An architect is designing a new vSphere environment with the following resources:
✑ 600 vCPU
✑ 5,760 GB RAM
Average resource usage is:
✑ 60 vCPU
✑ 1,152 GB RAM
The design must meet the following requirements:
✑ The environment has the ability to burst by 25%.
✑ Each host can schedule 36 vCPUs and has 512 GB RAM.
✑ Management overhead is 20%.
What is the minimum number of hosts required to meet the design requirements?
The architect needs to calculate both the CPU and RAM requirements considering the ability to burst and management overhead. The average usage is 60 vCPUs and 1152 GB RAM. Including the 25% burst requirement, the vCPU requirement becomes 75 (60 * 1.25) and the RAM requirement becomes 1440 GB (1152 * 1.25). Adding the 20% management overhead, the total vCPU and RAM requirements are 90 (75 * 1.2) and 1728 GB RAM (1440 * 1.2). Each host can schedule 36 vCPUs and has 512 GB RAM. Therefore, the minimum number of hosts required for vCPUs is 3 (90 / 36), and for RAM is 4 (1728 / 512). Since we need to satisfy both the CPU and RAM requirements, the higher number is chosen. Hence, 4 hosts are required to meet the design requirements.
I think it will be C (four hosts). Take for consideration "Management overhead is 20%." CPU: (60+60*0,25+60*0,2)/36 = ~ 2,4 (round up to 3) RAM: (1152+1152*0,25+1152*0,2)/512 = ~ 3,26 (round up to 4)
C - 4 hosts CPU - (60+60*0.45)/36 = 2.4 RAM: (1152+1152*0.45)/512 = 3.2 Taking the higher number, therefore 3.2 > 4 hosts
In order to meet all 3 requirements im going with C A is not correct as its not factoring in the management overhead of 20%
Kindly explain the math, 60+60 * 0.45 what is the reason for 60+60 and multiplied by .45?
20/100=0.2 25/100= 0.25 Distributive property 60 + 60 *0.25 + 60*0.2 = 60 + 60(0.25+0.2) = 60 + 60*0.45
Take the worst metric as guidance = RAM Average usage ✑ 1,152 GB RAM The design must meet the following requirements: ✑ The environment has the ability to burst by 25%. X = Average usage ✑ 1,152 GB RAM + 25% for burst = 1440 ✑ Each host can schedule 36 vCPUs and has 512 GB RAM. ✑ Management overhead is 20%. Y = VM usage + Burst capacity = 1140 Z= 1140 + 20% = 1728 N Hosts = 1728 / 512GB RAM = 3.375 N Hosts > 3 Hosts
C 4 HOSTS. Memory is the focus here. Vms use 1152gb ram + 25% burst = 1440gb + 20% mgmt = 1,728gb ram. Each host has 512gb ram so 4 hosts needed.
Since question is asking for the minimum, I would go with Answer A = 3 host. CPU: (60 * 1.25)/36 = ~ 2.08 (round up to 3) RAM: (1152 * 1.25)/512 = ~ 2.81 (round up to 3) Take the higher of the 2 figures - in this case RAM at 2.81 hosts, round it up to get 3 host. However if the question did not ask for the minimum, the industry best practice is to size it at 100% capacity to use 4 host.
How about management overhead?
Can you please explain the values why it is multiplied by 1.25?
Answer is A
How it is A? Can you please explain?
I change to C
Answer is C . 3 hosts
You mean A. 3 hosts
I think C?
The resource needs are 60 vCPU and 1152GB RAM. With a 25% burst, we will need up to 75 vCPU (60x1,25), and 1440 GB RAM (1152x1,25). For that 3 hosts could be sufficient (108 vCPU, 1536 GB RAM). But, as we have to take in account a 20% Management overhead, we will need 94 vCPU (75/0,8) and 1800 GB RAM (1440/0,8) so we need 4 hosts minimum to accommodate the necessary RAM.
can anyone provide a Link to explain how to calculate this equation?
CPU 60*1,25*1,2 = 90 /36 = 2,5 RAM 1152*1,25*1,2= 1728/512 = 3,37 Minimum can be 3 or 4 hosts. For best practices reasons i vote C (4 hosts).
Ans is 3 Avg vCPU usage = 60 Overhead - 20%, overhead for 60, = 60x20/100 = 12 Burst by 25%, Burst for 60=60x25/100=21 so, total vcpu utilization = 60+21+12 = 93 one host can give (schedule) 36 vCPU. So, how many host will provide 93 vCPU. 36 vCPUs given by hosts = 1 1 vCPUs given by hosts = 1/36 93 vCPUs given by hosts = (1/36)x93=2.58 let us round it to 3. So minimum 3 hosts are required. :)
I think you didn't consider memory in your calculations. Do the same for RAM and your answer will be 4 hosts.
Overhead will still be 20% even in burst. So 1152*1.25*1.20 would be more accurate. The worst situation. Whatever, C stay the answer.
Its 4 hosts for sure
4 Hosts CPU = 60*1.25*1.2/36 = 2.5 (3 hosts) RAM = 3.375 (4 hosts) Management overhead costs from the total number of employed resources, both in the case of an average and in the case of a burst consumption
I am not sure if between A and C Can anyone explain how this can be answered