Help with calculating cargo lots

[Varulv]

I need help interpreting the rules on calculating available freight shipment, CRB page 207-208.

If I roll 5 after DMs on the Freight Traffic table (p 208), the result is 2D lots. If I was checking for Major Cargo the rules states that “Major Cargo lots are composed of 1D x 10 tons of freight”. Do I interpret the rules correctly if I roll 2D, say 2 + 4 = 6 lots, and then roll 1D x 10 for each lot (e.g. 1 x 10, 2 x 10, 3 x 10, 4 x 10, 5 x 10, 6 x 10)?

If I’m correct, that would mean that every 1D lot of major cargo would result in 122.5 tons on average (range 10 to 360 ton). That’s more than a mouthful for a far trader.

 

[Anonymous]

200 ton traders are the smallest traders so it's no surprise they can't handle major lots. There are incidental and minor lots for the smaller ships.

edit: I get a different average but it doesn't change the answer.

 

[AnotherDilbert]

Do I interpret the rules correctly if I roll 2D, say 2 + 4 = 6 lots, and then roll 1D x 10 for each lot (e.g. 1 x 10, 2 x 10, 3 x 10, 4 x 10, 5 x 10, 6 x 10)?

Yes, that is my understanding.

 

[HalC]

While you are correct, what you may be missing is the fact that you may not subdivide those major cargo lots into something smaller. Let's say for the sake of argument, that you have a Far Trader ship. Your ship undertakes to carry a total of Five High Passengers (talk about LUCK!) and you find yourself with a market that has 1d6+2 minor freight lots, and 1d6+1 Major freight lots. Rolling 1d6+2, you end up with say, a 3+2 or 5 minor lots. You also roll a 4 for Major lots, becoming a 5 as well.

Rolling 1d6 five times, the rolls are 4,5,4,6, and 5. That's your minor lots. 6, 1, 6, 1, and 3 are your major lots.

Now, you start with 64 dTons of cargo capacity, but you lose 5 tons due to your High Passage passengers (High Passage passengers get 1 dTon of baggage allowance). Then you find yourself with 20, 25,20, 30, and 25 dTons of freight (minor) along with 60, 10, 60, 10, and 30 dtons of Freight (major).

64-5 = 59 dTons of cargo capacity. Best fit overall will be something like 30 tons (major) and 25 tons (Minor) for a total of 55 dTons of cargo capacity.

The only way to fill the cargo capacity that is less than 5 or less than 10 - is the purchase of speculative cargo, or finding "incidental freight" where you roll 1d6 for how many tons available.

Hope this helps.

 

[Varulv]

Thank you all!

(I calculated the average as average lot size 35 (3.5 x 10) and average number of lots on 1D as 3.5, i.e. 3.5 x 35 = 122.5 tons.)

 

[AnotherDilbert]

... and you find yourself with a market that has 1d6+2 minor freight lots, and 1d6+1 Major freight lots. Rolling 1d6+2, you end up with say, a 3+2 or 5 minor lots. You also roll a 4 for Major lots, becoming a 5 as well.

By RAW the mods apply to the roll on the table, not the roll for number of lots.


For each type of lots (Major, Minor, and Incidental):

1) Roll 2D + Mods on the Freight Traffic Table.

2) Roll nD (as specified in the table) for number of lots.

3) Roll 1D for each lot and multiply by lot size (10 Dt, 5 Dt, and 1 Dt respectively)

 

[Condottiere]

If you're undercapacitated, your broker should be opportunity costing the most profitable cargos.

On the other hand, if you can break up one of the cargoes, say a lot of socks, you could distribute the cartons around the ship, occupying every spare nook and cranny.

 

[Anonymous]

https://en.wikipedia.org/wiki/Knapsack_problem

The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible. It derives its name from the problem faced by someone who is constrained by a fixed-size knapsack and must fill it with the most valuable items.

The decision problem form of the knapsack problem (Can a value of at least V be achieved without exceeding the weight W?) is NP-complete, thus there isno known algorithm both correct and fast in all cases.

While the decision problem is NP-complete, the optimization problem is NP-hard, its resolution is at least as difficult as the decision problem, and there isno known polynomial algorithm which can tell, given a solution, whether it is optimal

A Free Trader isn't going to run into the limits above because there aren't that many boxes so worst come to the worst you can brute force all the permutations (or they have new mathematics at TL 9), but it's interesting as it's a problem that is simple to explain and still has potential for progress.

 

[HalC]

... and you find yourself with a market that has 1d6+2 minor freight lots, and 1d6+1 Major freight lots. Rolling 1d6+2, you end up with say, a 3+2 or 5 minor lots. You also roll a 4 for Major lots, becoming a 5 as well.

By RAW the mods apply to the roll on the table, not the roll for number of lots.


For each type of lots (Major, Minor, and Incidental):

1) Roll 2D + Mods on the Freight Traffic Table.

2) Roll nD (as specified in the table) for number of lots.

3) Roll 1D for each lot and multiply by lot size (10 Dt, 5 Dt, and 1 Dt respectively)

I'm a little confused here. I specified 1d6+2 minor lots, and 1d6+1 major lots as an example. I then rolled 1d+2 getting a 3 and adding the +2 resulted in a 5. 5 minor lots. Ditto with 1d6+1 with a roll of a 4 - also a final result of 5. Five rolls of 1d6 x 5 and Five rolls of 1d6*10.

Incidentally? The Example I chose was straight from the book for line Freight Traffic Value 5 on page 161 of First edition Main Rulebook. I didn't specify what the conditions were that led to selecting value five of the Freight Traffic Value - just the results inherent in it.

It appears that your "correction" seems to be no correction at all, but reiterating the very process that I used. Hence, my confusion...

???

 

[PsiTraveller]

I think the confusion Varulv had in the original post is that there is a lot of cargo available for a trader at every port. A ship with 64 tons of cargo space will have to pick and choose which lots to take to maximize the total amount of cargo carried.

The backpack challenge is to look at all the cargo available and pick which lots will come closest to filling the cargo hold without going over, since you have to take an entire lot and cannot break it up into smaller amounts.

SO my answer to Varulv is yes, there is a lot of cargo available, and in 2nd edition you get paid more to carry it (Yay!), and you have to pick the lots that let you fill your cargo hold to get paid the most.

 

[AnotherDilbert]

I'm a little confused here. I specified 1d6+2 minor lots, and 1d6+1 major lots as an example.

Sorry, I assumed 2nd edition.

I tried to point out that according to MgT2 Core, p208 you can't roll 1D+2 lots, only a straight 1D, 2D, 3D, ..., 10D (without mods).

 

[HalC]

I'm a little confused here. I specified 1d6+2 minor lots, and 1d6+1 major lots as an example.

Sorry, I assumed 2nd edition.

I tried to point out that according to MgT2 Core, p208 you can't roll 1D+2 lots, only a straight 1D, 2D, 3D, ..., 10D (without mods).

Interesting - a change in 2nd edition from first. NICE catch!

I have to wonder what other "little" things got changed overall. Thanks for the update by the way.