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Trying to look into the economics of SpaceX

Posted by: Ekkehard Augustin - Mon Aug 20, 2007 11:57 am
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Trying to look into the economics of SpaceX 
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Post    Posted on: Tue Sep 25, 2007 10:44 am
Before correcting and improving the equation system to incoprporate the
time structure all factors were applied to the same value of the labour
costs L. That value is L2006 now.And as can be seen by the values got for
L2007 and L2008 there is growth of L now.

This means that g, h and j have to be applied to higher values than i and
g and ha to a higher value than j. What about the effect on the launch
operations team? In principle the number of rockets is growing – this may
be a reason to assume that the launch operations team should grow also.

On the other hand the number of launches will be of meaning for the growth
of the launch operations team – and the number of launches per Quartal
seems to be constant to a very large degree. Peaks might be handeled via
temporaryly moving employees from the R&D-team to the launch operations
team for example.

From my point of view the table I posted based on the launch
manifest-versions Klaus posted is a reason to suppose NO growth of the
launch operations team – which menas groth of the manufacturing team or
perhaps one of the other two teams.

So what about growth of the R&D-team? They mainly will be developping
engines and it will be one engine after the other – not all at once. When
the first Merlin 1B is ready for a first launch they will work on
clustering/bundling them for the Falcon 9. And after that they will work
on using Falcon 9-first stages as boosters of the Falcon 9 Heavy. This
doesn’t sound as if any growth is to be expected.

And the Administration Team? Paperwork, phoning, hiring, ordering
repair-services and the like, selling launches – I don’t think that that
team will growth because of the rockets – they will grow when the business
grows.

There is one thing that has not been taken into account regarding the
growth of teams – the Dragon, The Dragon might cause growth of the R&D
Team and the Manufacturing Team – but first of all no numbers about the
Dragon are available that could be applied to handle this properly. And
next SpaceX is developing the Dragon with partners who are doing R&D and
manufacturing also. Last but not least it can be assumed here that the
majority of employees to be hired for the Dragon will be hired after 2008

So I consider the manufacturering-team to be the only one growing here.
Like I already said earlier I have in mind to apply alternative
numbers/values also to check what changes then and by how much it does
change.

This means that the structure of the salaries is changing by the years –
the share of the three not-growing teams is dimishing:

Code:
team                     2006   2007   2008

Launch Operations Team     25.     25     25
R&D                       51     51     51
Administration            22     22     22
non-growing               98     98     98
Manufacturing          157   243   335


Then the shares are

Code:
team            2006          2007          2008

non-growing     0 384313725   0,287390029   0.243792325
Manufacturing   0.615686275   0.712609971   0.756207675


These numbers include one correction and several roundings. These shares
now have to be multiplied by the factors and added on then – the shares of
the non-growing teams by 0.333... because of three rockets and equal
distribution of the salaries, the shares of the manufacturing team by the
other factor because of inequal distirbution.

Because of the application of g,h,i and j to salaries of different years I
now prefer to leave away the labour costs share- and auxiliary equations:

Falcon 9: 10 * X + f * 1/n9 * g * L2008 + pr9 + npr9 + p9 = F9 – (a + b)
Falcon 9 Heavy: 28 * X + f * 1/n9h * h * L 2008 + pr9h + npr9h + p9h = F9H
– (3 * a + b)
Falcon 1: 1 * Z + 1 * Y + f * 1/n1 * i * L2006 + pr1 + npr1 + p1 = F1 – (e
+ f)
Falcon 5: 6 * X + f * 1/n5 * j * L2007 + pr5 + npr5 + p5 = F5 - (c + d)
Similarity: Z = X
Cash Flow: n9 * pr9 + n9 * npr9 + n9h * pr9h + n9h * npr9h + n1 * pr1 + n1
* npr1 + n5 * pr5 + n5 * npr5 - r * I - np < n9 * F9 + n9h * F9H + n1 * F1
+ n5 * F5 – L
Investment: I = 1 * Z + 1 * Y + i1 * L2006 + j1 * L2006 + R



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Post    Posted on: Tue Oct 02, 2007 12:10 pm
The shares and factors result in the following values:

g = 0.333 * 0.243792325 + 0.667 * 0.756207675 = 0.58540255833333333333333333333333 rounded to 0.59
h = 0.333 * 0.243792325 + 0.333 * 0.756207675 = 0.333…

Since I already said a few posts earlier that I would apply the exact numbers of employees of that quarter of the year the first launch of a rocket is done in let’s think about it.

What I’ve said has to do with the circumstance that the number of employees is growing month y month and thus quarter by quarter. The number by which it grows is between 8 and 9 per month and thus 24 to 27 – 27 merely – per quarter.

This means that for the first quarter of a year the exact number is below the quarter of the average number of employees of that year while it is above it in the third and the last quarter of that year. In the second quarter it is equal to the average.

To do this properly the average per quarter must be calculated.

In the equation system L2006, L2007 and L2008 have to be replaced by L2006Q1 and so on in the four equations of the rockets an in the investment equation.

The effect of doing so would be increased values of X, Y and Z (at least) – which would mean safety margins included into the investment into each rocket. On the other hand doing NOT so would include safety margins into the variable costs – and thus the lower boundary of the launch and flight costs.

Because I delayed the calculations very much in between I prefer to apply both alternatives and start with keeping L2006 to L2008.

So let’s insert all known numbers now:

Falcon 9: 10 * X + 0.25 * 1/1 * 0.59 * 40889610.24 + 169130 + npr9 + p9 = 35 mio – (20958.4082353887 + 23151.7300274643)
Falcon 9 Heavy: 28 * X + 0.25 * 1/1 * 0.33 * 40889610.24 + 46553 + npr9h + p9h = 78 mio – (3 * 20958.4082353887 + 23151.7300274643)
Falcon 1: 1 * Z + 1 * Y + 0.25 * 1/2 * 0.29 * 24080486.40 + 16402,1 + npr1 + p1 = 6.7 mio – (5754,08494834638 + 306,884530578474)
Falcon 5: 6 * X + 0.25 * 1/1 * 0.71 * 32201748.48 + 91780 + npr5 + p5 = 18 mio - (26807.2663 + 11372.7797)
Similarity: Z = X
Cash Flow: 0 * pr9 + 0 * npr9 + 0 * pr9h + 0 * npr9h + n1 * pr1 + n1 * npr1 + 0 * pr5 + 0 * npr5 - r * I - np < 0 * 35 mio + 0 * 78 mio + 1 * 6.7 mio + 0 * 18 mio – 24080486.40
Investment: I = 1 * Z + 1 * Y + 0.1 * 24080486.40 + 0.52 * 24080486.40 + R

I didn’t insert numbers for n1 into the Cash Flow-equation because the three versions of the launch manifest list at least two different numbers of launches.

It also is looking as if the equations are adjusted to the seond version of the launch manifest at present because of applying L2008 for the Falcon 9 – I have to check that.

The Falcon 9 Heavy includes a share of the labour costs below the share included into the Falcon 9 – a bit more than half that share. This can be considered to be justified by the circumstance that two additional first stages are applied only while the second stage is left unchanged but newly constructed for the Falcon 9.

So after getting results from this L2008 has to be replaced by L2007 in the Falcon 9-equation(s) and two alternative results have to be calculated per case – four results to be expected if I haven’t forgotten something. Next additional four results might be got by replacing the L2006 etc. by L2006Q1 etc.



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Post    Posted on: Thu Oct 04, 2007 11:30 am
I forgot to insert the value for pr1 and for I into the Investment-equation into the Cash Flow-equation which I have done in this post now.

