How is the accuracy calculated in Cristian's method for synchronizing clocks in a distributed system?
In Cristian algorithm (https://en.wikipedia.org/wiki/Cristian%27s_algorithm) for clock synchronization, where S stands for server process, it's been mentioned in Distributed Systems Concept and Design book that:
The time by S’s clock when the reply message arrives is therefore in
the range [t + min, t + Tround - min ] . The width of this
range is Tround - 2min, so the accuracy is
+-(Tround/2 - min)
How is the accuracy calculated here, is it half of the width of the range and if yes why?
algorithm synchronization distributed-system
asked Nov 10 at 14:25
Rishabh
1491316
add a comment |
In Cristian algorithm (https://en.wikipedia.org/wiki/Cristian%27s_algorithm) for clock synchronization, where S stands for server process, it's been mentioned in Distributed Systems Concept and Design book that:
The time by S’s clock when the reply message arrives is therefore in
the range [t + min, t + Tround - min ] . The width of this
range is Tround - 2min, so the accuracy is
+-(Tround/2 - min)
How is the accuracy calculated here, is it half of the width of the range and if yes why?
algorithm synchronization distributed-system
asked Nov 10 at 14:25
Rishabh
1491316
add a comment |
In Cristian algorithm (https://en.wikipedia.org/wiki/Cristian%27s_algorithm) for clock synchronization, where S stands for server process, it's been mentioned in Distributed Systems Concept and Design book that:
The time by S’s clock when the reply message arrives is therefore in
the range [t + min, t + Tround - min ] . The width of this
range is Tround - 2min, so the accuracy is
+-(Tround/2 - min)
How is the accuracy calculated here, is it half of the width of the range and if yes why?
algorithm synchronization distributed-system
asked Nov 10 at 14:25
Rishabh
1491316
In Cristian algorithm (https://en.wikipedia.org/wiki/Cristian%27s_algorithm) for clock synchronization, where S stands for server process, it's been mentioned in Distributed Systems Concept and Design book that:
The time by S’s clock when the reply message arrives is therefore in
the range [t + min, t + Tround - min ] . The width of this
range is Tround - 2min, so the accuracy is
+-(Tround/2 - min)
How is the accuracy calculated here, is it half of the width of the range and if yes why?
algorithm synchronization distributed-system
algorithm synchronization distributed-system
asked Nov 10 at 14:25
Rishabh
1491316
asked Nov 10 at 14:25
Rishabh
1491316
asked Nov 10 at 14:25
Rishabh
1491316
asked Nov 10 at 14:25
Rishabh
1491316
asked Nov 10 at 14:25
Rishabh
1491316
1491316
add a comment |
add a comment |
1 Answer
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Yes, the accuracy is calculated as half of the width of the range.
In general, when an instrument or a device can provide output in a range, the accuracy is considered to be half the range. Because if you use the midpoint of the range as the answer, you can only be off by half the range to either side.
To put it in simpler terms, a range [l, l + h] can also be written as (l + h/2) ± h/2, illustrating the error value of h/2.
Of course, the important thing to remember is that the midpoint should be chosen as the answer for the above error analysis to be correct. In the algorithm you described, the range of possible time spans from T + min to T + RTT - min. The mid point of the range is T + RTT / 2, which is exactly what the algorithm picks to achieve minimal error.
answered Nov 10 at 15:56
merlyn
1,33511221
add a comment |
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1 Answer
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Yes, the accuracy is calculated as half of the width of the range.
In general, when an instrument or a device can provide output in a range, the accuracy is considered to be half the range. Because if you use the midpoint of the range as the answer, you can only be off by half the range to either side.
To put it in simpler terms, a range [l, l + h] can also be written as (l + h/2) ± h/2, illustrating the error value of h/2.
Of course, the important thing to remember is that the midpoint should be chosen as the answer for the above error analysis to be correct. In the algorithm you described, the range of possible time spans from T + min to T + RTT - min. The mid point of the range is T + RTT / 2, which is exactly what the algorithm picks to achieve minimal error.
answered Nov 10 at 15:56
merlyn
1,33511221
add a comment |
Yes, the accuracy is calculated as half of the width of the range.
In general, when an instrument or a device can provide output in a range, the accuracy is considered to be half the range. Because if you use the midpoint of the range as the answer, you can only be off by half the range to either side.
To put it in simpler terms, a range [l, l + h] can also be written as (l + h/2) ± h/2, illustrating the error value of h/2.
Of course, the important thing to remember is that the midpoint should be chosen as the answer for the above error analysis to be correct. In the algorithm you described, the range of possible time spans from T + min to T + RTT - min. The mid point of the range is T + RTT / 2, which is exactly what the algorithm picks to achieve minimal error.
answered Nov 10 at 15:56
merlyn
1,33511221
add a comment |
Yes, the accuracy is calculated as half of the width of the range.
In general, when an instrument or a device can provide output in a range, the accuracy is considered to be half the range. Because if you use the midpoint of the range as the answer, you can only be off by half the range to either side.
To put it in simpler terms, a range [l, l + h] can also be written as (l + h/2) ± h/2, illustrating the error value of h/2.
Of course, the important thing to remember is that the midpoint should be chosen as the answer for the above error analysis to be correct. In the algorithm you described, the range of possible time spans from T + min to T + RTT - min. The mid point of the range is T + RTT / 2, which is exactly what the algorithm picks to achieve minimal error.
answered Nov 10 at 15:56
merlyn
1,33511221
Yes, the accuracy is calculated as half of the width of the range.
In general, when an instrument or a device can provide output in a range, the accuracy is considered to be half the range. Because if you use the midpoint of the range as the answer, you can only be off by half the range to either side.
To put it in simpler terms, a range [l, l + h] can also be written as (l + h/2) ± h/2, illustrating the error value of h/2.
Of course, the important thing to remember is that the midpoint should be chosen as the answer for the above error analysis to be correct. In the algorithm you described, the range of possible time spans from T + min to T + RTT - min. The mid point of the range is T + RTT / 2, which is exactly what the algorithm picks to achieve minimal error.
answered Nov 10 at 15:56
merlyn
1,33511221
answered Nov 10 at 15:56
merlyn
1,33511221
answered Nov 10 at 15:56
merlyn
1,33511221
answered Nov 10 at 15:56
merlyn
1,33511221
1,33511221
add a comment |
add a comment |
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