Remote estimation over control area networks Aditya Mahajan McGill - - PowerPoint PPT Presentation
Remote estimation over control area networks Aditya Mahajan McGill University NetV Workshop, VTC 2017 24 Sep 2017 The road to self-driving cars . . . Car can handle dynamic driving tasks but still need human intervention Fully
Control either speed or steering Cruise control Automatic breaking
Control both speed or steering Automatic lane control (Tesla’s autopilot)
Car can handle “dynamic driving tasks” but still need human intervention . . .
Fully autonomous in certain situations . . .
Fully autonomous in all situations . . .
Control either speed or steering Cruise control Automatic breaking
Control both speed or steering Automatic lane control (Tesla’s autopilot)
Car can handle “dynamic driving tasks” but still need human intervention . . .
Fully autonomous in certain situations . . .
Fully autonomous in all situations . . .
Communication between the sensors, controllers, and actuators takes place over the Control Area Network (CAN) As the number of sensors increase, it is critical to ensure that the information exchange is effjcient.
Remote estimation over CAN–(Mahajan)
4
Remote estimation over CAN–(Mahajan)
4
Remote estimation over CAN–(Mahajan)
4
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload.
Remote estimation over CAN–(Mahajan)
4
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload.
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload. When the CAN bus is idle, all nodes start transmitting at the same time. Bitwise transmission can be dominant (high voltage) or recessive (low voltage) If any node transmits at a dominant level, the voltage of the bus is high.
Remote estimation over CAN–(Mahajan)
4
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload.
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload. When the CAN bus is idle, all nodes start transmitting at the same time. Bitwise transmission can be dominant (high voltage) or recessive (low voltage) If any node transmits at a dominant level, the voltage of the bus is high.
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload. When the CAN bus is idle, all nodes start transmitting at the same time. Bitwise transmission can be dominant (high voltage) or recessive (low voltage) If any node transmits at a dominant level, the voltage of the bus is high. Nodes monitor the voltage on the bus. If a node transmitting at a recessive level detects a dominant voltage on the bus, it immediately quits transmitting.
Remote estimation over CAN–(Mahajan)
4
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload.
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload. When the CAN bus is idle, all nodes start transmitting at the same time. Bitwise transmission can be dominant (high voltage) or recessive (low voltage) If any node transmits at a dominant level, the voltage of the bus is high.
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload. When the CAN bus is idle, all nodes start transmitting at the same time. Bitwise transmission can be dominant (high voltage) or recessive (low voltage) If any node transmits at a dominant level, the voltage of the bus is high. Nodes monitor the voltage on the bus. If a node transmitting at a recessive level detects a dominant voltage on the bus, it immediately quits transmitting.
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload. When the CAN bus is idle, all nodes start transmitting at the same time. Bitwise transmission can be dominant (high voltage) or recessive (low voltage) If any node transmits at a dominant level, the voltage of the bus is high. Nodes monitor the voltage on the bus. If a node transmitting at a recessive level detects a dominant voltage on the bus, it immediately quits transmitting.
Remote estimation over CAN–(Mahajan)
4
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload.
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload. When the CAN bus is idle, all nodes start transmitting at the same time. Bitwise transmission can be dominant (high voltage) or recessive (low voltage) If any node transmits at a dominant level, the voltage of the bus is high.
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload. When the CAN bus is idle, all nodes start transmitting at the same time. Bitwise transmission can be dominant (high voltage) or recessive (low voltage) If any node transmits at a dominant level, the voltage of the bus is high. Nodes monitor the voltage on the bus. If a node transmitting at a recessive level detects a dominant voltage on the bus, it immediately quits transmitting.
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload. When the CAN bus is idle, all nodes start transmitting at the same time. Bitwise transmission can be dominant (high voltage) or recessive (low voltage) If any node transmits at a dominant level, the voltage of the bus is high. Nodes monitor the voltage on the bus. If a node transmitting at a recessive level detects a dominant voltage on the bus, it immediately quits transmitting.
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload. When the CAN bus is idle, all nodes start transmitting at the same time. Bitwise transmission can be dominant (high voltage) or recessive (low voltage) If any node transmits at a dominant level, the voltage of the bus is high. Nodes monitor the voltage on the bus. If a node transmitting at a recessive level detects a dominant voltage on the bus, it immediately quits transmitting.
Suppose a sensor does not get access to the channel.
Remote estimation over CAN–(Mahajan)
4
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload.
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload. When the CAN bus is idle, all nodes start transmitting at the same time. Bitwise transmission can be dominant (high voltage) or recessive (low voltage) If any node transmits at a dominant level, the voltage of the bus is high.
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload. When the CAN bus is idle, all nodes start transmitting at the same time. Bitwise transmission can be dominant (high voltage) or recessive (low voltage) If any node transmits at a dominant level, the voltage of the bus is high. Nodes monitor the voltage on the bus. If a node transmitting at a recessive level detects a dominant voltage on the bus, it immediately quits transmitting.