Since all numbers are inserted now the equations can be modified. This is done in the appendix. The first three modifications result in the following version of the equation system:

Falcon 9: 10 * X + npr9 + p9 = 34786759.861737147 – 0.1475 * 40889610.24
Falcon 9 Heavy: 28 * X + npr9h + p9h = 77867420.0452663696 – 0.04125 * 40889610.24
Falcon 1: 1 * Z + 1 * Y + npr1 + p1 = 6677536.930521075146 – 0.03625 * 24080486.40
Falcon 5: 6 * X + npr5 + p5 = 17870039.954 – 0.1775 * 32201748.48
Similarity: Z – X = 0
Cash Flow: n1 * 16402.1 + n1 * npr1 - r * I - np < 1 * 6.7 mio – 24080486.40
Investment: 1 * Z + 1 * Y + R = 100 mio - 0.62 * 24080486.40

The unknowns here are

    X
    npr9
    p9
    npr9h
    p9h
    Z
    Y
    npr1
    p1
    npr5
    p5
    r
    np
    R


n1 simply isn’t inserted yet. So there are 14 unknowns at present and 7 equations. This is the consequence of the nprs. If they are combined with the profits p into one variable the number of unknowns is reduced to 10. I is NOT set here but is known – this is handled in the next modifications. r and np will be set and then a function is got. But let’s do more modifications first.

The next version of the equation system to be looked at is

Falcon 9: 10 * X + npr9 + p9 = 34786759.861737147 – 0.1475 * 40889610.24
Falcon 9 Heavy: 28 * X + npr9h + p9h = 77867420.0452663696 – 0.04125 * 40889610.24
Falcon 1: 1 * X + 1 * Y + npr1 + p1 = 6677536.930521075146 – 0.03625 * 24080486.40
Falcon 5: 6 * X + npr5 + p5 = 17870039.954 – 0.1775 * 32201748.48
Cash Flow: n1 * 16402.1 + n1 * npr1 - r * (1 * X + 1 * Y + R + 0.62 * 24080486.40) - np < 6.7 mio – 24080486.40
Investment: 1 * X + 1 * Y + R = 100 mio - 0.62 * 24080486.40

There is a known value on the left side of the Cash Flow-equation now because the rate of depreciation hasn’t been set yet. This has to be thought about in the next post as well as about np.

Now 13 unknowns now and 6 equations are left.

I’ll go on later.



Dipl.-Volkswirt (bdvb) Augustin (Political Economist)

Appendix

First 3 Modifications

1. Modification: Shift of knowns to the right sides and unknowns to the left sides

Falcon 9: 10 * X + npr9 + p9 = 35 mio – (20958.4082353887 + 23151.7300274643) – (0.25 * 1/1 * 0.59 * 40889610.24 + 169130)
Falcon 9 Heavy: 28 * X + npr9h + p9h = 78 mio – (3 * 20958.4082353887 + 23151.7300274643) – (0.25 * 1/1 * 0.33 * 40889610.24 + 46553)
Falcon 1: 1 * Z + 1 * Y + npr1 + p1 = 6.7 mio – (5754.08494834638 + 306.884530578474) – (0.25 * 1/2 * 0.29 * 24080486.40 + 16402.1)
Falcon 5: 6 * X + npr5 + p5 = 18 mio - (26807.2663 + 11372.7797) – (0.25 * 1/1 * 0.71 * 32201748.48 + 91780)
Similarity: Z – X = 0
Cash Flow: 0 * pr9 + 0 * npr9 + 0 * pr9h + 0 * npr9h + n1 * pr1 + n1 * npr1 + 0 * pr5 + 0 * npr5 - r * I - np < 0 * 35 mio + 0 * 78 mio + 6.7 mio + 0 * 18 mio – 24080486.40
Investment: 1 * Z + 1 * Y + R = 100 mio – (0.1 * 24080486.40 + 0.52 * 24080486.40)



2. Modification: Calculation of all terms completely consisting of knowns

Falcon 9: 10 * X + npr9 + p9 = 35 mio – 44110.138262853 – (0.1475 * 40889610.24 + 169130) = 35 mio – 44110.138262853 – 0.1475 * 40889610.24 - 169130 = 35 mio – 213240.138262853 – 0.1475 * 40889610.24 = 34786759.861737147 – 0.1475 * 40889610.24
Falcon 9 Heavy: 28 * X + npr9h + p9h = 78 mio – (62875.2247061661 + 23151.7300274643) – (0.04125 * 40889610.24 + 46553) = 78 mio – 86026,9547336304 – (0.04125 * 40889610.24 + 46553) = 78 mio – 86026.9547336304 – 0.04125 * 40889610.24 - 46553 = 78 mio – 132579.9547336304 – 0.04125 * 40889610.24 = 77867420.0452663696 – 0.04125 * 40889610.24
Falcon 1: 1 * Z + 1 * Y + npr1 + p1 = 6.7 mio – 6060.969478924854 – (0.03625 * 24080486.40 + 16402.1) = 6.7 mio – 6060.969478924854 – 0.03625 * 24080486.40 – 16402.1 = 6.7 mio – 22463,069478924854 – 0,03625 * 24080486.40 = 6677536.930521075146 – 0.03625 * 24080486.40
Falcon 5: 6 * X + npr5 + p5 = 18 mio – 38180.046 – (0.1775 * 32201748.48 + 91780) = 18 mio – 38180.046 – 0.1775 * 32201748.48 - 91780 = 18 mio – 129960.046 – 0.1775 * 32201748.48 = 17870039.954 – 0.1775 * 32201748.48
Similarity: Z – X = 0
Cash Flow: 0 * pr9 + 0 * npr9 + 0 * pr9h + 0 * npr9h + n1 * 16402.1 + n1 * npr1 + 0 * pr5 + 0 * npr5 - r * I - np < 0 * 35 mio + 0 * 78 mio + 6.7 mio + 0 * 18 mio – 24080486.40
Investment: I – (1 * Z + 1 * Y + R) = 0.62 * 24080486.40

Falcon 9: 10 * X + npr9 + p9 = 34786759.861737147 – 0.1475 * 40889610.24
Falcon 9 Heavy: 28 * X + npr9h + p9h = 77867420.0452663696 – 0.04125 * 40889610.24
Falcon 1: 1 * Z + 1 * Y + npr1 + p1 = 6677536.930521075146 – 0.03625 * 24080486.40
Falcon 5: 6 * X + npr5 + p5 = 17870039.954 – 0.1775 * 32201748.48
Similarity: Z – X = 0
Cash Flow: 0 * pr9 + 0 * npr9 + 0 * pr9h + 0 * npr9h + n1 * 16402.1 + n1 * npr1 + 0 * pr5 + 0 * npr5 - r * I - np < 0 * 35 mio + 0 * 78 mio + 6.7 mio + 0 * 18 mio – 24080486.40
Investment: 1 * Z + 1 * Y + R = 100 mio - 0.62 * 24080486.40

The labour cost-term is NOT calculated to keep it easier to replace the value labour cost.



