Answer / kangchuentat

ACCOUNTING AND FINANCIAL ENGINEERING - ANSWER 34.7 : (a) Purchase price = 0.95 x par bond value = 0.95 x $1000 = $950. (b) Annual discount = (total discount) / (25 years) = (0.05 x par bond value) / 25 = 0.05 x $1000 / 25 = $2. (c) Annual coupon rate payment = par value x coupon rate = $1000 x 0.08 = $80. (d) YTM = [ Answer in (c) + Answer in (b) ] x 200 / [ par bond value + Answer in (a) ] = (80 + 2) x 200 / (1000 + 950) = 8.41 %. 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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Question 71 - (a) The byte is the basic building block of computer data used in chemical engineering process simulation where 16 bits make a word, 4 bits make a nibble, 32 bits make a quad word and 8 bits make a byte. Then how many nibbles are there in a megabytes? (b) In computer data items, let : 1 bit - counts from 0 to 1, 8 bits - counts from 0 to 255, 16 bits - counts from 0 to A. What is the value of A? (c) In a binary system of 4 bits, if 1100 = 12, 1101 = 13, 1110 = 14, 1111 = 15, B = 16, then guess the value of B. (d) By using any form of tools, find the exact value of 2 power 64 or 2^64.

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Question 67 – In order to predict the wastewater production, the population number has to be understood. The population data is : 72000 (for year 1961 or P-1961), 85000 (for year 1971 or P-1971), 110500 (for year 1981 or P-1981). (a) Find the average population increase, or [ (P-1981 - P-1971) + (P-1971 - P-1961) ] / 2. (b) Find the average percentage population increase, or [ (P-1981 - P-1971) / P-1971 + (P-1971 - P-1961) / P-1961 ] / (2) X 100. (c) Find the incremental increase or P-1981 - 2 (P-1971) + P-1961. (d) Let Po = P-1981. After 2 decades or n = 2, the population is P-2001. By using arithmetical increase method, find P-2001 = Po + n (Answer for a). (e) By using incremental increase method, find P-2001 = (Answer of d) + n (n + 1) (Answer of c) / 2. (f) By using geometrical increase method, find P-2001 = Po [ 1 + (Answer of b) / 100 ] ^ n where ^ is power sign, or 1 ^ 2 = 1 x 1 = 1. (g) If the actual P-2001 = 184000, which method of estimation is more accurate, based on your answer in (d), (e) and (f)?

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