Answer:
Initial Investment, P = $100, 000
Recurring Cost, A=$6200
(a) Calculate the internal rate of return for infinite life,
A= P(i)
6200=100000(i)
i = 6200/100000
i =6.2%
(b) Calculate the internal rate of return for 100 years,
P = A(P/A, i, 100)
100000 = 6200 (P/A, i, 100)
(P/A, i, 100) = 16.129
For i = 6%
(P/A,6%,100) = 16.618
For i = 7%
(P/A ,7%,100) = 14.269
i = [(6 - 7) / (16.618 - 14.269)] (16.129 - 16.618) +6
i = 6.2%
Thus, IRR = 6.2%
(c) Calculate the internal rate of return for 50 years,
P = A(P/A ,i , 50)
100000 = 6200 (P/A, i, 50)
(P/A, i, 50) = 16.129
For i = 6%
(P/A, 6%, 100) = 15.762
For i = 5%
(P/A , 5%, 100) = 18.256
i = [(5 - 6) / (18.256 - 15.762)] (16.129-15.762) + 6
i = 5.853%
Thus, IRR = 5.853%
(d)
In all cases internal rate of return is greater than 4%, which is minimum interest rate that one can earn. So they should consider to install the pipeline.
Answer:
Work in process account= $76,680
Explanation:
Giving the following information:
Balance in work in process on May 1 $57,600
Direct material costs for May $89,200
Overhead applied at a rate of 120% of direct labor dollars
Direct labor= $76,500
Jobs completed during May and transferred to finished goods inventory was $242,420
Work in process account= Beginning work in process+direct materials + direct labor + manufacturing overhead - Jobs completed during May
Work in process account=57600 + 89200 + 76500 + (76500*1.20)- 242420
Work in process account= $76,680
You should maybe identify 1 or 2 of what you’d say your strongest qualifications are :)
