Problem 6.7 – NOT NEEDED
Humana Hospital Corporation installed a new MRI machine at a cost of $750,000 this year in its medical professional clinic in Cedar Park. This state-of-the-art system is expected to be used for 5 years and then sold for $125,000. Humana uses a return requirement of 25% per year for all of its medical diagnostic equipment. As a bioengineering student currently serving a coop semester on the management staff of Humana Hospital Corporation in Louisville, Kentucky, you are asked to determine the minimum revenue required each year to realize the expected recovery and return.
(a) What is your answer?
(b) If the AOC is expected to be $80,000 per year, what is the total revenue required to provide for recovery of capital, the 25% return, and the annual expenses?
Total revenue =$80000 + $263625 = $343625
(c) Write the spreadsheet functions to display your answers.
Excel function. 80000- PMT (25%, 5, 75000, 0) + PMT (25%,5,0,-125000)
Problem 6.8 White Oaks Properties builds strip shopping centers and small malls. The company plans to replace its refrigeration, cooking, and HVAC equipment with newer models in one entire center built 9 years ago. The original purchase price of the equipment was $638,000 nine years ago and the operating cost has averaged $240,000 per year. Determine the equivalent annual cost of the equipment if the company can now sell it for $184,000. The company’s MARR is 25% per year
The equivalent annual cost of the equipment if the company can now sell it for $184,000. The company’s MARR is 25% per year is -$417070
Chapter 6: Problems 6.3 using factors only, 6.5, 6.8, 6.15, 6.19, 6.26, 6.28 & 6.34
Problem 6.3: A machine used to shread cardboard boxes for composting has a first cost of $10,000, an AOC of $7000 per year, a 3-year life, and no appreciable salvage value. Assume you were told that the service provided by the machine will be needed for only 5 years, which entails a repurchase and retention for only 2 years.What salvage value, S, is required after 2 years in order to make the 2-year AW value the same as it is for its 3-year life cycle AW? Assume i = 10% per year. Solve using factors and, if assigned, using a spreadsheet and Goal Seek.
Solution:
-10,000(A/P,10%,3) – 7000 = -10,000(A/P,10%,2) – 7000 + S(A/F,10%,2)
Expand from above formula
-10,000(0.40211) – 7000 = -10,000(0.57619) – 7000 + S(0.47619)
-11,021 = -12,762 + 0.47619S
S = $3656
Problem 6.5 As discussed in Section 6.2, either of the following two equations can be applied to determine the amount necessary to recover invested capital and a required return:
CR1 = ?P(A?P,i,n) + S(A?F,i,n)
CR2 = ?[(P – S)(A/P,i,n) + S(i)]
For an alternative that has a first cost of $50,000 and a salvage value of $5,000 after its 5-year life, show that the capital recovery calculated using either of these equations is exactly the same. Use an interest rate of 10% per year.
CR1 = -50,000(A/P,10%,5) + 5,000(A/F,10%,5)
CR2 = – [(50,000 – 5,000) (A/P,10%,5) + 5,000(0.1)]
So we see the answer is same both formula we got the same capital inventory
Problem 6.8: White Oaks Properties builds strip shopping centers and small malls. The company plans to replace its refrigeration, cooking, and HVAC equipment with newer models in one entire center built 9 years ago. The original purchase price of the equipment was $638,000 nine years ago and the operating cost has averaged $240,000 per year. Determine the equivalent annual cost of the equipment if the company can now sell it for $184,000. The company’s MARR is 25% per year
Solution:
AW = -638,000(A/P,25%,9) – 240,000 + 184,000(A/F,25%,9)
The equivalent annual cost of the equipment if the company can now sell it for $184,000. The company’s MARR is 25% per year is -$417,097
Problem 6.15: Two methods can be used for producing solar panels for electric power generation. Method 1 will have an initial cost of $550,000, an AOC of $160,000 per year, and $125,000 salvage value after its 3-year life. Method 2 will cost $830,000 with an AOC of $120,000, and a $240,000 salvage value after its 5-year life. Assume your boss asked you to determine which method is better, but she wants the analysis done over a 3-year planning period. You estimate the salvage value of method 2 will be 35% higher after 3 years than it is after 5 years. If the MARR is 10% per year, which method should the company select?