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload. When the CAN bus is idle, all nodes start transmitting at the same time. Bitwise transmission can be dominant (high voltage) or recessive (low voltage) If any node transmits at a dominant level, the voltage of the bus is high. Nodes monitor the voltage on the bus. If a node transmitting at a recessive level detects a dominant voltage on the bus, it immediately quits transmitting.
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload. When the CAN bus is idle, all nodes start transmitting at the same time. Bitwise transmission can be dominant (high voltage) or recessive (low voltage) If any node transmits at a dominant level, the voltage of the bus is high. Nodes monitor the voltage on the bus. If a node transmitting at a recessive level detects a dominant voltage on the bus, it immediately quits transmitting.
Suppose a sensor does not get access to the channel.
Each data-frame consists of a 11 or 29 bit arbitrartion fjeld and payload. When the CAN bus is idle, all nodes start transmitting at the same time. Bitwise transmission can be dominant (high voltage) or recessive (low voltage) If any node transmits at a dominant level, the voltage of the bus is high. Nodes monitor the voltage on the bus. If a node transmitting at a recessive level detects a dominant voltage on the bus, it immediately quits transmitting.
Suppose a sensor does not get access to the channel. Then, it should simply discard the previous measurement rather than bufgering it. Transmit fresh measurement at the next transmission instant.
Remote estimation over CAN–(Mahajan)
6
Remote estimation over CAN–(Mahajan)
6
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
Remote estimation over CAN–(Mahajan)
6
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
Remote estimation over CAN–(Mahajan)
6
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Received packet Yi
t =
{ XI
t,
if sensor i has highest priority 𝔉,
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
Remote estimation over CAN–(Mahajan)
6
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Received packet Yi
t =
{ XI
t,
if sensor i has highest priority 𝔉,
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Received packet Yi
t =
{ XI
t,
if sensor i has highest priority 𝔉,
Estimate ˆ Xi
t =
{ Yi
t,
if Yi
t ≠ 𝔉
ai ˆ Xi
t−1,
if Yi
t = 𝔉
Distortion di(Xi
t − ˆ
Xi
t), where di(⋅) is an even and increasing function.
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
Remote estimation over CAN–(Mahajan)
6
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Received packet Yi
t =
{ XI
t,
if sensor i has highest priority 𝔉,
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Received packet Yi
t =
{ XI
t,
if sensor i has highest priority 𝔉,
Estimate ˆ Xi
t =
{ Yi
t,
if Yi
t ≠ 𝔉
ai ˆ Xi
t−1,
if Yi
t = 𝔉
Distortion di(Xi
t − ˆ
Xi
t), where di(⋅) is an even and increasing function.
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
St: Sensor with highest priority. All sensors observe St. Priority Assignment rule gi
t∶ (Xi 1:t, S1:t−1) ↦ Zi t.
St = arg max
i∈N Zi t
Remote estimation over CAN–(Mahajan)
6
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Received packet Yi
t =
{ XI
t,
if sensor i has highest priority 𝔉,
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Received packet Yi
t =
{ XI
t,
if sensor i has highest priority 𝔉,
Estimate ˆ Xi
t =
{ Yi
t,
if Yi
t ≠ 𝔉
ai ˆ Xi
t−1,
if Yi
t = 𝔉
Distortion di(Xi
t − ˆ
Xi
t), where di(⋅) is an even and increasing function.
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
St: Sensor with highest priority. All sensors observe St. Priority Assignment rule gi
t∶ (Xi 1:t, S1:t−1) ↦ Zi t.
St = arg max
i∈N Zi t
St: Sensor with highest priority. All sensors observe St. Priority Assignment rule gi
t∶ (Xi 1:t, S1:t−1) ↦ Zi t.
St = arg max
i∈N Zi t
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0
∑
i∈N i≠St
di(Xi
t − ˆ
Xi
t)
]
Remote estimation over CAN–(Mahajan)
6
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Received packet Yi
t =
{ XI
t,
if sensor i has highest priority 𝔉,
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Received packet Yi
t =
{ XI
t,
if sensor i has highest priority 𝔉,
Estimate ˆ Xi
t =
{ Yi
t,
if Yi
t ≠ 𝔉
ai ˆ Xi
t−1,
if Yi
t = 𝔉
Distortion di(Xi
t − ˆ
Xi
t), where di(⋅) is an even and increasing function.
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
St: Sensor with highest priority. All sensors observe St. Priority Assignment rule gi
t∶ (Xi 1:t, S1:t−1) ↦ Zi t.
St = arg max
i∈N Zi t
St: Sensor with highest priority. All sensors observe St. Priority Assignment rule gi
t∶ (Xi 1:t, S1:t−1) ↦ Zi t.
St = arg max
i∈N Zi t
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0
∑
i∈N i≠St
di(Xi
t − ˆ
Xi
t)
]
St: Sensor with highest priority. All sensors observe St. Priority Assignment rule gi
t∶ (Xi 1:t, S1:t−1) ↦ Zi t.
