Chapter 3 and 4 Book Problems Homework
docx
School
San Jose State University *
*We aren’t endorsed by this school
Course
200
Subject
Industrial Engineering
Date
Dec 6, 2023
Type
docx
Pages
4
Uploaded by AgentFlower9898
ISE 200 Chapter 3 and 4 Book Problems Homework Due Date : 03/23/2023 03/21/2023
Question 3-3 :
The amount of interest paid on the loan = $5350 - $5000 = $350
I = Prt
Substituting the known values : $350 = $5000*0.08*t
Simplifying:
t = $350/($5000*0.08) = 0.875 years
Converting years to months: t = 0.875*12 = 10.5 months
Question 3-32 :
i) If simple interest
R = 9%
T = 25 years
Amount = Q*25*9/100
= 2.25Q
ii) If compounded annually
Amount = Q[1 + 9/100]^25
= 8.623Q
Therefore, the amount available to help find an early retirement after 25 years of investment would be Q multiplied by 2.8908
Question 3-43 :
Effective annual interest rate = (1 + r/n)^n - 1
= ( 1 + 0.03/365)^365 - 1
= 0.0304 or 3.04%
Therefore, the effective annual interest rate for this bank account is 3.04%
Question 3-44 :
Effective interest rate = r
1 + r = (I + I/n)^n
Given r = 16.1 %
n = 12 since it is monthly compounded
i = nominal interest rate
1 + 16.1/100 = (1 + i/12)^12
1.161 = (1 + i/12)^12
Take log base 10 on both sides
log10(1.161) = 12log10(1 + i/12)
Solving for i we get i = 15%
Question 4-3 :
Annual deposits = $3500
Interest rate = 5%
Number of deposits = 40 (n = 40 years)
FV = $3500(F/A, 5%, 40)
= $3500(120.7998)
= $422799.3
He will have $422799.3 in his account after 40 deposits.
Question 4-11 :
Since present value is zero,
A*F/A*(r%,n) = F
2500*F/A(10%,n) = 61307
F/A(10%,n) = 24.5228
Since F/A(r%,n) = [(1 + r)^n - 1]/r
F/A(10%,n) = [(1.1)^n - 1/0.1
[(1.1)^n - 1]/0.1 = 24.5228
(1.1)^n - 1 = 2.4523 (1.1)^n = 3.4523
Taking natural logarithm on both sides,
n*ln1.1 = ln3.4523
n*0.0953 = 1.2390
n = 13
Question 4-19 :
Present worth of crane’s repair cost over its 15 year life:
Required present worth can be calculated as following :
PV = [A*{(1+i)^15-1/I(1+i)^15}] - [A*{(1+i)^5-1/i(1+I)^-10}]
= 5000[(1+0.1)^15-1/0.1(1+0.1)^15] - 5000[(1+0.1)^5-1/0.1(1+0.1)^5]
= 5000*7.6060 - 5000*3.7908
= 19076.50
Therefore, required Present worth of crane’s repair cost over it’s 15 year life is equal to 19076.50
Question 4-61 :
Question 4-76 :
Question 4-89:
Annuity = (R*PV)/[1-1/(1+R/N)^N]
93.12 = (R*1000)/[1-1/(1+R/12)^12]
A)Solving for R/12 we get R/12 = 1.7511%
B)R = 1.7511*12 =
21.0132%
C)EAR = (1 + 0.21032/12) - 1
= 23.16%
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
- Access to all documents
- Unlimited textbook solutions
- 24/7 expert homework help
EAR = 23.16%