Solution:
Method 1
AW = -550,000(A/P,10%,3) – 160,000 + 125,000(A/F,10%,3)
Method 2
AW = -830,000(A/P,10%,3) – 120,000 + 240,000(1.35) (A/F,10%,3)
Salvage value in 3 years=240,000*1.35=324000
So Method 1 is lower so it should be selected.
Problem 6.19: Two types of robots (Cartesian and Articulated) with the following estimates are under consideration for a dishwasher assembly process. Using an interest rate of 10% per year, determine which one should be selected on the basis of an annual worth analysis. Robot Cartesian Articulated First cost, $ ?300,000 ?430,000 AOC, $/year ?60,000 ?40,000 Salvage value, $ 70,000 95,000 Life, years 4 6
Solution:
Cartesian
AW Cartesian = -300,000(A/P,10%,4) – 60,000 + 70,000(A/F,10%,4)
Articulated
AWArt iculated = -430,000(A/P,10%,6) – 40,000 + 95,000(A/F,10%,6)
Based on annual worth analysis the Articulated is better as it has higher annual worth.
Problem 6.26 Calculate the perpetual equivalent annual cost (years 1 to ?) of $1,000,000 now and $1,000,000 three years from now at an interest rate of 10% per year
Annual equivalent cost = 1,751,300 x 0.10 = 175,130
Or
AW = 1,000,000 + 1,000,000(P/F,10%,3) * (0.10)
= 1,000,000 + 1,000,000(0.7513) * (0.10)
= $175,130
Problem 6.28 The State of Chiapas, Mexico, decided to fund a program for improving reading skills in elementary school students. The first cost is $300,000 now, and an update amount of $100,000 every 5years forever. Determine the perpetual equivalent annual cost at an interest rate of 10% per year
Solution:
AW = -300,000(0.10) – 100,000(A/F,10%,5)
Expand from above formula
= -30,000 – 100,000(0.16380)
= -$46,380
Problem 6.34 An international aerospace contractor has been asked by a municipal police department to estimate and analyze the life cycle costs for a proposed drone surveillance system to monitor traffic patterns and congestion within the central thoroughfares of the city. The list of items include the following general categories: R&D costs (R&D), nonrecurring investment costs (NRI), recurring investment costs (RI), scheduled and unscheduled maintenance costs (Maint), equipment usage costs (Equip), and disposal costs (Disp). The costs (in $ million units) for the 20-year life cycle have been estimated. Calculate the annual LCC at an interest rate of 7% per year.

Correct answer below, please expand in detail the below solution:
PW of LCC = -6.6 – 3.5(P/F,7%,1) – 2.5(P/F,7%,2) – 9.1(P/F,7%,3) – 18.6(P/F,7%,4)
– 21.6(P/F,7%,5) – 17(P/A,7%,5)(P/F,7%,5) – 14.2(P/A,7%,10)(P/F,7%,10)
– 2.7(P/A,7%,3)(P/F,7%,17)
= -6.6 – 3.5(0.9346) – 2.5(0.8734) – 9.1(0.8163) – 18.6(0.7629) – 21.6(0.7130)
– 17(4.1002)(0.7130) – 14.2(7.0236)(0.5083) – 2.7(2.6243)(0.3166)
= $-151,710,860
AW of LCC = -151,710,860(A/P,7%,20)
= -151,710,860(0.09439)
= $-14,319,988
Chapter 7: Problems 7.5 & 7.7 using factors only, 7.16, 7.25, 7.27 & 7.38
Problem 7.5 A small industrial contractor purchased a warehouse building for storing equipment and materials that are not immediately needed at construction job sites. The cost of the building was $100,000 and the contractor has just made an agreement with the seller to finance the purchase over a 5-year period. The agreement states that monthly payments will be made based on a 30-year repayment schedule of interest on the unrecovered balance of the principal; however, the total remaining balance of principal and interest at the end of year 5 must be paid in a lump-sum “balloon” payment. What is the size of the balloon payment, if the interest rate on the loan is 0.5% per month?
