Probability of long-run battery usage for cellphones
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There are two batteries in a cell phone, $B_1$ and $B_2$. Both are used simultaneously for avoiding battery shortage. The lifetimes of $B_1$ and $B_2$ are independent and exponentially distributed random variables with rates $lambda_1=3$ and $lambda_2=4$. When $B_1$ is empty it takes an exponential time to recharge it with rate $lambda_{R1}=5$ and if $B_2$ is empty it takes similar exponential time to recharge it with rate $lambda_{R2}=10$. The recharging time is independent between the batteries, as well as the discharge times.
Assume that both batteries are initially fully charged but just one of them is plugged in the cell phone.
The cell phone may begin with both batteries. For example, suppose cell phone begins with battery $B_1$, and $B_2$ is kept in reserve and is full. When $B_1$ finishes the owner team immediately begins to charge it, while $B_2$ immediately replaces $B_1$ to avoid shutting down the cell phone. Battery $B_2$ also immediately gets charged when it finishes.
If $B_1$ is charged by the time $B_2$ finishes, then $B_1$ replaces $B_2$ and the cell phone stops. Otherwise we say that the cell phone is dead and it remains dead until either one of the batteries is fixed.
What is the probability that $B_1$ is used in the cell phone in the long run?
This question is included in our lecture notes which i couldn't come up with an answer that I am sure. Can somebody please help me in finding the appropriate answer?
probability probability-distributions stochastic-processes
edited Dec 22 '18 at 0:00
alandella
339514
asked Dec 21 '18 at 23:35
Dennis96Dennis96
11
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add a comment |
$begingroup$
There are two batteries in a cell phone, $B_1$ and $B_2$. Both are used simultaneously for avoiding battery shortage. The lifetimes of $B_1$ and $B_2$ are independent and exponentially distributed random variables with rates $lambda_1=3$ and $lambda_2=4$. When $B_1$ is empty it takes an exponential time to recharge it with rate $lambda_{R1}=5$ and if $B_2$ is empty it takes similar exponential time to recharge it with rate $lambda_{R2}=10$. The recharging time is independent between the batteries, as well as the discharge times.
Assume that both batteries are initially fully charged but just one of them is plugged in the cell phone.
The cell phone may begin with both batteries. For example, suppose cell phone begins with battery $B_1$, and $B_2$ is kept in reserve and is full. When $B_1$ finishes the owner team immediately begins to charge it, while $B_2$ immediately replaces $B_1$ to avoid shutting down the cell phone. Battery $B_2$ also immediately gets charged when it finishes.
If $B_1$ is charged by the time $B_2$ finishes, then $B_1$ replaces $B_2$ and the cell phone stops. Otherwise we say that the cell phone is dead and it remains dead until either one of the batteries is fixed.
What is the probability that $B_1$ is used in the cell phone in the long run?
This question is included in our lecture notes which i couldn't come up with an answer that I am sure. Can somebody please help me in finding the appropriate answer?
probability probability-distributions stochastic-processes
edited Dec 22 '18 at 0:00
alandella
339514
asked Dec 21 '18 at 23:35
Dennis96Dennis96
11
$endgroup$
$begingroup$
By "the probability that $B_1$ is used in the cell phone" do you mean the probability that the first battery is depleted while the second battery is charging?
$endgroup$
– Math1000
Dec 22 '18 at 2:31
$begingroup$
Yes that means first battery is on duty. At the initial state the batteries are both accepted as full.
$endgroup$
– Dennis96
Dec 22 '18 at 5:55
$begingroup$
Some variation of this: Continuous time Markov Process. States 0,1 and 2 depending on whether both batteries are dead or which battery is in use. Set up $Q$-matrix of instantaneous transition rates (based on $lambda$s). Solve balance equations.// Seriously unrealistic situation because true cell-phone battery lives between changes and charging times are nowhere near exponential, but ignore that.
