Computer Cores Processing Power and Jump Control Software

[snrdg121408]

Morning all,

Looking over the details on MgT HG 2e PDF page 18 in the right column below Computer Core Table the first sentence is
"The Processing score for a computer core is in addition to the processing power needed for Jump Control programs and all Jump Control software is included in the price of the core."

A Computer Core/40 has a basic processing score of 40 without all of the Jump Control software bandwidth requirements.

The software list from CRB 2e PDF page 151 has the maximum bandwidth for the jump control software as 30 for a 6 parsec trip.

Would the total processing score for a Computer Core/40 be 70 since the lower Jump Control would be using only a small fraction of the processing score?

Another question concerning the Jump Control Software follows since HG 2e PDF page 14 allows a jump drive to make a maximum jump of 9 parsecs at TL 18.

Extending the Jump Control Software I believe the J7, J8, and J9 data would be:

Jump Control/7: TL 16; Bandwidths 35: Cr700,000
Jump Control/8: TL 17; Bandwidths 40: Cr800,000
Jump Control/9: TL 18; Bandwidths 45: Cr900,000

Are the numbers correct?

The maximum bandwdith, if I'm even close, would be 45 for a J9 drive which would give a Computer Core/40 a total processing score of 85.

How far out to lunch am I on this one?

 

[Condottiere]

Probably one of those issues that tends to get spread under the rug.

There is considerable difference between jump control one and six programmes, and technological level(s) availability might have been a natural cap on core computers' potential processing power; it may be that I had been lulled by previous editions that implied you don't have to pay for the jump control programmes, but I would have to look that up.

Since a clear value has been given to the processing power needed for each programme, and how much capacity all computers actually have, maybe you can make cores super bi.

 

[AnotherDilbert]

"The Processing score for a computer core is in addition to the processing power needed for Jump Control programs and all Jump Control software is included in the price of the core."
...
Would the total processing score for a Computer Core/40 be 70 since the lower Jump Control would be using only a small fraction of the processing score?

No, Jump Control is simply free for Core computers, both in cost and to run.

A Core/40 has processing/bandwidth 40, as specified.

Extending the Jump Control Software I believe the J7, J8, and J9 data would be:

Jump Control/7: TL 16; Bandwidths 35: Cr700,000
Jump Control/8: TL 17; Bandwidths 40: Cr800,000
Jump Control/9: TL 18; Bandwidths 45: Cr900,000

That would be logical, but J-7+ is not specified in MgT.

 

[Commander Zoom]

Would the total processing score for a Computer Core/40 be 70 since the lower Jump Control would be using only a small fraction of the processing score?

If you need a rational explanation for why the Core/40 only has a bandwidth of 40 even though it can run the jump software (30 bandwidth) for free, I'd say that the portion of the Core that runs the jump software consists of specialized hardware/processors/architecture that ONLY runs the jump software - one analogy is the bis systems, which have a specialized function to allow higher jumps.

Specialized processors exist in-game, so the Core computer systems might simply be another instance of such specialization.

 

[AndrewW]

Jump Control software could be hard coded into the Core computers and not require other resources such as memory.

 

[Condottiere]

Which would mean that a starship with a Core Computer factor Forty manufactured at technological level nine could perform a jump factor six, presuming they somehow got hold of a jump drive factor six.

 

[locarno24]

You definitely don't have to allow money or bandwidth but I'm not certain - it certainly doesn't say - that you can ignore Tech Level.

Nominally, you could claim an 'early prototype' (which has a x10 cost increase you don't have to pay) to field any piece of Jump Control software two TL early, but I'm not sure you should allow more than that.///

 

[snrdg121408]

Hello Condottiere,

Probably one of those issues that tends to get spread under the rug.

There is considerable difference between jump control one and six programmes, and technological level(s) availability might have been a natural cap on core computers' potential processing power; it may be that I had been lulled by previous editions that implied you don't have to pay for the jump control programmes, but I would have to look that up.

CT LBB 2 1977/1981 page 41 Computer Software List Routine Programs
Jump 1: Space = 1; MCr =0.1
Jump 2: Space = 2; MCr =0.3
Jump 3: Space = 3; MCr =0.4
Jump 4: Space = 4; MCr =0.5
Jump 5: Space = 5; MCr =0.6
Jump 6: Space = 6; MCr =0.7

MT and TNE I could not find a software list though there is mention for computer programing.

T4 Core Rulebook page 90 indicates that a ship's computer has all the required software installed and additional software can be found on page 105.

T20 The Traveller's Handbook 2002 3rd printing page 232
Jump 1: PP = 23; Cost = 100,000
Jump 2: PP = 30; Cost = 300,000
Jump 3: PP = 37; Cost = 400,000
Jump 4: PP = 41; Cost = 500,000
Jump 5: PP = 44; Cost = 600,000
Jump 6: PP = 52; Cost = 700,000

Since a clear value has been given to the processing power needed for each programme, and how much capacity all computers actually have, maybe you can make cores super bi.

The source material indicates that the jump programs are already integrated which is the case for the maneuver drive program.

 

[AnotherDilbert]

You definitely don't have to allow money or bandwidth but I'm not certain - it certainly doesn't say - that you can ignore Tech Level.

You need TL15 to develop Jump Control-6, since you need to understand Jump-6, but you don't need a TL15 computer to run it, just a computer with enough "bandwidth".

It you can build a Jump-6 drive, you can develop Jump Control-6, so I don't think it is much of a practical problem.

 

[snrdg121408]

Morning from the Pacific Northwest AnotherDilbert,

"The Processing score for a computer core is in addition to the processing power needed for Jump Control programs and all Jump Control software is included in the price of the core."
...
Would the total processing score for a Computer Core/40 be 70 since the lower Jump Control would be using only a small fraction of the processing score?

No, Jump Control is simply free for Core computers, both in cost and to run.

A Core/40 has processing/bandwidth 40, as specified.

The partial text of"...in addition to the processing power needed for Jump Control programs..." suggests to me that at TL 9 the Jump Control/1 software with a bandwidth of 5 is automatically installed on a TL 9 Computer Core/40 which has processing power of 5 plus the processing power capability of running number programs with maximum bandwidth of 40. A TL 9 Computer Core's total processing power is 45.

I agree that the cost of the Jump Control software is subsumed into the Computer Core's price.

]

Extending the Jump Control Software I believe the J7, J8, and J9 data would be:

Jump Control/7: TL 16; Bandwidths 35: Cr700,000
Jump Control/8: TL 17; Bandwidths 40: Cr800,000
Jump Control/9: TL 18; Bandwidths 45: Cr900,000

That would be logical, but J-7+ is not specified in MgT.

HG 2e Jump Potential Table lists J-7+ which in my opinion allows MgT the option to expand beyond established J-6 limit from the previous versions. The exception of GT which allows the development of jump drives that exceed J-6 using GURPS Vehicles.

 

[snrdg121408]

Hello Commander,

Thank you for your post.

Would the total processing score for a Computer Core/40 be 70 since the lower Jump Control would be using only a small fraction of the processing score?

If you need a rational explanation for why the Core/40 only has a bandwidth of 40 even though it can run the jump software (30 bandwidth) for free, I'd say that the portion of the Core that runs the jump software consists of specialized hardware/processors/architecture that ONLY runs the jump software - one analogy is the bis systems, which have a specialized function to allow higher jumps.

Specialized processors exist in-game, so the Core computer systems might simply be another instance of such specialization.

Thanks to AnotherDilbert I may have figured out the annotation for computer core processing power. The processing power for a computer core is that available bandwidth that can be processed after jump control software has been installed.

At TL 9 a star ship hull with TL 9 J-1 drive with a TL 9 Computer Core/40 has the ability to process a number of other programs requiring a total bandwidth of 40 plus/in addition to the installed Jump Control/1 software's bandwidth of 5.

 

[snrdg121408]

Howdy AndrewW,

Jump Control software could be hard coded into the Core computers and not require other resources such as memory.

The text "...is in addition to the processing power needed for Jump Control programs..." does not suggest to me that the software is hard coded or is not using memory.

 

[snrdg121408]

Hi locarno,

You definitely don't have to allow money or bandwidth but I'm not certain - it certainly doesn't say - that you can ignore Tech Level.

Nominally, you could claim an 'early prototype' (which has a x10 cost increase you don't have to pay) to field any piece of Jump Control software two TL early, but I'm not sure you should allow more than that.///

The HG 2e text of "...and all Jump Control software is included in the price of the core..." strongly suggests that the program costs are already included in a computer core's price.

The text of "The Processing score for a computer core is in addition to the processing power needed for Jump Control programs,..." suggests that the Jump Control programs do use bandwidth that a computer core's processing score has included plus the capability for processing a number of programs that require a maximum bandwidth of 40.

HG 2e Chapter 4 Primitive and Advanced Spacecraft page 48 on Prototype/Advanced Table shows that an Early Prototype has a cost modifier of +1000%.

 

[AnotherDilbert]

The partial text of"...in addition to the processing power needed for Jump Control programs..." suggests to me that at TL 9 the Jump Control/1 software with a bandwidth of 5 is automatically installed on a TL 9 Computer Core/40 which has processing power of 5 plus the processing power capability of running number programs with maximum bandwidth of 40. A TL 9 Computer Core's total processing power is 45.

Note that a Core/40 can be produced at any TL≥9 or presumably TL≥7 with Prototech (Core p107).

A TL15 Core/40 includes Jump Control-6. It still has bandwidth 40, regardless of whether Jump Control is running or not.

 

[snrdg121408]

Hello AnotherDilbert,

The partial text of"...in addition to the processing power needed for Jump Control programs..." suggests to me that at TL 9 the Jump Control/1 software with a bandwidth of 5 is automatically installed on a TL 9 Computer Core/40 which has processing power of 5 plus the processing power capability of running number programs with maximum bandwidth of 40. A TL 9 Computer Core's total processing power is 45.

Note that a Core/40 can be produced at any TL≥9 or presumably TL≥7 with Prototech (Core p107).

Thank you for the reference, which I forgot about, to the Computer Variants on CRB 2e page 107. I will have to add that to my spreadsheet.

CRB 2e page 107 Computer Variants
"A computer's optimum Technology Level is the level at which it reaches standard production and is no longer considered a prototype or experiment."

My understanding is that a Computer Core/40 is considered to be standard at TL 9. A Prototech model can be produced a TL 10 for 10x the cost and a TL 7 model would be 100x the cost.

Following the CRB 2e page 107 text "RetroTech: While computers may be built at, and rated as any TL above the optimum TL, a higher TL society may produce a lower rated TL System for reduced cost and weight. Any system may be built below the society's current level, but not below the optimum TL for the model. Each reduced level halves the cost and weight of the basic model.

If a TL 15 society produces a RetroTech TL 9 Computer Core/40 the 45,000,000 cost is reduced by half per optimum TL.
TL 15 to TL 14 = 45,000,000 x 0.5 = 22,500,000
TL 14 to TL 13 = 22,500,000 x 0.5 = 11,250,000
TL 13 to TL 12 = 11,250,000 x 0.5 = 5,625,000
TL 12 to TL 11 = 5,625,000 x 0.5 = 2,812,500
TL 11 to TL 10 = 2,812,500 x 0.5 = 1,406,250
TL 10 to TL 9 = 1,406,250 x 0.5 = 703,125

The cost to produce a RetroTech Computer Core/40 is Cr703,125.

A TL15 Core/40 includes Jump Control-6. It still has bandwidth 40, regardless of whether Jump Control is running or not.

A RetroTech TL 15 Computer Core/40 is programed to process the Jump Control (JC)-6 software Bandwidth of 30, which is also capable of running JC-1 through 5, in addition this core computer can install software that has a combined total of bandwidth of 40.

Does that mean when a Computer Core/40 has loaded programs having a total bandwidth of 40 that are not running the computer core still has a processing score of 40?