Correct Answer, please provide detailed caluculations
Monthly payment = 100,000(A/P,0.5%,360)
= 100,000(0.00600)
= $600
Balloon payment = 100,000(F/P,0.5%,60) – 600(F/A,0.5%,60)
= 100,000(1.3489) – 600(69.7700)
= $93,028
Problem 7.7 If a manufacturer of electronic devices invests $650,000 in equipment for making compact piezoelectric accelerometers for general purpose vibration measurement, estimate the rate of return from revenue of $225,000 per year for 10 years and $70,000 in salvage value from the used equipment sale in year 10. Solve by
(a) factors
0 = -650,000 + 225,000(P/A,i*,10) + 70,000(P/F,i*,10)
Trial and error method
Trying i= 30%: – 650,000 + 225,000(3.0915) + 70,000(0.0725) = $50,663 > 0 which is too low
Trying i= 35%: – 650,000 + 225,000(2.7150) + 70,000(0.0497) = $-35,646 < 0 which is too high (b) spreadsheet function below: Problem 7.16: A broadband service company borrowed $2 million for new equipment and repaid the loan in amounts of $200,000 in years 1 and 2 plus a lump sum amount of $2.2 million at the end of year 3. What was the interest rate on the loan? Solution: We substitute interest 9% the present value of the borrowing is less than the present value, thus interest rate further as 10% When substituting the interest rate as 10% present value of the borrowing amount is equal to the present value of repayment t, thus inters rate is 10% Problem 7.25: Solution: Conventional cash flow has only one change in direction of cash flow for during the project in the meanwhile unconventional cash flow can have more than one change in cash flow direction. So below cash flows can be categorized as: a) Conventional b) Non-conventional c) Conventional d) Non-conventional e) Conventional Problem 7.27 The annual revenues (in $1000 units) associated with several large apartment complexes are $0, $350, $290, $460, $150, and $320 for years 0, 1, 2, 3, 4, and 5, respectively. Determine (1) whether each cash flow series is conventional or nonconventional, (2) the maximum number of real-number roots, and (3) if there is one, positive real-number i* value. The cash flows for years 0, 1, 2, 3, 4, and 5, respectively, are: (a) $?1500, $?90, $?40, $?85, $?60, and $?90 (b) $?1500, $?450, $?300, $?400, $?125, and $?400 (c) $1500, $?450, $?300, $?500, $?200, $?400 (d) $?1500, $?450, $?300, $?400, $?125, and $?310 The cash flow is said to be conventional if only one sign change occur while cash flow is said to be unconventional if more than one sign change occur. Year 0 1 2 3 4 5 a) -1500 -90 -40 -85 -60 -90 b) -1500 -450 -300 -400 -125 -400 c) -1500 -450 -300 -500 -200 -400 d) -1500 -450 -300 -400 -125 -310 Revenue $ 0 350 290 460 150 320 Since, more than one sign change occur , the cash flow is unconventional. Year 0 1 2 3 4 5 No of sign Remark R+a -1500 260 250 375 90 230 1 Conventional R+b -1500 -100 -10 60 25 -80 2 Unconventional R+c -1500 -100 -10 -40 -50 -80 0 Conventional R+d -1500 -100 -10 60 25 10 1 Conventional Since , more than one sign change occur , the cash flow is unconventional. Year No of sign Max no of real roots 2 3 4 5 R+a 1 1 -300 -500 -200 -400 R+b 2 2 375 75 225 375 R+c 0 0 75 -425 25 -25 R+d 1 1 1575 1150 1175 1150 Since , more than one sign change occur , the cash flow is unconventional. Year Sign of first cash flow No of sign One + real number ? R+a - 1 Yes R+b - 2 No R+c - 0 No R+d - 1 Yes Solution: 7.27 – Please check the correct answer below and provide detailed explanation (a) Net cash flow: $-1500, $260, $250, $375, $90, and $230 Cum. cash flow: $-1500, $-1240, $-990, $-615, $-525, and $-295 Conventional; Descartes rule: 1; Norstrom’s criterion: inconclusive (b) Net cash flow: $-1500, $-100, $-10, $60, $25, and $-80 Cum cash flow: $-1500, $-1600, $-1610, $-1550, $-1525, and $-1605 Non-conventional; Descartes rule: 2; Norstrom’s criterion: inconclusive (c) Net cash flow: $1500, $-100, $-10, $-40, $-50, and $-80 Cum cash flow: $1500, $-1400, $1390, $1350, $1300, and $1220 Conventional; Descartes rule: 1; Norstrom’s criterion: S0 > 0, does not apply
(d) Net cash flow: $-1500, $-100, $-10, $60, $25, and $10
Cum cash flow: $-1500, $-1600, $-1610, $-1550, $-1525, and $-1515
