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Another algebra equation needed
Once again, I need a general equation that will solve for "A". Can someone
tell me what that would be please? K = -1000 R = 0.0083333 N=5 A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (A*(1+R)^-N) FYI: In this case I know A equals -40,554.920 Please note (don't know if it's important) that "A" can be negated one or more times.. TIA |
Another algebra equation needed
(-A*(1+R)^-3) + ... + (A*(1+R)^-N)
at what point in the above sequence does -A become +A -- Regards, Tom Ogilvy "Karl Thompson" wrote in message ... Once again, I need a general equation that will solve for "A". Can someone tell me what that would be please? K = -1000 R = 0.0083333 N=5 A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (A*(1+R)^-N) FYI: In this case I know A equals -40,554.920 Please note (don't know if it's important) that "A" can be negated one or more times.. TIA |
Another algebra equation needed
Oh, sorry.
There is only one occurance of -A. My assumption is though, that once I see how the equation is rewritten, that I'll be able to figure out how make the necessary modifications myself for 1) different values of N and 2) occassions when there may be more than one occurance of -A. "Tom Ogilvy" wrote in message ... (-A*(1+R)^-3) + ... + (A*(1+R)^-N) at what point in the above sequence does -A become +A -- Regards, Tom Ogilvy "Karl Thompson" wrote in message ... Once again, I need a general equation that will solve for "A". Can someone tell me what that would be please? K = -1000 R = 0.0083333 N=5 A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (A*(1+R)^-N) FYI: In this case I know A equals -40,554.920 Please note (don't know if it's important) that "A" can be negated one or more times.. TIA |
Another algebra equation needed
In the second half of the equation, is it A or K that is
being multiplied by (1+R) ^ -N? I can help if it is K.... -----Original Message----- Once again, I need a general equation that will solve for "A". Can someone tell me what that would be please? K = -1000 R = 0.0083333 N=5 A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (A*(1+R)^-N) FYI: In this case I know A equals -40,554.920 Please note (don't know if it's important) that "A" can be negated one or more times.. TIA . |
Another algebra equation needed
Thank you. It is "A".
However, maybe this example is a little better in that it's simplier(?): K = 1000 R = 0.0083333 A = (K*(1+R)^-1) + (-A*(1+R)^-2) in this case A = 499.98 "keyt" wrote in message ... In the second half of the equation, is it A or K that is being multiplied by (1+R) ^ -N? I can help if it is K.... -----Original Message----- Once again, I need a general equation that will solve for "A". Can someone tell me what that would be please? K = -1000 R = 0.0083333 N=5 A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (A*(1+R)^-N) FYI: In this case I know A equals -40,554.920 Please note (don't know if it's important) that "A" can be negated one or more times.. TIA . |
Another algebra equation needed
A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (-A*(1+R)^-N)
A = (K*(1+R)^-1) + (K*(1+R)^-2) + A*((-1*(1+R)^-3) + ... + (-1*(1+R)^-N)) A - A*((-1*(1+R)^-3) + ... + (-1*(1+R)^-N)) = (K*(1+R)^-1) + (K*(1+R)^-2) A * ( 1 - ((-1*(1+R)^-3) + ... + (-1*(1+R)^-N))) = (K*(1+R)^-1) + (K*(1+R)^-2) A = ((K*(1+R)^-1) + (K*(1+R)^-2)) / ( 1 - ((-1*(1+R)^-3) + ... + (-1*(1+R)^-N))) The answer for A with your numbers is -506.206107674368, checked by substituting back in the original. So if you know the answer is something else, then the formula must not be given correctly. -- Regards, Tom Ogilvy "Karl Thompson" wrote in message ... Oh, sorry. There is only one occurance of -A. My assumption is though, that once I see how the equation is rewritten, that I'll be able to figure out how make the necessary modifications myself for 1) different values of N and 2) occassions when there may be more than one occurance of -A. "Tom Ogilvy" wrote in message ... (-A*(1+R)^-3) + ... + (A*(1+R)^-N) at what point in the above sequence does -A become +A -- Regards, Tom Ogilvy "Karl Thompson" wrote in message ... Once again, I need a general equation that will solve for "A". Can someone tell me what that would be please? K = -1000 R = 0.0083333 N=5 A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (A*(1+R)^-N) FYI: In this case I know A equals -40,554.920 Please note (don't know if it's important) that "A" can be negated one or more times.. TIA |