$endgroup$
– BruceET
Dec 22 '18 at 8:13
add a comment |
$begingroup$
There are two batteries in a cell phone, $B_1$ and $B_2$. Both are used simultaneously for avoiding battery shortage. The lifetimes of $B_1$ and $B_2$ are independent and exponentially distributed random variables with rates $lambda_1=3$ and $lambda_2=4$. When $B_1$ is empty it takes an exponential time to recharge it with rate $lambda_{R1}=5$ and if $B_2$ is empty it takes similar exponential time to recharge it with rate $lambda_{R2}=10$. The recharging time is independent between the batteries, as well as the discharge times.
Assume that both batteries are initially fully charged but just one of them is plugged in the cell phone.
The cell phone may begin with both batteries. For example, suppose cell phone begins with battery $B_1$, and $B_2$ is kept in reserve and is full. When $B_1$ finishes the owner team immediately begins to charge it, while $B_2$ immediately replaces $B_1$ to avoid shutting down the cell phone. Battery $B_2$ also immediately gets charged when it finishes.
If $B_1$ is charged by the time $B_2$ finishes, then $B_1$ replaces $B_2$ and the cell phone stops. Otherwise we say that the cell phone is dead and it remains dead until either one of the batteries is fixed.
What is the probability that $B_1$ is used in the cell phone in the long run?
This question is included in our lecture notes which i couldn't come up with an answer that I am sure. Can somebody please help me in finding the appropriate answer?
probability probability-distributions stochastic-processes
edited Dec 22 '18 at 0:00
alandella
339514
asked Dec 21 '18 at 23:35
Dennis96Dennis96
11
$endgroup$
There are two batteries in a cell phone, $B_1$ and $B_2$. Both are used simultaneously for avoiding battery shortage. The lifetimes of $B_1$ and $B_2$ are independent and exponentially distributed random variables with rates $lambda_1=3$ and $lambda_2=4$. When $B_1$ is empty it takes an exponential time to recharge it with rate $lambda_{R1}=5$ and if $B_2$ is empty it takes similar exponential time to recharge it with rate $lambda_{R2}=10$. The recharging time is independent between the batteries, as well as the discharge times.
Assume that both batteries are initially fully charged but just one of them is plugged in the cell phone.
The cell phone may begin with both batteries. For example, suppose cell phone begins with battery $B_1$, and $B_2$ is kept in reserve and is full. When $B_1$ finishes the owner team immediately begins to charge it, while $B_2$ immediately replaces $B_1$ to avoid shutting down the cell phone. Battery $B_2$ also immediately gets charged when it finishes.
If $B_1$ is charged by the time $B_2$ finishes, then $B_1$ replaces $B_2$ and the cell phone stops. Otherwise we say that the cell phone is dead and it remains dead until either one of the batteries is fixed.
What is the probability that $B_1$ is used in the cell phone in the long run?
This question is included in our lecture notes which i couldn't come up with an answer that I am sure. Can somebody please help me in finding the appropriate answer?
probability probability-distributions stochastic-processes
probability probability-distributions stochastic-processes
edited Dec 22 '18 at 0:00
alandella
339514
asked Dec 21 '18 at 23:35
Dennis96Dennis96
11
edited Dec 22 '18 at 0:00
alandella
339514
asked Dec 21 '18 at 23:35
Dennis96Dennis96
11
edited Dec 22 '18 at 0:00
alandella
339514
edited Dec 22 '18 at 0:00
alandella
339514
edited Dec 22 '18 at 0:00
alandella
339514
339514
asked Dec 21 '18 at 23:35
Dennis96Dennis96
11
asked Dec 21 '18 at 23:35
Dennis96Dennis96
11
asked Dec 21 '18 at 23:35
Dennis96Dennis96
11
11
$begingroup$
By "the probability that $B_1$ is used in the cell phone" do you mean the probability that the first battery is depleted while the second battery is charging?
