the final answer should be expressed in 4 decimal places and follow the given format (given,solution, formula used) GENERAL INSTRUCTIONS: ENSC 108 ENGINEERING ECONOMY Read the questions carefully DATE: . . Use BLACK BALLPEN Write all your answers and solutions on the SOLUTION SHEET. SCORE . Write clearly and legibly, avoid making erasures. 160 COMPUTATION. Compute the following problems using four decimal places and box your final answer. Be sure to show your solution on the sheet. 1. A P100,000 item is purchased. Annual maintenance and operational costs are P18,000. Using 8%, what is the capitalized cost and annual cost of perpetual service? (6 points) 2. A manufacturing plant installed a new boiler at a total cost of P150,000 and is estimated to have a useful life of 10 years. It is estimated to have a scrap value at the end of its useful life of P5,000. If interest is 12% compounded annually, determine its capitalized cost and annual cost. (6 points) 3. A research foundation wishes to set up a trust fund earning 10% compounded annually to: a. Provide P2,000,000 for the lot and building and P1,000,000 for the initial equipment of a Structural Engineering and Materials Laboratory b. Pay P400,000 for the annual operating costs every year c. Pay P500,000 for the purchase of new equipment and replacement of some equipment every 5 years beginning 5 years from now How much money should be paid into the fund for the building and equipment and to pay for perpetual operation and equipment replacement? (3 points) 4. A heat exchanger is needed in a chemical process. If interest is 9% compounded annually. determine which of the following heat exchangers is cheaper by comparing the capitalized cost and annual cost. (15 points) Exchanger A costs P22,000 with a scrap value of P1,000 and a useful life of 7 years. Exchanger B costs P28,000 with a scrap value of P1.500 and a useful life of 10 years 5. Find the capitalized cost and annual cost of an asset whose cost is P100,000, salvage value is P10,000, life is 15 years at 5%. (6 points) 6. An asset was purchased for P100,000 and retired at the end of 15 years with a salvage value of P4,000. The annual operating cost was P18,000. Determine the capitalized cost and annual cost of the asset based on an interest rate of 8%. (6 points) 7. An item is purchased for P100,000. Annual cost is P18,000. Using interest rate of 8%, what is the capitalized cost of perpetual service? (3 points) 8. A bridge that was constructed at a cost of P75,000 is expected to last 30 years, at the end of which time its renewal cost will be P40,000. Annual repairs and maintenance are P3,000. What is the capitalized and annual cost of the bridge at an interest of 6%? (6 points)9. If a machine cost P3,000 with life 15 years and a final salvage value of P500, compute the capitalized cost money if money is worth 5%. (3 points) 10. A financial analysis of two types of bridges based on capitalized cost and on the following data is to be made: (6 points) Bridge A: Initial cost: P200,000 Cost of Renewal: P200,000 Salvage Value: PO Annual Maintenance: P1,000 Life: 30 years Bridge B: Initial cost: P240,000 Cost of Renewal: P240,000 Salvage Value: P20,000 . Annual Maintenance: None . Life: 40 years If the rate of interest is 8% compounded annually, determine the capitalized cost of each.follow this format: SAMPLE PROBLEMS Example 1: A machine costs P300,000 new, and must be replaced at the end of each 15 years. If the annual maintenance cost required is P5,000 find the capitalized cost, if money is worth 5% and the final salvage value is P50,000. Bulacan Agricultural State College Institute of Engineering and Applied Technology LECTURE 3: CAPITALIZED COST AND ANNUAL COST Agricultural and Biosystems EngineeringSAMPLE PROBLEMS Given: Solution: FC = P300, 000 i = 5% K = RC -SV -+ FC + MC (1+i)n -1 n = 15 years SV = P50,000 300,000 -50,009 + 300.000 + 5.000 MC = P5,000 (1+0.05)15 -1 0.05 RC = P300,000 = Php 631,711.438 Here are the formulas that will be used: WHAT IS ANNUAL COST? > It is defined by the formula AC = (RC -SV) + FC (i) + MC AC = K(i) (1+i)n -1 WHERE: K = Capitalized Cost FC = First Cost MC = Operation/Maintenance Cost i = Interest Rate RC = Replacement Cost ( if RC is not specified, use RC = FC ) SV = Salvage Value LECTURE 3: CAPITALIZED COST AND ANNUAL COSTWHAT IS CAPITALIZED COST? > It is defined by the formula K= RC -SV + FC + MC (1+i)n -1 i WHERE: K = Capitalized Cost FC = First Cost MC = Operation/Maintenance Cost i = Interest Rate RC = Replacement Cost ( if RC is not specified, use RC = FC ) SV = Salvage Value Bulacan Agricultural State College institute of Engineering and Applied Techno LECTURE 3: CAPITALIZED COST AND ANNUAL COST nibmal and