Cost planning assignment 6

docx

School

Centennial College *

*We aren’t endorsed by this school

Course

124165E

Subject

Finance

Date

Apr 3, 2024

Type

docx

Pages

13

Uploaded by MateCamel3667

Report
Q 4: If a project costs $100,000 and is expected to return $25,000 annually, how long does it take to recover the initial investment? What would be the discounted payback period at i=15%? A: Conventional payback period: $100000/$25000 = 4 years Discounted payback = 6 + [(10667.99203/25000) – 1] =6.57 years Q 5: Refer to Problem 5.4 , and answer the following questions: a. How long does it take to recover the investment? b. If the firm’s interest rate is 15% after taxes, what would be the discounted payback period for this project? Conventional payback period = 100000/25000 = 4 years Discounted payback = 6 + [(10667.99203/25000 – 1] = 6.57 years Q 7: Consider the investment projects in Table P5.7 , all of which have a four-year investment life.
CSM724-701 Assignment 3 a. What is the payback period of each project? b. What is the discounted payback period at an interest rate of 15% for each project? A: a. Payback period is when Net Cash flow becomes positive, Payback period of A = 3+(4-3)(2500/6200) = 3.4 years Payback period of B = 1 + (1900/2800) = 1.68 years Payback period at C = 3 + (1000/4500) = 3.22 years Payback period at D = 2 + (1400/2300) = 2.61 years Page 1 of 13
CSM724-701 Assignment 3 b. Discounted Payback period of A = 3 + (2500/3545.16) = 3.7 years Discounted Payback period of B = 1+ (2108.69/2117.08) = 1.99 years Discounted Payback period of C = 3 + (2087.02/2573.1) = 3.81 years Discounted payback period of D = 3 +(466.79/1372.32) = 3.34 years Page 2 of 13
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
CSM724-701 Assignment 3 Q 10: Your firm is considering purchasing an old office building with an estimated remaining service life of 25 years. Recently, the tenants signed a long-term lease, which leads you to believe that the current rental income of $250,000 per year will remain constant for the first five years. Then the rental income will increase by 10% for every five-year interval over the remaining life of the asset. That is, the annual rental income would be $275,000 for years 6 through 10, $302,500 for years 11 through 15, $332,750 for years 16 through 20, and $366,025 for years 21 through 25. You estimate that operating expenses, including income taxes, will be $85,000 for the first year and that they will increase by $5,000 each year thereafter. You also estimate that razing the building and selling the lot on which it stands will realize a net amount of $50,000 at the end of the 25-year period. If you had the opportunity to invest your money elsewhere and thereby earn interest at the rate of 12% per annum, what would be the maximum amount you would be willing to pay for the building and lot at the present time? A: Given: Estimated remaining service life = 25 years, current rental income = $250000 per year, O&M costs = $85000 for the first year increasing by $5000 thereafter, salvage value = $50000 and MARR = 12%. Let A 0 be the maximum investment required to break even. PW (12%) = -A 0 + [$250000(F/A,12%,25) + 25000(F/A,12%,20) + 27500(F/A,12%,15) + 30250(F/A,12%,10) + 33275(F/A,12%,5) + 50000(P/F,12%,25) – 85000(P/A,12%,25) – 5000(P/G,12%,25) = 0 Solving for A 0 = yields A 0 = $1,241,461 Q 11: Consider the following set of investment projects, all of which have a three-year investment life: Page 3 of 13
CSM724-701 Assignment 3 a. Compute the net present worth of each project at i=10%. b. Plot the present worth as a function of the interest rate (from 0% to 30%) for Project B . a. PW (10%) A = -1200 + 3000(P/F,10%,3) = -1200 + 3000(.7513) = -1200 + 2253.9 = 1053.9 PW (10%) B = -1800 + 600(P/F,10%,1) + 900(P/F,10%,2) + 1700(P/F,10%,3) =-1800 + 600(.9091) + 900(.8264) + 1700(.7513) = -1800 + 545.46 + 743.76 + 1277.21 = 766.43 PW (10%) C = -1000 -1200(P/F,10%,1) + 900(P/F,10%,2) + 3500(P/F,10%,3) = -1000 – 1200(.9091) + 900(.8264) + 3500(.7513) = -1000 – 1090.92 + 743.76 + 2629.55 = 1282.39 PW (10%) D = -6500 + 2500(P/F,10%,1) + 1900(P/F,10%,2) + 2800(P/F,10%,3) = -6500 + 2500(.9091) + 1900(.8264) + 2800(.7513) = -6500 + 2272.75 + 