I have 100 values and I want to find out how many iterations it takes for
each value to get to 0, each value decreases by 20% each month. Thanks.

If a value starts with a non-zero amount, then it can NEVER reach zero if it
decrease by 20% of its residual value.

If you decrease it 20 % of its original value it will take 5 Iterations:

100
80
60
40
20
0

--
Gary''s Student - gsnu200720

Sorry I forgot to write that if you start out at 100 how many iterations does
it take to go below 50...using the residual value

"Gary''s Student" wrote:

If a value starts with a non-zero amount, then it can NEVER reach zero if it
decrease by 20% of its residual value.

If you decrease it 20 % of its original value it will take 5 Iterations:

100
80
60
40
20
0

--
Gary''s Student - gsnu200720

In A1 put 100 and
In A2 thru A10 enter:

=A1*0.8
=A2*0.8
=A3*0.8
=A4*0.8
=A5*0.8
=A6*0.8
=A7*0.8
=A8*0.8
=A9*0.8
to see:

100
80
64
51.2
40.96
32.768
26.2144
20.97152
16.777216
13.4217728

and as you can see 4 iterations got use below 50.
--
Gary''s Student - gsnu200720


"Rick" wrote:

Sorry I forgot to write that if you start out at 100 how many iterations does
it take to go below 50...using the residual value

"Gary''s Student" wrote:

If a value starts with a non-zero amount, then it can NEVER reach zero if it
decrease by 20% of its residual value.

If you decrease it 20 % of its original value it will take 5 Iterations:

100
80
60
40
20
0

--
Gary''s Student - gsnu200720

Or something like this

# iterations =LOG(EndVal/BeginVal)/LOG(0.8)

"Gary''s Student" wrote:

In A1 put 100 and
In A2 thru A10 enter:

=A1*0.8
=A2*0.8
=A3*0.8
=A4*0.8
=A5*0.8
=A6*0.8
=A7*0.8
=A8*0.8
=A9*0.8
to see:

100
80
64
51.2
40.96
32.768
26.2144
20.97152
16.777216
13.4217728

and as you can see 4 iterations got use below 50.
--
Gary''s Student - gsnu200720


"Rick" wrote:

Sorry I forgot to write that if you start out at 100 how many iterations does
it take to go below 50...using the residual value

"Gary''s Student" wrote:

If a value starts with a non-zero amount, then it can NEVER reach zero if it
decrease by 20% of its residual value.

If you decrease it 20 % of its original value it will take 5 Iterations:

100
80
60
40
20
0

--
Gary''s Student - gsnu200720

Very good. Only a single cell and no need to iterate. Looks like thresholds
for exponential decay.
--
Gary''s Student - gsnu200720

I tried the log function below but it did not work...it gave me -12 something.

"Barb Reinhardt" wrote:

Or something like this

# iterations =LOG(EndVal/BeginVal)/LOG(0.8)

"Gary''s Student" wrote:

In A1 put 100 and
In A2 thru A10 enter:

=A1*0.8
=A2*0.8
=A3*0.8
=A4*0.8
=A5*0.8
=A6*0.8
=A7*0.8
=A8*0.8
=A9*0.8
to see:

100
80
64
51.2
40.96
32.768
26.2144
20.97152
16.777216
13.4217728

and as you can see 4 iterations got use below 50.
--
Gary''s Student - gsnu200720


"Rick" wrote:

Sorry I forgot to write that if you start out at 100 how many iterations does
it take to go below 50...using the residual value

"Gary''s Student" wrote:

If a value starts with a non-zero amount, then it can NEVER reach zero if it
decrease by 20% of its residual value.

If you decrease it 20 % of its original value it will take 5 Iterations:

100
80
60
40
20
0

--
Gary''s Student - gsnu200720

So what numbers did you plug in to the formula to get -12 something?

If EndVal is half of BeginVal (which is what you asked for), the answer is
3.106284
--
David Biddulph

"Rick" wrote in message
...
I tried the log function below but it did not work...it gave me -12
something.

"Barb Reinhardt" wrote:

Or something like this

# iterations =LOG(EndVal/BeginVal)/LOG(0.8)

"Gary''s Student" wrote:

In A1 put 100 and
In A2 thru A10 enter:

=A1*0.8
=A2*0.8
=A3*0.8
=A4*0.8
=A5*0.8
=A6*0.8
=A7*0.8
=A8*0.8
=A9*0.8
to see:

100
80
64
51.2
40.96
32.768
26.2144
20.97152
16.777216
13.4217728

and as you can see 4 iterations got use below 50.
--
Gary''s Student - gsnu200720


"Rick" wrote:

Sorry I forgot to write that if you start out at 100 how many
iterations does
it take to go below 50...using the residual value

"Gary''s Student" wrote:

If a value starts with a non-zero amount, then it can NEVER reach
zero if it
decrease by 20% of its residual value.

