Engineering Economy: Engr. Adonis C. Bibat

ENGINEERING ECONOMY ENGR. ADONIS C. BIBAT

Cash Flow Diagrams • May be drawn to help visualize and simplify problems having diverse receipts and disbursements.

Simple Interest • The amount of money to be paid for the use of money, which is proportional to the length of time the principal is used. • Ordinary simple interest is computed on the basis of one banker’s year. 1 banker’s year = 12 months 30 days each month = 360 days

Simple Interest Rate of Interest – the amount earned by one unit of principal during a unit time

I=Pin 𝑟 I=P (mt) 𝑚

I=Prt

where: I= interest P= principal r= rate of interest t= time in years i= interest rate per interest period m= number of interest period per year

Simple Interest Exact Interest for “t” days 365 days for ordinary year 366 days for leap year Future Worth after “t” days F=P+I F=P+Prt F=P(1+rt)

Simple Interest Discount

D=F-P Rate of Discount 𝑖 d= 𝑖+1

Effective Rate 𝐼 𝑟 m ER=𝑃=(1+𝑚)

-1

Simple Interest

1. If you borrowed money from your friend with simple interest of 12%, find the present worth of P50,000; which due at the end of 7 months.

Simple Interest 2. CE Board May 1997

A deposit of P110,000 was made for 31 days. The net interest after deducting 20% withholding tax is P890.36. Find the rate of return annually.

Simple Interest 3. CE Board May 2015 A businessman wishes to earn 7% on his capital after payment of taxes. If the income from an available investment will be taxed at an average rate of 42%, what minimum rate of return, before payment of taxes, must the investment offer to be justified?

Simple Interest 4. A man borrowed from a bank under a promissory note that he signed in the amount of P25,000 for a period of one year. He received only the amount of P21,915 after the bank collected the advance interest and an additional amount of P85.00 for notarial and inspection fees. What was the rate of interest that the bank collected in advance?

Simple Interest

5. Mr. J. de la Cruz borrowed money from a bank. He received from the bank P1,340.00 and promised to pay P1,500.00 at the end of 9 months. Determine the corresponding discount rate or often referred to as the “banker’s discount”.

Simple Interest

6. A bank loan of P2,000.00 was made at 8% simple interest. How long would it take in years for the amount of the loan and interest to equal P3,280.00

Simple Interest

7. A merchant is offered a 5% discount for immediate payment of bill which is due in 90 days. What is the largest simple interest rate at which he could afford to borrow in order to pay cash?

Simple Interest

8. If an investment of P15,000.00 earned an interest of P265 in 4 months, what is the annual rate of simple interest?

Simple Interest

9. If the rate of the simple interest is 8% per year, how long will it take for an investment of P6,000 to grow to P8,400?

Simple Interest

10. A man deposits a certain amount P which pays a simple interest rate of 8% per annum. Find the value of P when the principal plus interest at the end of 1.5 years is P6,720.

Simple Interest ANSWERS: 1. P 46,728.97 2. 11.75% 3. 12.1% 4. 13.64% 5. 13.73%

6. 8 years 7. 21% 8. 5.3% 9. 5 years 10. P 6,000

Compound Interest

If the interest due is added to the principal and thereafter earns interest, the sum by which the original principal has increased by the end of the term is called Compound Interest.

Compound Interest SINGLE PAYMENT:

n

F=P(1+i) where:

F= future worth of P after a period “n” P= present worth n= no. of periods (1+i)n= is called the single payment compound amount factor

Compound Interest Values i and n : the following examples show how to get the values of i and n Nominal interest rate, r=12% Number of years of investment, t=5 years » Compounded Annually (m=1) i=0.12/1=0.12 n=5(1)=5

Compound Interest •

Compounded Semi-annually (m=2)

•

i=0.12/2=0.06 n=5(2)=10 •

Compounded Quarterly (m=4) i=0.12/4=0.03 n=5(4)=20

Compounded Monthly (m=12) i=0.12/12=0.01 n=5(12)=60

•

Compounded Bi-Monthly (m=6) i=0.12/6=0.02 n=5(6)=30

Compound Interest

1. An investment of P200,000 earns 10% per annum compounded bi-monthly. How much will it earn in 6 years.

Compound Interest

2. An engineer invested his money at 16% per annum, net of taxes, compounded semi-annually after how many years will his money double?

