ,jrj ,i
�I
minE
subject to : ∑i
�βZi ,jri ,j
�+∑
i
�αZi ,jrj ,i
+
�1Hi ,j
exp (ri ,j/W )�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
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
exp (ri ,j/W )�1Hi ,j
�Zi ,jrj ,i
�I
minE
subject to : ∑i
�βZi ,jri ,j
�+∑
i
�αZi ,jrj ,i
+
�1Hi ,j
exp (ri ,j/W )�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
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
exp (ri ,j/W )�1Hi ,j
�Zi ,jrj ,i
�I
minE
subject to : ∑i
�βZi ,jri ,j
�+∑
i
�αZi ,jrj ,i
+
�1Hi ,j
exp (ri ,j/W )�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
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
exp (ri ,j/W )�1Hi ,j
�Zi ,jrj ,i
�I
minE
subject to : ∑i
�βZi ,jri ,j
�+∑
i
�αZi ,jrj ,i
+
�1Hi ,j
exp (ri ,j/W )�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
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
exp (ri ,j/W )�1Hi ,j
�Zi ,jrj ,i
�I
minE
subject to : ∑i
�βZi ,jri ,j
�+∑
i
�αZi ,jrj ,i
+
�1Hi ,j
exp (ri ,j/W )�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
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
exp (ri ,j/W )�1Hi ,j
�Zi ,jrj ,i
�
I
minE
subject to : ∑i
�βZi ,jri ,j
�+∑
i
�αZi ,jrj ,i
+
�1Hi ,j
exp (ri ,j/W )�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
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
exp (ri ,j/W )�1Hi ,j
�Zi ,jrj ,i
�I
minE
subject to : ∑i
�βZi ,jri ,j
�+∑
i
�αZi ,jrj ,i
+
�1Hi ,j
exp (ri ,j/W )�1Hi ,j
�Zi ,jrj ,i
�< E for all j
-
ConvexityI is
∑i
�βZi ,jri ,j
�+∑
i
�αZi ,jrj ,i
+
�1Hi ,j
exp (ri ,j/W )�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 + r2r3W 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
exp (ri ,j/W )�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 + r2r3W 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
exp (ri ,j/W )�1Hi ,j
�Zi ,jrj ,i
�convex? Why convex?
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 + r2r3W 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
exp (ri ,j/W )�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 convex
Iexp (r/W )
rddr� > 1
rWerW � 1
r2erW
Id2
dr2� >
�2r3� 2r2W
+1rW 2
�erW�
2W 2 � 2rW + r2r3W 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
exp (ri ,j/W )�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 + r2r3W 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
exp (ri ,j/W )�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 + r2r3W 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
exp (ri ,j/W )�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 + r2r3W 2
�erW
(r �W )2 +Wr3W 2
!erW
and(r �W )2 +W > 0
hence convex!I HW: solve such a problem with Matlab
-
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
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
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
I route data from sources to a sink
I 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
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
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 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
∑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
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 j
I 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
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
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
-
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
∑j is neighbor of i
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 � 4zAf = Z
-
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
∑j is neighbor of i
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 � 4zAf = Z
-
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
∑j is neighbor of i
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 � 4zAf = Z
-
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
∑j is neighbor of i
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 � 4zAf = Z
-
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
∑j is neighbor of i
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 � 4zAf = Z
-
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
∑j is neighbor of i
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 � 4zAf = Z
-