3. Modification: Elimination of multiplications by 0

Falcon 9: 10 * X + npr9 + p9 = 34786759.861737147 – 0.1475 * 40889610.24
Falcon 9 Heavy: 28 * X + npr9h + p9h = 77867420.0452663696 – 0.04125 * 40889610.24
Falcon 1: 1 * Z + 1 * Y + npr1 + p1 = 6677536.930521075146 – 0.03625 * 24080486.40
Falcon 5: 6 * X + npr5 + p5 = 17870039.954 – 0.1775 * 32201748.48
Similarity: Z – X = 0
Cash Flow: n1 * 16402.1 + n1 * npr1 - r * I - np < 6.7 mio – 24080486.40
Investment: 1 * Z + 1 * Y + R = 100 mio - 0.62 * 24080486.40




Modifications 4 and 5

4. Modification: Application of the Similarity-equation by insertion X for Z

Falcon 9: 10 * X + npr9 + p9 = 34786759.861737147 – 0.1475 * 40889610.24
Falcon 9 Heavy: 28 * X + npr9h + p9h = 77867420.0452663696 – 0.04125 * 40889610.24
Falcon 1: 1 * X + 1 * Y + npr1 + p1 = 6677536.930521075146 – 0.03625 * 24080486.40
Falcon 5: 6 * X + npr5 + p5 = 17870039.954 – 0.1775 * 32201748.48
Cash Flow: n1 * 16402.1 + n1 * npr1 - r * I - np < 6.7 mio – 24080486.40
Investment: 1 * X + 1 * Y + R = 100 mio - 0.62 * 24080486.40



5. Modification: Insertion of I = 1 * X + 1 * Y + R + 0.62 * 24080486.40 = 100 mio

Falcon 9: 10 * X + npr9 + p9 = 34786759.861737147 – 0.1475 * 40889610.24
Falcon 9 Heavy: 28 * X + npr9h + p9h = 77867420.0452663696 – 0.04125 * 40889610.24
Falcon 1: 1 * X + 1 * Y + npr1 + p1 = 6677536.930521075146 – 0.03625 * 24080486.40
Falcon 5: 6 * X + npr5 + p5 = 17870039.954 – 0.1775 * 32201748.48
Cash Flow: n1 * 16402.1 + n1 * npr1 - r * (1 * X + 1 * Y + R + 0.62 * 24080486.40) - np < 6.7 mio – 24080486.40
Investment: 1 * X + 1 * Y + R = 100 mio - 0.62 * 24080486.40



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Post    Posted on: Fri Oct 05, 2007 11:13 am
Regarding the new provisions np and the rate of depreciation r the launch manifests are of help – in particular for the new provisions they are the only data source.

But regarding the rate of depreciation there are more approaches:

    1. loss of the first Falcon 1
    2. loss of two Falcon 1s
    3. average including one of the other alternatives and the non-rocket-investments (buildings 30 years, machines 10 years)
    4. maximum number of launches of a rocket
    5. maximum number of launches over all rockets
    …

The alternative values of np are $ 13.4 mio in version1 and $ 20.1 mio in version 2 and 3.

The alternative values of r are

Code:
version 1 of launch manifest    33.333…%         by three Falcon 1-launches
version 2 of launch manifest    14.2857…%        by six Falcon 1-launches
version 2 of launch manifest     4.56%             by six Falcon 1-launches
                                               plus the Air Force-contract
version 3 of launch manifest    16.666…%         by six Falcon 1-launches

version 3 of launch manifest     4.75% to 5%   by six Falcon 1-launches
                                               plus the Air Force-contract
loss of first Falcon 1           5% and more
loss of two Falcon 1s           10% and more
average                          5.75% to 6%   15 years in average for
                                               buildings and machines
maximum number of launches      14.2857…%
maximum over all rockets        10%
maximum over all rockets         4%            plus the Air Force-contract


If the price of one Falcon 1-launch not accounting for reusability would be taken as depreciation then this would be a rate of depreciation of 6.7%. The average listed is 5.75% to 6% - which is less but close to it. The least rate is 4% and is based on rockets requiring an additional investment of $ 100 mio according to Elon Musk – so I don’t want to apply the 4%.

Because of all this I am going to apply 5% here.

Inserting the smaller np and the rate of depreciation chosen the equation system is

Falcon 9: 10 * X + npr9 + p9 = 34786759.861737147 – 0.1475 * 40889610.24
Falcon 9 Heavy: 28 * X + npr9h + p9h = 77867420.0452663696 – 0.04125 * 40889610.24
Falcon 1: 1 * X + 1 * Y + npr1 + p1 = 6677536.930521075146 – 0.03625 * 24080486.40
Falcon 5: 6 * X + npr5 + p5 = 17870039.954 – 0.1775 * 32201748.48
Cash Flow: n1 * 16402.1 + n1 * npr1 – 0.05 * (1 * X + 1 * Y + R + 0.62 * 24080486.40) – 13.4 mio < 6.7 mio – 24080486.40
Investment: 1 * X + 1 * Y + R = 100 mio - 0.62 * 24080486.40

6. Modification: Calculation of all terms completely consisting of knowns

Falcon 9: 10 * X + npr9 + p9 = 34786759.861737147 – 0.1475 * 40889610.24
Falcon 9 Heavy: 28 * X + npr9h + p9h = 77867420.0452663696 – 0.04125 * 40889610.24
Falcon 1: 1 * X + 1 * Y + npr1 + p1 = 6677536.930521075146 – 0.03625 * 24080486.40
Falcon 5: 6 * X + npr5 + p5 = 17870039.954 – 0.1775 * 32201748.48
Cash Flow: n1 * 16402.1 + n1 * npr1 – 0.05 * 1 * X + 0.05 * 1 * Y + 0.05 * R + 0.05 * 0.62 * 24080486.40 – 13.4 mio < 6.7 mio – 24080486.40
Investment: 1 * X + 1 * Y + R = 100 mio - 0.62 * 24080486.40

Falcon 9: 10 * X + npr9 + p9 = 34786759.861737147 – 0.1475 * 40889610.24
Falcon 9 Heavy: 28 * X + npr9h + p9h = 77867420.0452663696 – 0.04125 * 40889610.24
Falcon 1: 1 * X + 1 * Y + npr1 + p1 = 6677536.930521075146 – 0.03625 * 24080486.40
Falcon 5: 6 * X + npr5 + p5 = 17870039.954 – 0.1775 * 32201748.48
Cash Flow: n1 * 16402.1 + n1 * npr1 – 0.05 * X + 0.05 * Y + 0.05 * R + 0.031 * 24080486.40 – 13.4 mio < 6.7 mio – 24080486.40
Investment: 1 * X + 1 * Y + R = 100 mio - 0.62 * 24080486.40

7.Modification: Shift of knowns to the right sides and unknowns to the left sides

Falcon 9: 10 * X + npr9 + p9 = 34786759.861737147 – 0.1475 * 40889610.24
Falcon 9 Heavy: 28 * X + npr9h + p9h = 77867420.0452663696 – 0.04125 * 40889610.24
Falcon 1: 1 * X + 1 * Y + npr1 + p1 = 6677536.930521075146 – 0.03625 * 24080486.40
Falcon 5: 6 * X + npr5 + p5 = 17870039.954 – 0.1775 * 32201748.48
Cash Flow: n1 * 16402.1 + n1 * npr1 – 0.05 * X + 0.05 * Y + 0.05 * R < 6.7 mio – 24080486.40 – (0.031 * 24080486.40 – 13.4 mio)
Investment: 1 * X + 1 * Y + R = 100 mio - 0.62 * 24080486.40

6. Modification: Calculation of all terms completely consisting of knowns

Falcon 9: 10 * X + npr9 + p9 = 34786759.861737147 – 0.1475 * 40889610.24
Falcon 9 Heavy: 28 * X + npr9h + p9h = 77867420.0452663696 – 0.04125 * 40889610.24
Falcon 1: 1 * X + 1 * Y + npr1 + p1 = 6677536.930521075146 – 0.03625 * 24080486.40
Falcon 5: 6 * X + npr5 + p5 = 17870039.954 – 0.1775 * 32201748.48
Cash Flow: n1 * 16402.1 + n1 * npr1 – 0.05 * X + 0.05 * Y + 0.05 * R < 6.7 mio – 24080486.40 – 0.031 * 24080486.40 + 13.4 mio
Investment: 1 * X + 1 * Y + R = 100 mio - 0.62 * 24080486.40

Falcon 9: 10 * X + npr9 + p9 = 34786759.861737147 – 0.1475 * 40889610.24
Falcon 9 Heavy: 28 * X + npr9h + p9h = 77867420.0452663696 – 0.04125 * 40889610.24
Falcon 1: 1 * X + 1 * Y + npr1 + p1 = 6677536.930521075146 – 0.03625 * 24080486.40
Falcon 5: 6 * X + npr5 + p5 = 17870039.954 – 0.1775 * 32201748.48
Cash Flow: n1 * 16402.1 + n1 * npr1 – 0.05 * X + 0.05 * Y + 0.05 * R < 20.1 mio – 1.031 * 24080486.40
Investment: 1 * X + 1 * Y + R = 100 mio - 0.62 * 24080486.40

N1 isn’t inserted yet although the versions of the launch manifest are applied – the reason is that the number of possible launches is used while n1 is the actual number of launches.