St = arg max
i∈N Zi t
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0
∑
i∈N i≠St
di(Xi
t − ˆ
Xi
t)
]
Decentralized stochastic control problem. Finding optimal solution is notoriously diffjcult. Use a heuristic policy instead. Motivated by value of information in economics.
Remote estimation over CAN–(Mahajan)
9
Remote estimation over CAN–(Mahajan)
9
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i
Remote estimation over CAN–(Mahajan)
9
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i Called the error process because when St ≠ i, Xt − ˆ Xi
t = Ei
error can be written as
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0
∑
i∈N i≠St
di(Ei
t)
]
Remote estimation over CAN–(Mahajan)
9
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i Called the error process because when St ≠ i, Xt − ˆ Xi
t = Ei
error can be written as
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0
∑
i∈N i≠St
di(Ei
t)
]
The amount of money someone is willing to pay to access that information.
Remote estimation over CAN–(Mahajan)
9
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i Called the error process because when St ≠ i, Xt − ˆ Xi
t = Ei
error can be written as
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0
∑
i∈N i≠St
di(Ei
t)
]
The amount of money someone is willing to pay to access that information.
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel.
Remote estimation over CAN–(Mahajan)
9
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i Called the error process because when St ≠ i, Xt − ˆ Xi
t = Ei
error can be written as
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0
∑
i∈N i≠St
di(Ei
t)
]
The amount of money someone is willing to pay to access that information.
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel.
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
Remote estimation over CAN–(Mahajan)
9
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i Called the error process because when St ≠ i, Xt − ˆ Xi
t = Ei
error can be written as
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0
∑
i∈N i≠St
di(Ei
t)
]
The amount of money someone is willing to pay to access that information.
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel.
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0 [λiUi + (1 − Ui)di(Ei t)]]
Remote estimation over CAN–(Mahajan)
9
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i Called the error process because when St ≠ i, Xt − ˆ Xi
t = Ei
error can be written as
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0
∑
i∈N i≠St
di(Ei
t)
]
The amount of money someone is willing to pay to access that information.
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel.
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0 [λiUi + (1 − Ui)di(Ei t)]]
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming.
Remote estimation over CAN–(Mahajan)
9
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i Called the error process because when St ≠ i, Xt − ˆ Xi
t = Ei
error can be written as
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0
∑
i∈N i≠St
di(Ei
t)
]
The amount of money someone is willing to pay to access that information.
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel.
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0 [λiUi + (1 − Ui)di(Ei t)]]
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming.
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming. Let hi ∈ ℝ and vi∶ ℝ → ℝ satisfy the following dynamic program: for any e ∈ ℝ hi + vi(e) = min {λi + ∫
ℝ
φi(w)vi(w)dw, di(e) + ∫
ℝ
φi(w)vi(ae + w)dw}
Remote estimation over CAN–(Mahajan)
9
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i Called the error process because when St ≠ i, Xt − ˆ Xi
t = Ei
error can be written as
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0
∑
i∈N i≠St
di(Ei
t)
]
The amount of money someone is willing to pay to access that information.
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel.
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0 [λiUi + (1 − Ui)di(Ei t)]]
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming.
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming. Let hi ∈ ℝ and vi∶ ℝ → ℝ satisfy the following dynamic program: for any e ∈ ℝ hi + vi(e) = min {λi + ∫
ℝ
φi(w)vi(w)dw, di(e) + ∫
ℝ
φi(w)vi(ae + w)dw}
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming. Let hi ∈ ℝ and vi∶ ℝ → ℝ satisfy the following dynamic program: for any e ∈ ℝ hi + vi(e) = min {λi + ∫
ℝ
φi(w)vi(w)dw, di(e) + ∫
ℝ
φi(w)vi(ae + w)dw} Let fi
∗(e) = 0 if the fjrst term is smaller and fi ∗(e) = 1 if the second term is smaller.
Then, fi
∗(e) is the optimal action at state e.
Remote estimation over CAN–(Mahajan)
9
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i Called the error process because when St ≠ i, Xt − ˆ Xi
t = Ei
error can be written as
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0
∑
i∈N i≠St
di(Ei
t)
]
The amount of money someone is willing to pay to access that information.
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel.
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0 [λiUi + (1 − Ui)di(Ei t)]]
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming.
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming. Let hi ∈ ℝ and vi∶ ℝ → ℝ satisfy the following dynamic program: for any e ∈ ℝ hi + vi(e) = min {λi + ∫
ℝ
φi(w)vi(w)dw, di(e) + ∫
ℝ
φi(w)vi(ae + w)dw}
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming. Let hi ∈ ℝ and vi∶ ℝ → ℝ satisfy the following dynamic program: for any e ∈ ℝ hi + vi(e) = min {λi + ∫
ℝ
φi(w)vi(w)dw, di(e) + ∫
ℝ
φi(w)vi(ae + w)dw} Let fi
∗(e) = 0 if the fjrst term is smaller and fi ∗(e) = 1 if the second term is smaller.
