jnxPsuOutletPowerFactorValue

Power factor percentage of each PSU (2k/3k).
              
Algorithm for calculation of Power Factor is below.
For PowerOut values that fall in between 618.93W and 915.24W,
say 700W, the appropriate PF ranges from 0.910191 & 
0.917994. Following linear equation could help deduce a
fairly accurate input power value.
              
Linear equation y = mx + b (where m is the slope and b is 
the Y intercept)
              
Slope m = (y2 - y1) / (x2 - x1)
Y intercept  b = y - mx
              
Plugging it all together for our example:
              
m = (915.24 - 618.93) / (0.917994 - 0.910191) = 37973.86
b = 915.24 - (37973.86 * 0.917994) = -33944.5
              
for 700W (y), our efficiency (x) would then be:
              
x = (700 - (-33944.5)) / 37973.86 = 0.912326 = 91%
              
PowerIn = 700W /0.912326 = 767.26W

MIB

JUNIPER-POWER-SUPPLY-UNIT-MIB

OID

.1.3.6.1.4.1.2636.3.58.1.2.4.1.10

Type

column

Access

readonly

Status

current

Parent

jnxPsuOutletEntry