1. ## Linear programming question

DuPhos produces two chemicals: Formula S and Formula T.

These chemicals are produced via two manufacturing processes. Process 1 requires 2 hours of labour and 0.5 kg of raw material to produce 50 grams of Formula S and 25 grams of Formula T. Process 2 requires 3 hours of labour and 1 kg of raw material to produce 75 grams of Formula S and 50 grams of Formula T. Sixty hours of labour and 20 kg of raw material are available. Demand for Formula S is unlimited, but only 0.5 kg of Formula T can be sold. Formula S sells for $16/gram, and Formula T sells for$14/gram. Any Formula T that is unsold must be disposed of at a cost of $2/gram. Formulate and solve an LP which maximises DuPhos's revenue less disposal costs. (Hint: use five decision variables: x1 = units of process 1, x2 = units of process 2, x3 = grams of Formula S produced, x4 = grams of Formula T that are sold, and x5 = grams of Formula T that must be destroyed). I need to simply show this in algebraic form however i've hit a complete brick wall when it comes to the expression regarding the excess formula 2. ## Re: Linear programming question So far the best ive come up with is Maximse Z = 16x1 + 14x2 - 2x5 Subject to 2x1 + 3x2 <= 60 (labour) 0.5x1 + x2 <= 20 (materials) 0.05x1 + 0.075x2 >= 0 (Formula s produced) notsure about this Like i said it's the disposal amounts I can't work out how to incorporate into the next line. I'm thinkning something along the lines of: x4 - 2x5 <=500. 3. ## Re: Linear programming question nm, I misinterpreted the condition on T. So you just assume all of it will sell up to 500g and you have to destroy the rest. 4. ## Re: Linear programming question Originally Posted by Lanceharte0069 So far the best ive come up with is Maximse Z = 16x1 + 14x2 - 2x5 I think you made a careless error here. Dollars per gram times processing units is meaningless.$Z = 16x_3 + 14x_4 - 2x_5.$Subject to 2x1 + 3x2 <= 60 (labour) Looks good 0.5x1 + x2 <= 20 (materials) Looks good 0.05x1 + 0.075x2 >= 0 (Formula s produced) notsure about this This is not really necessary: see below. Like i said it's the disposal amounts I can't work out how to incorporate into the next line. I'm thinkning something along the lines of: x4 - 2x5 <=500. That is not QUITE right. Although they are obvious, you should show the non-negativity constraints on$x_1$and$x_2$, which will entail non-negativity on$x_3$and$x_4$.$0 \le x_1.0 \le x_2.$Now you need formulas to calculate$x_3$and$x_4$. You were almost 50% there:$x_3 = 50x_1 + 75x_2.$Notice that the definition immediately above and the two non-negativity conditions specified before that entail$x_3 \ge 0.$Now find a definition for$x_4.$Does the definition imply all relevant constraints on$x_4$?$x_5$IS a bit tricky. It is defined only by constraints. One is obvious. What is the other? 5. ## Re: Linear programming question Ah thanks for points out my initial error! I've been so focused on the other constraints the decision equation was not given proper thought!! So would a further constraint (for x3) once x3 = 50x1 + 75x2 is solved be: x3 - 50x1 - 50x2 >= 0? Any more tips for the wastage constraint? 6. ## Re: Linear programming question Originally Posted by Lanceharte0069 Ah thanks for points out my initial error! I've been so focused on the other constraints the decision equation was not given proper thought!! So would a further constraint (for x3) once x3 = 50x1 + 75x2 is solved be: This really is a definition, not a solution x3 - 50x1 - 50x2 >= 0? I explained before that this is unnecessary:$x_1 \ge 0\ and\ x_2 \ge 0 \implies 50x_1 + 75x_2 \ge 0.$Any more tips for the wastage constraint? Lance I am having a bit of trouble making sure that I am transcribing things correctly because I have to switch between different posts so I may make an error myself. Please read my posts with a critical eye, OK? Let's write down what he have so far. Writing things down clearly let's you concentrate your thoughts on the smallest number of things possible. And pay attention to units. Objective function:$Z = 16x_3 + 14x_4 - 2x_5.$Monetary units$x_1 \ge 0.$Non-negativity constraint. Process 1 units.$x_2 \ge 0.$Non-negativity constraint. Process 2 units$2x_1 + 3x_2 \le 60.$Labor hours constraint. Units are hours of labor. Note$x_1 \ge 0,\ and\ x_2 \ge 0 \implies 0 \le 2x_1 + 3x_2.0.5x_1 + x_2 \le 20.$Raw material constraint. Units are kilograms of raw material. Again non-negativity is entailed and need not be specified.$x_3 = 50x_1 + 75x_2.$Definition. Units are grams of formula S. Again non-negativity is entailed and need not be specified. Now as I said in my previous post, we need a definition for$x_4.$What is your best shot at that definition? What are its units? Does the non-negativity constraint need to be specified here? What are the units for$x_5$? There is a common sense definition implied for$x_5$. The problem does not give it explicitly. Take a few minutes to see if the light dawns. Is the non-negativity constraint entailed by that definition? 