Introduction to Quantitative Management - Assignment Example

Lecturer number My registration number is XXXX96. Then you take d = 96. So, b = d + 17 = 96 + 17 = 113. 2. From the Texxon data, the first step is completing the calculation of the exact values of all the parameters by replacing the missing value of “b = 113.” This is done as shown below: The profits realized per barrel of the blended 3 petrol grades A B C D E Profit Standard 14% 25% 22% 14% 25% 66 +0.02*b = 66 + 0.02*113 =66 + 2.26 = 68.26 Super 20% 20% 20% 18% 22% 80 Hi-Grade 25% 15% 15% 25% 23% 90 The Supply contraction for the 3 petrol grades Grade Order (barrels) Standard 20,000 Super 6,000 Hi-Grade 12,000 + b*20 = 12,000 + 113*20 =12,000 + 2260 =14,260 With the data now complete, we proceed to the next step. The contract for the supply of (12,000 + b*30) barrels of Hi-grade to be signed is equivalent to: 12,000 + b*30 where b = 113. So, 12,000 + b*30 = 12,000 + 113*30 = 12,000 + 3390 = 15,390 barrels of Hi-Grade petrol Report to Texxon Management: In order for Texxon to meet the current demand for the already contracted petrol grades, then, the production has to be met fully by the shipped crude oil types. However, for the company to maximize on its returns, it should cider undertaking the most profitable production. Below is a summary of the returns realized if the company produces only all units of any single petrol grade with the already ordered materials. I unit of crude material produces x units of grade petrol Barrels of crude Standard Super Hi-Grade A: 12,000 1/0.14 = 7.14286 1/0.20 = 5 1/0.25 = 4 B: 15,000 1/0.20 = 5 1/0.20 = 5 1/0.15 = 6.66667 C: 15,000 1/0.25 =4 1/0.20 = 5 1/0.12 = 8.33333 D: 12,000 1/0.14 = 7.14286 1/0.18 = 5.55556 1/0.25 = 4 E: 15,000 1/.25 = 4 1/0.22 = 4.54545 1/0.23 = 4.34783 So, from the table above, since the materials have to be used together, the maximum number of units each of the crude materials can produce is listed below. No. unit of blend grade petrol available material can produce Barrels of crude Standard Super Hi-Grade A: 12,000 7.14286*12,000 =85,714.32 5*12,000 = 60,000 4*12,000 = 48,000 B: 15,000 5*15,000 = 75,000 5*15,000 = 75,000 6.66667*15,000 =100,000.5 C: 15,000 4*15,000 = 60,000 5*15,000 = 75,000 8.33333*15,000 =124,999.5 D: 12,000 7.14286*12,000 = 85,714.32 5.55556*12,000 = 66,666.72 4*12,000 = 48,000 E: 15,000 4*14,000 = 60,000 4.54545*15,000 = 68,181.75 4.34783*15,000 = 65,217.45 Since the materials must be used together, the maximum numbers of individual blended units the materials available can make are as shown below i.e. the minimum number of units that available for all materials. Grade Petrol Number of grade oil that can be made Standard 60,000 Super 60,000 Hi-Grade 48,000 The profits table from processing of individual units of the blend alone is as shown below. Grade Petrol Number of materials that can be made Profits from the sales. Standard 60,000 60,000*68.26 = 4,095,600 Super 60,000 60,000*80 = 4,800,000 Hi-Grade 48,000 48,000*90 = 4,320,000 So, for maximum profits, processing of all units of super grade petrol would be the most profitable. i.e. $4,800,000 This is the maximum profit that can be gained: 4,800,000 Amounts of each type of crude oil unused: This is the number of units of oil above what is used in the processing of the minimum complete grade oil units. Crude Oil purchased Amount used Balance of material unused A: 12,000 48,000/5 = 9,600 12,000 – 9,600 = 2,400 B: 15,000 48,000/5 = 9,600 15,000 – 96,000 =5,400 C: 15,000 48,000/5 = 9,600 15,000 – 96,000 =5,400 D: 12,000 48,000/5.55556 = 8,639.99 12,000 - 8,639.99 = 3,360.01 E: 15,000 48,000/4.54545 = 10,560.01 15,000 - 10,560.01 = 4,439.99 Whether to sign the further deal Yes. Since on maximizing the profits the materials left can produce the required 14,260 barrels of Hi-Grade oil, then, the company should sign the deal. 1. Olympic Estates Construction From the information given, the net profit realized when the expert has been consulted is less $200,000 (= $0.2M). Consequently, in the tree decision tree, this is reflected by having its effect indicated on the last set of branches of the decision tree as shown below. Below is a summary of the returns realizable through the various strategies available to the company. Strategy Hire Expert Demand Favourability of outcome Probability Returns realizable (in $1 million) P1 Yes Strong Favourable 0.527*0.8 = 0.4216 2.2 P2 Yes Strong Favourable 0.527*0.8 = 0.4216 1.8 P3 Yes Strong Favourable 0.527*0.8 = 0.4216 7 P1 Yes Strong Not Favourable 0.527*0.2 = 0.1054 2.2 P2 Yes Strong Not Favourable 0.527*0.2 = 0.1054 1.8 P3 Yes Strong Not Favourable 0.527*0.2 = 0.1054 7 P1 No Weak Favourable 0.473 *0.3 = 0.1419 2.2 P2 No Weak Favourable 0.473 *0.3 = 0.1419 1.8 P3 No Weak Favourable 0.473 *0.3 = 0.1419 7 P1 No Weak Not Favourable 0.473 *0.7 = 0.3311 2.2 P2 No Weak Not Favourable 0.473 *0.7 = 0.3311 1.8 P3 No Weak Not Favourable 0.473 *0.7 = 0.3311 7 The table above further shows that hiring the expert offers indifferent returns. Therefore, considering the intermediate expected returns when the expert is not hired are higher as shown below. Strategy Demand Returns realizable (in $1 million) P1 Strong 2.4 P2 Strong 4.8 P3 Strong 7.2 P1 Weak 2 P2 Weak 1.2 P3 Weak -4.4 From the decision tree above, the optimal returns would be realized when the company hires the services of the expert in addition to undertaking construction of the large park (P3). This is evident from the high returns of $7.2 million expected to be realized and the high probability of 0.416 of the returns being realized. However, if the expert is hired, he/she would be paid $0.2 which would reduce the profits expected to be realized. For optimal returns, the company should consider commissioning the expert to undertake the survey for the development of a 290 unit park (P3). Reference Lawrence J. A. & Pasternack B. A. (2002) Applied Management Science, 2nd edition Read More