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Recovery, Payback, and Payout Periods

These measures estimate how long it takes to recover invested capital from profits or cash inflows.

$$Recovery\ Period=\frac{Total\ Investment}{Annual\ Profit}$$
$$Payback\ Period=\frac{Initial\ Investment}{Net\ Annual\ Profit}$$

Problem: Recovery Period

JG Summit invests P9.5 M and expects annual income P2.8 M. Annual depreciation P1 M, operating expenses P600,000, taxes and insurance P200,000, and miscellaneous P50,000. Find recovery period.

$$Annual\ profit=2,800,000-1,000,000-600,000-200,000-50,000=950,000$$
$$Recovery=\frac{9,500,000}{950,000}=10\text{ years}$$

Final answer: 10 years.

Problem: Payback From Uneven Cash Flow

Initial investment is $50,000. Cash inflows are $10,000, $12,000, $15,000, $18,000, and $20,000. Find payback period.

After 3 years, cumulative inflow is $37,000. After 4 years, it is $55,000. Remaining after year 3 is $13,000.

$$x=\frac{13000}{18000}=0.72$$

Final answer: $3.72$ years.

Problem: Payout Period

A plant requires fixed capital P8 M, working capital P1.5 M, annual profit P2 M, and depreciation 8% of fixed capital. Compute payout period.

$$Annual\ depreciation=0.08(8,000,000)=640,000$$
$$Payout=\frac{8,000,000}{2,000,000+640,000}=3.03\text{ years}$$

Final answer: 3.03 years.

Problem: Simple Payback with Uniform Savings

A project costs P480,000 and saves P80,000 per year. Find the simple payback period.

$$Payback=\frac{480000}{80000}=6\text{ years}$$

Answer: The simple payback period is 6 years.

Problem: Discounted Payback Check

An investment costs P100,000 and returns P60,000 at the end of each of years 1 and 2. At 10%, determine if discounted payback occurs within 2 years.

$$PW=60000(0.9091)+60000(0.8264)=104130$$

Answer: Yes. Discounted recovery exceeds P100,000 within 2 years.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q188

MSTE - Engineering Economy / Uniform Gradient Series / Engr. Janclyde Espinosa (Clidez)

An engineer is planning for a 15 year retirement. In order to supplement his pension and offset the anticipated effects of inflation, he intends to withdraw P5000 at the end of the first year, and to increase the withdrawal by P1000 at the end of each successive year. How much money must the engineer have in his savings account at the start of his retirement, if money earns 6% per year, compounded annually?
A/G (6%, 15) = 5.9260
P/A (6%, 15) = 9.7123
F/A (6%, 15) = 23.2760

Answer:

  1. 106116.59
  2. 101616.59
  3. 105116.95
  4. 115016.95
The withdrawal series is a uniform P5,000 plus an arithmetic gradient of P1,000:
$A = 5000 + G(A/G,6\%,15)$
$A = 5000 + 1000(5.9260)=10{,}926$
Present worth at retirement:
$P=A(P/A,6\%,15)=10{,}926(9.7123)=106{,}116.59$
$\boxed{P = 106116.59}$

Question Bank: q216

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

Determine the payback period for a proposed investment as follows:

q216

Answer:

  1. 3.72 years
  2. 3.27 years
  3. 3.5 years
  4. 3.33 years

Solution pending in psadquestions/q216.json.

Question Bank: t2016

MSTE - Engineering Economy / Simple Interest / BEMz

P400 is borrowed for 75 days at 16% per annum simple interes. How much wil be due at the end of 75 days?

  1. P4168.43
  2. P5124.54
  3. P4133.33
  4. P5625.43
The printed principal appears to be missing a zero; the keyed answer matches P4000, not P400. Using ordinary simple interest:
$F=P(1+rt)$
$F=4000\left[1+0.16\left(\frac{75}{360}\right)\right]$
$F=4000(1.03333)$
$\boxed{F=\text{P}4{,}133.33}$

Question Bank: t2019

MSTE - Engineering Economy / Simple Interest / BEMz

In buying a computer disk, the buyer was offered the options of paying P250 cash at the end of 30 days or P270 at the end of 120 days. At what rate is the buyer paying simple interest if he agree to pay at the end of 120 days.

  1. 32%
  2. 40%
  3. 28%
  4. 25%
Choosing P270 at 120 days instead of P250 at 30 days means paying P20 interest for an extra 90 days.
$I=Prt$
$20=250(r)\left(\frac{90}{360}\right)$
$20=62.5r$
$\boxed{r=32\%}$

Question Bank: w83

MSTE - Engineering Economy / Payback Period / MSTE November 2019

Rosenberg Engineering has offices in northern California, where a heat pump can be used for cooling in the summer and heating in the winter. Replacing the current system will cost $1,500 in May and $500 in June. Starting in July, it will save $200 per month for the summer months (June–August), $100 for the fall and spring months, and $150 for the winter months (November–March). What is the payback period for the heat pump?

  1. 12 months
  2. 18 months
  3. 14 months
  4. 16 months
Initial cost $= \$1{,}500 + \$500 = \$2{,}000$.
Annual savings:
• Summer (Jun–Aug): $\$200 \times 3 = \$600$
• Fall/spring (Sep–Oct): $\$100 \times 2 = \$200$
• Winter (Nov–Mar): $\$150 \times 5 = \$750$
Total $= \$1{,}550$ per year.
Payback $= \dfrac{2000}{1550} = 1.29$ yr $\times 12 \approx 15.5$ months
$\boxed{\approx 16\text{ months}}$