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Annuity Due

An annuity due has payments at the start of each period.

$$P=A\left(\frac{(1+i)^{n-1}-1}{(1+i)^{n-1}i}\right)+A$$
$$F=A\left(\frac{(1+i)^{n+1}-1}{i}\right)-A$$

Problem: Annuity Due Present Value

An engineer receives P25,000 at the beginning of each year for 18 years. Find the present value at 4% compounded annually at the time of the first payment.

$$P=25000\left(\frac{(1.04)^{17}-1}{(1.04)^{17}(0.04)}\right)+25000=329,142$$

Final answer: P329,142.

Problem: Quarterly Annuity Due

A student receives P3,000 at the beginning of each 3 months for 4 years. Find the sum at the end of 4 years at 6% compounded quarterly.

$$i=0.06/4=0.015, \quad n=16$$
$$F=3000\left(\frac{(1.015)^{17}-1}{0.015}\right)-3000=54,604$$

Final answer: P54,604.

Problem: Future Worth of Annuity Due

P6,000 is deposited at the beginning of each year for 4 years. Find the future worth at the end of year 4 if i = 8%.

$$F=6000(F/A,8\%,4)(1.08)=6000(4.5061)(1.08)=29200$$

Answer: The future worth is about P29,200.

Problem: Present Worth of Rent Paid in Advance

Rent of P18,000 is paid at the beginning of each year for 5 years. Find present worth at 10%.

$$P=18000(P/A,10\%,5)(1.10)=18000(3.7908)(1.10)=75058$$

Answer: The present worth is about P75,058.

Problem: Required Beginning-of-Year Deposit

How much must be deposited at the beginning of each year for 3 years to accumulate P50,000 at the end of year 3 at 7%?

$$A=\frac{50000}{(F/A,7\%,3)(1.07)}=\frac{50000}{3.2149(1.07)}=14533$$

Answer: Deposit about P14,533 at the beginning of each year.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q180

MSTE - Engineering Economy / Annuity Due / Engr. Janclyde Espinosa (Clidez)

An engineer is entitled to received P25000 at the beginning of each year for 18 yrs. What is the present value in pesos of this annuity at the time he is suppose to receive the first payment if the rate of interest is 4% compounded annually?

Answer:

  1. 329142
  2. 392142
  3. 324912
  4. 322419
This is an annuity due, valued at the time of the first payment:
$P=A[1+(P/A,4\%,17)]$
$P=25{,}000\left[1+\frac{1-(1.04)^{-17}}{0.04}\right]$
$\boxed{329142}$

Question Bank: q181

MSTE - Engineering Economy / Annuity Due / Engr. Janclyde Espinosa (Clidez)

A student will receive 3000 pesos at the beginning of each 3 months for 4 yrs. What is the sum of this annuity at the end of the 4th year if the interest rate is 6% compounded quarterly.

Answer:

  1. 54604
  2. 56404
  3. 54406
  4. 54064
Quarterly rate is $i=0.06/4=0.015$, with $n=16$ payments. Since payments are at the beginning of each quarter, use annuity due future worth:
$F=A(F/A,i,n)(1+i)$
$F=3000\left[\frac{(1.015)^{16}-1}{0.015}\right](1.015)$
$\boxed{54604}$

Question Bank: q223

MSTE - Engineering Economy / Perpetuity / Engr. Janclyde Espinosa (Clidez)

CE Board November 1995
Find the present value in pesos, of a perpetuity of P15,000 payable semi-annually if money is worth 8% compounded quarterly.

Answer:

  1. 371287.12
  2. 378271.12
  3. 327781.12
  4. 327718.12
Convert 8% compounded quarterly to effective semi-annual rate:
Quarterly rate $= 8\%/4 = 2\%$
Effective semi-annual $= (1.02)^2 - 1 = 0.0404 = 4.04\%$
PV of perpetuity: $P = \frac{A}{i} = \frac{15{,}000}{0.0404}$
$\boxed{\approx \text{P}371{,}287}$

Question Bank: q238

MSTE - Engineering Economy / Breakeven / Engr. Janclyde Espinosa (Clidez)

The Asian Transmission Co. makes and sells certain automotive parts. Present sales volume is 500,000 units a year at a selling price of P0.50 per unit. Fixed expenses total P80,000 per year. Determine the present total profit for a year.

