# NCEES Sample Problem #507 - Construction PM

### #1

Posted 05 December 2011 - 02:35 AM

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

Posted 05 December 2011 - 11:27 PM

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

### #3

Posted 05 December 2011 - 11:46 PM

### #4

Posted 06 December 2011 - 02:57 AM

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

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

Posted 06 December 2011 - 01:46 PM

### #7

Posted 03 April 2012 - 06:27 PM

### #8

Posted 04 April 2012 - 02:58 PM

### #9

Posted 05 April 2012 - 04:18 PM

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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