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 04-25-08, 11:23 AM #1 reich17 Thread Starter     Join Date: Jan 2005 Bikes: Posts: 414 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) For The Math Wizards And Other Intellectuals I am attempting to come up with a formula, that produces a number, that in some way, comes close to accurately reflecting the difficulty of assembling a product. I realize that the per piece value is misleading because a higher value does not necessarily mean a higher degree of difficulty to assemble but could represent a product that requires more (or more costly) material than labor. Anyway, based on square footage of material required per product, per product cost and number of the product for the order I wonder if there is a method of determining difficulty for reasons of scheduling production time.
 04-25-08, 11:43 AM #2 MTBLover But on the road more     Join Date: Jun 2006 Bikes: Bianchi Volpe '07 Posts: 864 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) If I read you correctly, it sounds like you're talking about a classical job shop scheduling problem. Google this- there's plenty of pointers out on how how to handle this computationally.
 04-25-08, 06:29 PM #3 cal_gundert05 My name is Mike, not Cal   Join Date: Jul 2006 Bikes: Posts: 474 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) What kinds of products are we talking about here? Regardless, I'm thinking you can quantify the difficulty of the steps in the assembly process, rather than the materials. For example, a bookshelf that requires 20 screws to attach 5 horizontal pieces to 2 vertical ones and 16 screws to attach the backing would be harder than a night stand that requires 8 screws to attach 2 horizontal pieces to 2 vertical pieces and no backing.
 04-25-08, 07:02 PM #4 stevesurf another cat...FAB!     Join Date: Jun 2005 Location: 1st star to the right... Bikes: Merlin Ti Build, Trek Y-50, Bianchi Titanium Build, Custom Cuevas Road bike Posts: 1,381 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) Making a Jig or assembly container will allow multiple subassemblies to be fastened. You really have to have an idea of how many assemblies you wish to produce at a sitting and then optimize your process for that. If it is a low volume (50 or less assemblies), the savings will be minimal in creating staging and assembly aids. However, if the qty grows and you have either just yourself or one other person doing this, it could be beneficial. In either case, the above fastener/unit labor approach will yield the most accurate estimate, as long as it is averaged over multiple assemblies. __________________ Change the course of the Epidemic | The Merlin | Merlin XLM | Bianchi Ti | CyclePeople Don't take it off until there's a cure LIVESTRONG | pay it forward | Cats | NEW Gallery 9
04-26-08, 07:10 PM   #5
DannoXYZ
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 Originally Posted by reich17 I am attempting to come up with a formula, that produces a number, that in some way, comes close to accurately reflecting the difficulty of assembling a product. I realize that the per piece value is misleading because a higher value does not necessarily mean a higher degree of difficulty to assemble but could represent a product that requires more (or more costly) material than labor. Anyway, based on square footage of material required per product, per product cost and number of the product for the order I wonder if there is a method of determining difficulty for reasons of scheduling production time.
Use a weighted average.

04-26-08, 07:16 PM   #6
ModoVincere
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 Originally Posted by DannoXYZ Use a weighted average.
That can lead to bad results when you are already at or near capacity, and would need to expand in order to take on the increased production.
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 04-26-08, 10:13 PM #7 fuzzbox Your imaginary friend.     Join Date: Jul 2007 Location: Wandering aimlessly. Bikes: A sweet Quamen ATL custom, GT Mach 2 Posts: 2,211 Mentioned: 0 Post(s) Tagged: 0 Thread(s) Quoted: 0 Post(s) You came to the right person, lvl 53 here. Read bio for more infor. __________________ Every passing minute is another chance to turn it all around.