Value Design of Yellowjacket - Case Study Example

VALUE DESIGN Yellowjacket, Inc., a large textile company, is trying to decide how long it should retain one ofits machines used in the sludge dewatering processes. The machine currently is estimated to have a $35,000 market value and a future market value of $18,000 next year, decreasing $1700 per year over its remaining maximum useful life of 8 years. The operating cost is expected to be $5500 next year, increasing by $450 each year thereafter. If the companys MARR is 15% per year, what is the economic service life of this asset? The economic service life of an asset is the remaining useful life of an asset that results in minimum annual equivalent cost. It is also the period of time during which a fixed asset competitively produces a good or service value. Economic life is calculated by Discounted Cash Flow (DCF) through calculating the total Annual Value (AV) of costs for the years the asset is in useful service. Present worth= $35,000 Worth 1 year from now=$18,000 (the salvage value) Yearly depreciating rate =$1700 Discount rate= 15% Operating cost=$5,500 and increases by $450 every year The economic asset cost of keeping the machine= PS (A/P, 15%, 1) +S (i) + OC =35,000(1.15) + 18,000 (0.15) +5,500 =$48,450 However, the operating cost increases by some margin and so is the yearly depreciation rate. Thus, using the discounting rate over the next 8 years, the economic service life of the machine will be calculated as follows, If N=1, then, AEC1= 35,000 (A/P, 15%, 1) + (18,000-1700)-(5,500+450) =52,050 Discounting to the future value after 8 years, EAC8=35,000 (A/P, 15%, 8) + (18,000-1700)-(5,500+450) =35,000-(16,300)-5050)/ (1+i)t =13,650/1.158 =$4,462 is the economic service value in the next 8 years. Lumberjack Power, operator of a nuclear power plant, is planning to replace its current equipment with some that is more environmentally friendly. The old equipment has annual operating expenses of $6750 and can be kept for 8 more years. The equipment will have a salvage value of $4000, if sold 8 years from now, and has a current market value of $24,000, if it is sold now. The new equipment has an initial cost of $62,000 and has estimated annual operating expenses of $6250 each year. The estimated market value of the new equipment is $19,000 after 8 years of operation. If the companys MARR is 16% per year, should the equipment be replaced? Use a study period of 8 years and the present worth method. The value of the defender (old equipment) after 8 years; Present worth= $24,000 Worth 8 year from now=$4,000 (the salvage value) Discount rate= 16% Annual Operating cost=$6,750 The future value of the defender= the discounted worth-the discounted costs/expense (24,000-4000)(1.16)8- 6,750(8) =$11,568.3 The value of the new equipment (challenger) Value after 8 years= $19,000 Discount rate= 16% Annual Operating cost=$6250 However, the initial cost of the machine =$62,000 The replacement analysis of the two equipments depends on the marginal costs and the expected value within the lap of 8 years. It also depends on three factors; the marginal cost and increasing, the marginal cost available but not increasing, and the no costs at all. The cost of the defender is $54,000 in 8 years whereas the cost of the challenger is $62,000 after 8 years. Thus, under the principle of keeping the one with the lowest cost, then old equipment should not be replaced. Defender Challenger Capital investment, $ 34,250, 5 years ago 38,250 Annual operating expenses, $ 5500 7800 Useful life, years 10 12 Estimated salvage value at the end of useful life 2500 5500 The Defender Present worth= $28,000 Salvage value of the useful life=$2,500 Useful life in years=5 Yearly depreciating rate =$1700 Discount rate= 5% Annual Operating cost=$5,500 The future value of the defender= (28,000-2,500) (1.05)5- 5,500(5) $5,045.18 The Challenger Present worth= $38,250 Salvage value of the useful life=$5,500 Useful life in years=12 Discount rate= 5% Annual Operating cost=$7800 = (38,250-5,500) (1.05)12- 7,800(5) =1,855.73 Under the circumstances of the next 5 years, the current vehicle should not be replaced now, but be replaced after 5 years. For the 1st year, k=1, therefore, salvage value=34000-3300=$30,700 Net annual revenue=5000+600=$5,600 AEC1=34,000(A/P,4%,1)+30,700+5600 =$71,460 AEC2=34,000(A/P,4%, 2)+30,700+5600 =$72,874.4 AEC3=74345.376 AEC4=75875.19 AEC5=77466.198 Considering that the actual value has been calculated using the net revenue earned from the equipment, the best time to abandon the equipment within the 5 years is when it has earned the highest possible value. Thus, the 5th year is the best time to abandon the equipment after maximizing the net revenue. The Inflation-adjustment interest rate= [(1+real interest rate)/1+ (inflation rate)]-1 =[1+7%)/ (1+36.75%)]-1 =[1.07/1.3675]-1 =21.75% An engineer wants to estimate the annual inflation-free cost of owning an aerobic digester system with a capacity of 195 million gallons per day (MGD) for the first 8 years of operation. Company records show that the cost of a similar system with a capacity of 75 MGD was $8 million five years ago. The equipment cost index has increased 26% per year since then, and the future general inflation rate is forecast to be 2% per year. He estimates