FANDOM


Standing gains with the 6 frenemy Syndicate factions (SM, AH, CS, PS, RV, NL) can be calculated using maths from linear algebra.

Uncapped Standing Gain

Acquiring standing direclty towards one syndicate (by wearing its sigil or redeeming syndicate medallions) will also add half of that to one allied syndicate, substract half to its opposed syndicate and substract the same amount to its enemy (we are not considering standing caps here).

Hence considering the final Standing Gain vector SG with its 6 components related to the standing to each syndicate, we can write SG as a sum of base vectors mutiplied by the standing directly acquired towards one faction, i.e.:

  • SG = a(SM) SM + a(AH) AH + a(CS) CS + a(PS) PS + a(RV) RV + a(NL) NL

where a(X) is the standing acquired directly towards syndicate X and the vector X is the 6 component vector representing the relationship of X to the syndicates (including itself). We can rewrite this in matrix form as :

  • SG = M a

where the vector a = (a(SM), a(AH), a(CS), a(PS), a(RV), a(NL)) represent the standing acquired directly to each syndicate and M is the 6x6 matrix composed by the 6 X vectors:

M SM AH CS PS RV NL
SM 1 0 0 -1 0.5 -0.5
AH 0 1 0.5 -0.5 -1 0
CS 0 0.5 1 0 -0.5 -1
PS -1 -0.5 0 1 0 0.5
RV 0.5 -1 -0.5 0 1 0
NL -0.5 0 -1 0.5 0 1

The order of the columns (or lines) is not important but it is rather useful to have the same order for the lines and columns. In that order, the M matrix is symmetric.

It is found that det(M) = 0 and rank(M) = 5 . That means M is not reversible, so we cannot direclty get a chosen standing gain vector SG and multiply it by M-1 to get the direct standing vector a. It also means that the system equation has one superfluous equation (and unknown). Indeed we can show that :

  • SM + AH + CS + PS + RV + NL = 0

so that, each column (or importantly line) is the opposite of the sums of the 5 others. This has 3 important consequences:

  • knowing the standing gain towards 5 syndicates is enough to know the standing gains towards the 6th.
  • the sum of final standing gain is always 0.
  • acquiring the same amount of direct standing to each syndicate result in a final standing gain of SG=0 (the kernel of M is the line whose base vector is k=(1, 1, 1, 1, 1, 1) )

Critically, that means the same final standing gain SG can be the result of (infinitely) many different direct standing combinations a (for example a and a+k give the same SG )

Therefore the maximum number of syndicate one can increase at the same time is 5. (Although this can be overcome when using the standing caps... see further below)

Inverting the equations

Generically we are now labeling the syndicate vectors as Xi with i= 1,2,.. 6, and the direct standing vector a=(a1,a2,.. a6)

As shown before, the sum of the 6 Xi vectors is 0. We can thus substitute X6 in the SG equation as :

  • SG = (a1-a6) X1 + (a2-a6) X2 + (a3-a6) X3 + (a4-a6) X4 + (a5-a6) X5

which is a 6x5 equation whose 6th line is irrelevant. We can thus reduce the problem to 5x5 sub-matrice M5 by removing the last line and column of matrix M which we can write as

  • SG5 = M5 b5

where SG5 is the 5 component vector (SG1,SG2,.. SG5), and b5=(a1-a6, a2-a6, .. a5-a6)

M5 is always reversible independantly of the chosen 6th syndicate, so for each final stading gain vector SG5 , we can find a unique input vector b5 = M5-1 SG5 .

It is however not direct to recover a from b5 since as we said earlier there are infinite solutions. Hence we need to impose another equation to recover a 6x6 matrix.

