Question

An investment pays $20,000 at the end of year 2, $30,000 at the end of year 4 and $X at the end of year 8. At an annual effective rate of 8.5%, the modified duration of this cash flow is 5.51704. Calculate X (round your answer to integer).

Ans should be one of these: Numerical answers (in random order) 6.955, 0.062, 4.5228, 6538.49, 0.04317, 2660.67, 50000, 400.36, 202.1, 3671.29, 1951.62, 4.4621.

please show all work and do not use excel or a financial calculator. The answer you get must be in the answer key posted at the bottom, thanks! I will thumb up if it's correct.

Answer 1

Modified Duration=Macaulay Duration/(1+yield)

=>5.51704=(2*20000/1.085^2+4*30000/1.085^4+8*X/1.085^8)/(20000/1.085^2+30000/1.085^4+X/1.085^8)*1/1.085

=>5.51704*1.085*(20000/1.085^2+30000/1.085^4+X/1.085^8)=(2*20000/1.085^2+4*30000/1.085^4+8*X/1.085^8)

=>5.51704*1.085*20000/1.085^2+5.51704*1.085*30000/1.085^4+5.51704*1.085*X/1.085^8=2*20000/1.085^2+4*30000/1.085^4+8*X/1.085^8

=>5.51704*1.085*20000/1.085^2+5.51704*1.085*30000/1.085^4=2*20000/1.085^2+4*30000/1.085^4+(8-5.51704*1.085)*X/1.085^8

=>5.51704*1.085*20000/1.085^2+5.51704*1.085*30000/1.085^4-2*20000/1.085^2-4*30000/1.085^4=(8-5.51704*1.085)*X/1.085^8

=>X=(5.51704*1.085*20000/1.085^2+5.51704*1.085*30000/1.085^4-2*20000/1.085^2-4*30000/1.085^4)*1.085^8/(8-5.51704*1.085)=105575

Alternatively,

Modified Duration=-1/P*dP/dy=-1/(20000/(1+y)^2+30000/(1+y)^4+X/(1+y)^8)*(-20000*2/(1+y)^3-4*30000/(1+y)^5-X*8/(1+y)^9)

-1/(20000/(1+8.5%)^2+30000/(1+8.5%)^4+X/(1+8.5%)^8)*(-20000*2/(1+8.5%)^3-4*30000/(1+8.5%)^5-X*8/(1+8.5%)^9)=5.51704

=>X=105575