Well, after spending nearly 40 hours this week on Ops Research, I have run into one last problem that is giving me a pain in the neck. I have a company that makes three components. The profit contributions for each component is $8, $6, & $9 respectively In order to make these components, I have two processes: Bending and folding. I have a total of 120 hours available on the bender, and 110 hours on the folder. It takes me 6 minutes on the bender and 4 minutes on the folder to make component 1 It takes me 4 minutes on the bender and 5 minutes on the folder to make component 2 It takes me 4 minutes on the bender and 2 minutes on the folder to make component 3 I need to make no less than 600 units of component 1 I need to make no more than 200 units of component 3 I can make up to 1000 units of component 1 and 2 How would I formulate this linear equation? I am so confused! Thanks! W
Sorry but I am a little rusty with my LP. Without looking at the book, I think this is how to go about solving this LP. When I get back to my office on Monday I will be happy to verify this result. x1 = component 1, x2 = component 2, ... If I were to program this into Lindo, the LP would look like this: MAX 8x1 + 6x2 + 9x3 ST (Subject To) 1) 6x1 + 4x2 + 4x3 <= 7200 2) 4x1 + 5x2 + 2x3 <= 6600 3) 10x1 + 9x2 + 6x3 <= 13800 4) x1 >= 600 5) x3 <= 200 6) x1 + x2 <= 1000 7) x1, x2, x3 >= 0 Explanation: The "MAX 8x1 + 6x2 + 9x3" maximizes your profit. 1&2) Since you only have 120h in the bender (7200 minutes), I restricted the amount of time in the bender. I did likewise with the folder. 3) Since you don't want to bend something you aren't going to fold, I combined the two processes to ensure that all components that get bent also get folded. Combine the processes for each component and you get 10 minutes to make x1, 9 minutes to make x2, 6 minutes to make x3, and 230h (or 13800 minutes) total time. 4,5,&6) These three constraints are self explanatory. 7) This ensures you don’t start making negative components. Like I said...I did this LP from my head and it has been a few years so it may not be 100% correct. I will verify this for you on Monday and let you know. LP problems are really a lot of fun once you get the hang of it.
PhD2B, I have done the problem myself independently, and I got identical results to yours except for constraint 3). Isn´t it redundant as that condition is already specified in 1) and 2)? Nevertheless, this subject is truly cool, and, yes, a lot fun. More problems, Mr.Engineer?
Since I didn't have my OR book handy I wasn't sure. I kind of thought it was redundant, but I was trying to figure out a way to ensure all components that get bent also get folded. Did it work without constraint #3? Did you get the same result using constraint #3 without constraint #s 1 and 2? Mr. Engineer…more problems please.
You want more work? wow - I wish my line maintenance staff would ask for more work (usually they are on the Internet looking for better jobs while inbetween assignments) OK - I completed this one but I cannot seem to get it to work right unless I maximize it (but the point is to minimize it). The objective is to minimize the total cost of the project. I have three carpenters working on a project that will take 150 hours. D1 makes $30 and hour D2 makes $25 D3 makes $18 D1 can work no more than 50 hours D3 has to work greater than 22.5 hours but no more than 25% of what D1 and D2 work combined D1 has to work at least 40% of what the combination of D1 and D2 work I formulated this so that D3 works >/= to 22.5 hours (15% of the 150 hour total). D1 and D2 together would work </= 127.5 which is the balance of 150 - 22.5 is. My problem is this: D1 </= 50 hours on project however, D1 has to work 40% of the hours worked by D1 and D2 together. That would mean that D1 >/= 51 hours. This won't work. Any suggestions?
MIN 30D1 + 25D2 + 18D3 st 1) D1 <= 50 2) D3 <= 22.5 3) D3 <= 0.25 (D1 + D2) Rearraging, - D1 - D2 + 4D3 <= 0 4) 0.4 (D1 + D2) <= D1 Reorganizing, -6D1 + 4D2 <= 0 Hopefully this coincides with PhD2B´s solution. Otherwise, with almost 100% certainty, I am the one who has it wrong.