Then, fi
∗(e) is the optimal action at state e.
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming. Let hi ∈ ℝ and vi∶ ℝ → ℝ satisfy the following dynamic program: for any e ∈ ℝ hi + vi(e) = min {λi + ∫
ℝ
φi(w)vi(w)dw, di(e) + ∫
ℝ
φi(w)vi(ae + w)dw} Let fi
∗(e) = 0 if the fjrst term is smaller and fi ∗(e) = 1 if the second term is smaller.
Then, fi
∗(e) is the optimal action at state e.
There exists a threshold ki(λi) such that the optimal policy is of the form fi
∗(e) =
{ 1, if |e| < ki(λi) 0,
Moreover, at k∗(λ∗), λi + ∫
ℝ
φi(w)vi(w)dw = di(e) + ∫
ℝ
φi(w)vi(ae + w)dw
Proof relies on stochastic monot-
submodularity.
Remote estimation over CAN–(Mahajan)
9
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i Called the error process because when St ≠ i, Xt − ˆ Xi
t = Ei
error can be written as
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0
∑
i∈N i≠St
di(Ei
t)
]
The amount of money someone is willing to pay to access that information.
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel.
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0 [λiUi + (1 − Ui)di(Ei t)]]
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming.
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming. Let hi ∈ ℝ and vi∶ ℝ → ℝ satisfy the following dynamic program: for any e ∈ ℝ hi + vi(e) = min {λi + ∫
ℝ
φi(w)vi(w)dw, di(e) + ∫
ℝ
φi(w)vi(ae + w)dw}
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming. Let hi ∈ ℝ and vi∶ ℝ → ℝ satisfy the following dynamic program: for any e ∈ ℝ hi + vi(e) = min {λi + ∫
ℝ
φi(w)vi(w)dw, di(e) + ∫
ℝ
φi(w)vi(ae + w)dw} Let fi
∗(e) = 0 if the fjrst term is smaller and fi ∗(e) = 1 if the second term is smaller.
Then, fi
∗(e) is the optimal action at state e.
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming. Let hi ∈ ℝ and vi∶ ℝ → ℝ satisfy the following dynamic program: for any e ∈ ℝ hi + vi(e) = min {λi + ∫
ℝ
φi(w)vi(w)dw, di(e) + ∫
ℝ
φi(w)vi(ae + w)dw} Let fi
∗(e) = 0 if the fjrst term is smaller and fi ∗(e) = 1 if the second term is smaller.
Then, fi
∗(e) is the optimal action at state e.
There exists a threshold ki(λi) such that the optimal policy is of the form fi
∗(e) =
{ 1, if |e| < ki(λi) 0,
Moreover, at k∗(λ∗), λi + ∫
ℝ
φi(w)vi(w)dw = di(e) + ∫
ℝ
φi(w)vi(ae + w)dw
Proof relies on stochastic monot-
submodularity.
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming. Let hi ∈ ℝ and vi∶ ℝ → ℝ satisfy the following dynamic program: for any e ∈ ℝ hi + vi(e) = min {λi + ∫
ℝ
φi(w)vi(w)dw, di(e) + ∫
ℝ
φi(w)vi(ae + w)dw} Let fi
∗(e) = 0 if the fjrst term is smaller and fi ∗(e) = 1 if the second term is smaller.
Then, fi
∗(e) is the optimal action at state e.
There exists a threshold ki(λi) such that the optimal policy is of the form fi
∗(e) =
{ 1, if |e| < ki(λi) 0,
Moreover, at k∗(λ∗), λi + ∫
ℝ
φi(w)vi(w)dw = di(e) + ∫
ℝ
φi(w)vi(ae + w)dw
Proof relies on stochastic monot-
submodularity.
Remote estimation over CAN–(Mahajan)
9
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i
Defjne Ei
0 = Xi 0 and for t ≥ 0,
Ei
t+1 =
{ Wi
t,
if St = i aiEi
t + Wi t
if St ≠ i Called the error process because when St ≠ i, Xt − ˆ Xi
t = Ei
error can be written as
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0
∑
i∈N i≠St
di(Ei
t)
]
The amount of money someone is willing to pay to access that information.
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel.
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0 [λiUi + (1 − Ui)di(Ei t)]]
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming.
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming. Let hi ∈ ℝ and vi∶ ℝ → ℝ satisfy the following dynamic program: for any e ∈ ℝ hi + vi(e) = min {λi + ∫
ℝ
φi(w)vi(w)dw, di(e) + ∫
ℝ
φi(w)vi(ae + w)dw}
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming. Let hi ∈ ℝ and vi∶ ℝ → ℝ satisfy the following dynamic program: for any e ∈ ℝ hi + vi(e) = min {λi + ∫
ℝ
φi(w)vi(w)dw, di(e) + ∫
ℝ
φi(w)vi(ae + w)dw} Let fi
∗(e) = 0 if the fjrst term is smaller and fi ∗(e) = 1 if the second term is smaller.
