Partial indicators method, Electrical Engineering

Partial Indicators Method

The partial indicators technique consists of calculating various measures of the financial, operating, commercial, and quality of a business's performance. Past performance can give information on improvements over time. These indicators account for the relationship of two easy measures, yielding indices of productivity, human capital development, or financial conditions, between others. Such performance indicators involve, for example, the number of workers per one thousand connections, training hours, the percentage of loss, etc.

The ultimate reasons of the benchmarking exercise are to bring the step forward improvement and align the several business processes along with the help of best practices to achieve the goal. Therefore it becomes more significant to select and tailor the set of indicators to the reasons and to the intended users.

Posted Date: 2/6/2013 2:37:25 AM | Location : United States







Related Discussions:- Partial indicators method, Assignment Help, Ask Question on Partial indicators method, Get Answer, Expert's Help, Partial indicators method Discussions

Write discussion on Partial indicators method
Your posts are moderated
Related Questions

what is experimental setup for Butterworth filter?

a sensorless brushless DC motor control with back emf (zero crossing point ) technique, using Arduino Microcontroller board control and describe in C. program language.

This is a synchronous motor that does not require a special start-up auxiliary motor. The  rotor  consists  of  stout  copper  (or aluminium)  conductors  arranged  in  the form of

why is delay line used in vertical deflection on systems of CRO?

Q. Three waveforms seen on an oscilloscope are shown in Figure. If the horizontal scale is set to 50 ms per division (500 ms for the entire screen width), and the vertical scale is


write a short notes of thermal compansation?

Q. Explain about Capital stock? Capital stock illustrates the amount of the owners' investment in the corporation.

Q. A 10-kW, 250-V dc shunt generator, having an armature resistance of 0.1and a field resistance of 250 , delivers full load at rated voltage and 800 r/min. The machine is now ru