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.
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:
106116.59
101616.59
105116.95
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: $P=A(P/A,6\%,15)=10{,}926(9.7123)=106{,}116.59$ $\boxed{P = 106116.59}$
Determine the payback period for a proposed investment as follows:
Answer:
3.72 years
3.27 years
3.5 years
3.33 years
### Payback period
The initial investment is $50{,}000$. Accumulate the undiscounted yearly cash flows:
| End of year | Cash flow (thousand dollars) | Cumulative cash flow (thousand dollars) |
| --- | ---: | ---: |
| 1 | 10 | 10 |
| 2 | 12 | 22 |
| 3 | 15 | 37 |
| 4 | 18 | 55 |
At the end of year 3, $50-37=13$ thousand dollars remains to be recovered. During year 4, the cash inflow is $18$ thousand dollars, so the fractional year is
$$\frac{13}{18}=0.722.$$
Therefore the payback period is
$$3+0.722=\boxed{3.72 \text{years}}.$$
P400 is borrowed for 75 days at 16% per annum simple interes. How much wil be due at the end of 75 days?
P4168.43
P5124.54
P4133.33
P5625.43
The printed principal appears to be missing a zero; the keyed answer matches P4000, not P400. Using ordinary simple interest: $F=P(1+rt)$ $F=4000\left[1+0.16\left(\frac{75}{360}\right)\right]$ $F=4000(1.03333)$ $\boxed{F=\text{P}4{,}133.33}$
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.
32%
40%
28%
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?