The unknowns left now are

    X
    npr9
    p9
    npr9h
    p9h
    Y
    npr1
    p1
    npr5
    p5
    R


Which are 11 unknowns at 6 equations. 4 unknowns can be combined with 4 other unknowns – resulting in 7 unknowns at 6 equations which means a function.

The unknowns to be combined will be looked at in the next post.

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Post    Posted on: Fri Oct 12, 2007 1:14 pm
The unknowns to be combined are

    npr9 with p9
    npr9h with p9h
    npr1 with p1
    npr5 with p5


The problem behind it is the question how the values of four of them could be got. The nprs are variable non-propellant costs per launch/flight similar to or like the propellant costs. There are no equations they could be calculated by – it would be unusual if such equations would exist. The ps are the profits per launch/flight and usually result from equations.

In so far both the non-propellant costs and the profits are very similar to each other – and the situation once more shows that the present approach is economically impossible and against the nature of the eight unknowns.

It seems that only values for combinations of each npr with the according p can be got. Then it could be thought or pondered about the distribution of the values over the two combined unknowns.

The nprs may include refurbishments. It is NOT clear however if they include labour costs of the refurbishments. Neither SpaceX nor Elon Musk have talked about refurbishments (not in any detail at least). At present I could imagine that the launch operation team also does the refurbishments and the rocket equations mean that the refurbishments are included into the prices – which is required to be able to cover those costs and to get into the profit zone one day.

So let’s add on the nprs and the ps to margins ms:

npr9 + p9 = m9
npr9h + p9h = m9h
npr1 + p1 = m1
npr5 + p5 = m5

There is one possible access to the value of a npr – npr1 is explicitly a part of the Cash Flow-equation. Perhaps it can be calculated directly. The other nprs may be got by assuming the ps to be zero.

Before going on it is required to recognize that there are errors in the mathematical operations in the previous post at modifications of the Cash Flow-equation – beginning at the 6. modification. I also detected that it would be more proper to NOT insert the Investment-equation into the Cash Flow-equation as 5. modification. Also the value for np must be removed again because it is linked to n1 directly. So I start again at the 5. modification:

5. Modification: Multiplying away factor = 1 in two equations

Falcon 9: 10 * X + m9 = 34786759.861737147 – 0.1475 * 40889610.24
Falcon 9 Heavy: 28 * X + m9h = 77867420.0452663696 – 0.04125 * 40889610.24
Falcon 1: X + Y + m1 = 6677536.930521075146 – 0.03625 * 24080486.40
Falcon 5: 6 * X + m5 = 17870039.954 – 0.1775 * 32201748.48
Cash Flow: n1 * 16402.1 + n1 * npr1 - r * I - np < 6.7 mio – 24080486.40
Investment: X + Y + R = 100 mio - 0.62 * 24080486.40

6. Modification: Solution for X from Falcon 9-equation – subtraction of m9

Falcon 9: 10 * X = 34786759.861737147 – 0.1475 * 40889610.24 – m9

7. Modification: Solution for X from Falcon 9-equation – division by 10

Falcon 9: X = 3478675.9861737147 – 0.01475 * 40889610.24 – 0.1 * m9 = 2875554.2351337147 – 0.1 * m9

At this point economical thoughts about the result should be done.

m9 consists of costs and a profit. So can m9 be negative?

A profit can be negative and thus be a loss – but can a launch price include a loss? This might be the case if launches of that rocket are driving the business of launching other rockets. If launches of the rocket the launch price of includes a loss is NOT offered the entrepreneur wouldn’t get customers for the other rockets. In THAT case a price might include a loss.

Negative costs would be revenues – and revenues are got via the price. So the costs included into m9 can’t be negative no way.

So it’s possible that m9 is negative because of business and then other rockets would cross-finance the loss. The method applied at present doesn’t account for that.

What are the boundaries of m9? The investment into the Merlin 1B (X) can’t be negative – meaning that m9 can’t be higher than $ 28,755,542.351337147. Then X would be zero.

How reasonable is the maximum value of m9? Assumed it is entirely npr9 then such a value is economically nonsense. The non-propellant costs not only would be a multiple of the propellant costs – they would be astronomical in comparison to the investment of 0. Assumed furthermore the non-propellant costs are refurbishment costs then the refurbishment would be astronomical in comparison to the investment itself which is 0.

Since the investment itself is higher than 0 that high refurbishment costs are unlikely. This also means that m9 can’t be entirely profit no way. So m9 will be no higher than the half of the maximum listed above.

Next a negative m9 is limited by the circumstance that the resulting investment into the Merlin (=X) effects the investment into and the profits of the other rockets. At least one of them will be intended to yield profits. The more negative m9 is the more cross-financing will be required and the higher the result of the equation got by the 7. modification.

At present the ONLY rocket available using a Merlin (=X) is the Falcon 1 – because of this I am doubting if the launch price of the Falcon 1 inlcudes a loss. Of course this doesn’t mean yet that m9 is not negative since it is not the margin of the price for a Falcon 1-launch.

Can a loss included into the launch price be as high as the investment into the rocket? I am doubting that very much – it would mean the loss of the complete investment into that rocket. The rocket would still be there because it is reusable but each launch of the rocket would reduce the capital left by a significant amount.

Of course it can’t be said for sure what portion of the investment the included loss might be but the upper boundary seems to be known now.

So m9 will be between $ – 28,755,542.351337147 and $ + 14,377,771.1756685735 – meaning that X will be between $ + 1,437,777.11756685735 and $ + 5,751,108.4702674294.

But it is possible to find it out in detail.

Let’s have a look to the results of the 7. Modification for the Falcon 9 Heavy and the Falcon 5:

Falcon 9 Heavy: 2,720,740.129388084 – 0.03571428… * m9h
Falcon 5: 2,025,704.9331333… - 0.1666… * m5

The earlier contradictional results show up again but completed by a term the value of isn’t known at present. This is what the equation system and the margins and their components are providing and what their purpose is (although it needs to be modified later because of economical impossibilities).

The Falcon 9 Heavy seems to allow for a larger range of values of m9h than is allowed for m9 while the Falcon 5 seems to allow for a smaller range of m5 than for m9. This I recognize by a look on the factors. But all the ranges are intercepting each other or they may form a hierarchy – this enables for approximations if required or wanted.

What about the Falcon 1? This rocket is the only one in the list where the two stages have different engines. So at this point the equation allows for a joint result only: 5,804,619.298521075 – m1. The range in this case is much more smaller than those for the other rockets. m1 can’t be larger than $ 2,902,309.649260537 and not smaller than $ - 6.7 mio. Then the both engines together require an investment of $ 2,902,309.649260537 minimum and 9,602,309.649260537 maximum.

But it can be found out more precisely and in more detail.



Dipl.-Volkswirt (bdvb) Augustin (Political Economist)


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Post    Posted on: Thu Oct 25, 2007 10:59 am
Like already mentioned the margins aren’t adjusted to each other up to now. But this can be handled now. The three alternative equations for X MUST have the same results since each Merlin of several instances of the same version have to have the same costs and price. This establishes functions by which m9h and m5 can be calculated from m9 – which I chose here because the Falcon 9 is the first of those three that will be launched.

