ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.27 : A biochemical engineering consultancy applies construction accounting in its finance. Its project began on 1 January 2010. Total revenue generated from the project was $9000. On 1 January 2011 as the budget, $2000 had been spent, with $6000 expected. However, the project cost increased latter, causing deviation from its initial budget on 1 January 2012, where $7000 had been spent, with $1400 expected. Let (estimated total cost) = (spent cost) + (expected cost to be spent), (percentage completion) = 100 (spent cost) / (estimated total costs), (total expected profits) = (total revenue) - (estimated total costs). Calculate : (a) total expected profits on 1 January 2011 and 1 January 2012; (b) estimated total cost as and not as the budget; (c) percentage completion of the project since the project began, in the first and second years.
ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.27 : On 1 January 2011 (as the budget), estimated total costs = $2000 + $6000 = $8000, total expected profits = $9000 - $8000 = $1000, percentage completion on first year = 100 ($2000 / $8000) = 25%. On 1 January 2012 (not as the budget), estimated total costs = $7000 + $1400 = $8400, total expected profits = $9000 - $8400 = $600, percentage completion on second year = 100 ($7000 / $8400) = 83.33%. Final answer : (a) Profits : $1000 (1 January 2011), $600 (1 January 2012). (b) Costs : $8000 (as the budget), $8400 (not as the budget). (c) Percentage completion : 25% (first year), 83.3% (second year). 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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Water flows through a pipe with circular cross sectional area at the rate of V / t = 80 L / s where V is the volume and t is time. Let Av = 80 L / s where A is cross sectional area and v is velocity of fluid. For point 1, the radius of the pipe is 16 cm. For point 2, the radius of the pipe is 8 cm. Find (a) the velocity at point 1; (b) the velocity at point 2; (c) the pressure at point 2 by using Bernoulli’s equation where P Rgy 0.5 RV = constant. P is the pressure, R = density of fluid, V = square of fluid’s velocity, g = gravitational constant of 9.81 N / kg and y = 2 m = difference of height at 2 points. The pressure of point 1 is 180 kPa.
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