We choose here the 6 components vector b = ( b5, T) where T = a1 + a2 + a3 + a4 + a5 + a6 is the total direct standing gain. The vector b is therefore given by :

  • b = K a

where K is the inversible matrix :

K
1 0 0 0 0 -1
0 1 0 0 0 -1
0 0 1 0 0 -1
0 0 0 1 0 -1
0 0 0 0 1 -1
1 1 1 1 1 1

We also construct the 6x6 matrix N as :

N
M5 0
0
0
0
0
0 0 0 0 0 1

and the corresponding 6 component vector NG = ( SG5, T). Hence we will use the total direct standing gain T as the last adjustable parameter.

As K and N are inversible, we can find the direct standing vector for a chosen end vector NG with :

  • a = K-1 N-1 NG = P NG

Note that since NG = ( SG5, T) we can access the infinite space of vectors a that will give the same SG by varying the parameter T. In practice, the vector a has some real-life constraints such as :

  • each component must be positive or nul (you cannot get negative direct standing)
  • T cannot be infinite (and should be minimal to maximize efficiency)

Hence, the optimal aopt to achieve the final standing gain SG can be found by minimizing T while keeping every component of aopt positive or nul.

Keeping in mind that k is a base vector of the kernel of M, we can find aopt using the value T = 0 and substracting the minimal value of the resulting a to the vector a, i.e. :

  • NG0 = ( SG5, 0)
  • a0 = K-1 N-1 NG0 = P NG0
  • aopt = a0 - min(a0)

The total directly standing is therefore Topt = sum(aopt), and the percentage of syndicate support is calculated as aopt x 100 / Topt

Practical case

We choose here the natural order of columns : (SM, AH, CS, PS, RV, NL) and we will therefore discard the NL vector. In this case the P = K-1 N-1 matrix is written as :

P
-15/6 10/6 3/6 -11/6 18/6 1/6
3/6 -14/6 -3/6 -5/6 -18/6 1/6
9/6 10/6 3/6 13/6 6/6 1/6
-15/6 -2/6 3/6 -11/6 6/6 1/6
15/6 -14/6 -3/6 7/6 -18/6 1/6
3/6 10/6 -3/6 7/6 6/6 1/6

As an example, let us assume we want to gain the following standing SG = (10, 0, 4, 2, -9, -7) which is valid as sum(SG) = 0 .

The constructed vector NG = (10, 0, 4, 2, -9, 0) gives the optimal direct standing vector :

  • aopt = P NG - min(P NG) = (0, 82, 66, 18, 106, 50)

Hence, to gain exactly the final standing SG = (10, 0, 4, 2, -9, -7) one needs to acquire a total of 322 direct standing dispatched as aopt = (0, 82, 66, 18, 106, 50)

Increasing 5 syndicates equally

It is possible to increase 5 syndicates equally at the same time, the 6th accumulating the negative standing. It is however very inefficient as most of the standing acquired is lost in balancing losses between ennemy Syndicates

Syndicate SG a percent T Eff
SM -5 33 32.04% 103 4.85%
AH 1 8 7.77%
CS 1 15 14.56%
PS 1 29 28.16%
RV 1 0 0.00%
NL 1 18 17.48%

Syndicate SG a percent T Eff
SM 1 0 0.00% 55 9.09%
AH -5 17 30.91%
CS 1 6 10.91%
PS 1 8 14.55%
RV 1 21 38.18%
NL 1 3 5.45%

Syndicate SG a percent T Eff
SM 1 4 10.81% 37 13.51%
AH 1 3 8.11%
CS -5 10 27.03%
PS 1 0 0.00%
RV 1 7 18.92%
NL 1 13 35.14%

Syndicate SG a percent T Eff
SM 1 21 38.18% 55 9.09%
AH 1 8 14.55%
CS 1 3 5.45%
PS -5 17 30.91%
RV 1 0 0.00%
NL 1 6 10.91%

Syndicate SG a percent T Eff
SM 1 0 0.00% 103 4.85%
AH 1 29 28.16%
CS 1 18 17.48%
PS 1 8 7.77%
RV -5 33 32.04%
NL 1 15 14.56%

Syndicate SG a percent T Eff
SM 1 7 18.92% 37 13.51%
AH 1 0 0.00%
CS 1 13 35.14%
PS 1 3 8.11%
RV 1 4 10.81%
NL -5 10 27.03%

  • the Efficiency Eff is calculated here as the sum of the positive final standing gains over the total acquired standing.