Another algebra equation needed
A = (K*(1+R)^-1) + (-A*(1+R)^-2)
A - (-A*(1+R)^-2) = (K*(1+R)^-1) A * ( 1 + (R+1)^-2) = (K*(1+R)^-1) A = (K*(1+R)^-1) / ( 1 + (R+1)^-2) That does solve to 499.982783099181 with your numbers. -- Regards, Tom Ogilvy "Karl Thompson" wrote in message ... Thank you. It is "A". However, maybe this example is a little better in that it's simplier(?): K = 1000 R = 0.0083333 A = (K*(1+R)^-1) + (-A*(1+R)^-2) in this case A = 499.98 "keyt" wrote in message ... In the second half of the equation, is it A or K that is being multiplied by (1+R) ^ -N? I can help if it is K.... -----Original Message----- Once again, I need a general equation that will solve for "A". Can someone tell me what that would be please? K = -1000 R = 0.0083333 N=5 A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (A*(1+R)^-N) FYI: In this case I know A equals -40,554.920 Please note (don't know if it's important) that "A" can be negated one or more times.. TIA . |
Another algebra equation needed
Thank you.
A with your numbers is -506.206107674368 I agree, except for one thing. Sometimes A is negated, and sometimes it isn't. I see if I modify your divisor to this (2 neg signs removed) then the result is the expected -40554.92: 1 - ( (-1*(1+R)^-3) + (1*(1+R)^-4) + (1*(1+R)^-5) ) This is great. Thanks. Now to write the VBA code to do this.... "Tom Ogilvy" wrote in message ... A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (-A*(1+R)^-N) A = (K*(1+R)^-1) + (K*(1+R)^-2) + A*((-1*(1+R)^-3) + ... + (-1*(1+R)^-N)) A - A*((-1*(1+R)^-3) + ... + (-1*(1+R)^-N)) = (K*(1+R)^-1) + (K*(1+R)^-2) A * ( 1 - ((-1*(1+R)^-3) + ... + (-1*(1+R)^-N))) = (K*(1+R)^-1) + (K*(1+R)^-2) A = ((K*(1+R)^-1) + (K*(1+R)^-2)) / ( 1 - ((-1*(1+R)^-3) + ... + (-1*(1+R)^-N))) The answer for A with your numbers is -506.206107674368, checked by substituting back in the original. So if you know the answer is something else, then the formula must not be given correctly. -- Regards, Tom Ogilvy "Karl Thompson" wrote in message ... Oh, sorry. There is only one occurance of -A. My assumption is though, that once I see how the equation is rewritten, that I'll be able to figure out how make the necessary modifications myself for 1) different values of N and 2) occassions when there may be more than one occurance of -A. "Tom Ogilvy" wrote in message ... (-A*(1+R)^-3) + ... + (A*(1+R)^-N) at what point in the above sequence does -A become +A -- Regards, Tom Ogilvy "Karl Thompson" wrote in message ... Once again, I need a general equation that will solve for "A". Can someone tell me what that would be please? K = -1000 R = 0.0083333 N=5 A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (A*(1+R)^-N) FYI: In this case I know A equals -40,554.920 Please note (don't know if it's important) that "A" can be negated one or more times.. TIA |
Another algebra equation needed
I guess that is what you meant by
