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Exam 2 Practice Problems
For Engineering Economy

Interest Rate Statements
There are 3 general ways to express interest rates as shown below: Interest Rate Statement Comment i = 12% per month i = 12% per year (1) When no compounding period is given, rate is effective i = 10% per year, comp’d semiannually i= 3% per quarter, comp’d monthly (2) When compounding period is given and it is not the same as period of interest rate, it is nominal When compounding period is given and rate is specified as effective, rate is effective over stated period i = effective 10%/yr, comp’d semiannually i = effective 4% per quarter, comp’d monthly (3)

Calculation of Effective Rates
Effective rates can be found from nominal rates for any time period longer than the compounding period by using the effective interest rate equation: i = (1 + r/m)m – 1 In using this equation, the time period on the left hand side of the equation for i must be exactly the same as the time period for the r, with m equal to the number of times interest is compounded over those time periods. For example, if the interest rate is given as nominal 24% per year compounded monthly, the effective interest rate per quarter would be determined as follows: i/quarter = [ 1 + r/quarter/no. times interest is Compounded, i.e., m] m – 1 = ( /3)3 – 1 = = 6.12%                                                                                                                              Note that the time period for i is exactly the same as the time period for r and that m is equal to 3 because interest is compounded 3 times in a quarter when interest is compounded monthly.

Answer: i/year = (1 + 0.12/4)4 – 1 = 12.55%
1. In calculating an effective semi-annual rate from an interest rate of 1.5 %  per month, the value of r and m in the effective interest rate equation are: Answer: 0.09 and 6 2. For an interest rate of 3% per quarter nominal, the effective annual rate is: Answer: i/year = ( /4)4 – 1          = 12.55% 3. For an interest rate of 9% per year, compounded monthly, the effective monthly rate is: Effective i/mo = 0.09/12                      = 0.75% per month

effective i /6 mos = (1 + 0.08/2)2 – 1 = 8.16%
4. For an interest rate of 16% per year, compounded quarterly, the effective i per six months is: effective i /6 mos = ( /2)2 – 1                                  = 8.16% 5. For an interest rate of effective 18% per year, compounded monthly, the effective monthly rate is Solve for nominal rate per year and then divide by 12:                   0.18 = ( 1 + r /12 )12 – 1 = (1+r/12)12             1 + r /12 =                    r /12 = r =   effective interest/month = 1.39% per month Note: r = 16.66% nominal rate/year

6. For an interest rate of 24% per year compounded quarterly, the nominal rate per six months is
Effective i/quarter = 0.24/4                                   = 6% per quarter       (effective)        r/6 mos = 0.06 * 2                     = 12% per six months                (nominal) 7. For an interest rate of 12% per year, compounded semiannually, the effective rate per two years is       r/2 years = 0.06 * 4                      = 24%       effective i/2 yrs = ( /4)4 - 1                                = 26.25%

8. An interest rate of 1% per month is the same as: (A)    Nominal 3% per quarter, compounded monthly (B)    Effective 3.03% per quarter, compounded monthly (C)    Effective 6.15% per six months, compounded monthly (D)    All of the above Answer A : i = (1 + r/m)m – 1 i/quarter = (1+0.03/3)3-1 = 3.03% Answer B : i = (1 + r/m)m – 1 i/6 mos. = (1+ .06/6)6 – 1 = 6.15% Answer C : i = (1 + r/m)m – 1 6.15%/6 mos. = (1+ r/6)6 – 1 r = .06 Answer: D

When using the single payment formulas (i. e
When using the single payment formulas (i.e. P/F or F/P), an effective interest rate over any time period can be used, as long as n is adjusted to equal the number of interest periods over the total time span between the P and F values. 9. For example, if the interest rate is 1% per month, an effective interest rate of 3.03% per quarter could be used in either the F/P or P/F factors for a 10 year period of time, and n would be 40 (i.e. 10 years x 4 quarters/year). Effective rates over other time periods could also be used, as long as n is adjusted accordingly. For example, if i of 1% per month is used, n would be 120.

On the other hand, when any of the uniform series or gradient formulas are used, only one value of n can be used and it is equal to the number of cash flow values in the cash flow series.  The interest rate must then be adjusted (if necessary) to an effective rate over the same time period as the n.     10. For example, if one wanted to know the amount of money that would be accumulated in 10 years from semiannual deposits of \$500 at an interest rate of 1% per month, the n would have to be 20 because of the semiannual deposits and, therefore, the i would have to be the effective semiannual interest rate (as calculated from the effective interest rate formula).  No other combination of n and i values could be used.