At TL 15 a computer core has the processing score of 30 to run the Jump Control/6 program's bandwidth 30 requirement plus a processing score of 100 to run any number of programs with a combined total bandwidth of 100.

 

[Condottiere]

I had developed an algorithm to calculate the cost of each computer at each technological level per bandwidth/capability, for the previous edition, but I recall I never managed to perfect it, since it had to fit with existing canon cost/capability at each technological level, and scale up and down in cost if you wanted more or less capability.

 

[locarno24]

You definitely don't have to allow money or bandwidth but I'm not certain - it certainly doesn't say - that you can ignore Tech Level.

You need TL15 to develop Jump Control-6, since you need to understand Jump-6, but you don't need a TL15 computer to run it, just a computer with enough "bandwidth".

It you can build a Jump-6 drive, you can develop Jump Control-6, so I don't think it is much of a practical problem.

That's pretty much what I was meaning - it's more for an issue for in-game narrative than out-of-game design mechanics - since if you're buying an off-world TL15 Jump-6 to attach for a ship, then yes, logically you must have access to TL15 'stuff'. With a core-type computer I agree you don't need to explicitly assign either cost or bandwidth to it provided you can say where you got it from.

 

[AnotherDilbert]

Does that mean when a Computer Core/40 has loaded programs having a total bandwidth of 40 that are not running the computer core still has a processing score of 40?

Yes, processing score describes how much software can be run concurrently.

All software has a Bandwidth score, which represents the processing power a computer requires in order to run it. At any one time, a computer can run a number of software packages whose combined Bandwidth does not exceed its Processing score.

Storage, how many programs you can have installed, is "big enough" to install as much software as you like:

Note that the storage capacity of computers is not used in Traveller as, beyond TL8, the capacity of even a modest computer is effectively unlimited.

The only anomaly is that Jump Control requires bandwidth 0 to run on a Core computer.

 

[snrdg121408]

Almost noon from the Pacific Northwest Another Dilbert,

Does that mean when a Computer Core/40 has loaded programs having a total bandwidth of 40 that are not running the computer core still has a processing score of 40?

Yes, processing score describes how much software can be run concurrently.

All software has a Bandwidth score, which represents the processing power a computer requires in order to run it. At any one time, a computer can run a number of software packages whose combined Bandwidth does not exceed its Processing score.

Storage, how many programs you can have installed, is "big enough" to install as much software as you like:

Note that the storage capacity of computers is not used in Traveller as, beyond TL8, the capacity of even a modest computer is effectively unlimited.

The only anomaly is that Jump Control requires bandwidth 0 to run on a Core computer.

Thank you for referencing the Core Rule Book (CRB) which I dug through after my last post, which would have answered my questions. Unfortunately, the only review I did when starting work on the Fessor was to compile a software list from the CRB 2e, HG 2e, and Central Supply Catalog (CSC).

There is no clear indication that the Jump Control programs have a bandwidth of zero.

HG 2e page 18 second column below the Computer Cores Table.

"The Processing score for a computer core is in addition to the processing power needed for Jump Control programs, and all Jump Control software is included in the price of the core. Other ship software must be added at extra cost as normal."

From the text the indication is that a computer core has processing power/score dedicated to run only the jump control software. The processing power/score listed for a particular computer core is available for use by any other software.

On my copy of CRB 2e PDF page 106 lists that the following software is included as basic package: Interface and Security/0. CRB 2e PDF page 151 adds two more to the list which are Maneuver/0 and Library. These four programs are clearly shown not to require bandwidth or have a cost in credits.

 

[snrdg121408]

Hi again Condottiere,

I had developed an algorithm to calculate the cost of each computer at each technological level per bandwidth/capability, for the previous edition, but I recall I never managed to perfect it, since it had to fit with existing canon cost/capability at each technological level, and scale up and down in cost if you wanted more or less capability.

I wished you had perfected the algorithm have you consulted with AnotherDilbert. AnotherDilbert has ship design spreadsheet over on the Citizens of the Imperium boards that make adjustments for various changes that are very good.