Answer:
First Expected Dividend will come in at the end of Year 3 or t=3 assuming current time is t=0.
D3 = $ 4.25, Growth Rate for year 4 and year 5 = 22.1 %
Therefore, D4 = D3 x 1.221 = 4.25 x 1.221 = $ 5.18925 and D5 = D4 x 1.221 = 5.18925 x 1.221 = $ 6.33607
Growth Rate post Year 5 = 4.08 %
D6 = D5 x 1.0408 = 6.33607 x 1.0408 = $ 6.59459
Required Return = 13.6 %
Therefore, Current Stock Price = Present Value of Expected Dividends = [6.59459 / (0.136-0.0408)] x [1/(1.136)^(5)] + 4.25 / (1.136)^(3) + 5.18925 / (1.136)^(4) + 6.33607 / (1.136)^(5) = $ 45.979 ~ $ 45.98
Price at the end of Year 2 = P2 = Present Value of Expected Dividends at the end of year 2 = [6.59459 / (0.136-0.0408)] x [1/(1.136)^(3)] + 4.25 / (1.136) + 5.18925 / (1.136)^(2) + 6.33607 / (1.136)^(3) = $ 59.3358 ~ $ 59.34
Dividend Yield at the end of year 3 = DY3 = D3 / P2 = 4.25 / 59.34 = 0.07612 or 7.612 %
Total Required Return = 14. 6 %
Therefore, Required Capital Gains Yield = 14.6 % - 7.612 % = 6.988 %
Answer:
Sales= $3,000,000
Explanation:
Giving the following information:
It expects to sell 10,000 mattresses in the current year and had 1,000 mattresses in finished goods inventory at the end of the previous year. Armando would like to complete operations in the current year with at least 1,250 completed mattresses in inventory. There is no ending work-in-process inventory. The mattresses sell for $300 each.
Production:
Sales= 10,000
Ending inventory= 1,250
Beginning inventory= (1,000)
Total= 10,250
Sales= 10,000*300= $3,000,000
Answer:
The profit for an investor who has $500,000 available to conduct locational arbitrage is $1,639.
Explanation:
Bank A has a ask rate of $0.305, so the investor can exchange his $500,000 at Bank A and get = $500,000/$.305 = MYR = 1,639,344
Bank B has a bid rate of $0.306, he can invest 1,639,344= 1,639,344 × $.306 = $501,639.
501,639 - $500,000 = $1,639.
Thus, the profit is $1,639.
Answer: 4 containers
Explanation:
The formula used to get the number of containers that are needed will be:
N = DT(1+X)/C
where,
N = total containers
D = planned usage rate used by the work center = 111 parts per hour
T = average waiting time = 100 minutes = 100/60 hours = 1.67 hours
X = inefficiency factor = 0.21
C = capacity of standard container = 5 dozens = 5 × 12 = 60 parts
N = DT(1+X)/C
N = (111 × 1.67)(1 + 0.21)/60
N = (185.37 × 1.21)/60
N = 224.2977/60
N = 3.738
N = 4 approximately
4 containers will be needed
