Department of Computer Science
College of Science and Mathematics
Montclair State University
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For the calculation of this number, please see the following:
11-22-13 | 11-24-13 | Temp. Difference | |
Avg Temp. | 54.66 | 26.08 | 28.58 |
Avg Low | 51.00 | 23.30 | 27.70 |
Avg High | 58.10 | 32.50 | 25.60 |
11-22-13 | 11-24-13 | Energy Difference | |
Total Power | 6375.69 | 5242.35 | 1133.34 |
IT Power | 2841.58 | 2850.38 | -8.80 |
PUE | 2.24 | 1.84 | 0.40 |
Where energy difference is Total Power on 11-22-13 minus Total Power on 11-24-13, and Temperature Difference is Avg Temp. on 11-22-13 minus Avg Temp. on 11-24-13.
\[ average\ kWh\ per\ degree\ Fahrenheit = \frac{6375.69-5242.35}{54.66-26.08} \] \[ average\ kWh\ per\ degree\ Fahrenheit = \frac{1133.34}{28.58} \] \[ average\ kWh\ per\ degree\ Fahrenheit = 39.65 \]In this situation, the data center would need to use 39.75 kWh of power to keep the internal data center temperature at the same level.
For the calculation of this number, please see the following:
11-14-13 | 11-21-13 | Temp. Difference | |
Avg Temp. | 40.99 | 46.72 | 5.73 |
Avg Low | 32.8 | 37 | 4.20 |
Avg High | 48.7 | 54 | 5.30 |
11-22-13 | 11-24-13 | Energy Difference | |
Total Power | 8102.35 | 5450.45 | 2651.90 |
IT Power | 2822.75 | 2789.10 | 33.65 |
PUE | 2.87 | 1.95 | 0.92 |
Where energy difference is Total Power on 11-14-13 minus Total Power on 11-21-13, and Temperature Difference is Avg Temp. on 11-14-13 minus Avg Temp. on 11-21-13.
\[ average\ kWh\ per\ degree\ Fahrenheit = \frac{2651.90}{5.73} \] \[ average\ kWh\ per\ degree\ Fahrenheit = 462.80 \]In this situation, the opposite occurs in that the data center would use 462.80 kWh of power for each degree cooler. While this situation is not logical, it demonstrates the variance in the data.
For the calculation of this number, please see the following:
Assuming that the cooling would be also cut in half and that the fan and pump would stay the same, the result is the following:
\[ CF = Total\ Energy\ Usage\ in\ kWh * 1.34 lbs/kWh*1 metric\ ton/2204.6lbs \]Where CF equals the Carbon Footprint, Total Energy (TE) equals the energy used by day for the cooling system and the PDUs, 1.34 lbs/kWh equals the national average for carbon emissions, and 1 metric ton equals 2,204.6 lbs
11-21-13 | Before | After |
Total Energy (TE) in kWh | 6375.69 | 3756.99 |
TE*1.34 | 8543.43 | 5034.37 |
Carbon footprint in metric tons per day | 3.88 | 2.28 |
For the calculation of this number, please see the following:
\[ \% \Delta CF = \left( \frac{CF\ Before - CF\ After}{CF\ Before} \right) * 100 \] \[ \% \Delta CF = \left( \frac{3.88-2.28}{3.88} \right) * 100 \] \[ \% \Delta CF = 41 \% \]For the calculation of this number, please see the following:
\[ PUE = \frac{Total Facility Energy}{IT Equipment Energy} \]Where Total Facility Energy equals the summation of PDU, AC, Pump and Fan energy usage, and IT Equipment Energy equals PDU energy usage.
11-21-13 | Before | After |
Total energy | 6375.69 | 3756.99 |
IT Energy | 2841.58 | 1420.79 |
PUE | 2.24 | 2.64 |
For the calculation of this number, please see the following:
\[ PUE = \frac{Total Facility Energy}{IT Equipment Energy} \]Where Total Facility Energy equals the summation of PDU, AC, Pump and Fan energy usage, and IT Equipment Energy equals PDU energy usage
\[ \% \Delta CF = \left( \frac{PUE\ After - PUE\ Before}{PUE\ Before} \right) * 100 \] \[ \% \Delta CF = \left( \frac{2.64 - 2.24}{2.24} \right) * 100 \] \[ \% \Delta CF = 17.85 \% \]For the calculation of this number, please see the following:
12-2-13 (Initial) | 12-4-13 (Lowered) | Change | |
Total Power in kWh | 5706.60 | 5671.58 | 35.02 kWh less power used |
IT Power in kWh | 2852.23 | 2841.10 | 11.12 kWh less power used |
PUE | 2.00 | 2.00 |
We have found that after adjusting the energy usage of the pumps and fans that are one of the primary sources of electrical consumption variance, that PUE in our data center is around 2.0. For a better visual representation of the data please see our website here.
Please see the Systems for the maximum system utilization. The maximum CPU-busy and Memory-busy values were used as input for an algorithm that aimed to find a reduced number of systems by consolidation. For example, if initially:
the two hosts could be consolidated in one with potential processor utilization of 90% and maximum memory utilization of 50%. The approach assumes similar properties among the systems. Additional parameters need to be employed if the systems are not equivalent.
Using data available for this server room, the following consolidation is possible: