Reuse business case

[georgesowers]

As I promised on the Q&A thread, I have posted a simple spread sheet that looks at reuse scenarios parametrically.  I have also posted a white paper that explains the input parameter and math behind the spreadsheet.  Using the tool, I compare two scenarios.  The first is my assessment of full booster stage reuse.  The second is ULA's approach of booster engine reuse.

The figure of merit I chose to evaluate is pure $/kg of the reusable system compared to $/kg of the same system used in expendable mode.  By keeping everything in terms of ratios, you don't need to know anything about actual costs.  I recognize that there are other factors like the quantization of performance and payloads, etc.  (I'll share some of those scenarios later).  But I think this assessment gives you a good intuitive feel for the basic problem and the parameters involved.

An example of where this pure case applies directly is the launch of a commodity like propellant to fill an orbiting stage or depot.  In that case every launcher would load as much as it could carry and $/kg is all that matters.

[docmordrid]

Another factor for the business case: it needs 10 flights/year

SFN....

ULA needs commercial business to close Vulcan rocket  business case
>
The Vulcan rocket must fly at least 10 times per year to keep factory and launch crews operating at the efficiencies needed to reach ULA’s price goal of $100 million per mission....
>

[Lar]

Dr. Sowers: I love the explanation of the spreadsheet (thanks for sharing the sheet and your exposition) but I was unclear as to the k values (k is "the fraction of the total cost of the expendable launch service represented by the production cost of the hardware to be reused." which is C(B)/ C(B) + C(~B) )

You stated for the SpaceX scenario 1 case that you are using .4 for k ("based on internet chatter" ) and that ULA is <.3 but in the ULA scenario 2 case you state k at .18

Could you clarify that? Are you saying that if ULA did a flyback booster you'd have a .3 or less but with the SMART proposal it's .18 or

Also if I understand k, it's the fraction of the total cost of the vehicle that is recovered because that part of the vehicle is reused... internet chatter on NSF seems to use .7 as the fraction of an F9 cost based on Elon saying the first stage is 70% of the cost.

Thanks for shedding light!

[RedLineTrain]

Interesting.  Thank you for posting.  Very sensitive to the performance ratio (p) and recovered production cost (k).

[NovaSilisko]

Lovely info. Hugely appreciate seeing some arithmetic behind the decisions here 

[georgesowers]

Dr. Sowers: I love the explanation of the spreadsheet (thanks for sharing the sheet and your exposition) but I was unclear as to the k values (k is "the fraction of the total cost of the expendable launch service represented by the production cost of the hardware to be reused." which is C(B)/ C(B) + C(~B) )

You stated for the SpaceX scenario 1 case that you are using .4 for k ("based on internet chatter" ) and that ULA is <.3 but in the ULA scenario 2 case you state k at .18

Could you clarify that? Are you saying that if ULA did a flyback booster you'd have a .3 or less but with the SMART proposal it's .18 or

Also if I understand k, it's the fraction of the total cost of the vehicle that is recovered because that part of the vehicle is reused... internet chatter on NSF seems to use .7 as the fraction of an F9 cost based on Elon saying the first stage is 70% of the cost.

Thanks for shedding light!

The booster is <0.3 of our total cost.  The engines, which we are recovering via SMART reuse are 0.18 of the total cost.  I don't think 0.7 can possibly be relative to the total cost including fixed cost and overhead.  It might be relative to the total production cost.  If you use that with some reasonable assumption about other non-hardware costs, 0.4 is appropriate. 

I have a detailed understanding of our cost structure.  I know how much is pads and factory and engineering and management and all those things other than pure hardware.  Even though I don't have any insight into details of SpX, I know they have more as many pads as we do, a factory like we do, a test site we don't have, more employees, etc.  All that has to be accounted for in the cost.

[a_langwich]

Also if I understand k, it's the fraction of the total cost of the vehicle that is recovered because that part of the vehicle is reused... internet chatter on NSF seems to use .7 as the fraction of an F9 cost based on Elon saying the first stage is 70% of the cost.

I think Elon Musk was saying the first stage is 70% of the manufacturing cost for F9.  k is the (manufacturing costs of the reusable portion)'s fraction of the total cost of the entire launch service, which the explanation PDF explicitly said was important to capture.

[Burninate]

This image was posted previously.

[joek]

Thank you very much Dr. Sowers.  This is hard to beat: interesting, informative and fun.  If I am interpreting this correctly, where I < 1 indicates the number of launches required for reuse to become cost effective?  (The figure below is from your scenarios.)  Thanks again

edit: update figure to clarify.

[sublimemarsupial]

"First is that when you reuse hardware, you lower the production rate on the hardware you are building, which results in a higher cost per unit produced."

What happens if this assumption is incorrect? In the case of the Scenario 1 company, their factory is readily set up to convert most all production resources from making the now reusable booster to the expendable second stage - everything from tankage, engines, avionics, tube bending, etc due to the commonality between the two stages. If the resuable booster allows them to increase their flight rate, they may actually end up reducing the overall production costs rather than increasing them.

[su27k]

Looks to me there're two issues with this calculation:
1. It didn't account for the value of the reusable core at the end of the calculation. After N expandable launches, you got nothing left; after N reusable launches, you still got a few cores, these have value. Another way to look at this is: you can expand the reusable core in its last launch, thus gets back some performance.
2. As I understand it, there're a lot of fixed overhead in the business that doesn't change much with regard to the # of launches. For example, assuming p = 2, I can try to launch the reusable twice to match the expandable performance, but then I got hit by twice the C(~B). I'm under the impression that the total cost wouldn't change much if your launch rate changes from once per year to twice per year.

Attached is my attempts to resolve these two issues. (PS: I'm on pain killers, so please forgive me for any arithmetic errors)

[Oli]

Some thoughts on scenario 2:

- I expected the recovered production cost to launch cost ratio to be bigger, but I guess that number is well known.
- A recovery and refurb ratio of 0.1 sounds rather optimistic to me...
- If I understand the F term correctly, 10 reuses (for example) will increase unit cost by a factor 1.42. So if only 1 engine is produced per year instead of 10 (for example), that one engine will only be 1.42x more expensive. Kind of hard to believe.

Other thoughts:

- The analysis assumes the same rocket configuration. However for an expendable rocket one could use arguably cheaper "technologies" like kerosene/solids instead of methane.
- Development cost are only quickly mentioned, but of course a decisive factor for profitability.
- It assumes launch reliability does not decrease with reuse.

Overall I must say that the spreadsheet convinced me that the hopes that are put into reusability in this forum are not justified. It seems to be just another potential cost reduction measure. Certainly not a decisive one.


Before I forget: Thank you very much for all this information, very much appreciated!!

[clongton]

Thank you Dr. Sowers for the spreadsheet and the white paper. Your numbers look promising but I wonder if they will actually come to fruition. Tory Bruno stated that the Vulcan will need to fly a minimum of 10 times per year in order to close the $100 million per flight business case for the Vulcan. Given the available launch market I wonder how you're going to get there. Less than 10 flights per year means that the cost will be over that price point. You will need to fight tooth and nail for commercial launches in the early years to get up to that number of annual flights. Does ULA plan to subsidize those early flights to keep that price point low until the volume reaches 10 per year where the price point can become self-sustaining?

One more thing if I may. In scenario 1 (booster stage re-use) the reuse index drops below 1 at 10 flights and in scenario 2 it drops below 1 at 2 flights. I know that scenario 2 is based on recovery of the Vulcan engine pod. Do I correctly assume that scenario 1 is based on ULA's estimate of recovering the entire Vulcan booster stage? What recovery method would Vulcan use in that scenario; RTLS with extendable legs or SRB-style splashdown & sea recovery? I would think that the RR_ratio would be calculated differently than scenario 2.

[Prober]

This image was posted previously.

you don't have the Delta IV costs/weights do you....the wheels are turning

[gosnold]

From the spreadsheet we can conclude that for Vulcan, the launch cost is 3.25x the cost of the first stage.

ref:
https://twitter.com/torybruno/status/588688390880530432
cost of engine=0.9*0.65*1st stage cost
cost of engine=0.18*launch cost

[a_langwich]

Do I correctly assume that scenario 1 is based on ULA's estimate of recovering the entire Vulcan booster stage?

Based on his quotes above, I think it's pretty clear scenario 1 was meant to track SpaceX recovering its entire first stage.  That is, he's using their stage production cost numbers, and then substituting his experience for the numbers that aren't public to get a ballpark estimate.

[joek]

A couple other thoughts based on this model; hope I got this about right...

1. Vulcan reuse becomes cost effective at two (2) launches; F9 reuse becomes cost effective at ten (10) launches.  That is for the same set of reused components.  E.g., the same engines are reused twice for Vulcan; the same first stage is used ten times for F9 before reuse becomes cost effective.

2. The above provides an approximation of the maximum production (edit: or acquisition) for reuse components.  E.g., for a flight rate of 20/yr over the long term:
- Vulcan reuse is cost effective if no more than 20 engines/yr are produced.
- F9 reuse is cost effective if no more than 2 first stages/yr are produced.
Exceeding those production quantities would incur cost with no benefit and put you below the reuse cost-benefit threshold.

3. The maximum benefit obtainable under these assumptions given virtually unlimited reuse: Vulcan ~10%; F9 ~4%.

4. The Vulcan scenario 2 parameters are presumably with no solids (?).  Adding solids will reduce performance impact (Pe/Pr closer to 1) but also reduce recoverable production cost.  The reuse index plot will be something between that shown for scenario 2 and 1.0.

5. The reuse index I is dimensionless and comparable if costs and cost structure are comparable.  Which reuse approach is most competitive in the market may change significantly based on actual costs of the competitors.  Not that I'm suggesting trying to incorporate that in this model; simply noting that as a competitive analysis this has limits.

6. $/kg is a good FOM (have to base this on something), but there is a step function which is difficult to quantify.  The degenerate case being that there is no performance penalty for payloads below a certain threshold.  The degenerate cases are shown in the figure below as Scenario 1-X and Scenario 2-X (Pe/Pr = 1.0 with all other parameters the same).  Reality is probably somewhere in the regions bounded by the plots for Scenario 1 and Scenario 1-X; and Scenario 2 and Scenario 2-X.

[Billium]

Thank you very much for posting this information. I’m just posting for discussion not asking for a reply.

I agree that item P (which I believe for Spacex is a 30% reduction in mass to orbit penalty for reuse) is appropriate to consider if what is being transported something like propellant, in other words if Spacex is being paid $x for each kg to orbit. However this calculation is not currently applicable to Spacex from a business perspective in my opinion because that is not how they are paid.

For Spacex they already designed their launch system to launch payloads with enough propellant for reuse. So whether or not they expend their launch system they still launch the same payload and still generate the same revenue.

So from a spacex business perspective, given the launcher they have in service, the financial hit that spacex takes for reuse is not 30% of the revenue but only the additional propellant, in other words filling the tank full instead of mostly full. In other words I think $/kg is not the correct measure for the business case here.

In conclusion I think from a spacex business perspective the P value should be 1.02 not 1.42. (I think I have the units right, this is meant to represent the performance loss due to reuse?)

I also agree with Lar that K should likely be closer to .7 not .4, I think the cost of the 1st stage is closer to 70% of the launch costs, not 40% of the launch costs.

