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Accounts Answers

Autor:   •  April 13, 2016  •  Coursework  •  734 Words (3 Pages)  •  764 Views

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Answer 1

i) Fixed bond = $100*0.06/2 = $3million

Value of fixed rate bond = 3e-0.05*3/12 + 3e-0.05*9/12 + 103e-0.05*15/12

                 = $102.61million    

ii) Floating bond = 0.07/2*$100 million + $100million

                             = $103.5million

Value of floating rate bond = $103.5e-0.07*3/12

                                             = $101.70million  

iii) Value of payment = (Notional principal) (Fixed rate %u2013 LIBOR) (Days/360)

                                    = ($100million) (0.06 %u2013 0.07) (90/360)

                                    = $0.25million.

iv) Value of payment = (Notional principal) (Fixed rate %u2013 LIBOR) (Days/360)

                                    = ($100million) (0.06 %u2013 0.07) (270/360)

                                    = $0.75million.

v) Value of payment = (Notional principal) (Fixed rate %u2013 LIBOR) (Days/360)

                                    = ($100million) (0.06 %u2013 0.07) (450/360)

                                    = $1.25million.

vi) Value of swap = Value of fixed bond %u2013 Value of floating bond

                              = $102.61million - $101.70million

                              = $0.91million

Answer 2

The fixed rate of interest can the company earn 2.7%

When the company invests at LIBOR minus 0.3% and then enters into a swap where it pays LIBOR and receives 3% it earns 2.7% per annum. Note that it is the bid rate that will apply to the swap.

Answer 3

The estimate is 100 x 0.2 x 1/52 = $2.77

Answer 4

  1. 1 x (10,000,000/100 x 500) = 200 contracts
  2. 9,500,000/ 100 x 200 = $475

475 since [9.5/10 - 1 – R] = 1 [ K/500 - 1 – R] implies K = 475. R is the redundant 6-month risk-free rate. Note that R is redundant if and only if the beta of the portfolio equals 1, because R cancels out from both sides of the equation.

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