8. Modification: Equalizing solutions for X

Falcon 9 and Falcon 9 Heavy: X = 2875554.2351337147 – 0.1 * m9 = 2720740.129388084 – 0.03571428… * m9h
Falcon 9 and Falcon 5: X = 2875554.2351337147 – 0.1 * m9 = 2025704.9331333… - 0.1666… * m5

9. Modification: Subtraction of the constant of the Falcon 9 Heavy-equation

Falcon 9 and Falcon 9 Heavy: 154814.1057456307 – 0.1 * m9 = 0.03571428… * m9h
Falcon 9 and Falcon 5: 849849.3020003814 – 0.1 * m9 = 2025704.9331333… - 0.1666… * m5

10. Modification: Division by the factors of m9h or m5

Falcon 9 and Falcon 9 Heavy: 4334795.6544449643 – 2.8000004480 * m9 = m9h
Falcon 9 and Falcon 5: 5099095.8120022882 – 0.6 * m9 = m5

At this point a better look at the boundaries already mentioned is possible since the values of the boundaries can be inserted to get real sets of values linked to each other. It also is possible to find additional extreme values where at least one of the margins is zero. This way it can be seen how the margins behave in concert – including the margins to be got for m9h and m5:

Code:
m9                       kind    m9h                     m5                     ruled out by

                       - invalid -

  28755542.3513378831    m9h=mn  - 76180735.8117840819   - 12154229.5988004417  m9
  28755542.3513368137    m5=min  - 76180735.8117810875   - 12154229.5987998     m9

                       - valid -

  14377771.1756685735    m9=max  - 35922970.0786685282   - 3527566.8933988559
   8498493.0200038136    m5 =0   - 19460988.6088905869           0
   1548141.0574563466    m9h=0            0                4170211.1775284802
         0               m9 =0      4334795.6544449643     5099095.8120022882
-  1630031.6456626863    m5=max     8898884.9925546633     6077114.7993999

                       - invalid -

- 12055559.5894844216    m9h=mx    38090367.905892041      12332430.9656929412  m5
– 28755542.351337147     m9=min    84850327.1206719492     22352421.2228045764  m9h


As can be seen in the table above the minimum values of m9h and m5 are invalid because the value of m9 would be far above the maximum valid value of m9 then. The minimum value of m9 is invalid also because m9 and m5 are above their maxima then. And the maximum value of m9h is invalid also because m5 is still above its maximum value then.

The valid values consist of three different groups – one where m9 is positive and at least m9h is negative, a second where none of the three is negative and a third where m9 is negative while both of the others are positive.

If you prefer to assume that all the margins are positive and none is negative then m9 is somewhere between $ 1,548,141 and $ 0 with m9h being between $ 0 and $ 4,334,795 and m5 between $ 4,170,211 and $ 5,099,095.

So let’s have a look at the values of X resulting now – by applying the equation Falcon 9: X = 2875554.2351337147 – 0.1 * m9:

Code:
m9                       kind     X                     ruled out by

                       - invalid -

  28755542.3513378831    m9h=mn   -       0.0000000736   m9
  28755542.3513368137    m5=min           0.0000000333   m9

                       - valid -

  14377771.1756685735    m9=max     1437777.1175668574
   8498493.0200038136    m5 =0      2025704.9331333333
   1548141.0574563466    m9h=0      2720740.1293880800
         0               m9 =0      2875554.2351337147
-  1630031.6456626863    m5=max     3038557.3996999833

                       - invalid -

- 12055559.5894844216    m9h=mx     4081110.19408215686   m5
– 28755542.351337147     m9=min     5751108.4702674294   m9h


Accoding to this table the investment into the Merlin (X) is between $ 1.434 mio and $ 3.038 mio at valid values for all three margins and between $ 2.721 mio and $ 2.875 mio if it is preferred to assume positive values for all three margins.

In so far reasonable and workable approximations are possible already and even values for the actual launch prices can be calculated.

In particular it seems to be possible to get values for X even if Z =X is invalid.

But let’s go on here.

m1 and Y haven’t been considered again yet. In the previous post a joint result for X and Y was possible only: X + Y = 5804619.298521075 – m1 with 2902309.649260537 >= m1 >= - 6.7 mio.

Applying the valid values of X this equation is turned into the following versions:

    1437777.1175668574 + Y = 5804619.298521075 – m1
    2025704.9331333333 + Y = 5804619.298521075 – m1
    2720740.1293880800 + Y = 5804619.298521075 – m1
    2875554.2351337147 + Y = 5804619.298521075 – m1
    3038557.3996999833 + Y = 5804619.298521075 – m1

The way to equations for Y is:

step 1

    Y = 5804619.298521075 - 1437777.1175668574 – m1
    Y = 5804619.298521075 - 2025704.9331333333 – m1
    Y = 5804619.298521075 - 2720740.1293880800 – m1
    Y = 5804619.298521075 - 2875554.2351337147 – m1
    Y = 5804619.298521075 - 3038557.3996999833 – m1

step 2

    Y = 4366842.1809542176 – m1
    Y = 3778914.3653877417 – m1
    Y = 3083879.169132995 – m1
    Y = 2929065.0633873603 – m1
    Y = 2766061.8988210917 – m1


The upper boundary and the lower boundary of m1 could be inserted to get alternative values of Y but it is not clear yet what the proper value or the proper range of values inside the range already list might be.

It’s a temptation to look at the boundaries of m1 by the second list – but that would be misleading because of the circumstance that there are several alternative equations.

So the unknowns not looked at yet – m1, Y, R – need to be found out by additional modifications to be done later.

Up to this point it is still looking as if simply a value has been found Andy Hill was assuming a longer time ago already. But the intention here is to look far beyond that value – to profits, refurbishments, also to the links between the margins etc. of the particular rockets and much more.



Dipl.-Volkswirt (bdvb) Augustin (Political Economist)


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Post    Posted on: Tue Nov 06, 2007 8:24 am
Before looking into the remaining unknowns systematically and in detail let’s think about m1 a bit. If all three other margins are supposed to be positive then it would be consequent to consider m1 to be also positive. But there is another essential aspect too – is it reasonable to include a loss into the launch price of the only reusable rocket SpaceX have at present? If so then it should be a marginal loss only because the whole initial investment of $ 100 mio would be lost relatively quickly. And this thought also holds if the one of or all other margins were supposed to be negative.

Another question again is if the Kestrel (Y) is cheaper or more expensive than the Merlin. Then from the previous post the following must be concluded: If the Kestrel is supposed to be cheaper and m9, m9h and m5 are considered to be positive then m1 logically must be positive – and it must be the more positive the lower the investment into the Merlin (X) is assumed to be. One argument seems to be speaking for the Kestrel to be cheaper – it is expendable and thus should be kept cheap the more.

To apply the equation system for getting systematical insights let’s consider Y and the three equations not applied yet now.

First of all there is the Investment-equation

Investment: X + Y + R = 100 mio - 0.62 * 24080486.40

The equation includes the Kestrel (Y). This seems – and really IS – an error I kept until now consciously.

It really is an error because the Kestrel is part of the expendable second stage of the Falcon 1 and thus is part of the variable costs – but it is not included into the variable non-propellant-costs. The reason causing the error was – among others – that development costs are investments of particular kind and that a portion of the investment of $ 100 mio must be considered to have gone into the development. Another reason or merely cause was that the second stage of the other three rockets explicitly will be reusable and NOT expended.

To keep the error enables to illustrate the economical effect of reusability.