1570.16 + 2103.64 = -553.45 Q.13 You are considering the purchase of a parking deck close to your office building. The parking deck is a 15-year-old structure with an estimated remaining service life of 25 years. The tenants have recently signed long-term leases, which leads you to believe that the current rental income of $250,000 per year will remain constant for the first five years. Then the rental income will increase by 10% for every five-year interval over the remaining asset life. Thus, the annual rental income would be $275,000 for years 6 through 10, $302,500 for years 11 through 15, $332,750 for years 16 through 20, and $366,025 for years 21 through 25. You Page 4 of 13
CSM724-701 Assignment 3 estimate that operating expenses, including income taxes, will be $65,000 for the first year and that they will increase by $6,000 each year thereafter. You estimate that razing the building and selling the lot on which it stands will realize a net amount of $200,000 at the end of the 25- year period. If you had the opportunity to invest your money elsewhere and thereby earn interest at the rate of 15% per annum, what would be the maximum amount you would be willing to pay for the parking deck and lot at the present time? Ans. Estimated Remaining service = 25 years Current rental income= $250,000 per year Estimated operating expenses including income tax= $65,000 for the first year and that they will increase by $6,000 each year thereafter. Salvage value=$200,000 MARR= 15% Let A 0 be the maximum investment required to break even. PW ( 15% ) = 250,000 ( P / A , 15% , 5 ) + 275,000 ( P / A , 15% , 5 ) ( P / F , 15% , 5 ) + 302,500 ( P / A , 15% , 5 ) ( P / F , 15% ¿ $ 1,116,775 Q.15 A university is trying to determine how much it should charge for tickets to basketball games to help offset the expenses of the new arena. The cost to build the arena including labor, materials, etc. was $92 million. Each year the maintenance cost is expected to increase by 5% as the building gets older. The maintenance cost for the first year is $150,000. Utilities are expected to average about $200,000 per year and labor costs $300,000. The average attendance at basketball games over the year is expected to be 100,000 people (or 100,000 tickets sold to events). Assuming the arena has no other source of income besides regular ticket sales (not including student tickets) for basketball games, what should the university charge so that it can recover at least 6% cost of borrowing on its investment? The university expects the arena to be used for 40 years and to have no appreciable salvage value. Page 5 of 13
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
CSM724-701 Assignment 3 Ans PriceofTicket ( P / A; 6% ; 40 ) = 92.000.000 + 150.000 ( P / A ; 5% ; 6 % ; 40 ) + ( 200.000 + 300.000 ) ( P / A ; 6 % ; 40 ) / 10 Q.19 A project has a service life of five years with the initial investment outlay of $200,000. If the discounted payback period occurs at the end of the project service life (say five years) at an interest rate of 10%, what can you say about the NFW of the project? Ans Since the initial investment plus interest had been fully recovered at the end of the project service life and after that, no amount of money was generated, the net future worth was 0, which means no profit. Q.20 Consider the following cash flows and present-worth profile. Table P5.20 Page 6 of 13
CSM724-701 Assignment 3 Figure P5.20 a. Determine the values of X and Y . Ans . From the project balance diagram, note that PW (24%)1 = 0 for project 1 PW (23%) 2 = 0 for project 2. PW (24%)1 = -$1, 000 + $400(P/F, 24%,1) + $800(P/F, 24%, 2) +X (P/F, 24%, 3) = 0 PW (23%)2 = -$1, 000 + $300(P/F, 23%,1) +Y(P/F, 23%, 2) + $800(P/F, 23%, 3) = 0 X= $299.58, Y= $493.49 b. Calculate the terminal project balance of project 1 at MARR=24% Ans. Since PW (24%)1= 0, FW (24%)1 = 0 c. Find the values of a, b , and C in the NPW plot. Ans. a= $593.49, b= $499.58, c= 17.91% Page 7 of 13
