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I have a few questions relating to linear programming which I could not understand from your lecture somehow.
1) What is the iso-contribution line and how to draw on the graph (i know we will not be asked in the exam but I’m learning that to draw)?
2) Is it correct that the optimal point is where we used all the limited resources available to produce maximum output and since it is where we produce maximum output therefore it is the point where we have a maximum contribution.
3) Optimum point is the point furthest away from the origin in the feasible area but we cannot simply identify that just by looking at the corners of the feasible area and see which point is the furthest away from the origin rather we need the iso-contribution line to see the angle/gradient/slope moving furthest towards either Point B or Point C?
Thanks again for your time.
1. The iso-contribution line is the line showing all combinations of the two products that give the same total contribution. It is only the engle of the line that we are concerned with because whatever the total contribution is, the angle of the line will be the same.
2. No. It is not where we use all of the limited resources and it is not necessarily where there is maximum output. Maximum output does not necessarily mean maximum contribution.
3. As I do, again, explain in my lecture, you can either check the contribution at all the corners of the feasible region (and the one giving the maximum contribution if the optimum solution), or (usually more efficiently) you can move out the iso-contribution line to find the corner that is furthest away from the origin.
1) This means that if the total contribution is $200 then products A and B would both be earning a $100 contribution each. You said in the lecture that we can take any contribution to draw the iso-contribution line because at this point we don’t know what the actual contribution will be. So we can choose any amount of contribution?
Let’s say the total contribution is $150 if we put this into objective function we would get:
C = 6S + 9E
S = ($150 / 6) = 25
E = ($150 / 9) = 16.67Now the iso-contribution line cannot be made for E because it would get over the demand line (which is at 10 E’s) and therefore feasible area (or it doesn’t matter?)
2) When looking for slack any line of a limited resource not touching the optimum point is considered to be not using all of the limited resources available (i.e. spare capacity).
That is why I said that the optimum point is where limited resources available are all used leaving no spare capacity such as material and labor lines are completely touching the optimum point but demand does not (i.e. slack of 5).
1. It doesn’t matter. All we need is the angle of the line.
2. The optimum point is not always where all of the limited resources are used. If there are three limited resources then it will be where two of them are used in full but there will be slack of the third.
Thanks. I appreciate your comment 🙂
(2) Yeah. That’s what I meant that we use all the limited resources of the constraints those lines are intersecting at the optimum point (but the other constraints those lines are not intersecting at this point then we are not using all their available resource – leaving spare capacity).
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