Compound Interest

3. An associations fund is placed in a time deposit that earns 11% per annum, net of deductions compounded quarterly. When will the fund triple?

Compound Interest

4. At what rate of interest, compounded monthly will an investment double in 3 years?

Compound Interest

5. A man buys a house and lot worth P2,000,000 if paid in cash. He agreed to pay a downpayment of P500,000 and two installments of P1,000,000 at the end of one year and the balance at the end of 3 years. Determine the amount of this balance if the interest rate is 22% compounded quarterly.

Compound Interest

6. A teachers fund is invested at 8% per annum, net of taxes, compounded quarterly. When will the fund quadruple?

Compound Interest

7. After how many years will an investment triple if invested at 10% per annum, net of deductions, compounded quarterly?

Compound Interest

8. How many years will be required for a given sum of money to triple, if it is deposited in a bank account that pays 6% per year, compounded annually?

Compound Interest 9. CE Board Nov. 2016

How long does it take an investment of P2000 to double if it is invested at 5% interest compounded quarterly?

Compound Interest 10. A Ford Mustang sports car costs $20,000 today. You can earn 12% tax-free on an “auto-purchase account”. If you expect the cost of the car to increase by 10% per year, the amount you would need to deposit in the account to provide for the purchase of the car five years from now is closest to:

Compound Interest ANSWERS: 1. P162,626.00 2. 4.5 years 3. 10.12 years 4. 23.3% 5. P1,317,015.55

6. 17.5 years 7. 11.12 years 8. 18.85 years 9. 14 years 10. $18,276.00

Compound Interest COMPOUNDED CONTINUOUSLY: rt

F=Pe

where: e=2.71828 F=future worth P=present worth r=interest rate compounded continuously t=no. of years

Compound Interest Effective Annual rate Compounded Continuously: er -1 = Effective Rate

e

rt

= Compound Amount Factor

1/ert = Present Worth Factor

Compound Interest 1. CE Board May 2015

What is the effective annual interest rate if the nominal rate of interest is 14% compounded continuously?

Compound Interest 2. CE Board Nov. 2015

If money is invested at a nominal rate of interest of 8% for a period of 4 years, what is the value of the compound amount factor if it is compounded continuously?

Compound Interest 3. CE Board May 2017

A savings ban offers long-term savings certificate at 7.5% per year, compounded continuously. If a 10-year certificate costs $1000, what will be its value at maturity?

Compound Interest

4. A man wished to have P40,000.00 in a certain fund at the end of 8 years that will pay a nominal rate of 6% compounded continuously. What is the value of the compound amount factor for this rate?

Compound Interest

5. Compute the interest for an amount of P200,000 for a period of 8 years if it was made at 16% compounded continuously.

Compound Interest

6. Compute the nominal interest rate with continuous compounding if the quarterly effective interest rate is 4%.

Compound Interest

7. The Bank of Philippine Island (BPI) is selling longterm savings certificates that pay interest at the rate of 7.5% per year compounded continuously. Compute the effective annual interest.

Compound Interest

8. The nominal interest of money is 9% for a period of 5 years. If it is compounded continuously, what is the value of the compounded amount factor?

Compound Interest

9. If the effective annual interest rate is 3.5%, find the equivalent nominal annual interest compounded continuously.

Compound Interest

10. A bank currently charges 10% interest compounded annually on business loans. If the bank were to change to continuous compounding, what would be the effective annual interest rate?