CSM724-701 Assignment 3 Q.21 Consider the project balances for a typical investment project with a service life of five years, as shown in Table P5.21 Table P5.21 Investment Project Balances a. Determine the interest rate used in the project balance calculation and compute the present worth of this project at the computed interest rate. -1500 = -2700 (1+i) + 1470 -2970 = -2700 (1+i) i = 10% -3000 (1+0,1) + a = -2700 a = 600 -1500 (1+0,10) + b = 0 b = 1650 -0 (1+0,10) + c = -300 c = 300 -300 (1+0,10) + 600 = d d = 270 PW (10%) = 270 (P/F;10%;5) = 270 (0,6209) = 167,64 Page 8 of 13
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
CSM724-701 Assignment 3 b. Construct the original cash flows of the project and the terminal balance and fill in the blanks in Table P5.21 . n An Project Balance 0 -$3000 -$3000 1 $600 -$2700 2 $1470 -$1500 3 $1650 $0 4 $300 -$300 5 $600 $270 Q 26. Consider an investment project, the cash flow pattern of which repeats itself every five years forever, as shown in the accompanying diagram. At an interest rate of 15%, compute the capitalized equivalent amount for this project. Ans) Given, i = 15%, number of years n= 5. The corresponding annual series A of the above cycle, A = [$100(P / A,15%, 2) + $40(P / A,15%, 2) (P / F,15%, 2) + $20(P / F,15%,5)] (A / P,15%, 5) = $66.13 And the Capitalized equivalent amount CE of 15% interest rates, CE (15%) = 66.13$ / 0.15 = 440.88$ Page 9 of 13
CSM724-701 Assignment 3 Q 27. A group of concerned citizens has established a trust fund that pays 5% interest, compounded monthly, to preserve a historical building by providing annual maintenance funds of $80,000 forever. Compute the capitalized equivalent amount for these building maintenance expenses. Ans) Given: r =5% compounded monthly, Maintenance cost = 80,000$ per year The rate of interest, i = {1+(0.05/12)} ^12 = 5.116% And the Capitalized equivalent amount CE of 5.116% interest rates, CE (5.116%) = 80,000$/0.06116 = 1,563,664$ Q 28. A newly constructed bridge costs $15,000,000. The same bridge is estimated to need renovation every 15 years at a cost of $3,000,000. Annual repairs and maintenance are estimated to be $1,000,000 per year. Comparing Mutually Exclusive Alternatives a. If the interest rate is 5%, determine the capitalized cost of the bridge. b. Suppose that in (a), the bridge must be renovated every 20 years, not every 15 years. What is the capitalized cost of the bridge? c. Repeat (a) and (b) with an interest rate of 10%. What can you say about the effect of interest on the results? Ans) Given: Construction cost = $15,000,000, renovation cost = $3,000,000 every 15 years, annual Operation and Maintenance costs = $1,000,000 and i = 5% per year. a) P1 = 15,000,000$ P2 = {3,000,000$ (A/F, 5%, 15)} / 0.05 = 2,780,537$ P3 = 1,000,000$ / 0.05 Page 10 of 13
CSM724-701 Assignment 3 = 20,000,000$ CE (5%) = P1+P2+P3 = 15,000,000$ + 2,780,537$ + 20,000,000$ = 37,780,537$ b) P1 = 15,000,000$ P2 = {3,000,000$(A/F, 5%, 20)} / 0.05 = 1,814,555$ CE (5%) = P1+P2+P3 = 36,814,555$ c) In 15 year cycle with 10 % interest: P1 = 15,000,000$ P2 = {3,000,000(A/F, 10%, 15)} / 0.1 = 944,213$ P3 = 1,000,000/0.1 = 10,000,000$ CE (10%) = P1+P2+P3 = 25,944,213$ In 20- year cycle with 10% of interest : P1 = 15,000,000$ P2 = {3,000,000(A/F, 10%, 20)} / 0.1 = 523,789$ P3 = 1,000,000 / 0.1 = 10,000,000$ CE (10%) = P1+P2+P3 = 25,523,789$ Hence, as the interest rate increases, CE value decreases. Q 29. To decrease the costs of operating a lock in a large river, a new system of operation is proposed. The system will cost $830,000 to design and build. It is estimated that it will have to be reworked every 10 years at a cost of $120,000. In addition, an expenditure of $80,000 will have to be made at the end of the fifth year for a new type of gear that will not be available Page 11 of 13
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
CSM724-701 Assignment 3 until then. Annual operating costs are expected to be $70,000 for the first 15 years and $100,000 a year thereafter. Compute the capitalized cost of perpetual service at i =7% Ans) Given: interest rate i=7%, Cost to design and build = $830,000, rework cost = $120,000 every 10 years, new type of gear = $80,000 at the end of 5th year, annual operating costs = $70,000 for the first 15 years and $100,000 thereafter CE (7%) = 830,000$ + {120,000$(A/F, 7%, 10)} / 0.07 + 80,000$ (P/F, 7%,5) + 70,000$ (P/A, 7%, 15) + {(100,000/0.07) (P/F, 7%, 15) = 2,166,448.62$ Page 12 of 13