decaying fractions and make them closer
$begingroup$
My problem is actually algorithmic. I have two request queues, each holding its average wait time. For each queue I am holding x = total_wait_time, y = number_of_requests.
Let a = x/y the average of queue one
and b = w/z the average of queue two
Every second I would like to decay the queue so old request will have less effect on the average wait time. One simple way to achieve this will be:
Let $0 < C < 1$
$x' = C * x$
$y' = C * y$
this will keep the same average (C is reduced), but in the same time when a new update happens it will have more affect because it was not multiplied by 0.9 yet.
In addition I would like the averages to go towards each other. This is done in case no requests will enter one of the queues for a long time. To achieve this I can do the following every second:
a' = $frac{a + b} {2}$
Two achive both I thought about doing 90% of the first and 10% of the second:
$a' = frac{x'}{y'} = 0.9 * frac{C*x}{C*y} + 0.1 * frac{a + b}{2} = frac{...}{2*c*y}$
To reduce the history effect I need y' < y. this only happens when C < 0.5, but I do not want to rid the history so rapidly. What am I missing? Why doesn't it act as I expect? How can I fix the formula?
I hope it was understandable and thank you
algebra-precalculus algorithms fractions
asked Dec 11 '18 at 9:24
EpicEpic
11
$endgroup$
add a comment |
$begingroup$
My problem is actually algorithmic. I have two request queues, each holding its average wait time. For each queue I am holding x = total_wait_time, y = number_of_requests.
Let a = x/y the average of queue one
and b = w/z the average of queue two
Every second I would like to decay the queue so old request will have less effect on the average wait time. One simple way to achieve this will be:
Let $0 < C < 1$
$x' = C * x$
$y' = C * y$
this will keep the same average (C is reduced), but in the same time when a new update happens it will have more affect because it was not multiplied by 0.9 yet.
In addition I would like the averages to go towards each other. This is done in case no requests will enter one of the queues for a long time. To achieve this I can do the following every second:
a' = $frac{a + b} {2}$
Two achive both I thought about doing 90% of the first and 10% of the second:
$a' = frac{x'}{y'} = 0.9 * frac{C*x}{C*y} + 0.1 * frac{a + b}{2} = frac{...}{2*c*y}$
To reduce the history effect I need y' < y. this only happens when C < 0.5, but I do not want to rid the history so rapidly. What am I missing? Why doesn't it act as I expect? How can I fix the formula?
I hope it was understandable and thank you
algebra-precalculus algorithms fractions
asked Dec 11 '18 at 9:24
EpicEpic
11
$endgroup$
add a comment |
$begingroup$
My problem is actually algorithmic. I have two request queues, each holding its average wait time. For each queue I am holding x = total_wait_time, y = number_of_requests.
Let a = x/y the average of queue one
and b = w/z the average of queue two
Every second I would like to decay the queue so old request will have less effect on the average wait time. One simple way to achieve this will be:
Let $0 < C < 1$
$x' = C * x$
$y' = C * y$
this will keep the same average (C is reduced), but in the same time when a new update happens it will have more affect because it was not multiplied by 0.9 yet.
In addition I would like the averages to go towards each other. This is done in case no requests will enter one of the queues for a long time. To achieve this I can do the following every second:
a' = $frac{a + b} {2}$
Two achive both I thought about doing 90% of the first and 10% of the second:
$a' = frac{x'}{y'} = 0.9 * frac{C*x}{C*y} + 0.1 * frac{a + b}{2} = frac{...}{2*c*y}$
To reduce the history effect I need y' < y. this only happens when C < 0.5, but I do not want to rid the history so rapidly. What am I missing? Why doesn't it act as I expect? How can I fix the formula?
I hope it was understandable and thank you
algebra-precalculus algorithms fractions
asked Dec 11 '18 at 9:24
EpicEpic
11
$endgroup$
My problem is actually algorithmic. I have two request queues, each holding its average wait time. For each queue I am holding x = total_wait_time, y = number_of_requests.
Let a = x/y the average of queue one
and b = w/z the average of queue two
Every second I would like to decay the queue so old request will have less effect on the average wait time. One simple way to achieve this will be:
Let $0 < C < 1$
$x' = C * x$
$y' = C * y$
this will keep the same average (C is reduced), but in the same time when a new update happens it will have more affect because it was not multiplied by 0.9 yet.
In addition I would like the averages to go towards each other. This is done in case no requests will enter one of the queues for a long time. To achieve this I can do the following every second:
a' = $frac{a + b} {2}$
Two achive both I thought about doing 90% of the first and 10% of the second:
$a' = frac{x'}{y'} = 0.9 * frac{C*x}{C*y} + 0.1 * frac{a + b}{2} = frac{...}{2*c*y}$
To reduce the history effect I need y' < y. this only happens when C < 0.5, but I do not want to rid the history so rapidly. What am I missing? Why doesn't it act as I expect? How can I fix the formula?
I hope it was understandable and thank you
algebra-precalculus algorithms fractions
algebra-precalculus algorithms fractions
asked Dec 11 '18 at 9:24
EpicEpic
11
asked Dec 11 '18 at 9:24
EpicEpic
11
asked Dec 11 '18 at 9:24
EpicEpic
11
asked Dec 11 '18 at 9:24
EpicEpic
11
asked Dec 11 '18 at 9:24
EpicEpic
11
11
add a comment |
add a comment |
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