So first here the two alternative versions – correct with Kestrel and second stage expendable and wrong with Kestrel and second stage reusable:

    a) expendable Y – corrected new version

    Falcon 1: X + Y + m1 = 6677536.930521075146 – 0.03625 * 24080486.40
    Cash Flow: n1 * 16402.1 + n1 * npr1 + n1 * Y - r * I - np < 6.7 mio – 24080486.40
    Investment: X + R = 100 mio - 0.62 * 24080486.40

    b) reusable Y – wrong original version

    Falcon 1: X + Y + m1 = 6677536.930521075146 – 0.03625 * 24080486.40
    Cash Flow: n1 * 16402.1 + n1 * npr1 - r * I - np < 6.7 mio – 24080486.40
    Investment: X + Y + R = 100 mio - 0.62 * 24080486.40

There are two alternative ways now – to insert $ 100 mio for I or to insert the Investment-equation for I. I insert $ 100 mio now – I keep the knowns uninserted but handle them as inserted:

6. modification

    a) expendable Y – corrected new version

    Falcon 1: X + Y + m1 = 6677536.930521075146 – 0.03625 * 24080486.40
    Cash Flow: n1 * 16402.1 + n1 * npr1 + n1 * Y - r * 100 mio - np < 6.7 mio – 24080486.40
    Investment: X + R = 100 mio - 0.62 * 24080486.40

    b) reusable Y – wrong original version

    Falcon 1: X + Y + m1 = 6677536.930521075146 – 0.03625 * 24080486.40
    Cash Flow: n1 * 16402.1 + n1 * npr1 - r * 100 mio - np < 6.7 mio – 24080486.40
    Investment: X + Y + R = 100 mio - 0.62 * 24080486.40



7. modification

    a) expendable Y – corrected new version

    Falcon 1: X = 6677536.930521075146 – 0.03625 * 24080486.40 - Y + m1
    Cash Flow: n1 * Y < 6.7 mio – 24080486.40 - n1 * 16402.1 - n1 * npr1 + r * 100 mio + np
    Investment: X = 100 mio - 0.62 * 24080486.40 - R

    b) reusable Y – wrong original version

    Falcon 1: X = 6677536.930521075146 – 0.03625 * 24080486.40 - Y - m1
    Cash Flow: n1 * npr1 < 6.7 mio – 24080486.40 - n1 * 16402.1 + r * 100 mio + np
    Investment: X = 100 mio - 0.62 * 24080486.40 - Y - R



8. modification

    a) expendable Y – corrected new version

    Falcon 1: X = 6677536.930521075146 – 0.03625 * 24080486.40 - Y + m1
    Cash Flow: Y < 6.7 mio/n1 – 24080486.40/n1 - 16402.1 - npr1 + r * 100 mio/n1 + np/n1
    Investment: X = 100 mio - 0.62 * 24080486.40 - R

    b) reusable Y – wrong original version

    Falcon 1: X = 6677536.930521075146 – 0.03625 * 24080486.40 - Y - m1
    Cash Flow: npr1 < 6.7 mio/n1 – 24080486.40/n1 - 16402.1 + r * 100 mio/n1 + np/n1
    Investment: X = 100 mio - 0.62 * 24080486.40 - Y - R

As can be seen in the 8. modification of the e quations not solved yet the two version prefer different results of the Cash Flow-equation – while the wrong version results in the value of npr1 the correct version seems to result in a value of the investment into a Kestrel (Y) while the npr1 is treated as if it is known.

To do like in the correct version is suggested by the circumstance that npr1 is part of m1 because of m1 = npr1 + p1. The wrong version on the other hand delivers a value for npr1 which seems to be of nearly no use but might be a hint to the value of npr1 in the correct version. The other equations of the wrong version seem to be of no use regarding the Kestrel.

So let’s concentrate on the Cash Flow-equation now – using r = 0.05 but alternatively 2 and 3 for n1 and alternatively 13.4 mio and 20.1 mio for np:

9. modification

    a) expendable Y – corrected new version

    aa) n1 = 2 np = 13.4 mio

    Cash Flow: Y < 6.7 mio/2 – 24080486.40/2 - 16402.1 - npr1 + 0.05 * 100 mio/2 + 13.4 mio/2

    ab) n1 = 3 np = 20.1 mio

    Cash Flow: Y < 6.7 mio/3 – 24080486.40/3 - 16402.1 - npr1 + 0.05 * 100 mio/3 + 20.1 mio/3



    b) reusable Y – wrong original version

    ba) n1 = 2 np = 13.4 mio

    Cash Flow: npr1 < 6.7 mio/2 – 24080486.40/2 - 16402.1 + 0.05 * 100 mio/2 + 13.4 mio/2

    bb) n1 = 3 np = 20.1 mio

    Cash Flow: npr1 < 6.7 mio/3 – 24080486.40/3 - 16402.1 + 0.05 * 100 mio/3 + 20.1 mio/3



10. modification

    a) expendable Y – corrected new version

    aa) n1 = 2 np = 13.4 mio

    Cash Flow: Y < 3.35 mio – 12040243.20 - 16402.1 - npr1 + 2.5 mio + 6.7 mio

    ab) n1 = 3 np = 20.1 mio

    Cash Flow: Y < 2.2333… mio – 8026828.80 - 16402.1 - npr1 + 1.666… mio + 6.7 mio



    b) reusable Y – wrong original version

    ba) n1 = 2 np = 13.4 mio

    Cash Flow: npr1 < 3.35 mio – 12040243.20 - 16402.1 + 2.5 mio + 6.7 mio

    bb) n1 = 3 np = 20.1 mio

    Cash Flow: npr1 < 2.2333… mio – 8026828.80 - 16402.1 + 1.666… mio + 6.7 mio




11. modification

    a) expendable Y – corrected new version

    aa) n1 = 2 np = 13.4 mio

    Cash Flow: Y < 5.85 mio – 12040243.20 - 16402.1 - npr1 + 6.7 mio

    ab) n1 = 3 np = 20.1 mio

    Cash Flow: Y < 3.9 mio – 8026828.80 - 16402.1 - npr1 + 6.7 mio



    b) reusable Y – wrong original version

    ba) n1 = 2 np = 13.4 mio

    Cash Flow: npr1 < 5.85 mio – 12040243.20 - 16402.1 + 6.7 mio

    bb) n1 = 3 np = 20.1 mio

    Cash Flow: npr1 < 3.9 mio – 8026828.80 - 16402.1 + 6.7 mio




11. modification

    a) expendable Y – corrected new version

    aa) n1 = 2 np = 13.4 mio

    Cash Flow: Y < 12.55 mio – 12056645.30 - npr1

    ab) n1 = 3 np = 20.1 mio

    Cash Flow: Y < 10.6 mio – 8043230.90 - npr1



    b) reusable Y – wrong original version

    ba) n1 = 2 np = 13.4 mio

    Cash Flow: npr1 < 12.55 mio – 12056645.30

    bb) n1 = 3 np = 20.1 mio

    Cash Flow: npr1 < 10.6 mio – 8043230.90
]



12. modification - results

    a) expendable Y – corrected new version

    aa) n1 = 2 np = 13.4 mio

    Cash Flow: Y < 493354.7 - npr1

    ab) n1 = 3 np = 20.1 mio

    Cash Flow: Y < 2556769.1 - npr1



    b) reusable Y – wrong original version

    ba) n1 = 2 np = 13.4 mio

    Cash Flow: npr1 < 493354.7

    bb) n1 = 3 np = 20.1 mio

    Cash Flow: npr1 < 2556769.1




It seems that the wrong version tells what must be concluded economically from the fact that Y can’t be negative because of being an investment or something having a price or at least production costs.

If n1 = 2 is applied then the Kestrel (Y) would cost less than $ 493,354.7 which is sounding very low in comparison to the Merlin (X) but might be possible under the aspect that the Kestrel is expendable.

