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This topic has 4 voices, contains 4 replies, and was last updated by
Chris Seider 310 days ago.
| Author | Posts |
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| Author | Posts |
| August 9, 2012 at 9:13 am #184609 | |
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Roger @foreverlearning Reputation - 15 Rank - Aluminum
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10 ct, 8 ct, 0 ct,7 ct,11 ct,3 ct,3 ct,3 ct,5 ct,10 ct…..what is the cycle time (ct) 80% of time? how did u get it? I used this equation below and i was told it was wrong….. do i need to sort the days? days ppl percent |
| August 10, 2012 at 4:24 am #184646 | |
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Prabhu V @prabhuvspj Reputation - 513 Rank - Copper
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Hi, Can you pls elaborate your query with what is the main objective and others….. |
| August 12, 2012 at 1:04 pm #184689 | |
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Eric Maass @poetengineer Reputation - 75 Rank - Aluminum
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Hi Roger, It sounds like you were given a set of individual cycle time measurements…and they want you to estimate the cycle time 80% of the time…Which is a rather vague request. |
| August 13, 2012 at 9:43 am #184708 | |
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manipage @manipage Reputation - 21 Rank - Aluminum
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Hi Roger, I think the 80% of the time the cycle time will be 3, If it is correct, then you can find this applying simple pareto. |
| August 13, 2012 at 10:57 am #184710 | |
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Chris Seider @cseider Reputation - 3120 Rank - Titanium
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If you want the maximum cycle time you’d expect 80% of the time, you can add 0.84 times the standard deviation and add to the mean cycle time–if you are comfortable with assuming a normal distribution. |
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