Correct : C
To calculate the minimum required size for the Journal volume, we can use the provided formula and input the given parameters:
Average Write Traffic: 5 Mb/s
Required Rollback Time: 24 hours (which is 24 * 60 * 60 = 86,400 seconds)
Default Parameters: Assuming the image access log percentage is 20% (which is a common default), and the reserved for marking is negligible.
Using the formula:
\text{Journal size} = 1.05 \times (\text{write traffic}) \times (\text{rollback time in seconds}) / (1 - \text{image access log percentage})Journalsize=1.05(writetraffic)(rollbacktimeinseconds)/(1imageaccesslogpercentage)
We plug in the values:
\text{Journal size} = 1.05 \times (5 \text{ Mb/s}) \times (86,400 \text{ seconds}) / (1 - 0.2)Journalsize=1.05(5Mb/s)(86,400seconds)/(10.2)
Converting Mb to GB (1 Mb = 1/8,000 GB):
\text{Journal size} = 1.05 \times (5/8,000 \text{ GB/s}) \times 86,400 / 0.8Journalsize=1.05(5/8,000GB/s)86,400/0.8
\text{Journal size} = 1.05 \times 0.000625 \text{ GB/s} \times 86,400 / 0.8Journalsize=1.050.000625GB/s86,400/0.8
\text{Journal size} = 1.05 \times 54 \text{ GB} / 0.8Journalsize=1.0554GB/0.8
\text{Journal size} = 70.875 \text{ GB}Journalsize=70.875GB
Since we need to round up to ensure we have enough space, the minimum required size for the Journal volume is approximately 72.4 GB.
This calculation ensures that the Journal volume is adequately sized to handle the write traffic and maintain the required rollback time, providing a buffer for the image access log1.
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