Then, fi
∗(e) is the optimal action at state e.
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming. Let hi ∈ ℝ and vi∶ ℝ → ℝ satisfy the following dynamic program: for any e ∈ ℝ hi + vi(e) = min {λi + ∫
ℝ
φi(w)vi(w)dw, di(e) + ∫
ℝ
φi(w)vi(ae + w)dw} Let fi
∗(e) = 0 if the fjrst term is smaller and fi ∗(e) = 1 if the second term is smaller.
Then, fi
∗(e) is the optimal action at state e.
There exists a threshold ki(λi) such that the optimal policy is of the form fi
∗(e) =
{ 1, if |e| < ki(λi) 0,
Moreover, at k∗(λ∗), λi + ∫
ℝ
φi(w)vi(w)dw = di(e) + ∫
ℝ
φi(w)vi(ae + w)dw
Proof relies on stochastic monot-
submodularity.
The objective is a single agent multi-stage optimization problem. Optimal solution is given by dynamic programming. Let hi ∈ ℝ and vi∶ ℝ → ℝ satisfy the following dynamic program: for any e ∈ ℝ hi + vi(e) = min {λi + ∫
ℝ
φi(w)vi(w)dw, di(e) + ∫
ℝ
φi(w)vi(ae + w)dw} Let fi
∗(e) = 0 if the fjrst term is smaller and fi ∗(e) = 1 if the second term is smaller.
Then, fi
∗(e) is the optimal action at state e.
There exists a threshold ki(λi) such that the optimal policy is of the form fi
∗(e) =
{ 1, if |e| < ki(λi) 0,
Moreover, at k∗(λ∗), λi + ∫
ℝ
φi(w)vi(w)dw = di(e) + ∫
ℝ
φi(w)vi(ae + w)dw
Proof relies on stochastic monot-
submodularity.
Remote estimation over CAN–(Mahajan)
13
Remote estimation over CAN–(Mahajan)
13
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Remote estimation over CAN–(Mahajan)
13
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
Remote estimation over CAN–(Mahajan)
13
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
Then, Ji(fi
k) = Di k + λiNi
k + λi∂kNi k = 0.
Therefore, VOIi(k) = −∂kDi
k
∂kNi
k
Remote estimation over CAN–(Mahajan)
13
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
Then, Ji(fi
k) = Di k + λiNi
k + λi∂kNi k = 0.
Therefore, VOIi(k) = −∂kDi
k
∂kNi
k
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Remote estimation over CAN–(Mahajan)
13
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
Then, Ji(fi
k) = Di k + λiNi
k + λi∂kNi k = 0.
Therefore, VOIi(k) = −∂kDi
k
∂kNi
k
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Remote estimation over CAN–(Mahajan)
13
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
Then, Ji(fi
k) = Di k + λiNi
k + λi∂kNi k = 0.
Therefore, VOIi(k) = −∂kDi
k
∂kNi
k
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Remote estimation over CAN–(Mahajan)
13
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
Then, Ji(fi
k) = Di k + λiNi
k + λi∂kNi k = 0.
Therefore, VOIi(k) = −∂kDi
k
∂kNi
k
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Need to compute Mi
k(0), Li k(0), ∂kMi k(0), ∂kLi k(0) to compute VOI.
Remote estimation over CAN–(Mahajan)
13
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
Then, Ji(fi
k) = Di k + λiNi
k + λi∂kNi k = 0.
Therefore, VOIi(k) = −∂kDi
k
∂kNi
k
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Need to compute Mi
k(0), Li k(0), ∂kMi k(0), ∂kLi k(0) to compute VOI.
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Remote estimation over CAN–(Mahajan)
13
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
Then, Ji(fi
k) = Di k + λiNi
k + λi∂kNi k = 0.
Therefore, VOIi(k) = −∂kDi
k
∂kNi
k
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Need to compute Mi
k(0), Li k(0), ∂kMi k(0), ∂kLi k(0) to compute VOI.
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Remote estimation over CAN–(Mahajan)
13
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
Then, Ji(fi
k) = Di k + λiNi
k + λi∂kNi k = 0.
Therefore, VOIi(k) = −∂kDi
k
∂kNi
k
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Need to compute Mi
k(0), Li k(0), ∂kMi k(0), ∂kLi k(0) to compute VOI.
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Or, in matrix form, 𝐌i = (𝐉 − 𝚾i)−1𝐞i
Remote estimation over CAN–(Mahajan)
13
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
Then, Ji(fi
k) = Di k + λiNi
k + λi∂kNi k = 0.
Therefore, VOIi(k) = −∂kDi
k
∂kNi
k
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Need to compute Mi
k(0), Li k(0), ∂kMi k(0), ∂kLi k(0) to compute VOI.
