Table: an asset

YEAR | BENEFIT |

0 | 30 |

1 | 40 |

2 | 50 |

3 | 60 |

4 | 70 |

QUESTION 1. How would you set up the calculation to find the value of this asset in period two?

(a) 30(P\G, I=0.1, 4) (1+0.1)^{2}

(b) (1+0.1) ^{3} (10(P\G, I=0.1, 5)+30(P\A, I=0.1,5))

(c) 30(P\G, I=0.1, 5)(1+0.1)^{2}

(d) (1+0.1) ^{2} (10(P\G,I=0.1, 4)+30(P\A, I=0.1, 4))

(e) (1+0.1)^{3} (10(P\G, I=0.1, 4)+30(P\A, I=0.1, 5))

# Answer:

The Asset can be divided into a constant serie "Asset(1)", and a gredient serie "Asset(2)", so Asset= Asset(1) + Asset(2)

Asset(1) | + | Asset(2) | ||

YEAR | BENEFIT | YEAR | BENEFIT | |

0 | 30 | 0 | 0 | |

1 | 30 | 1 | 10 | |

2 | 30 | 2 | 20 | |

3 | 30 | 3 | 30 | |

4 | 30 | 4 | 40 |

## Conditions:

P_{2}= P_{2}of Asset(1) + P_{2} of Asset(2)

P_{2}(1)= 30(P\A, I=0.1, 5) (1+0.1)^{3}

P_{2}(2)= 10(P\G, I=0.1, 5) (1+0.1)^{3}

## P_{2} equation:

P_{2}=P_{2}(1)+P_{2}(2)

=30(P\A, I=0.1, 5) (1+0.1)^{3} + 10(P\G, I=0.1, 5) (1+0.1)^{3}

=(1+0.1)^{3} [30(P\A, I=0.1, 5) + 10(P\G, I=0.1, 5)]

**So, The right answer is "b".**