Compound Interest ANSWERS 1. 15.03% 2. 1.377 3. $2,117.00 4. 1.616 5. P519,328.00

6. 15.68% 7. 7.79% 8. 1.57% 9. 3.44% 10. 10.517%

Ordinary Annuities One where equal payments are made at the end of each payment period starting from the first period.

Ordinary Annuities Present Worth P=A[

𝑛

1+𝑖 −1 𝑖 1+𝑖 𝑛

Future Worth F=A[

′

𝑛

1+𝑖 𝑖

′

−1

]

]

where: i= interest rate per payment n= number of interest periods n’= number of payments A= periodic payment

Ordinary Annuities Under Present Worth 1. Equal-payment-series-present-worth factor:

( i, n)=[ 𝑃 , 𝐴

𝑛

′

1+𝑖 −1 𝑖 1+𝑖 𝑛

]

2. Equal-payment-series-capital-recovery factor:

( i, n)=[ 𝐴 , 𝑃

𝑛

′

1+𝑖 −1 -1 𝑖 1+𝑖 𝑛

]

Ordinary Annuities Under Future Worth 1. Equal-payment-series-compound-amount factor:

( i, n)=[ 𝐹 , 𝐴

𝑛

1+𝑖

′

−1

𝑖

]

2. Equal-payment-sinking-fund factor:

(

𝐴 , 𝐹

i, n)=[

𝑛

′

]

1+𝑖 −1 -1 𝑖

Ordinary Annuities 1. CE Board May 2016 An engineer has just borrowed P41,000 from a local bank at the rate of 1% per month on the unpaid balance. His contract states that he must repay the loan in 35 equal monthly installments. How much money must be repaid each month?

Ordinary Annuities

2. A promissory note has outstanding payments of $650 at the end of each of the next five years. What market price would be paid for this note by an investor who requires a 12% yield on his investments, compounded quarterly?

Ordinary Annuities

3. An engineer constructs a house and sells it on monthly installments of 25,000 pesos. It will be fully paid in 7 years. If the rate of interest is 12% compounded monthly, what is the equivalent cash price of the house in thousands of pesos.

Ordinary Annuities

4. The cash price of a property is P4 million. The downpayment is P400,000 and the balance is payable on monthly installments of P85,643.74 for 5 years. What is the nominal rate of interest if it is compounded monthly?

Ordinary Annuities

5. An endowment today of one million pesos maybe converted into monthly pensions for 20 years. Determine the pension in pesos, considering that the rate of interest is 9% compounded monthly.

Ordinary Annuities

6. Compute the present value of a stream of monthly payments of P1,000 each over 10 years if the interest rate is 10% per annum compounding daily.

Ordinary Annuities

7. Hilario Amaba has P80,000 in a savings account that earns 7.25% per annum, compounded annually if he withdraws P120,000 at the end of each year. After how many years will the savings be exhausted?

Ordinary Annuities

8. In an ordinary annuity (uniform series of payments) if the nominal rate of interest is 8% compounded quarterly for 6 years. Compute the value of the sinking fund factor.

Ordinary Annuities

9. How many years will be required to accumulate P10,000 if P500 is deposited at the end of each year and interest is payable at 6 3Τ4 % per year, compounded annually?

Ordinary Annuities

10. The difference between uniform series capital recovery factor and the sinking fund factor is 0.06. Find the value of “n” if the capital recovery factor is 0.16104.

Ordinary Annuities ANSWERS 1. P1394.2 2. 2,311.5 3. 1416 thousands 4. 15% 5. P8,997.26

6. P75,543.22 7. 9.43 say 10 years 8. 0.03287 9. 13.08 say 14 years 10. 8

Deferred Annuity A deferred annuity is one where the payment of the first amount is deferred a certain number of periods after the first.

