What is the formula that can be derived using the values and formulas below to get the value of K43?
Here's what I have, so far, but no luck in getting the correct formula:
0 = 925191 - 119355 - (((39047 +y * (K3-J3))/((1+K5)^K1)) + ((38456+y*((L3-J3))/((1+K5)^L1)) + ((M32 + y * (M3 - J3)) / ((1 + K5)^M1)) + ((N32+y*(N3 - J3)) / ((1 + K5) ^ N1)) + ((O32+y*(O3-J3)) / ((1+ K5) ^ O1))) - (((((((39047 + (y * (K3 - J3))) + (K5 * 119355)) / ((119355 + ((39047(y * (K3 - J3))) + (K5 * 119355))* (1 - J24 ) + (K26*(K16*(1+K40)^K1))))) - K5) / (K5 - K37)) * (119355 + ((O32 + y * (O3 - J3)) + (K5 * J13)) * (1 - O24) + (O26* (K16 * ((1 + K40)) ^ 01)))))) / ((1 + K5) ^O1))
VALUES FOR SOME VARIABLES:
- J3 = 2015
- K3 = 2016
- J3 = 2015
- M3 = 2017
- N3 = 2018
- O3 = 2019
- K32 = 39047
- L32 = 38456
- M32 = 36685
- N32 = 32852
- 032 = 29019
- K5 = 0.09
- J13 = 119355
- K16 = 112
- K40 = 0.18
- K1 = 1
- L1 = 2
- M1 = 3
- N1 = 4
- O1 = 5
- P1 = 6
- J46 = 0
- J47 = 0
- J24 = 0.22
- K24 = 0.22
- L24 = 0.22
- M24 = 0.22
- O24 = 0.22
- P24 = 0.22
- J26 = 0
- K26 = 0
- L26 = 0
- M26 = 0
- N26 = 0
- O26 = 0
- P26 = 0
- K24 = 0.22
- K26 = 0
- P39 = 0.85
- O1 = 5
- K14 = 0
- K37 = 0.01
- K41 = 925191
Equations:
- K43 = ?
- K44 = K32+K43*(K3-J3)
- L44 = L32+K43*(L3-J3)
- M44 = M32+K43*(M3-J3)
- N44 = N32+K43*(N3-J3)
O44 = O32+K43*(O3-J3)
K45 = K5*K48
- L45 = K5*L48
- M45 = K5*M48
- N45 = K5*N48
O45 = K5*O48
K46 = K44 + K45
- L46 = L44 + L45
- M46 = M44 + M45
- N46 = N44 + N45
O46 = O44 + O45
K47 = K16*((1+K40)^K1)
- L47 = K16*((1+K40)^L1)
- M47 = K16*((1+K40)^M1)
- N47 = K16*((1+K40)^N1)
- O47 = K16*((1+K40)^O1)
P47 = K16*((1+K40)^P1)
K48 = J48+J46*(1-J24)+(J26*J47)
- L48 = K48+K46*(1-K24)+(K26*K47)
- M48 = L48+L46*(1-L24)+(L26*L47)
- N48 = M48+M46*(1-M24)+(M26*K47)
- O48 = N48+N46*(1-N24)+(N26*N47)
P48 = O48+O46*(1-O24)+(O26*O47)
K49 = K46/K48
- L49 = L46/L48
- M49 = M46/M48
- N49 = N46/N48
O49 = O46/O48
P50 = O49/P39
- K51 = K44/((1+K5)^(K1))
- L51 = L44/((1+K5)^(L1))
- M51 = M44/((1+K5)^(M1))
- N51 = N44/((1+K5)^(N1))
O51 = O44/((1+K5)^(O1))
P52 = K51 + L51 + M51 + N51+ O51
P53 = (((P50-K5)/(K5-K37))*P48)/((1+K5)^O1)
Q53 = K41-J13-P52-P53, WHERE Q53 IS ALWAYS 0
Below are the equations with the given values:
- K43 = ?
