In: Finance
Consider the following two cash flow series of payments: Series A is a geometric series increasing at a rate of 4% per year. The initial cash payment at the end of year 1 is $1,000. The payments occur annually for 5 years. Series B is a uniform series with payments of value X occurring annually at the end of years 1 through 5. You must make the payments in either Series A or Series B. Click here to access the TVM Factor Table Calculator Your answer is incorrect. Try again. Determine the value of X for which these two series are equivalent if your TVOM is i = 6%. $ Carry all interim calculations to 5 decimal places and then round your final answer to the nearest dollar. The tolerance is ±5. Your answer is incorrect. Try again. If your TVOM is 8%, would you be indifferent between these two series of payments? Enter the PW for each series to support this choice. PW, Series A: $ PW, Series B: $ Carry all interim calculations to 5 decimal places and then round your final answer to the nearest dollar. The tolerance is ±5%. Your answer is incorrect. Try again. If your TVOM is 5%, would you be indifferent between these two series of payments? Enter the PW for each series to support this choice. PW, Series A: $ PW, Series B: $ Carry all interim calculations to 5 decimal places and then round your final answer to the nearest dollar. The tolerance is ±5.
Series A | ||||||
Year | Geometric Series | TVOM F at 6% | TVOM at 6% | |||
C/f (n+1)=C/f( n)*1.04 | ||||||
1 | 2 | 3 | 4 | 5 | 6=3*5 | |
1 | 1000 | 1000 | 1/1.06^1= | 0.94340 | 943.39623 | |
2 | 1000*1.04= | 1040 | 1/1.06^2= | 0.89000 | 925.59630 | |
3 | 1040*1.04= | 1081.60000 | 1/1.06^3= | 0.83962 | 908.13222 | |
4 | 1081.6*1.04= | 1124.86400 | 1/1.06^4= | 0.79209 | 890.99765 | |
5.00000 | 1124.864*1.04= | 1169.85856 | 1/1.06^5= | 0.74726 | 874.18637 | |
Total PW of the series | 4.21236 | 4542.30876 | ||||
PW of Series B should be EQUAL to PW of the payments under geometric Series A | ||||||
Using P/A,i= 6%,n=5, 4.21236 | ||||||
4.21236*X=4542.30876 | ||||||
So, X=4542.30876/4.21236= | ||||||
1078.3287 | ||||||
So, | ||||||
the value of X for which these two series are equivalent if the TVOM is i = 6%= | ||||||
1078 | ||||||
2… | Series A | ||||
Year | Geometric Series | TVOM F at 8% | TVOM at 8% | ||
C/f (n+1)=C/f( n)*1.04 | |||||
1 | 2 | 3 | 4 | 5 | 6=3*5 |
1 | 1000 | 1000 | 1/1.08^1= | 0.92593 | 925.92593 |
2 | 1000*1.04= | 1040 | 1/1.08^2= | 0.85734 | 891.63237 |
3 | 1040*1.04= | 1081.60000 | 1/1.08^3= | 0.79383 | 858.60895 |
4 | 1081.6*1.04= | 1124.86400 | 1/1.08^4= | 0.73503 | 826.80862 |
5 | 1124.864*1.04= | 1169.85856 | 1/1.08^5= | 0.68058 | 796.18608 |
Total PW of the series | 3.99271 | 4299.16195 | |||
PW of Series A | 4299 | ||||
PW of Series B | 4304 | ||||
(1078*3.99271)= | |||||
4304.14138 | |||||
YES--given the tolerance is ±5% | |||||
3.. | Series A | ||||
Year | Geometric Series | TVOM F at 5% | TVOM at 5% | ||
C/f (n+1)=C/f( n)*1.04 | |||||
1 | 2 | 3 | 4 | 5 | 6=3*5 |
1 | 1000 | 1000 | 1/1.05^1= | 0.95238 | 952.38095 |
2 | 1000*1.04= | 1040 | 1/1.05^2= | 0.90703 | 943.31066 |
3 | 1040*1.04= | 1081.60000 | 1/1.05^3= | 0.86384 | 934.32675 |
4 | 1081.6*1.04= | 1124.86400 | 1/1.05^4= | 0.82270 | 925.42840 |
5 | 1124.864*1.04= | 1169.85856 | 1/1.05^5= | 0.78353 | 916.61479 |
Total PW of the series | 4.32948 | 4672.06155 | |||
PW of Series A | 4672 | ||||
PW of Series B | 4667 | ||||
(1078*4.32948)= | |||||
4667.17944 | |||||
YES--given the tolerance is ±5 | |||||