Deferred Annuity ➢Future worth F1=A[

𝑛

1+𝑖

′

−1

𝑖

𝑟 mt Fn=P(1+ ) 𝑚

] or

make F1=P 𝑟 mt Fn=F1(1+ ) 𝑚

Deferred Annuity ➢Present worth P1=A[

𝑛

′

1+𝑖 −1 𝑖 1+𝑖 𝑛

𝑟 mt Fn=P(1+ ) 𝑚

P=

𝑃1

𝑟 𝑚𝑡 1+ 𝑚

] or

make P1=F 𝑟 mt P1=P(1+ ) 𝑚

Deferred Annuity 1. CE Board May 2015 In 5 years, P1.8M will be needed to pay for the building renovation. In order to generate this sum, a sinking fund consisting of three annual payments is established now. For tax purposes, no further payments will be made after 3 years. What payments are necessary if money is worth 15% per annum?

Deferred Annuity 2. CE Board May 2004

A boy is entitled to 10 yearly endowments of P30,000 each starting at the end of the eleventh year from now. Using an interest rate of 8% compounded annually, what is the value of these endowment now?

Deferred Annuity 3. Find the value after 20 years in pesos of an annuity of P20,000 payable annually for 8 years with the first payment at the end of 2 years if money is worth 5%.

Deferred Annuity 4. Find the present value of an annuity of P25,000 payable annually for 8 years with the first payment at the end of 10 years if money is worth 5%.

Deferred Annuity 5. Ten annual installment of P60,000 each starting at the end of the eleventh year from now, is worth only P186,493.96. What is the rate of interest if it is compounded annually?

Deferred Annuity 6. A man invested P100,000 every end of the year for 10 years, then waited for another 10 years for his money to grow. If his investment earned 8%, after tax, compounded annually, what would be the sum of his investments and earnings at the end of the 20th year?

Deferred Annuity 7. P10,000 was deposited in a bank every end of the year for 10 consecutive years. Then the sum of all deposits and interest was left to earn. The same rate of interest of 8%, net of taxes, compounded annually. What would be the sum of all deposits and interests at the end of 15 years?

Deferred Annuity 8. On the day his son was born, a father decided to deposit a fund for his son’s college education. The father wants the son to be able to withdraw P4,000 from the fund on his 18th birthday and again on his 21st birthday. If the fund earns interest at 9% per year, compounded annually, how much should the father deposit at the end of each year, up through the 17th year?

Deferred Annuity 9. A father wants to set aside money for his 8 year old son’s future by making a monthly deposit to a bank account that pays 8% per year, compounded annually. What equal monthly deposits must the father make, the first 1 month after his 9th birthday and the last on his 17th birthday, in order for him to withdraw P4,000 on each of his next four birthdays, (the 18th through 21st)?

Deferred Annuity 10. San Miguel Brewery entered into a 10 year contract for raw materials which required a payment of P100,000 initially and P20,000 per year beginning at the end of the 5th year. The company made unexpected profits and ask that it be allowed to make a lump sum payment at the end of the 3rd year to payoff the remainder of the contract. What lump sum is necessary if the interest rate is 8%?

Deferred Annuity ANSWERS 1. P391,953.52 2. P93,241.98 3. 326,644.33 4. P104,156.11 5. 8%

6. P3,127,540.00 7. P212,855.12 8. P350.49 9. P100.38 10. P85,609.00

Annuity Due -

an equal series of payments paid at the beginning of each period.

Annuity Due ➢Present worth of an annuity due (1+𝑖)(𝑛−1) −1 P=A[ ]+ (𝑛−1) 𝑖(1+𝑖)

A

n=no. of payments A=amount of each installment

Annuity Due ➢Future worth of an annuity due

F=A[

(𝑛+1) (1+𝑖) −1

𝑖

]-A

Annuity Due ➢An engineer bought a dump truck costing $25,000 payable in 10 years, semi-annual payments, each installment payable at the beginning of each period. If the rate of interest is 12% compounded semi-annually, determine the amount of each installment. ➢Ans. $2,056.24

Annuity Due ➢Ben Abellana deposits P750 in a savings account in the Bank of Asia at the beginning of each year, starting now, for the next 10 years. If the bank pays 7% per year, compounded annually, how much money will Mr. Abellana have accumulated by the end of the 10th yr? ➢Ans. $11,087.70

Gradient Series • Is a series of annual payments in which each payment is greater than the previous one by the constant amount of G. ***Note: Gradient Series starts at year 2.