- K44 = 39047+K43*(2016-2015)
- L44 = 38456+K43*(2017-2015)
- M44 = 36685+K43*(2018-2015)
- N44 = 32852+K43*(2019-2015)
O44 = 29019+K43*(2020-2015)
K45 = 0.09*K48
- L45 = 0.09*L48
- M45 = 0.09*M48
- N45 = 0.09*N48
O45 = 0.09*O48
K46 = K44 + K45
- L46 = L44 + L45
- M46 = M44 + M45
- N46 = N44 + N45
O46 = O44 + O45
K47 = 112*((1+0.18)^1)
- L47 = 112*((1+0.18)^2)
- M47 = 112*((1+0.18)^3)
- N47 = 112*((1+0.18)^4)
- O47 = 112*((1+0.18)^5)
P47 = 112*((1+0.18)^6)
K48 = J48+0*(1-0.22)+(0*J47)
- L48 = K48+K46*(1-0.22)+(0*K47)
- M48 = L48+L46*(1-0.22)+(0*L47)
- N48 = M48+M46*(1-0.22)+(0*K47)
- O48 = N48+N46*(1-0.22)+(0*N47)
P48 = O48+O46*(1-0.22)+(0*O47)
K49 = K46/K48
- L49 = L46/L48
- M49 = M46/M48
- N49 = N46/N48
O49 = O46/O48
P50 = O49/0.85
- K51 = K44/((1+0.09)^(1))
- L51 = L44/((1+0.09)^(2))
- M51 = M44/((1+0.09)^(3))
- N51 = N44/((1+0.09)^(4))
O51 = O44/((1+0.09)^(5))
P52 = K51 + L51 + M51 + N51+ O51
P53 = (((P50-0.09)/(0.09-0.01))*P48)/((1+0.09)^5)
0 = 925191-119355-P52-P53, WHERE Q53 IS ALWAYS 0
Thank you so much for your time!
A partial answer:
Working backwards from what we want to find $K_{43}$. We have equations containing it that also contain $Var_{44}$ Where $Var$ is each of $K,L,M,N,O$. Each $Var_{44}$ is related $Var_{45}$ and $Var_{46}$ by given equations, and each $Var_{45}$ is related to $Var_{48}$. Thus we can rewrite the $Var_{44}$ equations to contain $Var_{46}$ and $Var_{48}$. Now the $Var_{48}$'s are also related to each other sequentially in equations that also contain $Var_{46}$'s.
Edit:
After examining the variable relationships, I think you can find what you need by starting with the final equation,
$0=925191-119355-P_{52}-P_{53}$
Rearrange so that vars are on the left and perform indicated subtraction to obtain:
$P_{52}+P{53}=805836$
From here substitute formulas for both vars to obtain an expression involving the vars $Var_{51}$ from subbing $P_{52}$ and $P_{50}$ and $P_{48}$ from subbing for $P_{53}$.
Simplify as much as you can, then sub for the $Var_{51}$'s replacing them expressions in terms of w/ $Var_{44}$'s, and simplify. Then sub for the $Var_{44}$'s which are in terms of $K_{43}$. (We have taken care of everything that came from $P_{52}$ now. Now to handle the stuff from $P_{53}$).
Simplify as much as you can, then sub for $P_{50}$ and $P_{48}$ which replaces those vars with expressions involving $O_{49},O_{48},O_{46}$. $O_{49}$ can be re-expressed in terms of $O_{48}$ and $O_{46}$ (reducing vars by 1). $O_{46}$ can be expressed as in terms of $O_{45}$ and $O_{44}$. Express $O_{44}$ by using formula in terms of $K_{43}$. Express $O_{45}$ in terms of $O_{48}$.
Now we are ready to re-express $O_{48}$ in terms of $N_{46}$ and $N_{48}$. Once this is done we will repeat the pattern that happened one we had expressions for $O_{46}$ and $O_{48}$, but with $N$ instead of $O$. This obtains expressions in terms of $M_{46}$, and $M_{48}$.
This process is looped, each time moving 1-letter backwards in the alphabet until we are working on $K$'s, the process goes through, except that this time, we don't get $J_{46}$'s, but only $J_{48}$'s. Do you know the value of $J_{48}$? If not, then you can't isolate $K_{43}$ as desired, but if you do, then at the end of all of this, you can finally obtain $K_{43}$. Note that I highly recommend simplifying the equations after each substitution to reduce complexity and help avoid mistakes (try to minimize the number of times a var to be substituted appears before subbing for it).