I also assume that while the .9 rate exponent may be appropriate for typical aerospace manufacturing, this number may not be the appropriate rate exponent for Spacex. Maybe someone could come up with another number for Spacex (considering the commonality of the upper stage, general manufacturing process and the fact Spacex might need the extra stages anyways for its small sat/mars projects) and then run all of these numbers and see how it looks.

I think the cost of recovery (drone ship ect.) and cost of refurbishment are unknown, it would nice to have that information but we don't have it.

[gosnold]

Thank you very much for posting this information. I’m just posting for discussion not asking for a reply.

I agree that item P (which I believe for Spacex is a 30% reduction in mass to orbit penalty for reuse) is appropriate to consider if what is being transported something like propellant, in other words if Spacex is being paid $x for each kg to orbit. However this calculation is not currently applicable to Spacex from a business perspective in my opinion because that is not how they are paid.

For Spacex they already designed their launch system to launch payloads with enough propellant for reuse. So whether or not they expend their launch system they still launch the same payload and still generate the same revenue.

So from a spacex business perspective, given the launcher they have in service, the financial hit that spacex takes for reuse is not 30% of the revenue but only the additional propellant, in other words filling the tank full instead of mostly full. In other words I think $/kg is not the correct measure for the business case here.

In conclusion I think from a spacex business perspective the P value should be 1.02 not 1.42. (I think I have the units right, this is meant to represent the performance loss due to reuse?)

 I think the spreadsheet is meant to help choose between reusability or not when designing a rocket from scratch, not to evaluate the competition. So from the ULA point of view, the question is do they want to use the SpaceX strategy and make a 30% bigger launcher, or do they want smart reuse?

I also agree with Lar that K should likely be closer to .7 not .4, I think the cost of the 1st stage is closer to 70% of the launch costs, not 40% of the launch costs.

1st stage is 70% of the rocket production cost I think (ie 2nd stage is 30%), but the launch costs include range costs, rocket assembly and checkout, etc...

[ChrisWilson68]

Dr. Sowers: I love the explanation of the spreadsheet (thanks for sharing the sheet and your exposition) but I was unclear as to the k values (k is "the fraction of the total cost of the expendable launch service represented by the production cost of the hardware to be reused." which is C(B)/ C(B) + C(~B) )

You stated for the SpaceX scenario 1 case that you are using .4 for k ("based on internet chatter" :) ) and that ULA is <.3 but in the ULA scenario 2 case you state k at .18

Could you clarify that? Are you saying that if ULA did a flyback booster you'd have a .3 or less but with the SMART proposal it's .18 or ???

Also if I understand k, it's the fraction of the total cost of the vehicle that is recovered because that part of the vehicle is reused... internet chatter on NSF seems to use .7 as the fraction of an F9 cost based on Elon saying the first stage is 70% of the cost.

Thanks for shedding light!

The booster is <0.3 of our total cost.  The engines, which we are recovering via SMART reuse are 0.18 of the total cost.  I don't think 0.7 can possibly be relative to the total cost including fixed cost and overhead.  It might be relative to the total production cost.  If you use that with some reasonable assumption about other non-hardware costs, 0.4 is appropriate. 

I have a detailed understanding of our cost structure.  I know how much is pads and factory and engineering and management and all those things other than pure hardware.  Even though I don't have any insight into details of SpX, I know they have more as many pads as we do, a factory like we do, a test site we don't have, more employees, etc.  All that has to be accounted for in the cost.

It's not valid to include fixed costs as part of k because the model assumes k is fixed even if the number of launches changes.  Since the model is a $/kg to orbit, putting fixed costs in k implies you are comparing $/kg to orbit on the same number of flights, not the same number of kg to orbit.

The $/kg model assumes something like propellant to a depot, where the number of kg is fixed.  For example, if you lose 50% performance for re-use, you would have twice as many flights, and by having fixed costs in k that assumes your fixed costs double also.

The model only makes sense if k includes only marginal costs, not fixed costs.

The conclusions of the model are badly skewed by this error.