There is only one occurance of -A. -- Regards, Tom Ogilvy Karl Thompson wrote in message ... Thank you. A with your numbers is -506.206107674368 I agree, except for one thing. Sometimes A is negated, and sometimes it isn't. I see if I modify your divisor to this (2 neg signs removed) then the result is the expected -40554.92: 1 - ( (-1*(1+R)^-3) + (1*(1+R)^-4) + (1*(1+R)^-5) ) This is great. Thanks. Now to write the VBA code to do this.... "Tom Ogilvy" wrote in message ... A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (-A*(1+R)^-N) A = (K*(1+R)^-1) + (K*(1+R)^-2) + A*((-1*(1+R)^-3) + ... + (-1*(1+R)^-N)) A - A*((-1*(1+R)^-3) + ... + (-1*(1+R)^-N)) = (K*(1+R)^-1) + (K*(1+R)^-2) A * ( 1 - ((-1*(1+R)^-3) + ... + (-1*(1+R)^-N))) = (K*(1+R)^-1) + (K*(1+R)^-2) A = ((K*(1+R)^-1) + (K*(1+R)^-2)) / ( 1 - ((-1*(1+R)^-3) + ... + (-1*(1+R)^-N))) The answer for A with your numbers is -506.206107674368, checked by substituting back in the original. So if you know the answer is something else, then the formula must not be given correctly. -- Regards, Tom Ogilvy "Karl Thompson" wrote in message ... Oh, sorry. There is only one occurance of -A. My assumption is though, that once I see how the equation is rewritten, that I'll be able to figure out how make the necessary modifications myself for 1) different values of N and 2) occassions when there may be more than one occurance of -A. "Tom Ogilvy" wrote in message ... (-A*(1+R)^-3) + ... + (A*(1+R)^-N) at what point in the above sequence does -A become +A -- Regards, Tom Ogilvy "Karl Thompson" wrote in message ... Once again, I need a general equation that will solve for "A". Can someone tell me what that would be please? K = -1000 R = 0.0083333 N=5 A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (A*(1+R)^-N) FYI: In this case I know A equals -40,554.920 Please note (don't know if it's important) that "A" can be negated one or more times.. TIA |
Another algebra equation needed
Here's my 2 cents. If the first two power terms use k, the 3rd power term
uses -A, and the 4th, 5th, and above use +A, then it appears to me that the equation may be something like this... A = (k*r*(1 + r)^(1 + n)*(2 + r))/((1 + r)^3 + (1 + r)^n*(-1 + r*(2 + r)*(1 + r + r^2))) With k = -1000, r=.0083333, and n=5, I get -40555.08210206875 If your only approximate number 'r' of .0083333 is really .008333333333333, (or more exactly 1/120), then I get your solution of -40554.92209880997 HTH -- Dana DeLouis Using Windows XP & Office XP = = = = = = = = = = = = = = = = = "Karl Thompson" wrote in message ... Thank you. A with your numbers is -506.206107674368 I agree, except for one thing. Sometimes A is negated, and sometimes it isn't. I see if I modify your divisor to this (2 neg signs removed) then the result is the expected -40554.92: 1 - ( (-1*(1+R)^-3) + (1*(1+R)^-4) + (1*(1+R)^-5) ) This is great. Thanks. Now to write the VBA code to do this.... "Tom Ogilvy" wrote in message ... A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (-A*(1+R)^-N) A = (K*(1+R)^-1) + (K*(1+R)^-2) + A*((-1*(1+R)^-3) + ... + (-1*(1+R)^-N)) A - A*((-1*(1+R)^-3) + ... + (-1*(1+R)^-N)) = (K*(1+R)^-1) + (K*(1+R)^-2) A * ( 1 - ((-1*(1+R)^-3) + ... + (-1*(1+R)^-N))) = (K*(1+R)^-1) + (K*(1+R)^-2) A = ((K*(1+R)^-1) + (K*(1+R)^-2)) / ( 1 - ((-1*(1+R)^-3) + ... + (-1*(1+R)^-N))) The answer for A with your numbers is -506.206107674368, checked by substituting back in the original. So if you know the answer is something else, then the formula must not be given correctly. -- Regards, Tom Ogilvy "Karl Thompson" wrote in message ... Oh, sorry. There is only one occurance of -A. My assumption is though, that once I see how the equation is rewritten, that I'll be able to figure out how make the necessary modifications myself for 1) different values of N and 2) occassions when there may be more than one occurance of -A. "Tom Ogilvy" wrote in message ... (-A*(1+R)^-3) + ... + (A*(1+R)^-N) at what point in the above sequence does -A become +A -- Regards, Tom Ogilvy "Karl Thompson" wrote in message ... Once again, I need a general equation that will solve for "A". Can someone tell me what that would be please? K = -1000 R = 0.0083333 N=5 A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (A*(1+R)^-N) FYI: In this case I know A equals -40,554.920 Please note (don't know if it's important) that "A" can be negated one or more times.. TIA |
Followup Questoin to Another algebra equation needed
Thanks Tom, Dana and Keyt for your help.
