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Improving operational efficiency begins with measuring it. Since operational efficiency is about the output to input ratio, it must be measured on both the input and output side. Quite often, company management is measuring primarily on the input side, e.g., the unit production cost or the man hours required to produce one unit.
Total effective equipment performance (TEEP) is a closely related measure which quantifies OEE against calendar hours rather than only against scheduled operating hours. A TEEP of 100% means that the operations have run with an OEE of 100% 24 hours a day and 365 days a year (100% loading). The term OEE was coined by Seiichi Nakajima. [2]
In statistics, efficiency is a measure of quality of an estimator, of an experimental design, [1] or of a hypothesis testing procedure. [2] Essentially, a more efficient estimator needs fewer input data or observations than a less efficient one to achieve the Cramér–Rao bound.
For example, the early UNIVAC I computer performed approximately 0.015 operations per watt-second (performing 1,905 operations per second (OPS), while consuming 125 kW). The Fujitsu FR-V VLIW / vector processor system on a chip in the 4 FR550 core variant released 2005 performs 51 Giga-OPS with 3 watts of power consumption resulting in 17 ...
Equipment is generally rated by the power it will deliver, for example, at the shaft of an electric or hydraulic motor. The power input to the equipment will be greater owing to the less than 100% efficiency of the device. [1] [2] [3] Efficiency of a device is often defined as the ratio of output power to the sum of output power and losses. In ...
If energy output and input are expressed in the same units, efficiency is a dimensionless number. [1] Where it is not customary or convenient to represent input and output energy in the same units, efficiency-like quantities have units associated with them. For example, the heat rate of a fossil fuel power plant may be expressed in BTU per ...
Outside of specific contexts, computer performance is estimated in terms of accuracy, efficiency and speed of executing computer program instructions. When it comes to high computer performance, one or more of the following factors might be involved: Short response time for a given piece of work. High throughput (rate of processing work tasks).
For a water electrolysis unit operating at a constant temperature of 25 °C without the input of any additional heat energy, electrical energy would have to be supplied at a rate equivalent of the enthalpy (heat) of reaction or 285.830 kJ (0.07940 kWh) per gram mol of water consumed. [6] It would operate at a cell voltage of 1.48 V.