Gradient Series

Convert into uniform series compounded → annually

𝐴1 1 = 𝐺 𝑖

−

𝑛 (1+𝑖)𝑛 −1

Since,

𝑛 (1+𝑖) −1

𝑃1 = 𝐴1 𝑖(1+𝑖)𝑛

Gradient Series

Uniform gradient series factor : (P/G , i , n)

Total Present Worth =𝑮 ∗

𝑃1 (1+𝑖)𝑛 −1 = 2 𝐺 𝑖 (1+𝑖)𝑛 𝑷 , 𝒊, 𝒏 𝑮

+ 𝑨𝟏 ∗

𝑷 , 𝒊, 𝒏 𝑨

−

𝑛 𝑖(1+𝑖)𝑛

Gradient Series

Uniform gradient series factor : (F/G , i , n)

Total Future Worth =𝑮 ∗

𝐹1 (1+𝑖)𝑛 −1 = 𝐺 𝑖2 𝑭 , 𝒊, 𝒏 𝑮

+ 𝑨𝟏 ∗

𝑭 , 𝒊, 𝒏 𝑨

−

𝑛 𝑖

Gradient Series • CE BOARD 2017

➢ Ans. P791

Gradient Series ➢A generator is expected to have a maintenance cost of P1,550 at the end of the first year and it is expected to increase P350 each year for the following seven years. What sum of money should be set aside now to pay the maintenance for the 8th year period? Assume rate of interest is 6%. ➢Ans. P16,569.9

Rate of Return • Rate of return =

𝑁𝑒𝑡 𝐴𝑛𝑛𝑢𝑎𝑙 𝑃𝑟𝑜𝑓𝑖𝑡 𝐹𝑖𝑥𝑒𝑑 𝐶𝑎𝑝𝑖𝑡𝑎𝑙+𝑊𝑜𝑟𝑘𝑖𝑛𝑔 𝐶𝑎𝑝𝑖𝑡𝑎𝑙

• Rate of return =

𝑁𝑒𝑡 𝐼𝑛𝑐𝑜𝑚𝑒 𝐶𝑎𝑝𝑖𝑡𝑎𝑙

• Net Income = gross income – capital – taxes • Net Annual Profit = Annual Profit – (annual depreciation in %)*(fixed capital)

Rate of Return • CE BOARD 2015 ➢A businessman wants to earn 7% on his capital after payment of taxes. If the income from an available investment will be taxed at an average rate of 42%, what minimum rate of return before payment of taxes must the investment offer to be justified? ➢Ans. 12.07%

Rate of Return ➢A fixed capital investment of P10,000,000 is required for a proposed manufacturing plant and an estimated working capital of P2,000,000. Annual depreciation is 10% of the fixed capital investment. If the annual profit is P2,500,000. What is the rate of return? ➢Ans. 12.5%

Benefit Cost Ratio •

𝑃𝑟𝑒𝑠𝑒𝑛𝑡 𝑤𝑜𝑟𝑡ℎ 𝑜𝑓 𝑎𝑛𝑛𝑢𝑎𝑙 𝑏𝑒𝑛𝑒𝑓𝑖𝑡𝑠 Benefit Cost Ratio or Present worth index = 𝐶𝑎𝑝𝑖𝑡𝑎𝑙 𝐼𝑛𝑣𝑒𝑠𝑡𝑚𝑒𝑛𝑡