If on the other hand n1 = 3 is applied the Kestrel (Y) would cost less than $ 2,556,769.1 which sounds reasonable – even in comparison to the Merlin(X).



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Post    Posted on: Mon Nov 12, 2007 11:48 am
Next the Falcon 1-equation can be solved. The wrong version of the Cash-Flow-equation and the Investment-equation wouldn’t result in something usefull but in an equation leading to functions where there are two independent variables instead of only one.

So I apply the correct version only from now on. Y can be inserted into the Falcon 1-equation now. Since the Cash Flow-equation is kept an inequation up to now the Falcon 1-equation is turned into an inequation also now. Since Y has a negative sign there the inequation-sign is reversed:

9. modification

a) expendable Y – corrected new version

aa) n1 = 2 np = 13.4 mio

Falcon 1: X > 6677536.930521075146 – 0.03625 * 24080486.40 – (493354.7 - npr1) + m1

ab) n1 = 3 np = 20.1 mio

Falcon 1: X > 6677536.930521075146 – 0.03625 * 24080486.40 – (2556769.1 - npr1) + m1



10. modification

a) expendable Y – corrected new version

aa) n1 = 2 np = 13.4 mio

Falcon 1: X > 6677536.930521075146 – 0.03625 * 24080486.40 – 493354.7 + npr1 + m1

ab) n1 = 3 np = 20.1 mio

Falcon 1: X > 6677536.930521075146 – 0.03625 * 24080486.40 – 2556769.1 + npr1 + m1



11. modification

a) expendable Y – corrected new version

aa) n1 = 2 np = 13.4 mio

Falcon 1: X > 6184182.230521075146 – 0.03625 * 24080486.40 + npr1 + m1

ab) n1 = 3 np = 20.1 mio

Falcon 1: X > 4120767.830521075146 – 0.03625 * 24080486.40 + npr1 + m1



12. modification

a) expendable Y – corrected new version

aa) n1 = 2 np = 13.4 mio

Falcon 1: X > 6184182.230521075146 – 872917.632 + npr1 + m1

ab) n1 = 3 np = 20.1 mio

Falcon 1: X > 4120767.830521075146 – 872917,632 + npr1 + m1



12. modification

a) expendable Y – corrected new version

aa) n1 = 2 np = 13.4 mio

Falcon 1: X > 5311264.598521075146 + npr1 + m1

ab) n1 = 3 np = 20.1 mio

Falcon 1: X > 3247850.198521075146 + npr1 + m1




Because of the preliminary results for X got earlier npr1 + m1 has to be negative it seems – and since npr1 as non-propellant variable costs including possible refurbishments can’t be negative it seems to mean that the profit p1 (from m1 = npr1 + p1) is negative.

But simply let’s insert the equation which calculates X from m9:

13. modification

a) expendable Y – corrected new version

aa) n1 = 2 np = 13.4 mio

Falcon 1: 2875554.2351337147 – 0.1 * m9 > 5311264.598521075146 + npr1 + m1

ab) n1 = 3 np = 20.1 mio

Falcon 1: 2875554.2351337147 – 0.1 * m9 > 3247850.198521075146 + npr1 + m1



14. modification

a) expendable Y – corrected new version

aa) n1 = 2 np = 13.4 mio

Falcon 1: – 0.1 * m9 > 2435710.363387360446 + npr1 + m1

ab) n1 = 3 np = 20.1 mio

Falcon 1: – 0.1 * m9 > 372295.963387360446 + npr1 + m1



15. modification – multiplication by -10 reverses the inequation-sign

a) expendable Y – corrected new version

aa) n1 = 2 np = 13.4 mio

Falcon 1: m9 < - 24357103.63387360446 - 10 * npr1 – 10 * m1

ab) n1 = 3 np = 20.1 mio

Falcon 1: m9 < - 3722959.63387360446 – 10 * npr1 – 10 * m1




As inequations these are telling anything only if assumptions about npr1 and m1 are applied. And since npr1 allways is positive it allows for a positive m9 only if m1 is negative which means that it includes a loss.

Because of this the Cash Flow-equation as well as the resulting equations above now must be turned into real equations very cautiously.

The Cash-Flow-INequation applied is

    a) expendable Y – corrected new version

    aa) n1 = 2 np = 13.4 mio

    Cash Flow: Y < 493354.7 - npr1

    ab) n1 = 3 np = 20.1 mio

    Cash Flow: Y < 2556769.1 - npr1

The Falcon 1-equation got is

    a) expendable Y – corrected new version

    aa) n1 = 2 np = 13.4 mio

    Falcon 1: m9 < - 24357103.63387360446 - 10 * npr1 – 10 * m1

    ab) n1 = 3 np = 20.1 mio

    Falcon 1: m9 < - 3722959.63387360446 – 10 * npr1 – 10 * m1

Y can’t be negative:

aa) 0 < Y < 493354.7 - npr1
ab) 0 < Y < 2556769.1 - npr1

So

aa) 0 < npr1 < 493354.7
ab) 0 < npr1 < 2556769.1

is valid.

Because of m1 = npr1 + p1 the Falcon 1-equation can be written as

    a) expendable Y – corrected new version

    aa) n1 = 2 np = 13.4 mio

    Falcon 1: m9 < - 24357103.63387360446 - 10 * npr1 – 10 * (npr1 + p1)

    ab) n1 = 3 np = 20.1 mio

    Falcon 1: m9 < - 3722959.63387360446 – 10 * npr1 – 10 * (npr1 + p1)



    a) expendable Y – corrected new version

    aa) n1 = 2 np = 13.4 mio

    Falcon 1: m9 < - 24357103.63387360446 - 10 * npr1 – 10 * npr1 – 10 * p1

    ab) n1 = 3 np = 20.1 mio

    Falcon 1: m9 < - 3722959.63387360446 – 10 * npr1 – 10 * npr1 – 10 * p1



    a) expendable Y – corrected new version

    aa) n1 = 2 np = 13.4 mio

    Falcon 1: m9 < - 24357103.63387360446 - 20 * npr1 – 10 * p1

    ab) n1 = 3 np = 20.1 mio

    Falcon 1: m9 < - 3722959.63387360446 – 20 * npr1 – 10 * p1

Insertion of npr1 results in

    a) expendable Y – corrected new version

    aa) n1 = 2 np = 13.4 mio

    Falcon 1: aaa) m9 < - 24357103.63387360446 - 20 * 0 – 10 * p1
    aab) < - 24357103.63387360446 - 20 * 493354.7 – 10 * p1

    ab) n1 = 3 np = 20.1 mio

    Falcon 1: aba) m9 < - 3722959.63387360446 – 20 * 0 – 10 * p1
    aba) m9 < - 3722959.63387360446 – 20 * 493354.7 – 10 * p1



    a) expendable Y – corrected new version

    aa) n1 = 2 np = 13.4 mio

    Falcon 1: aaa) m9 < - 24357103.63387360446 – 10 * p1
    aab) m9 < - 24357103.63387360446 - 9867094 – 10 * p1

    ab) n1 = 3 np = 20.1 mio

    Falcon 1: aba) m9 < - 3722959.63387360446 – 10 * p1
    aba) m9 < - 3722959.63387360446 – 51135382 – 10 * p1




    a) expendable Y – corrected new version

    aa) n1 = 2 np = 13.4 mio

    Falcon 1: aaa) m9 < - 24357103.63387360446 – 10 * p1
    aab) m9 < - 34224197.63387360446 – 10 * p1

    ab) n1 = 3 np = 20.1 mio

    Falcon 1: aba) m9 < - 3722959.63387360446 – 10 * p1
    abb) m9 < - 54858341.63387360446 – 10 * p1

To get an m9 of 0 in case of making the inequations equations the following results are got:

    a) expendable Y – corrected new version

    aa) n1 = 2 np = 13.4 mio

    Falcon 1: aaa) p1 = - 2435710.363387360446
    aab) p1 = - 3422419.763387360446

    ab) n1 = 3 np = 20.1 mio

    Falcon 1: aba) p1 = - 372295.963387360446
    abb) p1 = - 5485834.163387360446


So m9 = 0 requires a loss of between $ 373,000 and $ 5.5 mio per Falcon 1-launch. m9 = 0 is the case where a Merlin (X) requires an investment of $ 2,875,554.2351337147.

