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1. KDS Steel Ltd. has issued common stock at the market value Tk. 360 with flotation cost Tk. 15 at par. The company pays dividend per years Tk. 18 with 8% growth rate, what is the cost of the stock?
b) EDF 5 Housing Ltd. has issued bond at Tk. 1,000 at par at the rate of 12% coupon rate for 10 years. The flotation cost is Tk. 50 per bond and tax rate is 40%. If the bond is redeemable at 10% premium and issued (i) at par value (ii) at 10% premium and (iii) at 10% discount. What is the cost of the bond?
2.Gerald has taken out a loan of $100,000 today to start a business. He has agreed to repay the loan on the following terms: • Repayments will be made on a monthly basis. The first repayment will be made exactly one month from today. • The repayments for the first 5 years will cover interest only to help reduce the financial burden for Gerald's business at the start. • After the 5-year interest-only period, Gerald will make level monthly payments that will fully repay the loan after an additional 15 years (i.e. 20 years from today, the loan will be fully repaid). • The interest charged is 5% p.a. effective. Using this information, answer the following questions. a) Calculate the equivalent effective monthly rate on the loan. (1 mark) b) Calculate the size of the first repayment due exactly one month from now. (1 mark) c) Calculate the size of the level repayments that occur after the initial 5-year interest-only period. (2 marks)
3.Gerald has taken out a loan of $100,000 today to start a business. He has agreed to repay the loan on the following terms: • Repayments will be made on a monthly basis. The first repayment will be made exactly one month from today. • The repayments for the first 5 years will cover interest only to help reduce the financial burden for Gerald's business at the start. • After the 5-year interest-only period, Gerald will make level monthly payments that will fully repay the loan after an additional 15 years (.e. 20 years from today, the loan will be fully repaid). • The interest charged is 5% p.a. effective. Using this information, answer the following questions. a) Calculate the equivalent eective monthly rate on the loan. (1 mark) b) Calculate the size of the first repayment due exactly one month from now. (1 mark) c) Calculate the size of the level repayments that occur after the initial 5-year interest-only period. (2 marks) 10 years have passed, and Gerald's business is doing well. Further, he has made all the repayments on his loan so far as described above, and has just made the repayment due today. However, it has just been announced that the interest rate on Gerald's loan will go up to 5.5% p.a. compounding semi- annually. d) Calculate the new equivalent eective monthly rate on the loan. (1 mark) e) Calculate the current loan outstanding (again, it is 10 years after the loan was initially taken out). Note that the new interest rate only applies from today onwards. (2 marks) f) Because Gerald's business is doing well, he decides to repay a lump sum of $10,000 immediately. To further reduce the amount of interest he is paying to the bank, he will increase his monthly repayments to $1,000 per month. How many full repayments of $1,000 does Gerald have to make in order to fully repay this loan? (Note: Gerald may need to make a further, smaller payment in the subsequent month) (2 marks) g) Calculate the size of the smaller payment.
1.When a company pays dividends growing at a constant rate, we can calculate cost of equity using Gordon growth model as follows:
Cost of equity= {D1/(P-F)} + g
where,
D1= Expected dividend at the end of the year= Tk 18*(1+g)= Tk 18*(1+8%)= Tk 19.44
P= Issue price= Tk 360
F= Floation cost on issue of shares= Tk 15
g= constant growth rate= 8%
Therefore, replacing the above values in the said formula, we get as follows:
Cost of equity= {19.44/(360-15)} + 8% = (19.44/345) + 8% = 13.63%
2.Please use this google drive link to download the answer file.
https://drive.google.com/file/d/1wMXDibxxR0LC54uoZ6Fb6QDTpLOUCbfl/view?usp=sharing
Note: If you have any trouble in viewing/downloading the answer from the given link, please use this below guide to understand the whole process.
https://helpinhomework.org/blog/how-to-obtain-answer-through-google-drive-link
3.
Gerald's case
a) Equivalent effective monthly rate = (1+0.05)^(1/12)-1 = 0.0040741 or 0.40741%
b)1st repayment (interest only) = $100000 * 0.0040741 =$407.41
c) After 5 years , principal remaining wil be $100000 (as only interest has been paid in the 1st 5 years)
Level repayments (A) after 5 years is given by
A/0.0040741*(1-1/1.0040741^(15*12)) = 100000
=> A* 127.3854 =100000
=> A = $785.02
So, the level payments will be $785.02
d) Effective six monthly rate = 5.5%/2 = 0.0275
new equivalent effective monthly rate = (1+0.0275)^(1/6)-1 =0.0045317 or 0.45317%
e) Loan outstanding (5*12 =60 payments of $785.02 pending)
= 785.02/0.0040741*(1-1/1.0040741^60)
=$41711.17
f) After payment of $10000 immediately, principal remaining = $41711.17 - $10000 = $31711.17
No of payments (n) of $1000 required to pay the loan is given by
1000/0.0045317*(1-1/1.0045317^n) = 31711.12
=>1-1/1.0045317^n = 0.1437049
=> 1.0045317^n =1.1678217
Taking natural log of both sides
n = ln(1.1678217)/ln(1.0045317) = 34.312
So, Gerald will pay 34 full repayments of $1,000
g) The size of smaller payment = Future value of $31711.17 - Future value of 34 payments of $1000
= 31711.17*1.0045317^35 - 1000/0.00453147* (1.0045317^34-1)*1.0045317
=$312.58