Conventional; Descartes rule: 1; Norstrom’s criterion: inconclusive
Problem 7.38

Number of possible i = 5
Year Revenue Cost Net Cash Flow Return
0 0 -6000 -6000 -100%
1 25000 -30000 -5000 83%
2 19000 -7000 12000 200%
3 4000 -6000 -2000 33%
4 28000 -12000 16000 267%
B. IRR = 40%
Year Revenue Cost Net Cash Flow
0 0 -6000 -6000
1 25000 -30000 -5000
2 19000 -7000 12000
3 4000 -6000 -2000
4 28000 -12000 16000
C. External rate using MIRR = 28%
Reinvestment rate = 18%
Finance Rate – 10%
MIRR is the rate of return at which the net present value (NPV) of terminal inflows is equal to the investment (i.e. outflow)
Correct Answer :
First find net cash flow (NCF)
Year 0 1 2 3 4
Revenue, $ 0 25,000 19,000 4,000 28,000
Costs, $ -6,000 -30,000 -7,000 -6,000 -12,000
NCF, $ -6000 -5000 12,000 -2000 16,000
(a) Rule of signs test: up to 3 values
Cumulative CF sign test: 3 changes; no unique, positive i*
(b) in Excel Function: = IRR(B1:B5) displays i* = 39.9%
(c) PW0 = -6000 – 5000(P/F,10%,1) – 2000(P/F,10%,3)
= -6000 – 5000(0.9091) – 2000(0.7513)
= $-12,048
FW4 = 12,000(F/P,18%,2) + 16,000
= 12,000(1.3924) + 16,000
= $32,709
-12,048(F/P,i’,4) + 32,709 = 0
-12,048(1 + i’)4 + 32,709 = 0
i’ = (32,709/12,048)1/4 – 1
= 0.284 (28.4%)
Chapter 8: Problems 8.3, 8.5, 8.9, 8.17, 8.19, 8.27a, & 8.32
Problem 8.3: If the sum of the incremental cash flows is negative, what is known about the rate of return on the incremental investment?
Solution:
If the sum of the incremental cash flows is negative that means the incremental cash flow is not earning any return and it is deteriorating so the return will be less than 0%.
8.5 A food processing company is considering two types of moisture analyzers. Only one can be selected. The company expects an infrared model to yield a rate of return of 27% per year. A more expensive microwave model will yield a rate of return of 19% per year. If the company’s MARR is 19% per year, can you determine which model should be purchased solely on the basis of the overall rate of return information provided? Why or why not?
Incremental cash flow analysis is used to select better of mutually exclusive alternative, however the operation could be performed at only two alternatives at a time. The lowest initial cost alternatives are selected first for incremental cash flow analysis.
The overall rate of return analysis will not be helpful here because even though the infrared model is having higher rate of return than microwave model, still it has to be seen that incremental amount invested in infrared model should cover the cost of this model should also give returns higher than MARR, and if it is not doing so then there is no benefit of higher cost option. So, incremental cash flow analysis is needed to get the correct option.
Problem 8.9: A small construction company has $100,000 set aside in a capital improvement fund to purchase new equipment. If $30,000 is invested at 30%, $20,000 at 25%, and the remaining $50,000 at 20% per year, what is the overall rate of return on the entire $100,000?
24% per year
Problem 8.17 Lesco Chemical is considering two processes for making a cationic polymer. Process A has a first cost of $100,000 and an AOC of $60,000 per year. Process B’s first cost is $165,000. If both process will be adequate for 4 years and the rate of return on the increment between the alternatives is 25%, what is the amount of the AOC for process B?
For the first investment:
Initial amount = $ 100 000 Annual overhead cost = $ 60 000 Hence after 4 years at the rate of 25% interest the total cost will be calculated on the basis the the initial cost will incur an intensest for 4 years, the first year AOC for 3 year, the second year AOC for 2 years and so on. Hence the total cost will be:
For the second investment:
Let the AOC for the second investment be x. Hence we can write that:
Hence for the two investments to be adequate, the AOC for the second one should be $ 32476.237
Second was ok but the actual work is below please provide the below ans in detail
Correct ans: The incremental cash flow equation is 0 = -65,000 + x(P/A,25%,4), where x is
the difference in the AOC.