• Benefit Cost Ratio =

𝑃𝑟𝑒𝑠𝑒𝑛𝑡 𝑤𝑜𝑟𝑡ℎ 𝑜𝑓 𝑖𝑛𝑐𝑜𝑚𝑒 𝑃𝑟𝑒𝑠𝑒𝑛𝑡 𝑤𝑜𝑟𝑡ℎ 𝑜𝑓 𝑐𝑎𝑝𝑖𝑡𝑎𝑙

***Benefit cost ratio greater than 1 means the investment is good

Benefit Cost Ratio • CE BOARD 2015 ➢If the benefit cost ratio is to be 1.8, how much should he spend in a project that provides annual benefits of P120,000 for a period of 5 years without a salvage value. Cost of money is 12%. ➢Ans. P240,318

Benefit Cost Ratio • CE BOARD 2016 ➢ A small entrepreneur invested a capital of P80,000 for a buy and sell business. He estimated to have a gross income of P25,000 annually and on operating cost of P6,000 annually. It is assumed the business to have a life over 10 years. If the rate of interest is 12%, compute the benefit cost ratio. ➢ Ans. 1.34

Benefit Cost Ratio • CE BOARD 2017 ➢ Compute the present worth index of a certain investment of the Phil. Rock Corporation: Investment=P30,000 Annual net savings = P12,600 Useful life: n=9 yrs MARR = 12 % Present worth factor: (P/A, 12%, 9) = 5.3283 Compound amount factor: (F/A, 12%, 9) = 14.776 Capital recovery factor: (A/P, 12%, 9) = 0.1877 Uniform gradient present worth factor: (P/G, 12%, 9) = 8.226 ➢ Ans. 2.24

DEPRECIATION ➢The decrease in value of a property such as machinery, equipment, building or other structures due to passage of time. FC SV

→ First cost → Salvage cost

n m

→ Economic life → Any year before n

BVm

→ Book value after m years

DEPRECIATION BVm=FC - Dm

➢Straight-line: (mode-3-2) 𝑭𝑪−𝑺𝑽 d= 𝒏

Dm=d(m)

X(time)

Y(BV)

0

FC

n

SV

; d=depreciation per year ; Dm= total depreciation

DEPRECIATION BVm=FC - Dm

➢Sinking Fund: (𝟏+𝒊)𝒏 −𝟏 -1 d=(FC-SV)[ ] 𝒊 (𝟏+𝒊)𝒎 −𝟏 Dm=d[ ] 𝒊

; d=depreciation per year

; Dm= total depreciation

DEPRECIATION BVm=FC - Dm

➢Sum of the Years Digit: (mode-3-3) 𝒏−𝒎+𝟏 d=(FC-SV)[σ ] 𝒚𝒆𝒂𝒓𝒔

Note: σ 𝒚𝒆𝒂𝒓𝒔

𝑛 = (𝑛 + 1) 2 σ𝒏 𝒏−𝒎+𝟏 𝒙 σ𝒏 𝟏𝒙

Dm=(FC-SV)[

;

X(time)

Y(BV)

0

FC

n

SV

n+1

SV

d=depreciation per year

] ; Dm= total depreciation

DEPRECIATION

X(time)

Y(BV)

0

FC

n

SV

BVm=FC (1 - k)m

➢Declining Balance (Matheson): (mode-3-6) ➢Constant percentage method

SV = FC(𝟏 −

𝒏 𝒌)

Dm= FC – BVm

;

d=depreciation per year

;

Dm= total depreciation

DEPRECIATION BVm=FC (1 - k)m

➢Double Declining Balance:

k= 2/n Dm= FC – BVm

;

Dm= total depreciation

Dm= FC∗ 𝐤(𝟏 − 𝒌)𝒎−𝟏

DEPRECIATION ➢Service Output Method: 𝑭𝑪−𝑺𝑽 d= 𝑸𝒏

;

Qn= qty. produced during economic life

Dm= d(𝑸𝒎 ); Dm= total depreciation

DEPRECIATION ➢The first cost of a machine is P1,800,000 with a salvage value of P300,000 at the end of its life of 5 years. Determine the total depreciation after 3 years using straight line method. ➢Ans. P900,000

