Please refer to the question and to the reference. Please show the step-by-step solution in a clear format thanks.

Analyse and solve the following problems. 1. Two alternative designs are under consideration for a tapered fastening pin. The fastening pins are sold for Php.?[l each. Either design will serve cquall}T well and will involve the same material and manufacturing cost except for the lathe and drill operations. Design A will require IE hours of lathe time and 4.5 hours of drill time per 1.0% units. Design B will require '1" hours of lathe lime and 12 hours of drill time per Loco units. The variable operating cost of the lathe, including labor, is Php 18.61:) per hour. The variable operating cost of the drill, including labor, is Phpl per hour. Finally, there is a sunk. cost of Php for Design A and Php9,D for Design E due to obsolete tooling. a. 1Which design should be adopted if 125,0(10 units are sold each jl'ear'?I b. 1What is the annual saving over the other design? 5' SOLUTION: 2. A bicycle component manufacturer produces hubs for bike wheels. Two processes are possible for manufacturing, and the parameters of each process are as follows: -- - ducnn me _ visual ins - action Assume that the daily demand for hubs allows all defectfree hubs to be sold. Additionally, tested or rejected hubs cannot be sold. Find the process that maximizes prot per day if each part is made from th4- worth of material and can be sold for PhpSCICI. Both processes are fully automated, and variable overhead cost is charged at the rate of Php4-{l per hour. SOLUTION: 3. One method for developing a mine containing an estimated 100,000 tons of ore will result in the recovery of 62% of the available ore deposit and will cost Php23 per ton of material removed. A second method of development will recover only 50% of the ore deposit, but it will cost only Php15 per ton of materials removed. Subsequent processing of the removed ore recovers 300 pounds of metal from each ton of processed ore and costs Php40 per ton of ore processed. The recovered metal can be sold for Php0.80 per pound. Which method for developing the mine should be used if your objective is to maximize total profit from the mine? SOLUTION:PRESENT ECONOMY STUDIES When alternatives for accomplishing a specific task are being compared over one year or less and the influence of time on money can be ignored, engineering economic analyses are referred to as present economy studies. Several situations involving present economy studies are illustrated here. The rules, or criteria, will be used to select the preferred alternative when defect-free output (yield) is variable or constant among the alternatives being considered. When revenues and other economic benefits are present and vary among RULE 1: alternatives, choose the alternative that maximizes overall profitability based on the number of defect-free units of a product or service produced. When revenues and other economic benefits are not present or are RULE 2: constant among all alternatives, consider only the costs and select the alternative that minimizes the total cost per defect-free unit of product or service output. Present economy studies occur in the following situations: Example 2.4 (Selection of Material) A diesel engine uses Type A filter and high-grade lubricating oil costing Php5.50 per liter. With this filter, the oil and the filter have to be changed every 500 hours of operation, and 5 liters of oil have to be added every 100 hours. This filter costs Php148 a piece. Eighty liters of oil fill the engine. Another type, filter B, costing Php120 may be used with a lower grade of oil costing Php4.80 per liter. However, if this filter is used, the oil and filter have to be changed every 300 hours and 10 liters are added after each 150 hours the engine is used. Which type of filter and oil would you recommend? A SOLUTION: Set up a table to examine the total costs: Cost Items Type A Filter Type B Filter Cost of filter Php 148 Php 120 Initial Cost of oil 5.50/L x 80L = 440 4.80/L x 80L = 384 Additional Cost of oil 4 x 5.50/L x 5L = 110 1 x 4.80/L x 10L = 48 Total Cost Php 698 Phph 552 Lifespan (hours) 500 300 Cost per hour 698/500