solution of Assignment 2 - Solution of Assignment 2 for...

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Solution of Assignment 2 for SEG3590 Investment Science 1. (a) Monthly payment 12 30 12 30 (1 ) 0.1 12 (1 0.1 12) $100,000 $877.57 ) 1 (1 0.1 12) 1 n n rr P A r × × + × == = +− + Total Interest 877.57 12 30 100,000 $215,925.2 × = (b) Bi-weekly payment 877.57/ 2 $438.79 Let n = number of periods, then by ) ) 1 n n A r ′′ + ′ = P + 0.1 26 (1 0.1 26) 100,000 438.79 (1 0.1 26) 1 n n ×+ × = Hence 545 n ′ = or 545/26=20.95 years. Total Interest 438.79 545 100,000 $139,140.55 = Saving Interest or 35.6%. 877.57 12 30 438.79 545 $76,784.65 × −× = 2. A=(rP)/[1-1/(1+r) n ]=0.07/12×25,000/(1-1/(1+0.07.12) 84 )=377.32. 3. (a) 23 100 1.15 100 1.15 1100 1.15 885.84 A P =+ + = 50 1.15 50 1.15 1050 1.15 771.68 B P + = 3 1000 1.15 657.52 C P 1000 1.15 869.57 D P (b) 100 1.15 2 100 1.15 3 1100 1.15 2.72 885.84 A D 50 1.15 2 50 1.15 3 1050 1.15 2.84 771.68 B D 3 C D = 1 D D = (c) Bond C is the most sensitive to a change in yield.
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solution of Assignment 2 - Solution of Assignment 2 for...

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