DEPRECIATION ➢A Fuji Xerox machine was purchased 5 years ago at a cost of P120,000. The estimated salvage value at the end of 10 years is P10,000. If it is sold now for P30,000. Determine the sunk cost using straight line method in computing the depreciation. ➢Ans. P35,000

DEPRECIATION • CE BOARD 2015 ➢The first cost of a machine is P1,800,000 with a salvage value of P300,000 at the end of its life of 5 years. Determine the total depreciation after 3 years using constant percentage method. ➢Ans. P1,185,770

DEPRECIATION • CE BOARD 2004 ➢A machine has a first cost of P200,000. At the end of its economic life of 10 yrs., its scrap value is estimated to be P25,000, what will be its book value at the end of 8 yrs? Using double declining balance method. ➢Ans. P33,554

DEPRECIATION • CE BOARD 2003 ➢Cost of machine = P1,400,000 Useful life = 8 yrs Salvage value = P10,000 Determine the 4th year depreciation using the double declining balance method. ➢Ans. P147,656

DEPRECIATION ➢A machine costs P100,000 with a salvage value of P20,000 at the end of its life of 8 years. What gives the depreciation at the 6th year using sinking fund method if money is worth 6% annually? Determine the book value at the 6th yr? ➢Ans. P56,380.63, P43,619.37

DEPRECIATION • CE BOARD 2015 ➢An equipment costs P1,200,000. At the end of its economic life of 5 years its salvage value is P300,000. Using sum of years digit method, what will be its book value on the 3rd year. ➢Ans. P480,000

DEPRECIATION • CE BOARD 2015 ➢A machine costs P1,500,000. It has a salvage value of P600,000 at the end of its economic life. Using the sum of the years digit method, the book value at the end of 2yrs is estimated to be P870,000. What is the machines economic life in years? ➢Ans. 4 yrs

Perpetuity ➢Is an annuity where the payment periods extend forever or in which the periodic payments continue indefinitely.

Present Worth =

Perpetuity=

𝐴 𝑖

𝐴 𝑖

Perpetuity • CE BOARD 2016 ➢Determine the appropriate sized of an annual payment needed to retire P70,000,000 in bonds issued by a city to build a dam. The bonds must be repaid over a 50-yr period and they can earn interest at an annual rate of 6% compounded annually. ➢Ans. P4,200,000

Perpetuity • CE BOARD 1995 ➢Find the present value in pesos, of a perpetuity of P15,000 payable semi-annually if money is worth 8% compounded quarterly. ➢Ans. P371,287.12

***important ➢Classification of cost 1.

Fixed cost per unit = power consumption per unit + maintenance per unit + labor cost per unit 2. Variable cost per unit = (depreciation cost per annum + cost of housing the machine per annum)/(annual production units) 3. Total production cost per unit = fixed cost per unit + variable cost per unit ***Total production cost per unit is the Total expenses per unit.

***important • CE BOARD 2015 ➢ A machine costing P1,000,000 is expected to produce 10,000 units of steel pipe during the entire life before being replaced. At the end of its life, it will have a scrap value of P50,000. The cost of housing the machine is P25,000 a year. The power consumption per unit is P9 and the maintenance of the machine per unit will be P7. Labor costs P35 per unit. If depreciation of the machine is P198,772.71 per year, determine the total production cost per unit. The annual production is 2,500 units. ➢ Ans. P140.51

***important ➢ Break-Even Analysis 1. Break-Even Point- is the value of an independent variable such that two alternatives are equally attractive. 2. (Present sales volume of units per annum)*(Present break-even point cost)= Fixed expenses per annum 3. Total sales = total expenses 4. Selling price per unit = (total income of sold units)/ (total # of units sold)

***important • CE BOARD 2015 ➢Toyota Motors makes and sells certain automobile parts. Present sales volume is 500,000 units a year at a selling price of P0.50 per unit. Fixed expenses are P80,000 per year. Compute the present break even point in pesos. ➢Ans. P0.16

***important ➢A firm has the capacity to produce 1,000,000 units of production per year. At present, it is able to produce and sell 600,000 units yearly at a total income of P720,000. Annual fixed costs are P250,000 and the variable costs per unit is P0.70. Determine the number of units that should be sold annually to break-even. ➢Ans. 500,000 units

***important ➢CAPITALIZED COST ✓The sum of money such that the annual interest on it at the rate i% is equal to the annual investment of the asset.

