# NCEES Sample Problem #507 - Construction PM

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### #1 Lucky1

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Posted 05 December 2011 - 02:35 AM

From the NCEES Civil Sample Questions & Solutions

Can someone explain why the 5 minute delay is subtracted from the numerator versus added to the trip time in the denominator when computing trips per hour?

Thanks.

### #2 treyjay

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Posted 05 December 2011 - 11:27 PM

Do you mean Problem #515?

If so, you are trying to find "cycles/hr" .........(1 cycle/17 min)(55 min/hr) = 3.24 cycles/hr

### #3 treyjay

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Posted 05 December 2011 - 11:46 PM

The cycle is 17 min with no delays. Since there is a 5 minute delay/hr, the working time is 55 min/hr.

### #4 Lucky1

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Posted 06 December 2011 - 02:57 AM

Problem #515 is the one in question.
Apparently it's in the way the problem is worded if the delay time is added into the cycle time or subtracted from 60 minutes when calculating cycles/hr.

### #5 treyjay

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Posted 06 December 2011 - 03:20 AM

Problem #515 is the one in question.
Apparently it's in the way the problem is worded if the delay time is added into the cycle time or subtracted from 60 minutes when calculating cycles/hr.

I guess I can see the confusion, but these problems always work with a cycle time with no delay. The delay is taken out in the hour portion. This allows for things like worker breaks, etc. The trick part of Problem #515 is recognizing the weight limit in how much the truck can haul per cycle.

### #6 Lucky1

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Posted 06 December 2011 - 01:46 PM

The weight limit issue was easy and I let the easy part (the delay and cycle time) throw me. Thanks for the clarification on the delay.

### #7 evrick1952

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Posted 03 April 2012 - 06:27 PM

Usually considered to be the efficiency factor when calculating CY (or whatever) per hour. For a five minute delay the efficiency factor is 0.91667 (55/60)

### #8 Lucky1

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Posted 04 April 2012 - 02:58 PM

I had never read it explained in that context. The old "understanding the concept" light bulb went on again. Thanks for this explanation.

### #9 Jayman_PE

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Posted 05 April 2012 - 04:18 PM

Yeah, the way I approached it was we are looking for Cyles per hour. Without a 5-min delay we have 17min/60min.cycle/1hr = 0.28 hrs/cycle. The reciprocal gives 3.52 cycles/hr. OK.
Now, if we subtract 5 min from the demoninator we have 17 min/55min.cycle/1hr = 0.31 hrs/cyle. The reciprocal gives 3.23 cycles/hr < 3.52 (above). Therefore I've now convinced myself the 5 minute delay slows production. The frist time I worked this problem it bothered me subtracting 5 out of the denominator...seemed like a contradiction. But if you check the units it makes sense.... Good problem.

Edited by Jayman_10x, 05 April 2012 - 04:18 PM.

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