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 fJ
I
∑j is neighbor of i
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 � 4zAf = Z
-
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
∑j is neighbor of i
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 � 4zAf = Z
-
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
∑j is neighbor of i
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 � 4zAf = Z
-
Linear Programming can be solved E¢ ciently
I
x+y=C
x+2y=1
2x+y=1
-
Multiow 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
-
Multiow 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
-
Multiow 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
-
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
exp (ri ,j/W )�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
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
exp (ri ,j/W )�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
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
exp (ri ,j/W )�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
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
exp (ri ,j/W )�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
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
exp (ri ,j/W )�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
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
exp (ri ,j/W )�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
-
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 f ki ,jri ,j
!+∑
i
α
∑k f kj ,irj ,i
+ ESS1ff kj ,i>0g + pj ,i∑k f kj ,irj ,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 f ki ,jri ,j
!+∑
i
α
∑k f kj ,irj ,i
+ ESS1ff kj ,i>0g + pj ,i∑k f kj ,irj ,i
!+∑
j∑k
Nki ,j +∑j
∑k
Nkj ,i
f ki ,j � Nki ,j � 100000Nki ,j 2 f0, 1g
-
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 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 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 f ki ,jri ,j
!+∑
i
α
∑k f kj ,irj ,i
+ ESS1ff kj ,i>0g + pj ,i∑k f kj ,irj ,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 f ki ,jri ,j
!+∑
i
α
∑k f kj ,irj ,i
+ ESS1ff kj ,i>0g + pj ,i∑k f kj ,irj ,i
!+∑
j∑k
Nki ,j +∑j
∑k
Nkj ,i
f ki ,j � Nki ,j � 100000Nki ,j 2 f0, 1g
-
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 f ki ,jri ,j
!+∑
i
α
∑k f kj ,irj ,i
+ ESS1ff kj ,i>0g + pj ,i∑k f kj ,irj ,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 f ki ,jri ,j
!+∑
i
α
∑k f kj ,irj ,i
+ ESS1ff kj ,i>0g + pj ,i∑k f kj ,irj ,i
!+∑
j∑k
Nki ,j +∑j
∑k
Nkj ,i
f ki ,j � Nki ,j � 100000Nki ,j 2 f0, 1g
-
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 f ki ,jri ,j
!+∑
i
α
∑k f kj ,irj ,i
+ ESS1ff kj ,i>0g + pj ,i∑k f kj ,irj ,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 f ki ,jri ,j
!+∑
i
α
∑k f kj ,irj ,i
+ ESS1ff kj ,i>0g + pj ,i∑k f kj ,irj ,i
!+∑
j∑k
Nki ,j +∑j
∑k
Nkj ,i
f ki ,j � Nki ,j � 100000Nki ,j 2 f0, 1g
-
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 f ki ,jri ,j
!+∑
i
α
∑k f kj ,irj ,i
+ ESS1ff kj ,i>0g + pj ,i∑k f kj ,irj ,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 f ki ,jri ,j
!+∑
i
α
∑k f kj ,irj ,i
+ ESS1ff kj ,i>0g + pj ,i∑k f kj ,irj ,i
!+∑
j∑k
Nki ,j +∑j
∑k
Nkj ,i
f ki ,j � Nki ,j � 100000Nki ,j 2 f0, 1g
-
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 f ki ,jri ,j
!+∑
i
α
∑k f kj ,irj ,i
+ ESS1ff kj ,i>0g + pj ,i∑k f kj ,irj ,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 f ki ,jri ,j
!+∑
i
α
∑k f kj ,irj ,i
+ ESS1ff kj ,i>0g + pj ,i∑k f kj ,irj ,i
!+∑
j∑k
Nki ,j +∑j
∑k
Nkj ,i
f ki ,j � Nki ,j � 100000Nki ,j 2 f0, 1g
-
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 f ki ,jri ,j
!+∑
i
α
∑k f kj ,irj ,i
+ ESS1ff kj ,i>0g + pj ,i∑k f kj ,irj ,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 f ki ,jri ,j
!+∑
i
α
∑k f kj ,irj ,i
+ ESS1ff kj ,i>0g + pj ,i∑k f kj ,irj ,i
!+∑
j∑k
Nki ,j +∑j
∑k
Nkj ,i
f ki ,j � Nki ,j � 100000Nki ,j 2 f0, 1g
-
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 f ki ,jri ,j
!+∑
i
α
∑k f kj ,irj ,i
+ ESS1ff kj ,i>0g + pj ,i∑k f kj ,irj ,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 f ki ,jri ,j
!+∑
i
α
∑k f kj ,irj ,i
+ ESS1ff kj ,i>0g + pj ,i∑k f kj ,irj ,i
!+∑
j∑k
Nki ,j +∑j
∑k
Nkj ,i
f ki ,j � Nki ,j � 100000Nki ,j 2 f0, 1g