# A01_Tan - Group Tan everyone did equal share Brendon...

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Unformatted text preview: Group Tan : everyone did equal share Brendon Beede-33.3% Mark Fitchett- 33.3% Trisha McCuen-33% A. The Production facility cost of setting up would need to be three times what it is right now in order for th grandfather clocks cease to be profitable. This is because before when the setup cost was 50,000 the brea even cost was 100. Thus when the setup cost is 100,000 the break even cost is 200, and so when the setu is 150,000 the setup cost is 300. 300 units is equal to the units that they are able to sell. Here is an image represent what I did in Excel format. B. The Marginal Production cost can be \$733(733.33) before the grandfather clocks cease to be profitable. T because that they will not be making enough profit after that point to cover fixed cost. Here is the equation 9 733= 167, 167*300=50,100. fixed cost for production is 50,000. If the company were to go ahead and prod these clocks at 733 they would only be making \$100 profit. here is a photo of the marginal cost just one dollar over. This shows they are losing money on it. C.Unit revenue can be as less as \$566.67(or \$567) before the grandfather clock production ceases to be prof This is because when Unit Revenue- Marginal Cost = 126 , then you must times that by how many products here is a photo of the marginal cost just one dollar over. This shows they are losing money on it. C.Unit revenue can be as less as \$566.67(or \$567) before the grandfather clock production ceases to be prof This is because when Unit Revenue- Marginal Cost = 126 , then you must times that by how many products ght now in order for the st was 50,000 the break, and so when the setup cost sell. Here is an image to ease to be profitable. This is Here is the equation 900e to go ahead and produce oney on it. ction ceases to be profitable. y how many products you oney on it. ction ceases to be profitable. y how many products you A. J ennifer needs to make the following decision: How many units to ship from plant A to both R etail O utlet 1 and 2 Also, she needs to find how many units to ship from plant B to both retail outlet 1 and 2. T must be done by his minimizing shipping costs while staying within the supply limits of each plant and meeting the customers’ demands. T can be done by finding the following variables: P his roducts_S hipped, T otal_C ost, T otal _R eceived and T otal_S hipped BT . otal S hipping C can be represented mathematically by: =S ost UMP O UC (B 4,P R D T 3:C roducts_S hipped) or T ota S hipping C osts = (\$700*0)+(\$400 *30)+(\$800*40)+(\$600*10) CT . otal_S hipped = 80, T otal_R ecieved = 80 and P roducts_S hipped >= 0 D Here is a cut and paste of the mathematical model from excel: . E I feel J . ennifer’s shipping plan should include maximizing the production of P lant B in order to ship more units to R etail O utlet 2. Also, she should maximize the units shipped out of P lant A to the R etail O utlet 2. D oing this will help her increase sales by shipping over the minimum of 25 units to R etail O utlet 2, and will increase her profit margin since the shipping costs are almost half the price of shipping to R etail O utlet 1. S hipping P lan: R etail O utlet 1 will have 0 P roducts_S hipped from P lant A, and 40 P roducts_S hipped from P lant B . R etail O utlet 2 will have 30 P roducts_S hipped from P lant A and 10 P roducts_S hipped from plant B . S hipping P lan: R etail O utlet 1 will have 0 P roducts_S hipped from P lant A, and 40 P roducts_S hipped from P lant B . R etail O utlet 2 will have 30 P roducts_S hipped from P lant A and 10 P roducts_S hipped from plant B . oth R etail O utlet 1 and 2. T must be done by his g t he customers’ C ost, T otal _R eceived cts_S hipped) or T otal er to ship more units to utlet 2. D oing this will will increase her profit P lant B . m plant B . P lant B . m plant B . 2.10 (a) Decision: How many units of each product to produce so as to maximize profit? Constraints: Number of units of frame parts is limited to 200 and number of electrical components is limited to 300 available. Measure of Performance: Total Profit (b) Maximize Profit = \$1(Units of Product 1) + [\$2(Units of Product 2) if Product 2 < 60, otherwise \$0] Product 1 = 1(frame parts) + 2(electrical components) Product 2 = 3(frame parts) + 2(electrical components) where frame parts < 200 and electrical components < 300 and frame parts > 0 and electrical components > 0 (c) WorldLight Company Unit Profit Product 1 \$1 Product 2 \$2 Resources Available 200 300 Total Profit \$175 Frames Electrical 1 2 Product 1 3 2 Product 2 25 ≤ ≤ Units Produced 125 Cells F5:F6 C7:D8 F11 C4:D4 C11:D11 Range Name ResourcesAvailable ResourcesUsedPerUnitProduced TotalProfit UnitProfit UnitsProduced (d) Maximize Profit: Z = (x1) + 2(x2) (x1) + 3(x2) < 200 2(x1) + 2(x2) < 300 and x1 > 0 x2 > 0 (e) I used the Graphical Linear Programming and Sensitivity Analysis in the Interactive Management Science Module to solve the problem. The resulting profit (Z) is 175. Z = 175 with x1 = 125 and x2 = 25 ...
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