$endgroup$
– Math1000
Dec 22 '18 at 2:31
$begingroup$
Yes that means first battery is on duty. At the initial state the batteries are both accepted as full.
$endgroup$
– Dennis96
Dec 22 '18 at 5:55
$begingroup$
Some variation of this: Continuous time Markov Process. States 0,1 and 2 depending on whether both batteries are dead or which battery is in use. Set up $Q$-matrix of instantaneous transition rates (based on $lambda$s). Solve balance equations.// Seriously unrealistic situation because true cell-phone battery lives between changes and charging times are nowhere near exponential, but ignore that.
$endgroup$
– BruceET
Dec 22 '18 at 8:13
add a comment |
$begingroup$
By "the probability that $B_1$ is used in the cell phone" do you mean the probability that the first battery is depleted while the second battery is charging?
$endgroup$
– Math1000
Dec 22 '18 at 2:31
$begingroup$
Yes that means first battery is on duty. At the initial state the batteries are both accepted as full.
$endgroup$
– Dennis96
Dec 22 '18 at 5:55
$begingroup$
Some variation of this: Continuous time Markov Process. States 0,1 and 2 depending on whether both batteries are dead or which battery is in use. Set up $Q$-matrix of instantaneous transition rates (based on $lambda$s). Solve balance equations.// Seriously unrealistic situation because true cell-phone battery lives between changes and charging times are nowhere near exponential, but ignore that.
$endgroup$
– BruceET
Dec 22 '18 at 8:13
$begingroup$
By "the probability that $B_1$ is used in the cell phone" do you mean the probability that the first battery is depleted while the second battery is charging?
$endgroup$
– Math1000
Dec 22 '18 at 2:31
$begingroup$
By "the probability that $B_1$ is used in the cell phone" do you mean the probability that the first battery is depleted while the second battery is charging?
$endgroup$
– Math1000
Dec 22 '18 at 2:31
$begingroup$
Yes that means first battery is on duty. At the initial state the batteries are both accepted as full.
$endgroup$
– Dennis96
Dec 22 '18 at 5:55
$begingroup$
Yes that means first battery is on duty. At the initial state the batteries are both accepted as full.
$endgroup$
– Dennis96
Dec 22 '18 at 5:55
$begingroup$
Some variation of this: Continuous time Markov Process. States 0,1 and 2 depending on whether both batteries are dead or which battery is in use. Set up $Q$-matrix of instantaneous transition rates (based on $lambda$s). Solve balance equations.// Seriously unrealistic situation because true cell-phone battery lives between changes and charging times are nowhere near exponential, but ignore that.
$endgroup$
– BruceET
Dec 22 '18 at 8:13
$begingroup$
Some variation of this: Continuous time Markov Process. States 0,1 and 2 depending on whether both batteries are dead or which battery is in use. Set up $Q$-matrix of instantaneous transition rates (based on $lambda$s). Solve balance equations.// Seriously unrealistic situation because true cell-phone battery lives between changes and charging times are nowhere near exponential, but ignore that.
$endgroup$
– BruceET
Dec 22 '18 at 8:13
add a comment |
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$begingroup$
By "the probability that $B_1$ is used in the cell phone" do you mean the probability that the first battery is depleted while the second battery is charging?
$endgroup$
– Math1000
Dec 22 '18 at 2:31
$begingroup$
Yes that means first battery is on duty. At the initial state the batteries are both accepted as full.
$endgroup$
– Dennis96
Dec 22 '18 at 5:55
$begingroup$
Some variation of this: Continuous time Markov Process. States 0,1 and 2 depending on whether both batteries are dead or which battery is in use. Set up $Q$-matrix of instantaneous transition rates (based on $lambda$s). Solve balance equations.// Seriously unrealistic situation because true cell-phone battery lives between changes and charging times are nowhere near exponential, but ignore that.
$endgroup$
– BruceET
Dec 22 '18 at 8:13