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Or, in matrix form, 𝐌i = (𝐉 − 𝚾i)−1𝐞i
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Or, in matrix form, 𝐌i = (𝐉 − 𝚾i)−1𝐞i and 𝐍i = (𝐉 − 𝚾i)−1𝟐
Remote estimation over CAN–(Mahajan)
13
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
Then, Ji(fi
k) = Di k + λiNi
k + λi∂kNi k = 0.
Therefore, VOIi(k) = −∂kDi
k
∂kNi
k
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Need to compute Mi
k(0), Li k(0), ∂kMi k(0), ∂kLi k(0) to compute VOI.
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Or, in matrix form, 𝐌i = (𝐉 − 𝚾i)−1𝐞i
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Or, in matrix form, 𝐌i = (𝐉 − 𝚾i)−1𝐞i and 𝐍i = (𝐉 − 𝚾i)−1𝟐
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
Remote estimation over CAN–(Mahajan)
13
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
Then, Ji(fi
k) = Di k + λiNi
k + λi∂kNi k = 0.
Therefore, VOIi(k) = −∂kDi
k
∂kNi
k
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Need to compute Mi
k(0), Li k(0), ∂kMi k(0), ∂kLi k(0) to compute VOI.
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Or, in matrix form, 𝐌i = (𝐉 − 𝚾i)−1𝐞i
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Or, in matrix form, 𝐌i = (𝐉 − 𝚾i)−1𝐞i and 𝐍i = (𝐉 − 𝚾i)−1𝟐
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
Using Leibniz rule ∂kLi
k(x) = φi(x − ak)Li k(x) + φi(x + ak)Li k(−k) + ∫ k −k
φi(x − ay)Li
k(y)dy
∂kMi
k(x) = φi(x − ak)Mi k(x) + φi(x + ak)Mi k(−k) + ∫ k −k
φi(x − ay)Mi
k(y)dy
Remote estimation over CAN–(Mahajan)
13
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
Then, Ji(fi
k) = Di k + λiNi
k + λi∂kNi k = 0.
Therefore, VOIi(k) = −∂kDi
k
∂kNi
k
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Need to compute Mi
k(0), Li k(0), ∂kMi k(0), ∂kLi k(0) to compute VOI.
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Or, in matrix form, 𝐌i = (𝐉 − 𝚾i)−1𝐞i
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Or, in matrix form, 𝐌i = (𝐉 − 𝚾i)−1𝐞i and 𝐍i = (𝐉 − 𝚾i)−1𝟐
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
Using Leibniz rule ∂kLi
k(x) = φi(x − ak)Li k(x) + φi(x + ak)Li k(−k) + ∫ k −k
φi(x − ay)Li
k(y)dy
∂kMi
k(x) = φi(x − ak)Mi k(x) + φi(x + ak)Mi k(−k) + ∫ k −k
φi(x − ay)Mi
k(y)dy
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
Using Leibniz rule ∂kLi
k(x) = φi(x − ak)Li k(x) + φi(x + ak)Li k(−k) + ∫ k −k
φi(x − ay)Li
k(y)dy
∂kMi
k(x) = φi(x − ak)Mi k(x) + φi(x + ak)Mi k(−k) + ∫ k −k
φi(x − ay)Mi
k(y)dy
Taking ratios, we get ∂kLi
k(x)
∂kMi
k(x) = Li k(k)
Mi
k(k)
Remote estimation over CAN–(Mahajan)
13
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
Then, Ji(fi
k) = Di k + λiNi
k + λi∂kNi k = 0.
Therefore, VOIi(k) = −∂kDi
k
∂kNi
k
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Need to compute Mi
k(0), Li k(0), ∂kMi k(0), ∂kLi k(0) to compute VOI.
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Or, in matrix form, 𝐌i = (𝐉 − 𝚾i)−1𝐞i
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Or, in matrix form, 𝐌i = (𝐉 − 𝚾i)−1𝐞i and 𝐍i = (𝐉 − 𝚾i)−1𝟐
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
Using Leibniz rule ∂kLi
k(x) = φi(x − ak)Li k(x) + φi(x + ak)Li k(−k) + ∫ k −k
φi(x − ay)Li
k(y)dy
∂kMi
k(x) = φi(x − ak)Mi k(x) + φi(x + ak)Mi k(−k) + ∫ k −k
φi(x − ay)Mi
k(y)dy
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
Using Leibniz rule ∂kLi
k(x) = φi(x − ak)Li k(x) + φi(x + ak)Li k(−k) + ∫ k −k
φi(x − ay)Li
k(y)dy
∂kMi
k(x) = φi(x − ak)Mi k(x) + φi(x + ak)Mi k(−k) + ∫ k −k
φi(x − ay)Mi
k(y)dy
Taking ratios, we get ∂kLi
k(x)
∂kMi
k(x) = Li k(k)
Mi
k(k)
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
Using Leibniz rule ∂kLi
k(x) = φi(x − ak)Li k(x) + φi(x + ak)Li k(−k) + ∫ k −k
φi(x − ay)Li
k(y)dy
∂kMi
k(x) = φi(x − ak)Mi k(x) + φi(x + ak)Mi k(−k) + ∫ k −k
φi(x − ay)Mi
k(y)dy
Taking ratios, we get ∂kLi
k(x)
∂kMi
k(x) = Li k(k)
Mi
k(k)
VOIi(k) = 𝐍i 𝐌i
m
𝐍i
m
− 𝐌i
0,
where 𝐌i = (𝐉 − 𝚾i)−1𝐞i and 𝐍i = (𝐉 − 𝚾i)−1𝟐 .
