A customer wants to store PSM recordings for 100 days and estimates they will have 10 Windows sessions per day for 100 minutes each.
What is the minimum storage required for the Vault and PAReplicate for the PSM recordings?
A customer wants to store PSM recordings for 100 days and estimates they will have 10 Windows sessions per day for 100 minutes each.
What is the minimum storage required for the Vault and PAReplicate for the PSM recordings?
To calculate the minimum storage required for the Vault and PAReplicate for PSM recordings, we need to consider the number of recording minutes and the storage requirement per minute. The customer will have 10 sessions per day, each lasting 100 minutes, over 100 days. This results in 10 sessions/day * 100 minutes/session * 100 days = 100,000 minutes. PSM recordings require approximately 250 KB per minute. Therefore, the total storage needed is 100,000 minutes * 250 KB/minute = 25,000,000 KB, which is equivalent to 25 GB. Hence, 25 GB is the accurate calculation for the storage required.
The answer is A The estimated storage requirement for PSM recordings is approximately 50-250 KB for each minute of a recording session. The recording size is affected by the type of session recording (console vs. GUI recording) as well as by the type and number of activities that are performed during the session. In this case, the customer wants to store PSM recordings for 100 days and estimates they will have 10 Windows sessions per day for 100 minutes each. This means that the total number of minutes of recording will be 100 days * 10 sessions/day * 100 minutes/session = 100,000 minutes. At the high end of the estimated storage requirement, this would require 100,000 minutes * 250 KB/minute = 25 GB of storage. However, the customer may be able to get away with less storage if they use a lower quality recording setting or if they only record certain types of sessions.
[[(100 min)*(10 sessions)]*(100 days)]*250(kb/min) = 25000000kb = 25 GB. The answer should be A.
((100 days) x (10 sessions/day) x (100 minutes/session) x (250 kb/min)) = 25 GB
it's (π_πππ )=(πΆ_π ππ π πππ )(π‘_π ππ π πππ )(π _(π ππ π πππ πππππππππ) )+20πΊπ΅ so is 25 gb + 20gb = 45 gb. So minimum is 250 GB
Ans 25GB Recording 250KB/Min And exp 1Day -> 10*100= 1000Min/Day 100Days - > 1000*100= 100000 Min 1MB = 1000KB 1GB = 1000MB= 1000*1000KB= 1000000KB 1KB = 1/1000000 GB Total GB = 100000*250/1000000= 25GB
Another trick question, based on the formula provided by CyberArk: 100(retention) x 10(# of sessions) x 100(minutes each) x 250 (kb/min) + 20GB = 45GB therefore the next smallest size is 250GB so I'd be inclined to agree that the answer is in fact B. https://docs.cyberark.com/PAS/13.2/en/Content/PAS%20INST/Considerations-Before-Installing-PSM.htm?tocpath=Installation%7CInstall%20PAM%20-%20Self-Hosted%7CInstall%20PSM%7C_____1#:~:text=of%20your%20implementation.-,Planning%20capacity,-Determine%20the%20amount
[[(100 min)*(10 sessions)]*(100 days)]*250(kb/min) https://docs.cyberark.com/Product-Doc/OnlineHelp/PAS/13.0/en/Content/PAS%20INST/Considerations-Before-Installing-PSM.htm?tocpath=Installation%7CInstall%20PAM%20-%20Self-Hosted%7CInstall%20PSM%7C_____1#Planningcapacity
[[(100 min)*(10 sessions)]*(100 days)]*250(kb/min) = 25000000kb = 25 GB. So according to the way you calculated, the answer should be A.
(100 days) x (10 sessions/day) x (100minutes/session)+ 20GB = 100 020 (100 GB) minimum is 250 GB
*100 GB + 20 Gb = 120 GB
the correct calculation: it's (π_πππ )=(πΆ_π ππ π πππ )(π‘_π ππ π πππ )(π _(π ππ π πππ πππππππππ) )+20πΊπ΅ so is 25 gb + 20gb = 45 gb. So minimum is 250 GB
It is option B
After exausting calculation, the best answer is the B. First of all, we need to understand the minimum quality possible it is 100 KB/s 100 x 10 x 100 x 100 KB + 20 GB. after we need to convert the value of session KB to GB: 100x10x100x0,0001+20 GB= 30 GB