# MGST391 Tutorial Solution for Formulation and Graph

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University of Calgary **We aren't endorsed by this school
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MGST 391
Subject
Business
Date
Oct 17, 2023
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5
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UNIVERSITY OF CALGARY HASKAYNE SCHOOL OF BUSINESS MGST391 Online Tutorial: Formulation & Graph 1. Cattle Feed This problem must be typed (in Word). Ensure to align the decision variables in both the objective function and constraints. You should only type the LP problem; do not show details of calculations. Nor should you do it in Excel or Excel Solver. Because the decision variables are given, you do not have to state them explicitly in the question. Two types of winter beef cattle feed are manufactured by a company, Regular and Premium, with the following characteristics where DE is digestible energy: Regular | Premium Selling price (\$/kg) \$4.40 \$5.65 Maximum kg demanded 65,000 38,000 Protein quality per kg minimum 14% 16% DE (Mcal/kg) minimum 2.9 32 This cattle feed is made by combining two ingredients, Alfalfa and Barley. Alfalfa is a plant of the pea family grown primarily for forage. Barley is a cereal crop. They have the following characteristics: Alfalfa | Barley Cost (\$/kg) \$1.90 | \$2.50 Availability (kg) 88,500 | 100,000 Protein quality per kg 17.7% | 12.3% DE per kg 2.60 3.65 Use the following decision variables to formulate the linear programming model with 9 constraints to maximize profit. Ag = kilograms of Alfalfa in Regular Ap = kilograms of Alfalfa in Premium Br = kilograms of Barley in Regular Bp = kilograms of Barley in Premium Hints: To formulate the OFC (Profit = Price ā Cost) you need only consider the cost of the Alfalfa and Barley put into either of the two different types of feed. Ar + Bgis the kilograms of regular feed. Ar + Apis the kilograms of Alfalfa. Percentages can be converted to proportions in the constraints.
SOLUTION For the OFC take the selling price of the cattle feed and subtracf: the cost of the ingredient (everything is in kilograms). Students do not need to show this work: ' Ag = kilograms of Alfalfa in Regular \$4.40 - \$1.90 = \$2.50 As = kilograms of Alfalfa in Premium \$5.65 - \$1.90 = \$3.75 Br = kilograms of Barley in Regular \$4.40 - \$2.50 = \$1.90 Bs = kilograms of Barley in Premium \$5.65 - \$2.50 = \$3.15 Another helpful thing to realize is that (Ar + Br) = Regular (kg) and (Ap+ Bp) = Premium (kg) cattle feed MAXIMIZE 2.50 Ag + 3.75 Ap + 1.90 Bg +3.15 Bp subject to Max Demand Regular (kg) Ar + A Br < 65,000 Max Demand Premium (kg) Ap+ Bp < 38,000 Min Protein in Regular (kg) 0.177 Ar+ 0.123 Bg > 0.14 *( Ag+ Bg) Min Protein in Premium (kg) 0.177 Ap + 0.123 Bp > 0.16 * (Ap+ Bp) Min DE in Regular (Mcal/kg) 2.6 Ar + 3.65 Bg > 2.9 *(Ar+ Bg) Min DE in Premium (Mcal/kg) 2.6 Ap + 3.65 Bp > 3.2 * (Ap+Bp) Availability Alfalfa (kg) Ar + Ap < 88,500 Availability Barley (kg) Br + Bp < 100,000 Non-negativity Ar, Ap,Bg, Bp >0 ā_āā MGST 391 (Fall 2019) Ass't #1 Question 2 ) Page 2
2. Antique Lamps R Us For question , calculations are not to be shown. Please submit answers on the following single-page template (use a ruler). You can include the problem (and print out this entire page). Solve the following Linear Programming problem graphically, using both an iso-profit line of \$3000 and the corner point method. Ensure that you label the constraints and feasible region clearly. Highlight the feasible region. Explicitly indicate the optimal solution (on the graph or chart) and state the optimal solution in words. Antique Lamps R Us want to determine the number to assemble of their two types of lamps: B(rass) and P(latinum) to maximum profit: Maximize 300B + 500P Subject to Assembly (hours) 3B + 4P < 48 Wiring (hours) B + 2P < 20 Max Brass Demand B < 13 Max Platinum Demand P < 8 Non-negativity BP > 0 Co-ordinates (B,P) | Profit (\$) (0,0) 0 20 P(latinum) 16 12 8 4 0 | 4 8 12 16 20 B(rass) NN MGST 391 Tutorial for Formulation and Graph Page 2
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