In question D, what is the range, and why is 2 to 6 within range and 1 to 3 not
ID: 388915 • Letter: I
Question
In question D, what is the range, and why is 2 to 6 within range and 1 to 3 not in range? ctive function coefficient for A changes from 3 to 5. Does the optimal solution change? Use the graphical solution procedure to find the new optimal solution. The same extreme point remains optimal V Optimal solution 41 c. Assume that the objective function coefficient for A remains 3, but the objective function coefficient for B changes from 2 to 4. Does the optimal solution change? Use the graphical solution procedure to find the new optimal solution. A new extreme point becomes optimal optimal solution 30 d. The sensitivity report for the linear program in part (a) provides the following objective coefficient range information: Objective Coefficient 3.00000 2.00000 Allowable Increase 3.00000 1.00000 Allowable Decrease 1.00000 1.00000 Variable A. Use this objective coefficient range information to answer parts (b) and (c). V to V so the optimal solution will now change in part (b) The objective coefficient range for A is from because the new objective coefficient is in this range. The objective coefficient range for B is from the new objective coefficient is not in this range. 2 6 3 Vso the optimal solution will change in part (e) because Check My Work NextExplanation / Answer
We have objective function as
Max 3A + 2B.. Here the coefficeint of A is 3 and coefficeint of B is 2.
Now note that in part D it has been provided that A can have coefficeint that can increase by 3.0000 from the coeffeiceint which is given in objective function. That is the coefficeint of A can be at maximum 3 (coeffiecint in objective function) + 3 ( alowable increase). Hence the maximum limit is 6.
Similarly the coefficeint of A can be lowered by 1.0000 from original coeffiecient that is 3 in objective function. Thus it can be at least 3 (coefficeint in objective function) - 1 (allowable decrease) = 3-1 = 2.
Hence the minium limit is 2 and maximum limit is 6 for the coefficeint of A. This will make range 2 to 6 for A' coeffiecint.
Similarly,
in part D it has been provided that B can have coefficeint that can increase by 1.0000 from the coeffeiceint which is given in objective function. That is the coefficeint of B can be at maximum 2(coeffiecint in objective function) + 1 ( alowable increase). Hence the maximum limit is 2+1 = 3..
Similarly the coefficeint of B can be lowered by 1.0000 from original coeffiecient that is 2 in objective function. Thus it can be at least 2 (coefficeint in objective function) - 1 (allowable decrease) = 2-1 = 1.
Hence the minium limit is 1 and maximum limit is 3 for the coefficeint of B. This will make range of 1 to 3 for B' coeffiecint.
Therefore, For A coefficeint range is 2 t0 6
and for B coeffiecint the range is 1 to 3.
Now,
In part b the optimal solution is provided as A = 7 and B = 3. It is also provided that coeffiecint of A changes from 3 to 5. Notice that in part D we calculated that the range for A is 2 to 6. Hence the new coeffiecint in part B which is 5 is in range of 2 to 6. As 5 comes between 2 and 6. it is in range and the solution will not change.
On other hand, in part c the coeffeicint of B cahnges from 2 to 4. Now notice that in part D we calculated that range for B coeffiecint is 1 to 3 and hence 4 does not come between 1 and 3, it is not in range and the solution will change.
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