There turns out to be one more variation to this that I'm hoping one of you can help me with. Given this: K = -1000 K2 = 500.0011 R = 1/120 = 0.0083333333.. N=5 K2 = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + (+A*(1+R)^-4) + (+A*(1+R)^-N) Again, how do I solve for A? In this case A equals 2,602.01 TIA, Karl "Karl Thompson" wrote in message ... Once again, I need a general equation that will solve for "A". Can someone tell me what that would be please? K = -1000 R = 0.0083333 N=5 A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (A*(1+R)^-N) FYI: In this case I know A equals -40,554.920 Please note (don't know if it's important) that "A" can be negated one or more times.. TIA |
Followup Questoin to Another algebra equation needed
K2 = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + (+A*(1+R)^-4) +
(+A*(1+R)^-5) K2 = (K*(1+R)^-1) + (K*(1+R)^-2) + ((-A*(1+R)^-3) + (+A*(1+R)^-4) + (+A*(1+R)^-5)) K2 - ((-A*(1+R)^-3) + (+A*(1+R)^-4) + (+A*(1+R)^-N)) = (K*(1+R)^-1) + (K*(1+R)^-2) K2 - A * ((-(1+R)^-3) + ((1+R)^-4) + ((1+R)^-5)) = (K*(1+R)^-1) + (K*(1+R)^-2) A * ((+(1+R)^-3) - ((1+R)^-4) - ((1+R)^-5)) = -(K*(1+R)^-1) - (K*(1+R)^-2) + K2 A = ( -(K*(1+R)^-1) - (K*(1+R)^-2) + K2)/((+(1+R)^-3) - ((1+R)^-4) - ((1+R)^-5)) -- Regards, Tom Ogilvy Karl wrote in message ... Thanks Tom, Dana and Keyt for your help. There turns out to be one more variation to this that I'm hoping one of you can help me with. Given this: K = -1000 K2 = 500.0011 R = 1/120 = 0.0083333333.. N=5 K2 = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + (+A*(1+R)^-4) + (+A*(1+R)^-N) Again, how do I solve for A? In this case A equals 2,602.01 TIA, Karl "Karl Thompson" wrote in message ... Once again, I need a general equation that will solve for "A". Can someone tell me what that would be please? K = -1000 R = 0.0083333 N=5 A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (A*(1+R)^-N) FYI: In this case I know A equals -40,554.920 Please note (don't know if it's important) that "A" can be negated one or more times.. TIA |
Followup Questoin to Another algebra equation needed
If n could be 5,10, 20, etc...(with +a being used for the 4th power term and
above) then here's another idea... Sub Demo() Dim k, k2, z, n, answer k = -1000 k2 = 500.0011 'z=1+r z = 1 + (1 / 120) n = 5 answer = ((z - 1) * z ^ (n + 1) * (k2 * z ^ 2 - k * (z + 1))) / ((z - 2) * z ^ n + z ^ 3) Debug.Print answer End Sub returns: -2602.00995082074 HTH. -- Dana DeLouis Using Windows XP & Office XP = = = = = = = = = = = = = = = = = "Karl" wrote in message ... Thanks Tom, Dana and Keyt for your help. There turns out to be one more variation to this that I'm hoping one of you can help me with. Given this: K = -1000 K2 = 500.0011 R = 1/120 = 0.0083333333.. N=5 K2 = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + (+A*(1+R)^-4) + (+A*(1+R)^-N) Again, how do I solve for A? In this case A equals 2,602.01 TIA, Karl "Karl Thompson" wrote in message ... Once again, I need a general equation that will solve for "A". Can someone tell me what that would be please? K = -1000 R = 0.0083333 N=5 A = (K*(1+R)^-1) + (K*(1+R)^-2) + (-A*(1+R)^-3) + ... + (A*(1+R)^-N) FYI: In this case I know A equals -40,554.920 Please note (don't know if it's important) that "A" can be negated one or more times.. TIA |
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