ACCOUNTING AND FINANCIAL ENGINEERING - EXAMPLE 34.5 : In an American style option for share market, 2 persons - A and B agree to the following : B is required to sell 100 shares of IBMS to A for $85 per share anytime that A wants in the next 8 months. A will pay B $2 per share up front, non-refundable for this option. IBMS involves in petrochemical processing. IBMS stock is currently selling for $80 per share. (a) If A did not buy the share of IBMS from B after 8 months, how much will B earn? (b) If the share of IBMS goes up to $100 / share in 6 months later : (i) how much should A pay B for 100 shares according to their optional agreement? (ii) how much will A earn from 100 shares purchased from B when all the 100 shares are sold to the open market? (iii) how much net profit will A earn for selling 100 shares to the open market?
ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.5 : (a) Non-refundable payment = $2 / share x 100 shares = $200. (b) (i) Payment of A to B = $85 / share x 100 shares = $8500. (ii) Payment of open market to A = $100 / share x 100 shares = $10000. (iii) Net profit = answer b (ii) - answer b (i) - answer a = $10000 - $8500 - $200 = $1300. 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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QUANTUM COMPUTING - EXAMPLE 32.8 : In quantum computing, a quantum state is given by S = a | 00 > + b | 01 > + g | 10 > + d | 11 >. (a) Find S in term of | 0 > and | 1 > etc. (b) The probability of getting x is P(x). For S = 0.5 | 00 > + 0.5 | 01 > + 0.5 | 10 > + 0.5 | 11 >, find P(0) and P(1). Hint : P(00) + P(01) = P(0) = a x a + b x b, P(10) + P(11) = P(1) = g x g + d x d.
QUANTUM COMPUTING - EXAMPLE 32.10 : In quantum computing, the conversion of Control Not (CNOT) gate in two input quantum bit gate could be decribed as : | 00 > --> | 00 >, | 01 > --> | 01 >, | 10 > --> | 11 >, | 11 > --> | 10 >. If | P > = 0.707 ( | 01 > - | 11 > ), find the value of CNOT | P >.
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Question 48 - Let a ^ 2 = a x a and a ^ 3 = a x a x a where ^ is power function. Niobium is a metal with a body-centered cubic structure. The length of the unit cell structure is b = 0.3349 nm. (a) Find the volume for a unit cell structure for niobium. (b) There are 2 atoms per unit cell structure of niobium. The metal has a molar mass of 92.9 g / mol. One mole of the metal consists of 6.02 x 10 ^ 23 atoms. Find the mass of niobium per unit cell and the density of niobium.
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Question 45 - According to Raoult’s law for ideal liquid, x (PSAT) = yP where x is mole fraction of component in liquid, y is mole fraction of component in vapor, P is overall pressure and PSAT is saturation pressure. A liquid with 60 mole % component 1 and 40 mole % component 2 is flashed to 1210 kPa. The saturation pressure for component 1 is ln (PSAT) = 15 - 3010 / (T + 250) and for component 2 is ln (PSAT) = 14 - 2700 / (T + 205) where PSAT is in kPa and T is in degree Celsius. By assuming the liquid is ideal, calculate (a) the fraction of the effluent that is liquid; (b) the compositions of the liquid and vapor phases. The outlet T is 150 degree Celsius.
CHEMICAL MATERIAL BALANCE - EXAMPLE 2.5 : In a non-dilute absorber, the inlet gas stream consists of 8 mol % carbon dioxide in nitrogen. By contact with room temperature water at atmospheric pressure, 65 % of the carbon dioxide from a gas stream has been removed. (a) Find the mole ratio of carbon dioxide and nitrogen gases at inlet and outlet gas streams. (b) The Henry's Law provides y = 1640 x for carbon dioxide in water. Find the mole ratio when x = 0.0000427. Mole ratio is y / (1 - y) for y.
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