TD 發問時間: 社會與文化語言 · 5 年前

生產作業管理的Reliability問題~

The guidance system of a ship is controlled by a computer that has three major modules. In order for the computer to function properly, all three modules must function. two of the modules have reliabilities of 0.97 and the other has a reliability of 0.99

a. What is the reliability of the computer?

b. A backup computer identical to the one being used will be installed to improve overall reliability. Assuming the new computer automatically functions if the main one fails, determine the resulting reliability.

c. If the backup computer must be activated by a switch in the event that the first computer fails, and the switch has a reliability of 0.98, what is the overall reliability of the system? (Both the switch and the backup computer must function in order for the backup to take over.)

1 個解答

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  • 5 年前
    最佳解答

    a. What is the reliability of the computer?

    Reliability = (reliability of 1st module) x (reliability of 2nd module) x (reliability of 3rd module)

    = 0.97 x 0.97 x 0.99

    = 0.9315

    b. A backup computer identical to the one being used will be installed to improve overall reliability. Assuming the new computer automatically functions if the main one fails, determine the resulting reliability.

    the system reliability

    = reliability of first computer + (1 - reliability of first computer)(reliability of second computer)

    = (0.9315) + (1 - 0.9315)(0.9315)

    = 0.9953

    c. If the backup computer must be activated by a switch in the event that the first computer fails, and the switch has a reliability of 0.98, what is the overall reliability of the system? (Both the switch and the backup computer must function in order for the backup to take over.)

    This addition of switch changes the reliability of 2nd computer to:

    (reliability of switch) x (reliability of first module) x reliability of second module) x (reliability of third module)

    = 0.98 x 0.97 x 0.97 x 0.99

    = 0.9129

    the overall reliability of the system

    = reliability of first computer + (1 - reliability of first computer)x(reliability of second computer)

    = 0.9315 + (1 - 0.9315)(0.9129)

    = 0.9940

    2014-10-23 09:56:11 補充:

    Please double check the multiplication!

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