If you decrease it 20 % of its original value it will take 5
Iterations:

100
80
60
40
20
0

--
Gary''s Student - gsnu200720

This is what I typed =LOG(50,100)/LOG(0.8)
and I get -8.76571

"David Biddulph" wrote:

So what numbers did you plug in to the formula to get -12 something?

If EndVal is half of BeginVal (which is what you asked for), the answer is
3.106284
--
David Biddulph

"Rick" wrote in message
...
I tried the log function below but it did not work...it gave me -12
something.

"Barb Reinhardt" wrote:

Or something like this

# iterations =LOG(EndVal/BeginVal)/LOG(0.8)

"Gary''s Student" wrote:

In A1 put 100 and
In A2 thru A10 enter:

=A1*0.8
=A2*0.8
=A3*0.8
=A4*0.8
=A5*0.8
=A6*0.8
=A7*0.8
=A8*0.8
=A9*0.8
to see:

100
80
64
51.2
40.96
32.768
26.2144
20.97152
16.777216
13.4217728

and as you can see 4 iterations got use below 50.
--
Gary''s Student - gsnu200720


"Rick" wrote:

Sorry I forgot to write that if you start out at 100 how many
iterations does
it take to go below 50...using the residual value

"Gary''s Student" wrote:

If a value starts with a non-zero amount, then it can NEVER reach
zero if it
decrease by 20% of its residual value.

If you decrease it 20 % of its original value it will take 5
Iterations:

100
80
60
40
20
0

--
Gary''s Student - gsnu200720

I suggest you read again the formula given by Barb.

If you get a different answer from what you're expecting, it's always worth
checking whether you've typed the formula wrongly. It's usually safest to
copy and paste rather than to retype, then you can copy that again and
change the relevant parameters.
--
David Biddulph

"Rick" wrote in message
...
This is what I typed =LOG(50,100)/LOG(0.8)
and I get -8.76571

"David Biddulph" wrote:

So what numbers did you plug in to the formula to get -12 something?

If EndVal is half of BeginVal (which is what you asked for), the answer
is
3.106284

"Rick" wrote in message
...
I tried the log function below but it did not work...it gave me -12
something.

"Barb Reinhardt" wrote:

Or something like this

# iterations =LOG(EndVal/BeginVal)/LOG(0.8)

"Gary''s Student" wrote:

In A1 put 100 and
In A2 thru A10 enter:

=A1*0.8
=A2*0.8
=A3*0.8
=A4*0.8
=A5*0.8
=A6*0.8
=A7*0.8
=A8*0.8
=A9*0.8
to see:

100
80
64
51.2
40.96
32.768
26.2144
20.97152
16.777216
13.4217728

and as you can see 4 iterations got use below 50.
--
Gary''s Student - gsnu200720


"Rick" wrote:

Sorry I forgot to write that if you start out at 100 how many
iterations does
it take to go below 50...using the residual value

"Gary''s Student" wrote:

If a value starts with a non-zero amount, then it can NEVER
reach
zero if it
decrease by 20% of its residual value.

If you decrease it 20 % of its original value it will take 5
Iterations:

100
80
60
40
20
0

--
Gary''s Student - gsnu200720

Its not LOG...its the natural log. =LN(50/100)/LN(0.8) Did you get it using
LOG?

"David Biddulph" wrote:

I suggest you read again the formula given by Barb.

If you get a different answer from what you're expecting, it's always worth
checking whether you've typed the formula wrongly. It's usually safest to
copy and paste rather than to retype, then you can copy that again and
change the relevant parameters.
--
David Biddulph

"Rick" wrote in message
...
This is what I typed =LOG(50,100)/LOG(0.8)
and I get -8.76571

"David Biddulph" wrote:

So what numbers did you plug in to the formula to get -12 something?

If EndVal is half of BeginVal (which is what you asked for), the answer
is
3.106284

"Rick" wrote in message
...
I tried the log function below but it did not work...it gave me -12
something.

"Barb Reinhardt" wrote:

Or something like this

# iterations =LOG(EndVal/BeginVal)/LOG(0.8)

"Gary''s Student" wrote:

In A1 put 100 and
In A2 thru A10 enter:

=A1*0.8
=A2*0.8
=A3*0.8
=A4*0.8
=A5*0.8
=A6*0.8
=A7*0.8
=A8*0.8
=A9*0.8
to see:

100
80
64
51.2
40.96
32.768
26.2144
20.97152
16.777216
13.4217728

and as you can see 4 iterations got use below 50.
--
Gary''s Student - gsnu200720


"Rick" wrote:

Sorry I forgot to write that if you start out at 100 how many
iterations does
it take to go below 50...using the residual value

"Gary''s Student" wrote:

If a value starts with a non-zero amount, then it can NEVER
reach
zero if it
decrease by 20% of its residual value.