It's nice to see that we are thinking similarily. MIN 30D1 + 25D2 + 18D3 ST D1 + D2 + D3 = 150 D1 <= 50 D3 >= 22.5 D3 <= .25D1 + .25D2 or D3 - .25D1 - .25D2 <= 0 D1 >= .4D1 + .4D2 or .6D1 - .4D2 >= 0 D1,D2 >= 0 Solution: Have D1 work 48h, D2 work 72h, and D3 work 30h
I got wrong D3 >= 22.5. The sign should be in the opposite direction. Just a careless mistake. Nevertheless, yes, it is nice to see we are thinking alike. This is a very nice subject. I hope Mr. Engineer keeps on sending problems . Just by reading your conversations I got a couple of ideas I could use in my job. I could try to use, I should have said. Regards
Well, let's see here. Since I did quite abysmally in my undergrad Finite Math class and in my grad Applied Statistical Processes, darned if I know how to write out the equation that would prove this. But, here's the "conceptual math" answer: 1. It seems that we are absolutely required to make 600 units of component 1. This will burn up 600 x 6 = 3600 minutes (60 hours) on the bender and 600 x 4 = 2400 minutes (40 hours) on the folder and will produce 600 x $8 = $4800 profit contributions. We now have 60 hours left on the bender and 70 hours left on the folder. 2. Component 3 has the highest profit contribution @ $9 and we are allowed to make up to 200 units thereof. This will burn up 200 x 4 = 800 minutes (13 1/3 hours) on the bender and 200 x 2 = 400 minutes (6 2/3 hours) on the folder and will produce 200 x $9 = $1800 profit contribution. We are now left with 46 2/3 hours on the bender and 63 1/3 hours left on the folder. 3. We can now make up to 400 units of some combination of components 1 and 2. If there was enough time on the machines to produce an additional 400 units of component 1, this would produce an additional $3200 of profit contribution, while 400 units of component of component 2 would produce only $2400 in profit contribution. Producing an additional 400 units of component 1 will burn up 400 x 6 = 2400 minutes (40 hours) on the bender and 400 x 4 = 1600 minutes (26 2/3 hours) on the folder and will produce 400 x $8 = $3200 in profit contribution. We still, by the way, have 6 2/3 hours available on the bender and 36 2/3 hours left on the folder.
You are welcome. Does your course cover blending problems, non-linear programming, or stochastics? If so, let me know; I'll start brushing up on those topics. It was a real joy to do LP problems again! I thought it was especially fun to compare results with JLV. It made me realize how much I miss doing these kinds of problems. Hmmm...Are there any online colleges out there, with courses that involve operations research, that need an adjunct professor?
I am sure most of the students in the class would welcome both you and JLV as the Professor. The one we have replies almost sarcastically (like we are peons - his typical answer is "read the question" -- I was thinking about the same type of nonsensical reply, but the is the dean of the school)
One more just for kicks. I was in the process of doing all of the problems in the chapter -- I think I have most of this down (at least for this portion) -- however, I cannot seem to get this one to work. I am developing two products. A wafer and a widget. I have two facilities, one in Nevada and one in Oregon. I have contract to manufacture 200,000 wafers and 75,000 widgets. My Nevada facility can produce a maximum of 120,000 units total (widgets and wafers), and my Oregon facility can produce up to 180,000 total units. The costs for making each product is as follows Nevada Oregon Wafer $5.25 4.95 Widgets $5.45 $5.70 Using a linear programming model, how would you determine a production model to minimize total production cost?
Re: One more just for kicks. x1 = # Nevada Wafers x2 = # Oregon Wafers x3 = # Nevada Widgets x4 = # Oregon Widgets min z = 5.25x1 + 4.95x2 + 5.45x3 + 5.70x4 s.t. x1 + x2 = 200,000 (Wafer Constraint) x3 + x4 = 75,000 (Widget Constraint) x1 + x3 <= 120,000 (Nevada Facility Constraint) x2 + x4 <= 180,000 (Oregon Facility Constraint) x1,x2,x3,x4 >= 0 (Sign Restriction)
