Question

Question 5 A company expects to receive the following payments from a customer: • £10,000 on 1.4.2017 • £4,500 on 1.7.2018 • £3,200 on 1.1.2022 The current date is 1 July 2016. This date represents t = 0, and hence 1 July 2017 ist 1. Assume that the discount function v(t) runs from time to time t and is given by: (O) = 1- 2 (a) Find the value of each payment as at 1 January 2020 (5 marks) (b) Using the results from (a), or otherwise, find the value of all payments at 1 July 2023. [2 marks) (c) Let p(t) represent a continuous payment of 200 per unit time. This is paid betweent1 and t = 4. Now let (s) be the force of interest and is represented by 8(s) = 0.1 + Bs. Write down an integral that represents the accumulated value at t = b (where b>4) for this continuous payment. [3 marks) [Total: 10 marks]

Can someone please help me with this question. I legit got no clue.

Answer 1

The question is based on time value of money which is a fundamental concept in finance.

On 1.1.20 $10,000 on 1.4.17 and $4500 on 1.7.18 has already been received . So the value will remain  the same . No change

However $3200 will be received after 2 years so value = 1- 2^2/60 = 56/60*3200 = $2986.67.is the present value

By Jul 1'2023 the 3rd payment of $3200 will be received. So the values will have no change i.e. $10000,$4200,$3200.

So total value on 1.7.23 = $17700 ( am using $sign as pound is unavailable on my key board)

The 3rd question is out of scope as it involves integral calculus