6 gradient cash flow
TRANSCRIPT
GRADIENT CASH FLOW
When cash incomes and/or cash expenditures increase or decrease according to gradient named as;
gradient cash flows.
increase or decrease amont is;
G gradient
Gradient series;
• uniform (linear) gradient series
• geometric gradient series
• escalating gradient series
A1
A1+G
G
A1+(n-1)G
0 1 2 3 4 5 ....... ..... n
uniform gradient series present worth factor
P present value calculates with (P/G,i,n) factors each payment periods G amount increase or decrease cash flows with i interest rate and after n periods
P=G{(1-(1+ni)(1+i)-n)/i2}
P=G(P/G,i,n)
uniform gradient series future worth factor
F future value calculates with (F/G, i,n) factors each payment periods G amount increase or decrease cash flows with i interest rate and after n periods
F= G/i{((1+i)n-1/i)-n}
F = G(F/G, i,n)
uniform gradient annual worth factor
A annual worth calculates with (A/G, i, n) factors each payment periods G amount increase or decrease cash flows with i interest rate and after n periods
A = G { 1 – n } i (1+i)n -1 A = G(A/G, i, n)
0 1 2 3 4 5 n
G=100 TL, i= %5
Example
G= 100 TL (increase each payment)
i= %5 (yaerly)
n= 5 years
P= ?
P =G(P/G,i,n)
P =G(P/G,5,5)=100x8,237
P =823,70 TL
Example
G= 100 TL (increase each payment)
i= %5 (yearly)
n= 5 years
F= ?
F =G(F/G,i,n)
F =G(F/G,5,5)=100x10,513
F =1.051,30 TL
Example
G= 100 TL (increase each payment)
i= %5 (yearly)
n= 5 years
A= ?
A =G(A/G,i,n)
A =G(A/G,5,5)=100x1,903
A =190,30 TL
P =
A=100 TL
n=5, i=5
P= PA+/- PG
n=5, i=5
G=50 TL
Example
A=100 TL (first payment)
G= 50 TL (increase each payment)
i= %5 (yearly)
n=10 years
P= ?
P= PA+ PG
PA = A(P/A,i,n)
PA = 100x(P/A,5,10)=100x7,7217
PA = 772,17 TL
PG =G(P/G,i,n)
PG =50(P/G,5,10)=50x31,652
PG =1.582,60 TL
P= PA+ PG
P= 772,17 + 1.582,60 = 2.354,77 TL
Example
A=100 TL (first payment)
G= 50 TL (increase each payment)
i= %5 (yearly)
n=10 years
F= ?
F= FA+ FG
FA = A(F/A,i,n)
FA = 100x(F/A,5,10)=100x12,578
FA = 1.257,80 TL
FG =G(F/G,i,n)
FG =50(F/G,5,10)=50x51,558
FG =2.577,90 TL
F= FA+ FG
F= 1.257,8 + 2.577,90 =3.835,70 TL
Example
A1=100 TL (first payment)
G= 50 TL (increase each payment)
i= %5 (yearly)
n=10 years
A= ? (equal uniform series to gradient series)
A= A1+ AG
A1 = 100 TL
AG =G(A/G,i,n)
AG =50(A/G,5,10)=50x4,099
AG =204,95 TLA= A1+ AG A= 100 + 204,95 = 304,95 TL