Single cycle material handling operations in warehouse layout design

  • Thread starter Amaelle
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Amaelle
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Good day all,
I m currently studying how to determine "the optimal layout" in INDUSTRIAL PLANT & SAFETY WAREHOUSES course, and I was stucked with the following section:
Material Handling operation

Single cycle:
• it is an handling operation of input or output of a SKU
• consists of a round trip transfer between the point I/O and the generic location of storage
• in the input cycle the way back is "unloaded“
• in the output cycle the way to is "unloaded“
and then the slide goes to the following section:
Slide titled Determining the optimal layout listing assumptions and formulas r = 2 Cs = 4(U/4 + V/2) and r = 4(U/4 + A/(2U)); a rectangle of area A = U*V shows an I/O point at the middle of the bottom side and a dashed path to a storage location.

I really need to understand :
  1. what do they mean by single cycle?
  2. what are they trying to do in the picture i'have posted bellow? I really tried the whole day to find a hint but uselessly
Many thanks in advance!
 

Discussion

You simply want to minimize the average path length. Take ## \frac{dr}{dU}=0 ##. Have you had any calculus yet? After that, there's just a little algebra to see how ## U ## and ## V ## are related for optimal conditions.
 
Thanks for your pr ompt reply,
my issue is not calculus related, i need to understand what do they mean by average path!? what is the idea behind this why is the input out put point different from allocation?
many thanks in advance!
 
I didn't look at the diagram very carefully, but they give you the answer=formula here. I don't know how they arrived at their first formula=if you study their diagram, maybe it makes perfect sense, but it really doesn't need to. They give you the formula for ## r ##, and that is the one that they expect you to use.
One item you should recognize is area ## A=UV ##. (That's how they got the second formula for ## r ## from the first one).