Remote estimation over CAN–(Mahajan)
13
n sensors, each observing a Gauss-Markov process. Scenario A 50 homogeneous sensors with (ai, σi) = (1, 1). Scenario B 25 sensors with (ai, σi) = (1, 1) and 25 sensors with (ai, σi) = (1, 5). Scenario C 20 sensors (ai, σi) = (1, 1); 15 with (1, 5); 15 with (1, 10).
Remote estimation over CAN–(Mahajan)
13
n sensors, each observing a Gauss-Markov process. Scenario A 50 homogeneous sensors with (ai, σi) = (1, 1). Scenario B 25 sensors with (ai, σi) = (1, 1) and 25 sensors with (ai, σi) = (1, 5). Scenario C 20 sensors (ai, σi) = (1, 1); 15 with (1, 5); 15 with (1, 10).
TDMA Sensors transmit periodically ERR Sensor with highest error transmits VOI Sensor with the highest VOI transmits
Remote estimation over CAN–(Mahajan)
13
n sensors, each observing a Gauss-Markov process. Scenario A 50 homogeneous sensors with (ai, σi) = (1, 1). Scenario B 25 sensors with (ai, σi) = (1, 1) and 25 sensors with (ai, σi) = (1, 5). Scenario C 20 sensors (ai, σi) = (1, 1); 15 with (1, 5); 15 with (1, 10).
TDMA Sensors transmit periodically ERR Sensor with highest error transmits VOI Sensor with the highest VOI transmits ERR VOI 2 4 6 8 ERR VOI 20 40 60 80 100 ERR VOI 100 200 Scenario A Scenario B Scenario C
Remote estimation over CAN–(Mahajan)
14
Remote estimation over CAN–(Mahajan)
14
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Received packet Yi
t =
{ XI
t,
if sensor i has highest priority 𝔉,
Estimate ˆ Xi
t =
{ Yi
t,
if Yi
t ≠ 𝔉
ai ˆ Xi
t−1,
if Yi
t = 𝔉
Distortion di(Xi
t − ˆ
Xi
t), where di(⋅) is an even and increasing function.
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
Remote estimation over CAN–(Mahajan)
14
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Received packet Yi
t =
{ XI
t,
if sensor i has highest priority 𝔉,
Estimate ˆ Xi
t =
{ Yi
t,
if Yi
t ≠ 𝔉
ai ˆ Xi
t−1,
if Yi
t = 𝔉
Distortion di(Xi
t − ˆ
Xi
t), where di(⋅) is an even and increasing function.
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0 [λiUi + (1 − Ui)di(Ei t)]]
Remote estimation over CAN–(Mahajan)
14
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Received packet Yi
t =
{ XI
t,
if sensor i has highest priority 𝔉,
Estimate ˆ Xi
t =
{ Yi
t,
if Yi
t ≠ 𝔉
ai ˆ Xi
t−1,
if Yi
t = 𝔉
Distortion di(Xi
t − ˆ
Xi
t), where di(⋅) is an even and increasing function.
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0 [λiUi + (1 − Ui)di(Ei t)]]
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
Then, Ji(fi
k) = Di k + λiNi
k + λi∂kNi k = 0.
Therefore, VOIi(k) = −∂kDi
k
∂kNi
k
Remote estimation over CAN–(Mahajan)
14
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Received packet Yi
t =
{ XI
t,
if sensor i has highest priority 𝔉,
Estimate ˆ Xi
t =
{ Yi
t,
if Yi
t ≠ 𝔉
ai ˆ Xi
t−1,
if Yi
t = 𝔉
Distortion di(Xi
t − ˆ
Xi
t), where di(⋅) is an even and increasing function.
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0 [λiUi + (1 − Ui)di(Ei t)]]
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
Then, Ji(fi
k) = Di k + λiNi
k + λi∂kNi k = 0.
Therefore, VOIi(k) = −∂kDi
k
∂kNi
k
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Need to compute Mi
k(0), Li k(0), ∂kMi k(0), ∂kLi k(0) to compute VOI.