Are such losses likely? The least one is limited but still would mean a reduction of the capital invested into SpaceX launch by launch of a Falcon 1 – the highest one would mean a quick destruction of SpaceX if there aren’t soon and permanent launches of the other rockets. If a positive m9 is supposed the losses per Falcon 1-launch will be higher by up to $ 1,548,141.0574563466 if all of m9, m9h and m5 are supposed to be NOT negative. But the launches all together may sum up to a profit. Under this aspect the losses might be correct.

At this point it should be mentioned what the upper boundary of the investment into a Merlin (X) under the assumption that all m9, m9h and m5 are positive results in – an investment into one Falcon 9 Heavy of $ 80,515,518.5837440116 without the aluminium. This is by $ 2.5 mio higher than the launch price of $ 78 mio listed last year. So there is something wrong. The reason may be

1) erroneously applied/inserted wrong number
2) undetected error during modifications
3) too high salaries assumed
4) the application of economical impossibilities
5) improper employee ratios

May be that the yet to be calculated alternative sets of numbers will lead to more proper results.

The results here also don’t accord to what has been said at the beginning of the previous post – except for the result aba) where the loss per Falcon 1-launch is less than $ 400,000 and thus marginal in comparison to the investment into the Merlin (X).

It’s looking a bit as if conclusions to the salaries are possible.

On the other hand it might be perhaps that the idea that all margins must be positive really is not correct. In particular it might be that m9 can or will be negative – because then m1 can be positive. This would mean that Falcon 1-launches yield a profit. Then the investment into the Merlin (X) would be higher (although invalid here because of something wrong or improper). At an m9 < 0 m9h would be positive. Please refer to tables about links between the margins and links between m9 and X (Merlin) posted earlier.

But at present the view is within the equation system which among others is based on Elon Musk’s issue that the present prices do NOT account for reusability although he is out on reusability. This means that a wider view must be applied for a short moment now. The prices don’t account for reusability but the rockets are intended to be reusable. If the reusability-technology works then the prices depreciate the Merlin (X) by one single flight – which means that beginning at the second flight that portion of the price that is caused by not-accounting-for-reusability really is a revenue. And this revenue may well include a profit!

If a margin is negative then it simply must be subtracted from that revenue/profit. Then at the minimum p1 and npr1 close to zero there will be a profit of close to $ 2 mio per Falcon 1-launch. Next at that minimum p1 and npr1 below $ 2 mio there still will be a profit per Falcon 1-launch.

This means that perhaps the first flight only might yield a loss.

Another important point also is that the larger npr1 is supposed to be and the larger the refurbishment-portion of this large npr1 (< 2556769.1) is supposed the more this portion is turned into a revenue if the rocket fails and is lost. Of course this does NOT mean a profit – but since it will not be spent for refurbishments in such a case it can be used to finance the replacement of the lost rocket after subtraction of the loss included in the price. This contributes to keeping a positive Cash Flow even in the case of the loss of the rocket.

But before looking into details applying the wider view first some critical tasks need to be done within the narrower view – during the thread the following tasks have been listed:

Quote:
So I consider the manufacturering-team to be the only one growing here.
Like I already said earlier I have in mind to apply alternative
numbers/values also to check what changes then and by how much it does
change.


Quote:
It also is looking as if the equations are adjusted to the seond version of the launch manifest at present because of applying L2008 for the Falcon 9 – I have to check that.


Quote:
So after getting results from this L2008 has to be replaced by L2007 in the Falcon 9-equation(s) and two alternative results have to be calculated per case – four results to be expected if I haven’t forgotten something. Next additional four results might be got by replacing the L2006 etc. by L2006Q1 etc.


…
…
…

Finally it seems to be unclear yet what the relation or function between X and Y is in the calculations done up to now. The Falcon 1-equation is

Falcon 1: X = 6677536.930521075146 – 0.03625 * 24080486.40 - Y + m1
= 5804619.298521075146 – Y + m1

.
So it’s clear that the Merlin (X) and the Kestrel (Y) are negatively linked – the higher the investment into the Merlin the smaller that into the Kestrel. The result for the Kestrel (Y) was got from the Cash Flow-equation and is

Cash Flow: Y < 2556769.1 - npr1

.

So the profit p1 obviously doesn’t have no impact on the investment into the Kestrel. The maximum investment into it is $ 2,556,769.1 – then the minimum investment into the Merlin (X) is got by inserting the maximum Y:

Falcon 1: X = 5804619.298521075146 – 2556769.1 + m1
= 3247850.198521075146 + m1

.

At the maximum investment into the Kestrel (Y) npr1 would be 0. Above it turned out that p1 must be a loss and thus negative because of npr1 = 0 p1 = m1 is valid – and the minimum X is higher than the maximum Y. In other words the Kestrel always is a smaller investment than the Merlin – if the current calculations and results are kept including the errors etc.

May be that later much more can be seen when better tables are available via Excel.

What I didn’t think about yet are npr9, npr9h and npr5. They might be concluded from npr1. I am not sure if the number of engines or the surface or a combination of them would be the best to be applied. It seem to be clear though that the higher npr1 the worse the number of engines as criterion because a built-in loss might be got that never might be counteracted by reusability – I have in mind to so calculations of this but don’t know if I will do them in the next post or later.

The question is what needs refurbishment – the engine(s) or the surface. This moment I prefer the surface as criterion. The surfaces of the Falcons are

Code:
Falcon           surface   stages   ratio to Falcon 1   npri

Falcon 1           80.11      1        1                npr1  =  1   * npr1
Falcon 5          531.56      2        6.6              npr5  =  6.6 * npr1
Falcon 9          599.42      2        7.5              npr9  =  7.5 * npr1
Falcon 9 Heavy   1197.70      4       15                npr9h = 15   * npr1


This list seems to add another constraint having an effect on earlier results because of the upper boundary of npr1 – I will consider it later.



Dipl.-Volkswirt (bdvb) Augustin (Political Economist)


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Post    Posted on: Tue Nov 13, 2007 12:29 pm
Something I forgot to post..

SpaceX updated their homepage with new performance data from Cape Canaveral for Falcon 9, Falcon 9 Heavy.

Image

http://www.spacex.com/falcon9.php
http://www.spacex.com/falcon9_heavy.php

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Post    Posted on: Tue Nov 13, 2007 12:40 pm
Hello, Klaus,

I am working on the Excel spreadsheet already and provide the possibility to insert alternative data.

So the data you are linking to can be applied easyly. I up to now apply the data from 2005/2006 but will apply the actual data later I think at present.

Regarding my previous post one remark seems to required - the variable non-propellant costs npr NOT NECESSARYLY are refurbishment costs and if they are really not it should be doubted if the aurface area or the number of engines are proper criterions to conclude from npr1 to npr9, npr9h and npr5.

As long as the prices don't account for reusability I am doubting much if the nprs got here include refurbishment costs because the non-accounting for reusabilty means that a complete replacement of the engines is supposed to apply a pricing that includes self-insurance. If the engines are NOT lost then the investment into the engines included in the prices is turned into refurbishment costs to any degree. ...



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