7. ## Re: Linear programming question Hi Jeffm Given the units for x4 and x5 are both grams of Formula t. Would the equation to deal with wastage be: 14x4 -2x5 <=.5? Sorry I'm really having difficulties with this one!! I've tried starting off with x5= ...... and working it from there but keep on getting turned around! I'd imagine that the non-negativity constraint was assumed with all the equations, I intended to indicate x1, x2, x3, x4, x5 >= 0 as the final non-negativity constraint. You're help has been great though, I honestly appreciate it. 8. ## Re: Linear programming question given x1 and x2 how much T do you make? You sell x4 of it so how much is left over to destroy? x4 is bounded by what? 9. ## Re: Linear programming question Originally Posted by Lanceharte0069 Hi Jeffm Given the units for x4 and x5 are both grams of Formula t. YES Would the equation to deal with wastage be: 14x4 -2x5 <=.5? Does this make sense in terms of units? Do you know dimensional analysis? Sorry I'm really having difficulties with this one!! I've tried starting off with x5= ...... and working it from there but keep on getting turned around! I'd imagine that the non-negativity constraint was assumed with all the equations, I intended to indicate x1, x2, x3, x4, x5 >= 0 as the final non-negativity constraint. You're help has been great though, I honestly appreciate it. It is certainly possible to just get all the non-negativity constraints out of the way in one line by saying$x_1,\ x_1,\ x_1,\ x_1,\ x_1 \ge 0$even though some of those are redundant in this particular problem. You still have not attempted to set up a defining equation for$x_4.$As Romsek has pointed out, the amount of S produced equuals the amount sold, but the amount of T produced may not equal what is sold. What is sold cannot exceed what is produced. But what is produced may exceed what can be sold. Let$x_6$be the grams of T produced. What is the definition of$x_6$? What is the relationship among$x_4$,$x_5$, and$x_6$? Notice that these three variables are all in grams of T. 10. ## Re: Linear programming question So given x4 can only <= 0.5. 0.025x1 + 0.05x2 <=0.5x4 0.5x4 - 0.025x1 - 0.05x2 >= 0 (given this amount cannot be less than 0) 11. ## Re: Linear programming question Originally Posted by Lanceharte0069 So given x4 can only <= 0.5. 0.025x1 + 0.05x2 <=0.5x4 0.5x4 - 0.025x1 - 0.05x2 >= 0 (given this amount cannot be less than 0) You are making this much much more difficult than it needs to be.$grams\ of\ T\ produced = x_6 = 25x_1 + 50x_2.$Each unit of process 1 produces 25 grams of T and each unit of process 2 produces 75 grams.$grams\ of\ T\ sold = x_4 \le 500.$The variable$x_4$has units in grams rather than kilograms, remember. And you can sell only 500 grams.$grams\ of\ T\ to\ be\ disposed\ of = x_5 = x_6 - x_4,\ subject\ to\ x_5 \ge 0.$I know you are studying linear programming right now, but I strongly advise watching some tutorials on dimensional analysis. As I think I said earlier, it won't think for you, but it will prevent errors. Most of the youtube videos on it are geared toward chemistry students (which is where I learned it back in neolithic times), but it is useful in many more ways. 12. ## Re: Linear programming question I think i'm a little confused (again) by the introduction of x6. My professor has stated the variabled only go up to x5 as stated in the question. Sorry i'm completely lost now and have no idea what the answer is to be honest...... 13. ## Re: Linear programming question Originally Posted by Lanceharte0069 I think i'm a little confused (again) by the introduction of x6. My professor has stated the variabled only go up to x5 as stated in the question. Sorry i'm completely lost now and have no idea what the answer is to be honest...... Yes, I know he set up five variables, but that does not preclude you from helping yourself out by introducing a new one. How many grams of T do you sell? That is what$x_4$is defined as, correct? How many grams of T do you produce? That =$25x_1 + 50x_2.$Do you see where that came from? How many grams of T do you sell? Can you sell more than you produce? No matter how much you produce, what is the maximum you can sell? So$25x_1 + 50x_2 > 500 \implies x_4 = 500.$And$25x_1 + 50x_2 \le 500 \implies x_4 = 25x_1 + 50x_2.$So what are your constraining statements? If you produce more than 500, you can sell only 500. If you produce not more than 500, you can sell what you produce. How many grams of T must you dispose of? If you sell all you produce, it is zero. If you produce more than you sell, that means you produced more than 500 and sold only 500. So$25x_1 + 50x_2 > 500 \implies x_5 = (25x_1 + 50x_2)- 500.$And$25x_1 + 50x_2 \le 500 \implies x_5 = 0.$So what are your constraining statements? 14. ## Re: Linear programming question Thanks for your help but I still have no idea. I'll put this one in the too hard basket for now and come back to it when I practise these a bit more. Thanks again for your attempts at helping this lost soul! Lance 15. ## Re: Linear programming question Does this help you?$\displaystyle \\x_1:\mbox{units of process 1}\\x_2:\mbox{units of process 2}\\x_3:\mbox{grams of Formula S produced}\\x_4:\mbox{grams of Formula T produced}\\x_5:\mbox{grams of Formula T to be disposed}\\ \\ \mbox{Labor amount: } 2x_1+3x_2 \leq 60 \\ \mbox{Material amount: } 500x_1+1000x_2 \leq 20000 \\ \\ \mbox{Formula S produced: } x_3=50x_1+75x_2\\ \mbox{Formula T produced: } x_4=25x_1+50x_2 \\ \mbox{Formula T to be disposed: } x_5 =\begin{cases}0 & x_4 \leq 500\\x_4-500 & x_4 > 500\end{cases} \\ \mbox{Revenue: } Z=16x_3+14x_4-16x_5$Loss from amount disposed would be:$\displaystyle 2x_5 $from disposal plus$\displaystyle 14x_5\$ (cost of opportunity, i.e loss of what we could sell from formula T if we could sell more than 500 grams). As we have consider sales of all formula T in the revenue formula we have to subtract cost of opportunity from the revenue.

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