bohacek/SensorAndDataWirelessNets_2010/En… · Bit-rate Selection over a Single Hop I Recall that...

97
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 where P t is the transmit power, H is the channel, and N is the noise.

Transcript of bohacek/SensorAndDataWirelessNets_2010/En… · Bit-rate Selection over a Single Hop I Recall that...

  • 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