Yioop - PHP Search Engine

Nov 4 In-Class Exercise Thread.

Post your solutions to the Nov 4 In-Class Exercise to this thread.

Best, Chris

(Edited: 2020-11-03)

Post your solutions to the Nov 4 In-Class Exercise to this thread. Best, Chris

-- Nov 4 In-Class Exercise Thread

 Parameters that could be affected or considered - size, read/write speeds, and connection speeds of hard drive

 We have 5 disks => probability that 1 of the disks fail in a given year = 5/10 = 0.5 or 1 disks could fail in 2
 Replacement takes 5 hours at max => 5/24 of a day = 1/252 year.
 Odds of failure for 1 disk = 1/10 * 1/252 = 1 in 2520
 Considering 5 disks => Mean time to data loss = 2 * 2520 = 5040 years

(Edited: 2020-11-04)

Parameters that could be affected or considered - size, read/write speeds, and connection speeds of hard drive We have 5 disks => probability that 1 of the disks fail in a given year = 5/10 = 0.5 or 1 disks could fail in 2 Replacement takes 5 hours at max => 5/24 of a day = 1/252 year. Odds of failure for 1 disk = 1/10 * 1/252 = 1 in 2520 Considering 5 disks => Mean time to data loss = 2 * 2520 = 5040 years

-- Nov 4 In-Class Exercise Thread

	Concrete parameters:
	Setting the RTO to be 5 hours would enforce a type of hard disk that needs to be use.
	Essentially, we need to make sure to use a hard drive less than a certain size to satisfy the RTO condition.

	RTO objective = replacement process time = 5 hours = 5/24 days = 5/8760 years

	The odds that a disk will fail during the replacement process is:

	1 failure / (MTTF for 1 disk = 10 years) * 5/8760 years = 5/87600 years = 1/17520 years

	If we have 5 disks (each with an MTTF being 10 years) then the MTTF for 5 disks is 10/5 = 2 years.
	Therefore, the mean time to data loss is 2 * 17520 = 35,040 years.

(Edited: 2020-11-04)

Concrete parameters: Setting the RTO to be 5 hours would enforce a type of hard disk that needs to be use. Essentially, we need to make sure to use a hard drive less than a certain size to satisfy the RTO condition. RTO objective = replacement process time = 5 hours = 5/24 days = 5/8760 years The odds that a disk will fail during the replacement process is: 1 failure / (MTTF for 1 disk = 10 years) * 5/8760 years = 5/87600 years = 1/17520 years If we have 5 disks (each with an MTTF being 10 years) then the MTTF for 5 disks is 10/5 = 2 years. Therefore, the mean time to data loss is 2 * 17520 = 35,040 years.

-- Nov 4 In-Class Exercise Thread

Concrete parameters -

 1. RTO dictates the type of disk used in the particular setting. 
 2. Another factor is the network used to connect storage to the server. 
 3. Another very trivial factor can be operating system used to take the backup which will dictate the speed of backup.

Concrete parameters - 1. RTO dictates the type of disk used in the particular setting. 2. Another factor is the network used to connect storage to the server. 3. Another very trivial factor can be operating system used to take the backup which will dictate the speed of backup.

-- Nov 4 In-Class Exercise Thread

 parameters that affect hardrive based on RTO:
 - size and type of hard disk used.
 - network connecting hard drive to node server.
 
 RTO is of 5 hours so recovery process takes 5/24 days = 5/8760 years
 
 MTTF of single hard drive = 1/10 * 5/8760 = 1/17520 so 1 in 17520 failures.
 
 MTTF for 5 disks = 5/10 = 1/2 i.e. 2 yrs.

 Mean time to data loss: 2*17520 = 35040 yrs

(Edited: 2020-11-04)

parameters that affect hardrive based on RTO: - size and type of hard disk used. - network connecting hard drive to node server. RTO is of 5 hours so recovery process takes 5/24 days = 5/8760 years MTTF of single hard drive = 1/10 * 5/8760 = 1/17520 so 1 in 17520 failures. MTTF for 5 disks = 5/10 = 1/2 i.e. 2 yrs. Mean time to data loss: 2*17520 = 35040 yrs

-- Nov 4 In-Class Exercise Thread

 RTO is 5 hours which is 5/24 of a day or 5/8760 of a year
 1/10 chance of failure * 5/8760 = 1/17520 failures
  
 5/10 or 1/2 is the odds 1 of the five disks fail in a given year.
 Mean time to failure would be 2 * 17520 = 35040 years

(Edited: 2020-11-08)

RTO is 5 hours which is 5/24 of a day or 5/8760 of a year 1/10 chance of failure * 5/8760 = 1/17520 failures 5/10 or 1/2 is the odds 1 of the five disks fail in a given year. Mean time to failure would be 2 * 17520 = 35040 years

-- Nov 4 In-Class Exercise Thread

 Concrete Parameters : capacity and type of hard drives.
 
 RTO is 5 hours = 5/24 day = 5/(24*365) = 1/1752 year
 MTTF for 1 hard drive is 10 years, so chances of failure for one hard drive = 1/10 
 * 1/1752  = 1/17520

 For 5 hard drives odds of failures = 5/10 * 1/17520 = 1/35040
 Thus Mean time to data loss = 35040 years.

Concrete Parameters : capacity and type of hard drives. RTO is 5 hours = 5/24 day = 5/(24*365) = 1/1752 year MTTF for 1 hard drive is 10 years, so chances of failure for one hard drive = 1/10 * 1/1752 = 1/17520 For 5 hard drives odds of failures = 5/10 * 1/17520 = 1/35040 Thus Mean time to data loss = 35040 years.

-- Nov 4 In-Class Exercise Thread

 Hard drive parameters that could be affected: network connection speeds, size.

 Mean time to data loss:
 RTO = 5 hours = 5/8760 of a year
 1/10 MTTF * 5/8760 = 1/17520 failures
  
 5 disks each with MTTF of 10 years = 10/5 = 2 years MTTF
 Thus, mean time to data loss = 2 * 17520 = 35040 years

Hard drive parameters that could be affected: network connection speeds, size. Mean time to data loss: RTO = 5 hours = 5/8760 of a year 1/10 MTTF * 5/8760 = 1/17520 failures 5 disks each with MTTF of 10 years = 10/5 = 2 years MTTF Thus, mean time to data loss = 2 * 17520 = 35040 years