Answer:

  1. 170000
  2. 250000
  3. 225000
  4. 190000
Annual revenue is:
$500{,}000(0.50)=250{,}000$
Profit after fixed expenses:
$250{,}000-80{,}000=170{,}000$
$\boxed{170000}$

Question Bank: t1009

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

You have a choice to borrow P5,000 from a friend or from a bank. The bank charges 20% per year while your friend charges 1.6% per month. If the loan is payable in one year, where will you borrow to get the best deal and why?

  1. From the bank; I will save P1049
  2. From the bank; I will save P49.00
  3. From the friend; I will save P49.00
  4. From a friend; I will save P1049
Bank: 20%/yr simple interest on P5,000:
$I_{\text{bank}} = 5{,}000 \times 0.20 = $ P1,000
Friend: 1.6%/month compounded monthly for 1 year:
$F = 5{,}000(1.016)^{12} = 5{,}000 \times 1.2090 \approx $ P6,045
$I_{\text{friend}} \approx $ P1,045
Bank interest is lower by $\approx 1{,}045 - 1{,}000$
$\boxed{\approx \text{P}49 \text{ savings; borrow from bank}}$

Question Bank: t1013

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

A house and lot costing P2 million was bought at a down payment of P500,000 and P1 million after one year. The remaining balance will be paid at the end of the fourth year. If the interest rate is 24% compounded semi-annually, what is the required payment?

  1. P1,740,122
  2. P1,568,345
  3. P1,632,547
  4. P1,704,586
Effective annual rate: $i = (1 + 0.24/2)^2 - 1 = (1.12)^2 - 1 = 25.44\%$
FV of price at year 4: $2{,}000{,}000(1.2544)^4 = 2{,}000{,}000 \times (1.12)^8 = 2{,}000{,}000 \times 2.47596 = $ P4,951,920
FV of down payment at year 4: $500{,}000(1.12)^8 = 500{,}000 \times 2.47596 = $ P1,237,980
FV of year-1 payment at year 4: $1{,}000{,}000(1.12)^6 = 1{,}000{,}000 \times 1.97382 = $ P1,973,820
Balance = $4{,}951{,}920 - 1{,}237{,}980 - 1{,}973{,}820$
$\boxed{= \text{P}1{,}740{,}120 \approx \text{P}1{,}740{,}122}$

Question Bank: t1028

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

Twenty-one equal end-of-year deposits are made into a savings account that pays 4% interest. Compute the amount of each deposit that will permit withdrawals of P20,000 at the ends of the eighteenth through twenty-first years, leaving the account empty.

  1. P2657
  2. P3215
  3. P2875
  4. P4523
21 deposits at $i=4\%$; 4 withdrawals of P20,000 at end of years 18, 19, 20, 21.
FW of withdrawals at year 21:
$FW_W = 20{,}000[(1.04)^3+(1.04)^2+(1.04)^1+1] = 20{,}000 \times 4.2465 = $ P84,929
FW of deposits at year 21:
$(F/A,4\%,21) = \frac{(1.04)^{21}-1}{0.04} = \frac{2.27877-1}{0.04} = 31.969$
Setting FW equal: $A \times 31.969 = 84{,}929$
$A = \frac{84{,}929}{31.969}$
$\boxed{\approx \text{P}2{,}657}$

Question Bank: t1058

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

A civil engineer is considering the purchase of a dump truck. He made a 10% down payment of P120,000 and the balance paid in 48 equal monthly payments with interest at 1% per month. The payments are due at the end of each month. What is the monthly payment?

  1. P31452
  2. P30574
  3. P29124
  4. P28441

Solution pending in psadquestions/t1058.json.

Question Bank: t1083

MSTE - Engineering Economy / Engineering Economy / Gemini mapped Chapter 7 to 10

How large a contribution is required to endow perpetually a research laboratory which requires P500,000 for original construction, P200,000 per year for operating expenses, and P100,000 every three years for new and replacement equipment? Interest is 4%.

  1. P5.9M
  2. P6.8M
  3. P6.3M
  4. P7.2M
Capitalized cost = construction + perpetual operating + perpetual equipment replacement.
Annual operating (perpetual): $200{,}000/0.04 = 5{,}000{,}000$.
Equipment every 3 yr: $\dfrac{100{,}000}{(1.04)^3 - 1} = \dfrac{100{,}000}{0.12486} = 800{,}871$.
$$CC = 500{,}000 + 5{,}000{,}000 + 800{,}871$$
$$\boxed{CC \approx \text{P}6.3\text{M}}$$