If you decrease it 20 % of its original value it will take 5
Iterations:

100
80
60
40
20
0

--
Gary''s Student - gsnu200720

OK, Rick, let's take this a step at a time:
What answer do you get from =LN(50/100)/LN(0.8) ?
What answer do you get from =LOG(50/100)/LOG(0.8) ?

You may wish to remind yourself of the conversion between logs to different
bases.
--
David Biddulph

"Rick" wrote in message
...
Its not LOG...its the natural log. =LN(50/100)/LN(0.8) Did you get it
using
LOG?

"David Biddulph" wrote:

I suggest you read again the formula given by Barb.

If you get a different answer from what you're expecting, it's always
worth
checking whether you've typed the formula wrongly. It's usually safest
to
copy and paste rather than to retype, then you can copy that again and
change the relevant parameters.
--
David Biddulph

"Rick" wrote in message
...
This is what I typed =LOG(50,100)/LOG(0.8)
and I get -8.76571

"David Biddulph" wrote:

So what numbers did you plug in to the formula to get -12 something?

If EndVal is half of BeginVal (which is what you asked for), the
answer
is
3.106284

"Rick" wrote in message
...
I tried the log function below but it did not work...it gave me -12
something.

"Barb Reinhardt" wrote:

Or something like this

# iterations =LOG(EndVal/BeginVal)/LOG(0.8)

"Gary''s Student" wrote:

In A1 put 100 and
In A2 thru A10 enter:

=A1*0.8
=A2*0.8
=A3*0.8
=A4*0.8
=A5*0.8
=A6*0.8
=A7*0.8
=A8*0.8
=A9*0.8
to see:

100
80
64
51.2
40.96
32.768
26.2144
20.97152
16.777216
13.4217728

and as you can see 4 iterations got use below 50.
--
Gary''s Student - gsnu200720


"Rick" wrote:

Sorry I forgot to write that if you start out at 100 how many
iterations does
it take to go below 50...using the residual value

"Gary''s Student" wrote:

If a value starts with a non-zero amount, then it can NEVER
reach
zero if it
decrease by 20% of its residual value.

If you decrease it 20 % of its original value it will take 5
Iterations:

100
80
60
40
20
0

--
Gary''s Student - gsnu200720

I got it now. I am an idiot. I was writing =LOG(50,100) instead of (50/100).
You were right I should have looked how to type it.

Thanks.

"David Biddulph" wrote:

OK, Rick, let's take this a step at a time:
What answer do you get from =LN(50/100)/LN(0.8) ?
What answer do you get from =LOG(50/100)/LOG(0.8) ?

You may wish to remind yourself of the conversion between logs to different
bases.
--
David Biddulph

"Rick" wrote in message
...
Its not LOG...its the natural log. =LN(50/100)/LN(0.8) Did you get it
using
LOG?

"David Biddulph" wrote:

I suggest you read again the formula given by Barb.

If you get a different answer from what you're expecting, it's always
worth
checking whether you've typed the formula wrongly. It's usually safest
to
copy and paste rather than to retype, then you can copy that again and
change the relevant parameters.
--
David Biddulph

"Rick" wrote in message
...
This is what I typed =LOG(50,100)/LOG(0.8)
and I get -8.76571

"David Biddulph" wrote:

So what numbers did you plug in to the formula to get -12 something?

If EndVal is half of BeginVal (which is what you asked for), the
answer
is
3.106284

"Rick" wrote in message
...
I tried the log function below but it did not work...it gave me -12
something.

"Barb Reinhardt" wrote:

Or something like this

# iterations =LOG(EndVal/BeginVal)/LOG(0.8)

"Gary''s Student" wrote:

In A1 put 100 and
In A2 thru A10 enter:

=A1*0.8
=A2*0.8
=A3*0.8
=A4*0.8
=A5*0.8
=A6*0.8
=A7*0.8
=A8*0.8
=A9*0.8
to see:

100
80
64
51.2
40.96
32.768
26.2144
20.97152
16.777216
13.4217728

and as you can see 4 iterations got use below 50.
--
Gary''s Student - gsnu200720


"Rick" wrote:

Sorry I forgot to write that if you start out at 100 how many
iterations does
it take to go below 50...using the residual value

"Gary''s Student" wrote:

If a value starts with a non-zero amount, then it can NEVER
reach
zero if it
decrease by 20% of its residual value.

If you decrease it 20 % of its original value it will take 5
Iterations:

100
80
60
40
20
0

--
Gary''s Student - gsnu200720