11. If deposits of \$500 are made quarterly into a savings account for 5 years at an interest rate of 12% per year, compounded monthly, the correct i and n values to use in the F/A equation are what? effective i/quarter = ( /3)3 - 1                                   = 3.03%                               n = 5 * 4                                  = 20 12. An engineer working for a small chemical company estimates that the cost for improving the efficiency of a certain process now would be \$100,000. The company president wants to know the equivalent cost for doing it 4 years from now at an interest rate of 12% per year, compounded quarterly. The correct values of i and n to use in the F/P equation are Answer: 3% and 16

13. Energy costs for a certain process are expected to be \$3000 in year 1, \$3,200 in year 2, and amounts increasing by \$200 per year thru year 10 at an interest rate of 12% per year, compounded monthly, the correct values of i and n to use in the P/A and P/G equations are: The cash flow occurs at yearly intervals. Therefore, an effective interest rate per year is required with n equal to the number of years. 12.68% and 10

When using the single payment formulas, any combination of i and n values can be used wherein i is an effective interest rate over any time period and n is the number of those interest rate periods between the P and F values. For example, for P and F values 6 years apart and interest compounded monthly, some of the possible i and n values that could be used are: i/mo and n = 72, i/quarter and n = 24, i/6 mos and n = 12, etc. The i must be an effective rate. 14. A garage door manufacturing company wants to make a single deposit now so that it can replace a small production line in 5 years. If the company wants to have \$1.2 million available and it can earn interest at 14% per year compounded semiannually, the amount of the deposit is Solution: The easiest effective interest rate to obtain is the semiannual rate of 7%. Using this rate,             P = 1.2 (P/F, 7%, 10)                = 1.2 (0.5083)                = \$609,960

Example – PP > CP The present worth of semiannual cash flows of \$500 for 8 years, beginning 6 months from now at an interest rate of 12% per year, compounded monthly is closest to: Solution: The payment period (i.e. 6 mos.) is greater than the CP (i.e. 1 month). Therefore, n is equal to the number of A values, which is 16 in this case. The only interest rate that can be used is an effective rate per semiannual period because n is in semiannual periods. The effective rate per 6 months must be calculated from the effective interest rate formula as follows: i /6 mos = ( / 6 )6 - 1              = 6.15% Now,         P = 500 (P/A, 6.15%, 16) =    \$5, (PV function in Excel)

A manufacturing company invests \$250,000 in a new warehouse facility
A manufacturing company invests \$250,000 in a new warehouse facility. At an interest rate of 12% per year, compounded quarterly, the equivalent cost per year over a 5 year recovery period would be closest to: Since cash flow is for yearly time periods, must use an effective interest rate per year:        i/year = ( /4)4 – 1                 = 12.55%              A = 250,000 (A/P, 12.55%, 5)                 = 250,000 [ ( )5]  /   [ ( )5 – 1 ]                 = 250,000 (0.2812)                 = \$70,300

An engineer deposits \$100 per month (beginning one month from now) into an account for 10 years which earns interest at 6% per year compounded semiannually. The amount in the account immediately after the last deposit is closest to: Move deposits to end of compounding period and then use effective i of 3% per six months in F/A factor:         F = 600 (F/A, 3%, 20)            = 600 ( )            = \$16,122

Example - Continuous Compounding of Interest The present worth of payments of \$1000 per quarter for 5 years starting 3 months from now at an interest rate of 12% per year compounded continuously is closest to _____________. Solution: Since PP>CP, n is equal to the number of payments, which is 20 in this case. Since the time period for n is quarters, i must be an effective rate per quarter. Using the effective interest rate equation,         i/quarter = e                       = 3.045% Now,         P = 1,000 (P/A, 3.045%, 20)            = \$14, (from PV function in Excel)

Varying Interest Rates
When interest rates vary over time, cash flow that is moved through periods having different rates should be adjusted with each period’s rate. For example, if the interest rates in years 1, 2, and 3 were 10%, 11%, and 12%, respectively, the equation for finding the present worth of a \$3000 amount in year 3 would be P = 3000 (P/F, 12%, 1) (P/F, 11%, 1) (P/F, 10%, 1)

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