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Arithmetic Gradient Series

An arithmetic gradient changes by a constant amount each period.

$$A_{equiv}=G(A/G,i,n)$$

Problem: CE Board May 2022

Maintenance is $155 at the end of year 1 and increases by $35 each year for 8 years. Find the present worth at 6%.

$$P_1=155(P/A,6\%,8)=155(6.2098)=962.50$$
$$P_2=35(P/G,6\%,8)=35(19.845)=694.58$$

Final answer: $P=P_1+P_2=1,657$.

Problem: Retirement Gradient

Withdrawals start at $5,000 and increase by $1,000 yearly for 15 years. At 6%, find required savings at retirement.

$$A_2=1000(A/G,6\%,15)=1000(5.9260)=5,926$$
$$A_3=5000+5926=10,926$$
$$P=10926(P/A,6\%,15)=106,116.59$$

Final answer: $106,116.59.

Problem: Present Worth of Arithmetic Gradient

Maintenance starts at P10,000 in year 1 and increases by P2,000 each year through year 5. At 10%, find the present worth.

$$P=10000(P/A,10\%,5)+2000(P/G,10\%,5)$$
$$P=10000(3.7908)+2000(6.8618)=51632$$

Answer: Present worth is about P51,632.

Problem: Equivalent Annual Arithmetic Gradient

A cost series has base A = P20,000 and arithmetic gradient G = P3,000 for 6 years at 8%. Find the equivalent annual cost.

$$A_{eq}=20000+3000(A/G,8\%,6)$$
$$(A/G)=\frac{1}{0.08}-\frac{6}{(1.08)^6-1}=2.350$$

Answer: $A_{eq}=20000+3000(2.350)=P27,050$.

Problem: Find the Last Payment

A series starts at P5,000 and increases by P750 each year for 8 years. Find the payment in year 8.

$$A_8=A_1+(8-1)G=5000+7(750)=10250$$

Answer: The year-8 payment is P10,250.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: t1014

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

If money is worth 6% compounded annually, what payment 12 years from now is equivalent to a payment of P125,000, nine years from now?

  1. P156321
  2. P148877
  3. P149876
  4. P145628
Move P125,000 from year 9 to year 12 (3 years forward) at $i = 6\%$:
$F = 125{,}000 \times (1.06)^3 = 125{,}000 \times 1.19102$
$\boxed{= \text{P}148{,}877}$

Question Bank: t1049

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

The cost of an asphalt pavement is as follows: Initial cost........................................... P20M. Annual maintenance.......................... P1.5M. Additional maintenance every 5 years... P3 M. Cost of money...................................... 8.0%.

What is the present worth of all additional maintenance costs for 25 years?

  1. P5,653,254
  2. P6,547,125
  3. P5,458,753
  4. P4,852,365

What is the present worth of all the costs for 25 years?

  1. P18,256,365
  2. P21,470,938
  3. P20,235,653
  4. P19,256,365

What is the equivalent annual cost of this pavement?

  1. P4,125,568
  2. P3,874,562
  3. P3,125,745
  4. P3,611,369

Solution pending in psadquestions/t1049.json.

Question Bank: t1086

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

An investor is considering buying a 20-year corporate bond. The bond has a face value of P1M and pays 6% interest per year in two semiannual payments. Thus the purchaser of the bond would receive P30,000 every six months and in addition he would receive P1M at the end of 20 years, along with the last P30,000 interest payment. If the investor thought he should receive 8% interest, compounded semiannually, how much would he be willing to pay for the bond?

  1. P802,072
  2. P523,114
  3. P851,120
  4. P658,147
Per period (semiannual): yield $i = 8\%/2 = 4\%$; coupon $= 30{,}000$; $n = 40$ periods; redemption $= 1{,}000{,}000$.
$$P = 30{,}000\,(P/A,4\%,40) + 1{,}000{,}000\,(P/F,4\%,40)$$
$$P = 30{,}000(19.793) + 1{,}000{,}000(0.2083)$$
$$\boxed{P \approx \text{P}802{,}072}$$