***important ➢CAPITALIZED COST Capitalized Cost =

(𝐹𝐶−𝑆𝑉) (1+𝑖)𝑛 −1

+ FC +

𝑂𝐶 𝑖

Annual Cost = Capitalized Cost*i = Capital recovery cost + Operating cost or maintenance cost FC=first cost or replacement cost SV=salvage cost or disposal value (FC – SV) = rebuilt cost n=economic life OC=operating cost or maintenance cost

***important ➢CAPITAL RECOVERY COST ➢An asset whose series of cash flows consist of just two terms: an original cost P and an actual or estimated salvage value.

***important ➢CAPITAL RECOVERY COST

CRC=

(𝐹𝐶−𝑆𝑉)(1+𝑖)𝑛 𝑖 (1+𝑖)𝑛 −1

+ SV 𝑖

Annual savings = Capital recovery (CRC) + annual operating cost

***important ➢ A new bridge connecting Dumaguete and Santander with a 100-yr life is expected to have an initial cost of $20 M. This bridge must be resurfaced every five years at a cost of $1 M. The annual inspection and operating costs are estimated to be $50,000. Determine the present worth cost of the bridge using the capitalized equivalent approach (that is the life of the bridge as infinite), if the interest rate is 10% per year, compounded annually. ➢ Ans. $22,137,974.81

***important • CE BOARD 1999 ➢The first cost of a certain equipment is P324,000 and a salvage value of P50,000 at the end of its life of 4 years. Find the capitalized cost if money is worth 6% compounded annually. ➢Ans. P1,367,901.15

***important • CE BOARD 2016 ➢A machine which costs P50,000 when new has a 10-year lifetime and a salvage value equal to 10% of its original value. Determine the capital recovery based upon an interest rate of 8% per year, compounded annually. ➢Ans. P7,105

***important ➢ Pasay City is considering adding new buses for its current mass-transit system that links NAIA III to major city destinations on nonstop basis to decongest the terrible traffic. The total investment package is worth P8,000,000 and is expected to last 10 years with a salvage value of P750,000. The annual operating and maintenance costs for buses would be P2,000,000. If the system is used for 600,000 trips per year, what would be the fare price to charge per trip. Assume rate of interest is 5% compounded annually. ➢ Ans. P4.96 per trip

***important • CE BOARD 2015 ➢ A construction equipment is badly needed for a certain project so as to shorten the time of completion of the project. The equipment cost P1.2M and has a life of 5 years with a salvage value of P200,000 at the end of its life. The machine can be bought with money borrowed at an interest rate of 20% per annum. Annual operating cost is P10,000. Find the annual investment cost of using the equipment. ➢ Ans. P384,380

***important • CE BOARD 2015 ➢ A telephone company purchased a microwave equipment for P6 M with a salvage value of P600,000, over a period of 5 years and pay a lump sum of P400,000 for the maintenance cost. Minimum attractive rate of return is 16% annually. Compute the annual cost of investment of purchasing the microwave equipment. ➢ Ans. P1,867,374.7

***important • CE BOARD 2016 ➢ Maintenance expenditures for the Mananga Bridge in Tabunok, Talisay with a 20-yr life will come as periodic outlays of P1000 at the end of the 5th yr, P2000 at the end of the 10th yr. and P3000 at the end of the 15th yr. With interest at 10%, what is the equivalent uniform annual cost of maintenance for the 20-yr period? ➢ Ans. P248