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Chapter 6 Homework, Day 2
Exercises 6-6, 7, 9, 13; Problems 6-1, 3
E6-6 (Future Value and Present Value Problems) Presented below are three unrelated situations. a. Ron Stein Company recently signed a lease for a new office building, for a lease period of 10 years. Under the lease agreement, a security deposit of $12,000 is made, with the deposit to be returned at the expiration of the lease, with interest compounded at 10% per year. What amount will the company receive at the time the lease expires? Future Value of 1 factor (n = 10, i = 10%) is 2.59374 $12,000 * 2.59374 = $31,124. b. Kate Greenway Corporation, having recently issued a $20 million, 15-year bond issue, is committed to make annual sinking fund deposits of $620,000. The deposits are made on the last day of each year and yield a return of 10%. Will the fund at the end of 15 years be sufficient to retire the bonds? If not, what will the deficiency be? Future Value of an Ordinary Annuity of 1 factor (n = 15, i = 10%) is 31. $620,000 * 31.77248 = $19,698,937.60; the fund will not be sufficient to retire the bonds – it will end up being short $301, 062.40 of its goal. c. Under the terms of his salary agreement, president Juan Rivera has an option of receiving either an immediate bonus of $40,000, or a deferred bonus of $75,000 payable in 10 years. Ignoring tax considerations, and assuming a relevant interest rate of 8%, which form of settlement should Rivera accept? Present Value of 1 Factor (n = 10, i = 8%) is. $75,000 * .46319 = $34,739. Future Value of 1 Factor (n = 10, i = 8%) is 2. $40,000 * 2.15892 = $86,356. Rivera should accept the immediate bonus of $40,000, because it is worth more in the present as well as the future, and he will also then have the money immediately to invest or do as he chooses with it instead of delaying the satisfaction for ten years.
E6-7 (Computation of Bond Prices) What would you pay for a $100,000 debenture bond that matures in 15 years and pays $10,000 a year in interest if you wanted to earn a yield of: a. 8%? Present Value of 1 factor (n = 15, i = 8%) is .31254 $100,000 * .31524 = $31,524. Present Value of an Ordinary Annuity of 1 (n = 15, i = 8%) factor is 8. $10,000 * 8.55948 = $85,594.80 $31, 524.00 + $85,594.80 = $117,118. b. 10%? Present Value of 1 factor (n = 15, i = 10%) is .23939 $100,000 * .23939 = $23,939. Present Value of an Ordinary Annuity of 1 (n = 15, i = 10%) factor is 7. $10,000 * 7.60608= $76,060.80 $23, 939.00 + $76,060.80 = $99,999. c. 12%? Present Value of 1 factor (n = 15, i = 12%) is .18270 $100,000 * .18270 = $18,270. Present Value of an Ordinary Annuity of 1 (n = 15, i = 12%) factor is 6. $10,000 * 6.81086 = $68,108.60 $18,270.00 + $68, 108.60 = $86,378. E6-9 (Unknown Rate) Kross Company purchased a machine at a price of $100,000 by signing a note payable, which requires a single payment of $118,810 in 2 years. Assuming annual compounding of interest, what rate of interest is being paid on the loan? $118, 810 = 100,000 * Future Value Factor (n = 2, i = ?) Future Value Factor = $118,810 / $100,000; read across n = 2 row to find interest rate. Here, rate is 9%. E6-13 (Computation of Bond Liability) Messier Inc. manufactures cycling equipment. Recently, the vice president of operations of the company has requested construction of a new plant to meet the increasing demand for the company’s bikes. After a careful evaluation of the request, the board of directors has decided to raise funds for the new plant by issuing $3,000,000 of 11% term corporate bonds on March 1, 2012, due on March 1, 2027, with interest payable each March 1 and September 1. At the time of the issuance, the market interest rate for similar financial instruments is 10%.
c. Fishbone Corporation bought a new machine and agreed to pay for it in equal annual installments of $4,000 at the end of each of the next 10 years. Assuming that a prevailing interest rate of 8% applies to this contract, how much should Fishbone record as the cost of the machine? Principal = $4,000.00 n = 10 i = 8% Present Value of an Ordinary Annuity of 1 factor (n = 10, i = 8%) is 6. $4,000.00 * 6.71008 = $26,840. d. Fishbone Corporation purchased a special tractor on December 31, 2012. The purchase agreement stipulated that Fishbone should pay $20,000 at the time of purchase and $5,000 at the end of each of the next 8 years. The tractor should be recorded on December 31, 2012, at what amount, assuming an appropriate interest rate of 12%? Principle = $5,000.00 n = 8 i = 12% Present Value of an Ordinary Annuity of 1 factor (n = 8, i = 12%) is 4. $5,000.00 * 4.96764 = $24, 838.20 $24,838.20 + $20,000.00 = $44, 838. e. Fishbone Corporation wants to withdraw $120,000 (including principal) from an investment fund at the end of each year for 9 years. What should be the required initial investment at the beginning of the first year if the funds earn 11%? Principal = $120,000.00 n = 9 i = 11% Present Value of an Ordinary Annuity of 1 factor (n = 9, i = 11%) is 5. $120,000.00 * 5.53705 = $664,446. P6-3 (Analysis of Alternatives) Assume that Wal-Mart Stores, Inc. has decided to surface and maintain for 10 years a vacant lot next to one of its stores to serve as a parking lot for customers. Management is considering the following bids involving two different qualities of surfacing for a parking area of 12,000 square yards. Bid A: A surface that costs $5.75 per square yard to install. This surface will have to be replaced at the end of 5 years. The annual maintenance cost on this surface is estimated at 25 cents per square yard for each year except the last year of its service. The replacement surface will be similar to the initial surface. Cost to install: 12,000 * 5.75 = $69,000.00 Present Value of 1 (n = 5, i = 9%) figure is. Cost to re-install: $69,000.00 * .64993 = $44,845. Total Installation Cost: $69,000.00 + $44,845.17 = $113,845. Present Value of an Ordinary Annuity of 1 factor (n = 4, i = 9%) is 3.
Maintenance Cost: 12,000 * $.25 = $3,000.00 $3,000.00 * 3.23972 = $9,719.16 (Years 1-4) Present Value of an Ordinary Annuity of 1 factor (n = 9-5, i = 9%) = 5.99525 – 3.88965 = 2. $3000.00 * 2.1056 = $6,316.80 (Years 6-9) Total Maintenance Cost: $9,719.16 + $6,316.80 = $16,035. Total Cost: $113,845.17 + $16,035.96 = $129,881. Bid B: A surface that costs $10.50 per square yard to install. This surface has a probable useful life of 10 years and will require annual maintenance in each year except the last year, at an estimated cost of 9 cents per square yard. Cost to install: 12,000 * $10.50 = $126,000. Present Value of an Ordinary Annuity of 1, figure (n = 9, i = 9%) is 5. Maintenance Cost: 12,000 * $.09 = $1,080.00 $1,080.00 * 5.99525 = $6,474. Total Cost: $126,000.00 + $6,474.87 = $132,474. Instructions: Prepare computations showing which bid should be accepted by Wal-Mart. You may assume that the cost of capital is 9%, that the annual maintenance expenditures are incurred at the end of each year, and that prices are not expected to change during the next 10 years. Bid A should be accepted because its present value lower; the total estimated cost of the project for Bid A comes to only $129,881.13 (in today’s terms) while Bid B’s total estimated costs are $132,474. (in today’s terms).