that the annual operating expenses of the new system would be $440,000 per year for the first two years and increase to $441,500 per year thereafter, due to the increase in maintenance costs. Calculate the annual inflation-free cost of owning the 195-MGD system for the first 8 years. Use a cost-capacity exponent of 0.14 for the system and a market-based MARR of 8% per year. Cost capacity equation relates to the cost ratio of the cost of something now and its cost at sometimes in the past. The cost of a system with capacity of 75MGD =$8 million, valued at 5 years ago. Cost index increased by 26% pa by now, and future inflation forecasted at 2% pa. Ct= C0 (It/I0) $8million (26%/2%) 0.13*8m $1.04million Cost capacity= C1 (Q2/Q1) x C195-MGD=$1.04 (441,500/440,000)0.14 =$1.44049 million Trans-Atlantic Petroleum Corp. plans to seek two additional production licenses from the Romanian government. The company estimates the costs to drill, acquire the seismic data, and conduct technical studies for these new fields will be $14.2 billion, 7 years from now. The company plans to establish a fund in a Romanian bank. The fund earns a rate of return of 5% per year (a rate relative to the Romanian New Lei- RON). How much will the company have to set aside now in U.S. dollars if it is estimated that U.S. dollars will be devalued at an average of 2.25% per year and the present exchange rate is 2.57 RON per U.S. dollar? Cost estimate / (1+r)n= $14.2 billion/(1+5%)7 =14.2 billion/ (1.05)7 =14.2 billion/1.4071 =10.0917 billion Cost now=14.2-10.0917= $4.1083 billion The funds are to earn return at the rate of 5% p.a. Interest =4.1083*5%*1 =$0.205414 billion Devaluation is done at an average of 2.25% 0.205414*2.25% = $0.004621815 billion =0.011870646 billion RON A Caribbean cruise line has purchased a new cruise ship for $35 million and expects to realize a net revenue of $210,000 each year for the next 10 years. The estimated salvage value of the ship at the end of its useful life of 10 years is $134,000. Assume an effective federal tax rate of 40%, a state income tax rate of 7% per year, and an after-tax inflation-free MARR of 8% per year. Calculate the actual dollar present worth of ATCF if straight-line depreciation is used and the average rate of inflation is 4% per year. The salvage value =p (1-i) y $134,000= $35 million (1-i) 10 =$261.194 (1-i) 10 Net revenue=$210,000 every year Federal Tax rate=40% State income tax rate=7% After-tax inflation free MARR=8% Actual dollar present=$2.1 million*63%*10 years 2.1*0.63*10 =$13.23 million A dentist is deciding between two X-ray machines for his new office. Estimated costs for each machine are given below. Machine A B Installed cost $50,000 $52,000 Annual maintenance cost $1500 $1200 Market value at year 10 $5750 $6250 Life, years 10 10 Which machine should be recommended based on the annual worth method and an actual-dollar analysis? Use an inflation-free MARR of 14%, an inflation rate of 6.5% per year, and a study period of 10 years. 1+MARR combined= (1+inflation) (1+MARR real) PWA= (5750*21.5%)1+(1500+50000) =1236.25+6500 =$7736.25 PWB=(6250*21.5%)1+(1200+52000) =1343.75+53200 =$54543.75 Based on the annual worth method and an actual dollar analysis, the best machine is machine B due to its higher value than machine A. Cougar Telemarketing is considering establishing a call center. The initial cost will be $2,750,000 with a $27,500 market value any time within a 13-year period. The fixed cost of the center will be $830,000 per year with an average variable cost of $3.00 per call. Cougar expects to generate revenue of $5.25 per call with a capacity of 110,000 calls for the first year. The company also expects to increase the capacity uniformly each year. At an interest rate of 2% per year, determine the uniform amount the capacity must increase each year so that the company can recover its investment in 3 years. The initial cost=$2,750,000 Within the 13 year period, the market value=$27,500 Total operational costs= [830,000+(3*110,000)] =$1,160,000 the first year The revenue per call=$5.25 Thus, out of the 110,000 calls, the revenue=$557,500 In order to gain the initial investment after three years, the company need to recover ; Cost=830000+990,000=1829000 Revenue=1672500 [27,500+1672500]-1829000*2% =$16622200 Two machines are under consideration for a new production line. Machine X costs $50,000 and is expected to have a salvage value of $6500 at the end of its useful life of 5 years. It will have a fixed cost of $16,000 per year and a variable cost of $55 per unit per year. On the other hand, machine Y costs $55,000 and is expected to have a salvage value of $7000 at the end of its useful life of 7 years. It will have a fixed cost of $14,500 per year and a variable cost of $58 per unit per year. Determine the quantity that must be produced for the two machines to break even at an interest rate of 3% per year. Machine X =50,000-6500)+(1.03)5- 16275 =$27,226.159 Machine Y =55,000-7000]+(1.03)7-14906 =$33,095.2298 Work cited Brigham, Eugene F, and Joel F. Houston. Fundamentals of Financial Management. Mason, Ohio: Thomson/South-Western, 2004. Print. Goetzmann, William N, and Roger G. Ibbotson. The Equity Risk Premium: Essays and Explorations. Oxford: Oxford University Press, USA, 2006. Internet resource. Read More