Transcript
Engineering Economics – Chapter 2
Alternate method when g > i
If g > i, then g1 will be negative. The factor could still be calculated with the formula but it cannot be obtained from the table. A different form can be used in this case.
Define a fictitious growth rate:
(11)
Note: Although the factor is F/A, here it is used to calculate a present value.
Example: A series of 10 transactions grows by 9.2% each time, starting at $100. If i = 5% per period, find PV (Present Value).
(a) Method 1 - eqn. (10):
(b) Method 2 eqn. (11):
Note: One method has (1 + g) in the denominator, whereas the other has (1 + i) in the denominator.
Example: A person's salary is increasing at 5%/year and the interest rate is 5%/year. With an initial after-tax salary of $20,000 received in one year, what is the present value of the person's earnings if he/she works for 25 years.
= $20,000, i = 0.05, g = 0.05, n = 25
case: i = g P = 20,000(25)/1.05 = $476,190
Examples of equivalence calculations
178308060960F
P
A
G
g
00F
P
A
G
g
288798015240000
Composite Cash flows:
- can apply factors to subsets of transactions
1. Negative Linear gradient
An investor can make 3 annual end-of-year payments, starting with $15,000 but reduced by $1,000 each year thereafter. The project will generate receipts of $10,000 at the end of year 4 which will increase annually by $2,500 for the following 4 years. If the investor can earn a rate of return of 10 percent annually, is this alternative attractive?
PW = [-15,000 – (-1000)(A/G, 10, 3)](P/A, 10, 3) + [10,000 + 2500(A/G, 10, 5)](P/A, 10, 5)(P/F, 10, 3)
2164080463550040005003873500
35433004635500
30784805397500
195072052070$10,000
00$10,000
238506052070$12,500
00$12,500
284988052070$15,000
00$15,000
377190052070$20,000
00$20,000
330708052070$17,500
00$17,500
26212806159500
2141220768350021717006921500
533400101600$15,000
00$15,000
1005840101600$14,000
00$14,000
1409700133985$13,000
00$13,000
39700201377958
008
35128201377957
007
30480001377956
006
25984201377955
005
21412201377954
004
16687801454153
003
12039601454152
002
7391401454151
001
22860016827500124968076835001706880768350077724076835003276607683500
169164092075007772409207500
77724010731500
Example: An ambitious saver plans to deposit $2,000 in a money market account starting 1 year from now and wants to increase annual deposit by $1,000 each year for the following 6 years. Assuming that deposits earn 9 percent annually, determine what equal-payment annuity would accumulate the same amount over the 7-year period.
45605701454150037661851454150029718001454150021640801454150013639801454150056578514541500536829014541500
4114806159500502920615950
000
2927985615953
003
2114550615952
002
3703320615954
004
4526280615955
005
5349240615956
006
1303020330201
001
3876675101600022688556350030822906350046901106350055035456350029375101054103G
003G
2124075387352G
002G
1341120-8890G
00G
565785635005562601016000
2268855825500308229099695004535805996955G
005G
371284582554G
004G
38766751587500469011097790005358765158756G
006G
55130702349500
A = + G (A/G, i, N)
= $2,000 + $1,000 (A/G, 9, 7) = $2,000 + $1,000(2.6574) = $2,000 + $2,657 = $4,654
=> seven equal payments of $4657 are equivalent to seven payments increasing by $1,000 from $2,000 for the first one to $8,000 for the last one.
PW = [-15,000 + 1,000(0.9366)] x 2.4896 + [10,000 +2,500(1.8101)] x 3.7908 x 0.7513 = 6,356.08
Conclusion: positive PW, therefore invest.
2. Negative geometric gradient:
You win a lottery which pays $20,000 and 20 end-of-year payments for an amount equal to 5% LESS than the previous year. If you save all of the money, and the interest rate is 7% how much will you have at the end of 20 years.
Let
F = [20,000 +A/(1+g)*(P/A, 12.63, 20)] x (F/P, 7, 20) = [20,000 + 19,000/(0.95) x (7.184)] x (3.8697)
F = $633,392
Would you rather have $150,000 now?
Can this problem be done without separating the initial receipt of $20,000 from the rest of the cash flow?
3. Composite cash flow
An 18-year old student will need to withdraw $5,000 per year for 6 years for her education starting in 1 year (Master's degree included). Her rich uncle will contribute $3,000 into the account now, increasing by a fixed amount each year until the end of the 5th year. What is the required annual gradient to exactly satisfy the student's requirements? i = 12%/year.
Set the PW equal
5000(P/A, 12, 6) = [3000(P/A, 12, 6) + G(P/G, 12, 6)](F/P, 12, 6)
5000(4.1114) = [3000(4.1114) + G(8.9302)](1.12)
G = $674.15/year
Categories of unknowns
For most calculations involving factors, there are 4 variables of which 3 are known and the fourth one can be solved for. For example, given: P, i, N solve F. Sometimes i or N is the unknown.
Example: Invest $1,000 for 6 years to attain $1,600. What rate of return i is required?
F = P(F/P, i, N) (F/P, i, 6) = F/P = 1.6 = (1 + i)6 1 + i = (1.6)1/6 = 1.08148 i = 0.08148 = 8.15%/yr
or by interpolation in the interest tables.
(F/P, 8, 6) = 1.5868 and (F/P, 9, 6) = 1.6770 i = 0.08 + 0.01 = 0.08146
IF THE EXPRESSION IS TOO COMPLEX TO ISOLATE i, THEN TRIAL & ERROR AND INTERPLATION MUST BE USED (OR A SUITABLE COMPUTER PROGRAM).
EQUIVALENCE
1013460120650Each Payment = $317.70
00Each Payment = $317.70
502729559690006934202159000502920029210003185160292100045643803492500411480027305003649980273050031927803492500
47548806540500477012080645$1,791
00$1,791
175260501650019050065405$1,000
00$1,000
276606011303000731520292100044958014922500
3139440144145Each Payment = $237.40
00Each Payment = $237.40
534924052705i = 6%
00i = 6%
4343406794500272034073025002278380730250018059407302500135636073025008915407302500
4495801270000388620660400
000
8972551270000836295755651
001
1301115755652
002
135445512700001765935755653
003
180784512700002238375679454
004
22783801270000272224512700002695575679455
005
3145155679456
006
318897012700003609975679457
007
365569512700004067175679458
008
411289512700004518660660409
009
456438012700005027295127000049682406604010
0010
457209652000
2423160-137795Time in Years
00Time in Years
$1,000 today is equivalent to $1,791 received 10 years from now.
$1,000 today is equivalent to $237.40 received at the end of each year for the next five years.
$1,000 today is equivalent to $317.70 received at the end of years 6, 7, 8, 9 and 10.
167068534099500
110299529273500
4.2124 0.7473
$237.40 received at the end of each year for the next 5 years is equivalent to a lump sum of $1,791 received 10 years from now.
$317.70 received at the end of years 6, 7, 8, 9 and 10 is equivalent to $1,791 in 10 years form now.
$237.40 received at the end of each year for the next 5 years is equivalent to $317.70 at the end of years 6, 7, 8, 9 and 10.
Example: Given the values of F and PA below, if i = 10%, find N?
P = 1000 F = 1600 i = 10% N = ?
(F/P, 10, N) = F/P = 1.6 = (1 + i)N = 1.1N ln(1.6) = Nln(1.1) years.
or by interpolation in the interest tables.
(F/P, 10, 4) = 1.4641 and (F/P, 10, 5) = 1.6105 N = 4 + 1 years.
