Md. Imrul Kaes - OEE Problem

download Md. Imrul Kaes - OEE Problem

of 5

Transcript of Md. Imrul Kaes - OEE Problem

  • 8/22/2019 Md. Imrul Kaes - OEE Problem

    1/5

    OEE Formulas:

    Availability = Operating Time Planned Production Time

    Performance = Ideal cycle Time Actual Cycle Time

    Actual Cycle Time = Operating Time Total pieces produced

    Ideal run time = 1 Ideal cycle time

    Quality = Good pieces Total pieces produced

    OEE = Availability x Performance x Quality x 100%

    Example 1:

    Shift length: 8 hours

    Short breaks: 2 @ 15 minutes

    Meal break: 1 @ 30 minutes

    Downtime: 47 minutes

    Ideal runtime: 60 pcs per minute

    Total pcs produced: 19,271 pcs

    Reject pcs: 423 pcs

    Calculate OEE of that shift.

    Solution:

  • 8/22/2019 Md. Imrul Kaes - OEE Problem

    2/5

    Here, sift length = Plant operating time = 8 x 60 mins =

    480 mins

    Planned shut down time = 2 x 15 + 30 mins = 60 mins

    Planned production time = plant operating time Planned shutdown time

    = 480 60 mins = 420 mins

    Down time loss = 47 mins

    Operating time = Planned production time Down time loss

    = 420 47 mins = 373 mins

    Good pcs = Total pcs Rejected pcs

    = 19271 423 pcs

    = 18848 pcs

    Availability = Operating Time Planned Production

    Time

    = 373 mins 420 mins = 0.88

    Ideal cycle time = 1 Ideal run time

    = 1 60 pcs per min = 0.0166 min per pcs

    Actual Cycle Time = Operating Time Total pieces

    produced

    = 373 mins 19271 pcs = 0.0193 min per pc

    Performance = Ideal cycle Time Actual Cycle Time

    = 0.0166 0.0193 = 0.8576

    Quality = Good pieces Total pieces produced

    =18848 pcs 19271 pcs = 0.9780

    OEE = Availability x Performance x Quality x 100%

    =0.88 x 0.8576 x 0.9780 x 100%

  • 8/22/2019 Md. Imrul Kaes - OEE Problem

    3/5

    =0.7381 x 100% =73.81% (ANS)

    Example 2:

    Shift length: 10 hours

    Short breaks: 2 @ 10 minutes

    Meal break: 1 @ 1 hour

    Downtime: 69 minutes

    Ideal runtime: 120 pcs per hour = 120 60 = 2 pcs per min

    Total pcs produced: 771 pcs

    Reject pcs: 70 pcs

    Calculate OEE of that shift.

  • 8/22/2019 Md. Imrul Kaes - OEE Problem

    4/5

    Solution:

    Here, Sift length = Plant operating time = 10 x 60 mins =

    600 mins

    Planned shut down time = 2 x 10 + 60 mins = 80 mins

    Planned production time = plant operating time Planned shut

    down time

    = 600 80 mins = 520 mins

    Down time loss = 69 mins

    Operating time = Planned production time Down time loss

    = 520 69 mins = 451 mins

    Good pcs = Total pcs Rejected pcs

    = 771 70 pcs

    = 701 pcs

    Availability = Operating Time Planned Production

    Time

    = 451 mins 520 mins = 0.867

    Ideal cycle time = 1 Ideal run time

    = 1 2 pcs per min = 0.50 min per pcs

    Actual Cycle Time = Operating Time Total pieces

    produced

    = 451 mins 771 pcs = 0.5849 min per pc

    Performance = Ideal cycle Time Actual Cycle Time

    = 0.50 0.5849 = 0.8548

    Quality = Good pieces Total pieces produced

    =701 pcs 771 pcs = 0.909

    OEE = Availability x Performance x Quality x 100%

  • 8/22/2019 Md. Imrul Kaes - OEE Problem

    5/5

    =0.867 x 0.8548 x 0.909 x 100%

    =0.676 x 100% =67.62% (ANS)