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- LPP Model Maximize P = 12x + 10y Subject to : 4x + 3y < 480 2x + 3y < 360 X, y 2 0 Which of the following points (x, y) is feasible? A) ( 120, 10) B ( 30, 100 ) c) ( 60, 90 ) D) ( 10, 120 )Consider the following LP problem: Min 6X+ 27Y Subject to : 2 X + 9Y => 25, and X + Y <= 75. Pick a suitable statement for this problem: a. X=37.5, Y=37.5 is the only optimal solution. b. Optimal Obj. function value is 75 c. X = 0, Y = 0 is the only optimal solution. d. Optimal Obj. function value is 0Problem 7-19 eBook Given the linear program Max 3A +48 s.t. Y -1A+ 1A + 2A + s.t. 28 ≤ 8 2B ≤ 12 18 ≤ 16 Α, Β 2 0 a. Write the problem in standard form. For those boxes in which you must enter subtractive or negative numbers use a minus sign. (Example: -300) Al+ A+ A+C A+ B+ B B+ B B b. Select the correct graph that shows the optimal solution for the problem. S1 S1 + + S₂ + S2 + S3 53 A, B, S1, S2, S3 A Q☆
- Briefly explain these terms:a. Basic variableb. Shadow pricec. Range of feasibilityd. Range of optimalityDetermine the pivot element in the simplex tableau. (If there is more than one correct pivot element, choose the element with the smaller row number.) X1 X2 X3 S1 S2 3 4 2 1 15 1 20 -8 -3 10 1 row column N O OB.) Convert the following function to standared form Min Z = 2x₂ + 5x3 S.t: x₁-x₂ ≥2 2x₁ + x₂ + 6x3 ≤ 6 x₁ - x₂ + 3x3 = 4 X1, X2, X3 20
- STAR Co. provides paper to smaller companies whose volumes are not large enough to warran paper rolls from the mill and cuts the rolls into smaller rolls of widths 12, 15, and 30 feet. The cutting patterns have been established: 1 2 Pattern 12ft. 15ft. 30ft. Trim Loss 0 4 1 10 ft. 3 0 7 ft. 8 0 0 4 ft. 2 1 2 1 ft. 5 2 3 1 1 ft. Trim loss is the leftover paper from a pattern (e.g., for pattern 4, 2(12)+1(15) + 2(30) = 99 hand for the coming week are 5,670 12-foot rolls, 1,680 15-foot rolls, and 3,350 30-foot rolls. hand will be sold on the open market at the selling price. No inventory is held. Number of: 31. If constraint has a shadow price of $6, Right-Hand-Side (RHS) is 12, allowable increase is 2, allowable decrease is 4. How would objective function change if the RHS of this constrains changes from 12 to 9? Answer___________Suppose we are solving a maximization problem andthe variable xr is about to leave the basis.a What is the coefficient of xr in the current row 0?b Show that after the current pivot is performed, thecoefficient of xr in row 0 cannot be less than zero.c Explain why a variable that has left the basis on agiven pivot cannot re-enter the basis on the next pivot.
- Find the Ma X Z = Xit X2 optimal solution by the large M X+ X2 3• ---- S.to メ4 >1 (3) X2 < 2 14) X1 30Q2. Solve the given LP problem on the right by (LP): Max Z = 2X1 + 4X2 %3D using The Graphical Solution Method. a) Find the optimal solution, determine the solution type. b) Find the optimality range for the changes in the objective coefficient c2. c) Find the feasibility range for the changes in the Right Hand Side (RHS) of one st. 3X1 + 2X2 < 12 Xị + 2X2 s 8 2X1 + X2 2 2 X1, X2 2 0 of the binding constraints.L.P. Model: Maximize Subject to: Z= 1X + 10Y 4X + 3Y ≤ 36 2X+4Y ≤ 40 1Y27 X,Y 20 (C₁) (C₂) (C3) 1.) Plot and label the constraints C₁, C₂ and C3 (using the line drawing tool) on the provided graph. 2.) Using the point drawing tool, plot the point that maximizes the objective function. The optimum solution is: X = (round your response to two decimal places). Y = (round your response to two decimal places). Optimal solution value Z = (round your response to two decimal places). C Y 22- 20- 18- 16- 14- 12- 10- 8- co 6+ 4- 2- ó -~ 0 2 -4 4 -6 8 10 12 X . . 14 16 18 20 22 Ly