Remote estimation over CAN–(Mahajan)
14
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Received packet Yi
t =
{ XI
t,
if sensor i has highest priority 𝔉,
Estimate ˆ Xi
t =
{ Yi
t,
if Yi
t ≠ 𝔉
ai ˆ Xi
t−1,
if Yi
t = 𝔉
Distortion di(Xi
t − ˆ
Xi
t), where di(⋅) is an even and increasing function.
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0 [λiUi + (1 − Ui)di(Ei t)]]
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
Then, Ji(fi
k) = Di k + λiNi
k + λi∂kNi k = 0.
Therefore, VOIi(k) = −∂kDi
k
∂kNi
k
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Need to compute Mi
k(0), Li k(0), ∂kMi k(0), ∂kLi k(0) to compute VOI.
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Or, in matrix form, 𝐌i = (𝐉 − 𝚾i)−1𝐞i and 𝐍i = (𝐉 − 𝚾i)−1𝟐
Remote estimation over CAN–(Mahajan)
14
n sensors indexed by N = {1, . . . , n}. Xi
t+1 = aiXi t + Wi t,
ai, Xi
t, Wi t ∈ ℝ,
Wi
t ∼ φi(⋅).
The observation processes across sensors are independent. The noise process is independent across time (and independent of initial state) The density φi(⋅) is even and unimodal.
Received packet Yi
t =
{ XI
t,
if sensor i has highest priority 𝔉,
Estimate ˆ Xi
t =
{ Yi
t,
if Yi
t ≠ 𝔉
ai ˆ Xi
t−1,
if Yi
t = 𝔉
Distortion di(Xi
t − ˆ
Xi
t), where di(⋅) is an even and increasing function.
Control Area Network Control Area Network ⋮ ⋮ Receiver 1 Receiver n Sensor 1 Sensor n (Z1
t, X1 t)
(Zn
t , Xn t )
Y1
t
Yn
t
ˆ X1
t
ˆ Xn
t
X1
t
Xn
t
The amount of money someone is willing to pay to access that information.
Suppose that there is a single sensor, say i, and a dedicated communication channel is available. The sensor has to pay an access fee λi each time it uses the channel. Let Ui
t ∈ {0, 1} denote the sensor’s decision.
Then, the error process is Ei
t+1 =
{ Wi
t,
if Ui
t = 1
aiEi
t + Wi t
if Ui
t = 0
min lim
T→∞
1 T 𝔽 [
T−1
∑
t=0 [λiUi + (1 − Ui)di(Ei t)]]
For a given λi, fjnd ki(λi) by numerically solving the dynamic program. VOIi(e) can be computed by doing a binary search of λi until ki(λi) = |e|. This method is extremely ineffjcient because solving DP is hard.
Let fi
k denote the threshold policy with threshold k.
Defjne Di
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
(1 − Ui
t)di(Ei t)
] and Ni
k = lim T→∞
1 T 𝔽 [
T−1
∑
t=0
Ui
t]
Then, Ji(fi
k) = Di k + λiNi
k + λi∂kNi k = 0.
Therefore, VOIi(k) = −∂kDi
k
∂kNi
k
Let τ denote the stopping time of the fjrst transmission. Defjne Li
k(x) = 𝔽
[
τ−1
∑
t=0
d(Ei
t)
| Ei
0 = x
] and Mi
k(x) = 𝔽 [τ | Ei 0 = x].
Then, from renewal theory: Di
k = Li k(0)
Mi
k(0) and Ni k =
1 Mi
k(0) .
Therefore, VOIi(k) = Mi
k(0) ∂kLi k(0)
∂kMi
k(0) − Li k(0)
Need to compute Mi
k(0), Li k(0), ∂kMi k(0), ∂kLi k(0) to compute VOI.
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
(Fredholm integral eqn of the 2nd kind)
Let {w−m, . . . , wm} and {x−m, . . . , xm} be the weights and abscissas for any quadrature rule of 2m + 1 points over [−k, k]. Then the above integral equation can be approximated as Li
k(xp) ≈ di(xp) + m
∑
q=−m
wqφi(xp − axq)Li
k(xq)
Or, in matrix form, 𝐌i = (𝐉 − 𝚾i)−1𝐞i and 𝐍i = (𝐉 − 𝚾i)−1𝟐
Li
k(x) = di k(x) + ∫ k −k
φi(y − ax)Li
k(y)dy
Using Leibniz rule ∂kLi
k(x) = φi(x − ak)Li k(x) + φi(x + ak)Li k(−k) + ∫ k −k
φi(x − ay)Li
k(y)dy
∂kMi
k(x) = φi(x − ak)Mi k(x) + φi(x + ak)Mi k(−k) + ∫ k −k
φi(x − ay)Mi
k(y)dy
Taking ratios, we get ∂kLi
k(x)
∂kMi
k(x) = Li k(k)
Mi
k(k)
VOIi(k) = 𝐍i 𝐌i
m
𝐍i
m
− 𝐌i
0,
where 𝐌i = (𝐉 − 𝚾i)−1𝐞i and 𝐍i = (𝐉 − 𝚾i)−1𝟐 .