Due to floods overseas, there is a cut in the supply of the mineral iron ore and its price increases dramatically. An Australian iron ore mining company therefore expects a large but temporary increase in its profit and cash flows. The mining company does not have any positive NPV projects to begin, so what should it do? Select the most correct answer.
A 90-day Bank Accepted Bill (BAB) has a face value of $1,000,000. The simple interest rate is 10% pa and there are 365 days in the year. What is its price now?
In mid 2009 the listed mining company Rio Tinto announced a 21-for-40 renounceable rights issue. Below is the chronology of events:
- 04/06/2009. Share price opens at $69.00 and closes at $66.90.
- 05/06/2009. 21-for-40 rights issue announced at a subscription price of $28.29.
- 16/06/2009. Last day that shares trade cum-rights. Share price opens at $76.40 and closes at $75.50.
- 17/06/2009. Shares trade ex-rights. Rights trading commences.
All things remaining equal, what would you expect Rio Tinto's stock price to open at on the first day that it trades ex-rights (17/6/2009)? Ignore the time value of money since time is negligibly short. Also ignore taxes.
In the dividend discount model:
###P_0 = \dfrac{C_1}{r-g}###
The return ##r## is supposed to be the:
The below screenshot of Commonwealth Bank of Australia's (CBA) details were taken from the Google Finance website on 7 Nov 2014. Some information has been deliberately blanked out.
What was CBA's approximate payout ratio over the 2014 financial year?
Note that the firm's interim and final dividends were $1.83 and $2.18 respectively over the 2014 financial year.
Question 513 stock split, reverse stock split, stock dividend, bonus issue, rights issue
Which of the following statements is NOT correct?
Question 794 option, Black-Scholes-Merton option pricing, option delta, no explanation
Which of the following quantities from the Black-Scholes-Merton option pricing formula gives the Delta of a European call option?
Where:
###d_1=\dfrac{\ln[S_0/K]+(r+\sigma^2/2).T)}{\sigma.\sqrt{T}}### ###d_2=d_1-\sigma.\sqrt{T}=\dfrac{\ln[S_0/K]+(r-\sigma^2/2).T)}{\sigma.\sqrt{T}}###