Capitalized Equivalent Cost For extremely long projects lives. (40 years and more)
ENGR 301 Lecture 18
(1 + i ) N − 1 1 lim ( P / A, i , N ) = lim N = N→∞ i(1 + i ) i
N→∞
PE(i) = A(P/A, i, N Application of Economic Evaluation Techniques
∞) = A/i
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ENGR 301 Lecture 18
S. El-Omari
Capitalized Equivalent Cost
ENGR 301 Lecture 18
Capitalized Equivalent Cost
A major University has just completed a new engineering complex worth $50 million dollars. A campaign, targeting alumni, is planned to raise for future maintenance costs, which are estimated at $2 million per year. Any unforeseen costs above 2$ million per year would be obtained by raising tuition. Assuming that the university can create a trust fund that earns 8% annually, how much has to be raised now to cover the perpetual string of 2$ million annual costs?
Given: A = $2 million, i = 8% per year, N = ∞ Find: CE(8%) Capitalized cost CE(i) = A/i = $2,000,000 / 0.08 = $25,000,000 After one year $25 million x 8% = $2 million After N years $25 million x 8% = $2 million
S. El-Omari
S. El-Omari
ENGR 301 Lecture 18
Unit Profit /Cost Calculation To obtain a unit profit (or cost) of operating an asset the following steps should be performed; 1. Determine the number of units to be produced each year. 2. Identify the cash flow series associated with production or service over the life of the asset. 3. Calculate PE of the project cash flow then determine AE. 4. Divide AE by the number of units to be produced or service and if it varies, you may need to convert them into equivalent units. S. El-Omari
ENGR 301 Lecture 18
ENGR 301 Lecture 18
Unit Profit /Cost Calculation Example: Equal Annual operating hours Tiger machine tool company is considering the acquisition of a new metal cutting machine. The required initial investment of $75,000 and the projected cash benefits over the 3-year’s project are as follows End of year Net cash flow 0 1 2 3 S. El-Omari
-$75,000 24,400 27,340 55,760
ENGR 301 Lecture 18
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Unit Profit /Cost Calculation
Unit Profit /Cost Calculation
55,760 27,340
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Given: PE(15%) = $3553, N= 3 years, i = 15% 2000 machine hours / year Find: Equivalent savings per machine hour S. El-Omari
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ENGR 301 Lecture 18
AE(15%) = $3553(A/P, 15%, 3) AE(15%) = $1556 Savings per machine hour = $1556 / 2000 hours = $0.78/hour S. El-Omari
Unit Profit /Cost Calculation
ENGR 301 Lecture 18
Unit Profit /Cost Calculation
Example: Unequal Annual operating hours C(1500)
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Given: PE(15%) = $3553, N= 3 years, i = 15% 1500 machine hours / year for 1st year 2500 for 2nd year and 2000 for 3rd year Find: Equivalent savings per machine hour S. El-Omari
ENGR 301 Lecture 18
Make-or-Buy Decision A company is considering either producing or buying empty videocassette. The production rate of cassettes is 79,815 units per week for 48 weeks of operation per year. The planning horizon is 7 years. Find the unit cost for each option and which one is preferred at i = 14%
Equivalent annual savings=[C(1500)(P/F,15%,1) +C(2500)(P/F,15%,2) + C(2000)(P/F,15%,3) ] (A/P,15%,3)= 1975.16 C = $1556 C = $1556 / 1975.16 = $0.79 / hour S. El-Omari
ENGR 301 Lecture 18
•Make option (annual costs): Labor Materials Incremental overhead Total cost / year •Buy option: New loading machine Salvage value at end of 5 years Annual operating costs Labor Purchasing empty cassette ($0.85/unit) Incremental overhead Total Annual operating costs
S. El-Omari
ENGR 301 Lecture 18
S. El-Omari
$1,445,633 $2,048,511 $1,088,110 $4,582,254 $405,000 $45,000 $251,956 $3,256,452 $822,719 $4,331,127
ENGR 301 Lecture 18
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Make-or-Buy Decision
Make-or-Buy Decision
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AE(14%) make = $4,582,254
$405,000 $4,331,127 Production volume = 79,815 units / 48 weeks = 3,831,120 units / year S. El-Omari
ENGR 301 Lecture 18
S. El-Omari
Make-or-Buy Decision Buy Option 0
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ENGR 301 Lecture 18
Make-or-Buy Decision $45,000 7
$405,000 $4,331,127
• Make option: Unit cost = $4,582,254 / 3,831,120= $1.2/unit • Buy option Unit cost = $4,421,376 / 3,831,120= $1.15/unit
Capital recovery cost CR(14%) = $405,000(A/P,14%,7) -45,000(A/F,14%,7) = $90,249 AE(14%) buy = $4,331,127 + $90,249 = $4,421,376
Buying the empty cassette then loading the tape inhouse will save 5 cents / cassette
S. El-Omari
S. El-Omari
ENGR 301 Lecture 18
ENGR 301 Lecture 18
Break–even Point: Cost Reimbursement
Break–even Point: Cost Reimbursement
Is the process of calculating the cost of equipment that correspond to the unit of use of that equipment. Ex. The reimbursement of costs from a company to an employee for the use of his car. Those costs include: Gasoline, oil, tires, and also ownership costs like depreciation, insurance, finance charges, registration, maintenance fee, garage ….etc.
•
S. El-Omari
ENGR 301 Lecture 18
Depreciation is the loss in value of the vehicle during the time it’s owned due to: 1. The passage of time 2. Its mechanical and physical condition 3. The number of kilometers it is driven
• There two types of depreciation 1. Economic depreciation 2. Accounting depreciation (details next week) S. El-Omari
ENGR 301 Lecture 18
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Example: Sam is a sales engineer and owns two cars. One of them is entirely dedicated to his work use. His car is 1998 subcompact automobile purchased with personal savings for $11,000. Sam expects to drive 14,500, 13,000 and 11,500 business kilometers for the next three years. If his interest rate is 6%, what should be the reimbursement rate per kilometers so that Sam can break even. He estimated the costs owning of operating his business car for the first three years as follows
Year Depreciation Scheduled maintenance Insurance Registration fee Total ownership cost Nonscheduled repairs accessories Gasoline and taxes Oil Parking and tolls Total operating cost Total of all costs
S. El-Omari
S. El-Omari
Break–even Point: Cost Reimbursement
ENGR 301 Lecture 18
1st Year 2879 100 635 78 $3692 70 15 688 80 135 $988 $4680
2nd year 1776 132 635 78 $2621 115 13 650 100 125 $1003 $3624
3rd year 1545 172 635 78 $2450 227 12 522 100 110 $971 $3421
ENGR 301 Lecture 18
Break–even Point: Cost Reimbursement Year 1 2 3
Total kilometers driven Reimbursement 14,500 (X)(14,500) = 14,500 X 13,000 (X)(13,000) = 13,000 X 11,500 (X)(11,500) = 11,500 X
Annual equivalent reimbursement [14,500X(P/F,6%,1) +13,000X(P/F,6%,2) +11,500X(P/F,6%,3)] (A/P,6%,3)= $13,038X Annual equivalent costs [4680(P/F,6%,1) +3624(P/F,6%,2) +3421(P/F,6%,3)] (A/P,6%,3)= $3933 13,058X = $3933 X = 30.12 cents per kilometer S. El-Omari
ENGR 301 Lecture 18
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