0 = -65,000 + x(2.3616)
x = 65,000/2.3616
= $27,524
AOCB = 60,000 – 27,524
= $32,476
Incremental ROR Comparison (Two Alternatives)
Problem 8.19 A consulting engineering firm’s CFO wants to purchase either Ford Explorers or Toyota 4Runners for company principals. The two models under consideration cost $30,900 for the Ford and $36,400 for the Toyota. When considering life-cycle costs, the AOC of the Explorer is expected to be $600 per year more than that of the 4Runner. The trade-in values after 3 years are estimated to be 50% of the first cost for the Explorer and 60% for the 4Runner. (a) What is the incremental ROR between the two vehicles? (b) Provided the firm’s MARR is 18% per year, which vehicle should it buy?
I = 15.5
Use excel please
8.19 (a) Find rate of return on incremental cash flow
0 = -5500 + 600(P/A,?i*,3) + 6390(P/F,?i*,3)
?i* = 15.5% (spreadsheet)
b)
As incremental IRR is less than 18% MARR ,Ford Explorer should be bought
Problem 8.27 Alternative R has a first cost of $100,000, annual M&O costs of $50,000, and a $20,000 salvage value after 5 years. Alternative S has a first cost of $175,000 and a $40,000 salvage value after 5 years, but its annual M&O costs are not known. Determine the M&O costs for alternative S that would yield a required incremental rate of return of 20%. Solve (a) by hand.
Incremental investment = $175,000 – $100,000= $75,000
Incremental salvage value = $40,000 – $20,000 = $20,000
Present value of incremental salvage value (20%,5 years) = $20,000 * 0.4019= $8,038
Net incremental cost = Incremental investment – PV of incremental salvage value
= $75,000 – $8,038= $66,962
Annual savings in incremental M&O costs = Net incremental costs / annuity factor
= $66,962 / 2.9906= $22,390.8245
Annual M&O cost for alternative S = Annual M&O for alternative R – Savings
= $50,000 – $22,390.8245= $27,609.1754
Therefore the Annual M&O cost for alternative S is $27,609.1754
Ans: 8.27 By hand: Let x = M & O costs. Perform an incremental cash flow analysis.
0 = -75,000 + (-x + 50,000)(P/A,20%,5) + 20,000(P/F,20%,5)
0 = -75,000 + (-x + 50,000)(2.9906) + 20,000(0.4019)
x = $27,609
M & O cost for S = $27,609
8.32 Old Southwest Canning Co. has determined that any one of four machines can be used in its chilicanning operation. The cost of the machines are estimated below, and all machines have a 5-year life. If the minimum attractive rate of return is 25% per year, determine which machine should be selected on the basis of a rate of return analysis. Machine First Cost, $ AOC, $ 1 ?28,000 ?20,000 2 ?51,000 ?12,000 3 ?32,000 ?19,000 4 ?33,000 ?18,000
Solution:
MARR = 25%
We need to do incremental IRR analysis
In order to calculate incremental IRR, first, we arrange the options in the increasing value of their initial cost
We need to find incremental cash flow between lowest and second lowest option first, then use the formula of IRR in excel on the incremental cash flow, if incremental IRR is greater than MARR then select the second lowest option if it is less than the MARR then select the lowest option.
Keep repeating the above until the last option is evaluated
Machine Incremental IRR Analysis
Year 1 3 4 2 3-1 4-1 2-4
0 -28000 -32000 -33000 -51000 -4000 -5000 -18000
1 -20000 -19000 -18000 -12000 1000 2000 6000
2 -20000 -19000 -18000 -12000 1000 2000 6000
3 -20000 -19000 -18000 -12000 1000 2000 6000
4 -20000 -19000 -18000 -12000 1000 2000 6000
5 -20000 -19000 -18000 -12000 1000 2000 6000
Incremental IRR 7.93% 28.65% 19.86%
We compare first Machine 3 and machine 1 since machine 1 has the lowest cost, and the second lowest option is machine 3, incremental IRR is less than MARR so we select machine 1, and machine 3
Now we find incremental cash flow between machine 4 & machine 1, we find incremental IRR>MARR, so we select machine 4 and reject machine 1
Now we find incremental cash flow between machine 2 & machine 4, we find incremental IRR