EXAMPLE 34.14 : For a formula of y = [ 1 - 1 / (1 + r) ^ n ] / r, let y = present value, r = interest rate / year, n = number of years to future value of 1 : (a) Find a simple mathematical relationship of y as a function of r when n = 1 million years. (b) What is the present value of $10000, if the annual discount rate is 10 %, forever?
(a) When n = 1 million years, it could be approximated as n = infinity. Then y = [ 1 - 1 / (1 + r) ^ n ] / r = [ 1 - 1 / (1 + r) ^ (infinity) ] / r = [ 1 - 1 / (infinity) ] / r = (1 - 0) / r = 1 / r. (b) For future value of $10000 and n = infinity forever, use the equation of y = 1 / r for future value of 1 and interest rate r. As r = 10 % = 0.1, then y = x / r = $10000 / 0.1 = $100000, where x = future value. The answer is given by Kang Chuen Tat; PO Box 6263, Dandenong, Victoria VIC 3175, Australia; SMS +61405421706; chuentat@hotmail.com; http://kangchuentat.wordpress.com.
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ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.8 : Cash flow agreement where the sign + or - of the amounts are start negative and at a point switch to positive and stay positive is referred to has an investment. The cash flow could be denoted by the symbol of [a, b, c ...]. Consider the following two cases in a biochemical factory : (a) You pay $1 today. You receive $1.05, one year from now. (b) You pay $3 today. You receive $5, one year from today. You pay $2, two years from today. Determine if cases (a) and (b) are an investment or a business transaction with cash flow symbol. Find the interest rate r in the cash flow in both cases (a) and (b).
Question 68 – Biochemical Oxygen Demand (BOD) could be calculated using the formula BOD = (DOi - DOf) (Vb / Vs) where Vb = Volume of bottle in ml, Vs = Volume of sample in ml, DOi = Initial dissolved oxygen in mg / L, DOf = Final dissolved oxygen in mg / L. (a) By using a bottle of Vb = 300 ml with sample Vs = 30 ml, find the BOD if DOi = 8.8 mg / L and DOf = 5.9 mg / L. (b) By using a bottle Vb = 600 mL with sample Vs = 100 mL, find the BOD if DOi = 8.8 mg / L and DOf = 4.2 mg / L. (c) Find the average BOD = [ Answer of (a) Answer of (b) ] / 2. (d) If the BOD-5 test for (a) - (c) is run on a secondary effluent using a nitrification inhibitor, find the nitrogenous BOD (NBOD) = TBOD - CBOD. Let TBOD = 45 mg / L and CBOD = Answer of (c).
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