When using the erasure coding technique in an object-based storage system, data is divided into 12 data segments and 4 coding segments. What is the maximum number of disk drive failures against which data is protected by the configuration?
In a typical erasure coded storage system, a set of n disks is divided into m disks to hold data and k disks to hold coding information, where n, m, and k are integers. The coding information is calculated from the data. If up to k of the n disks fail, their contents can be recomputed from the surviving disks.
In this example k=4, so the correct answer is B.
B. 4
"Provides space-optimal data redundancy to protect data loss against multiple drive failures
– A set of n disks is divided into m disks to hold data and k disks to hold coding information
– Coding information is calculated from data
The figure illustrates an example of dividing a data into nine data segments (m = 9) and three coding fragments (k = 3). The maximum number of drive failure supported in this example is three."
In erasure coding, data is divided into data segments and coding segments. The total number of segments is the sum of data segments and coding segments. In this configuration, there are 12 data segments and 4 coding segments, making a total of 16 segments.
The maximum number of disk drive failures the system can tolerate is equal to the number of coding segments, which is 4 in this case. This means that data can still be reconstructed even if up to 4 disks fail simultaneously.
The figure illustrates an example of dividing a data into nine data segments (m = 9) and three coding fragments (k = 3). The maximum number of drive failure supported in this example is three.
Erasure coding offers higher fault tolerance (tolerates k faults) than replication with less storage cost.
"Provides space-optimal data redundancy to protect data loss against multiple drive failures
– A set of n disks is divided into m disks to hold data and k disks to hold coding information
– Coding information is calculated from data
The figure illustrates an example of dividing a data into nine data segments (m = 9) and three coding fragments (k = 3). The maximum number of drive failure supported in this example is three."
I think the correct is A.
Because the data segments is 12 and the coding segments is 4 then:
M/K=9/4= 3
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