Increasing 1 syndicate with equally spread losses

It may also be beneficial in some cases to increase standing with 1 syndicate but spread the standing losses evenly amongst the other 5. Technically it is the oppposite operation of the previous paragraph, however the direct standing distributions are quite different. It is also very inefficient.

Syndicate SG a percent T Eff
SM 5 0 0.00% 95 5.26%
AH -1 25 26.32%
CS -1 18 18.95%
PS -1 4 4.21%
RV -1 33 34.74%
NL -1 15 15.79%

Syndicate SG a percent T Eff
SM -1 21 29.58% 71 7.04%
AH 5 4 5.63%
CS -1 15 21.13%
PS -1 13 18.31%
RV -1 0 0.00%
NL -1 18 25.35%

Syndicate SG a percent T Eff
SM -1 9 21.95% 41 12.20%
AH -1 10 24.39%
CS 5 3 7.32%
PS -1 13 31.71%
RV -1 6 14.63%
NL -1 0 0.00%

Syndicate SG a percent T Eff
SM -1 0 0.00% 71 7.04%
AH -1 13 18.31%
CS -1 18 25.35%
PS 5 4 5.63%
RV -1 21 29.58%
NL -1 15 21.13%

Syndicate SG a percent T Eff
SM -1 33 34.74% 95 5.26%
AH -1 4 4.21%
CS -1 15 15.79%
PS -1 25 26.32%
RV 5 0 0.00%
NL -1 18 18.95%

Syndicate SG a percent T Eff
SM -1 6 14.63% 41 12.20%
AH -1 13 31.71%
CS -1 0 0.00%
PS -1 10 24.39%
RV -1 9 21.95%
NL 5 3 7.32%


Capped Standing Gain : getting 6 syndicates to max rank

As stated above, it is possible to gain standing to 5 syndicates at the same time, while the 6th syndicate absorb all standing losses. However it is possible to get all 6 syndicates to rank 5 by using the standing caps.

In the table below we see that the maximum cumulative standing is 372,000 and the minimum is -71,000. Since the minium cap is way below the maximum (in absolute value) it is possible to accumulate positive total standing as the negative standing resulting from a direct gain may be aborbed by the negative cap.

Rank Minimum Standing Maximum Standing Cumulative standing
5 0 132,000 372,000
4 0 99,000 240,000
3 0 70,000 141,000
2 0 44,000 71,000
1 0 22,000 27,000
0 -5,000 5,000 0
-1 -22,000 0 -27,000
-2 -44,000 0 -71,000

Hence the absolute total syndicate standing one can accumluate over the 6 syndicates is 5 x 372,000 - 71,000 = 1,789,000 considering 5 syndicates are at the maximum of ank 5 and 1 at the bottom of rank -2.

Starting from that situatoin, increasing the 6th syndicate while dispatching evenly the losses amongst the others as described above will ultimately spread the total stadning amongst the 6 syndicates to 1,789,000/6 = 298,166.67 (*). In the process the 6th syndicate will gain 227,166,67 standing, while the other 5 will lose 73,833.33 standing, leading all 6 to have 58,166.67 standing in rank 5.

This situation is however not stable, for the following reasons :

  • Any purchase from syndicate offerings will lead to a loss of total standing
  • Any direct standing gain (coming from sigils or medaillions) will leave the total standing unchanged

Hence the total standing can only go down as offerings are bought from the syndicates. It follows that once the total standing has gone down, getting back to the situation with every syndicates at the same standing of 58,166.67 at rank 5 can only be done by gaining positive total standing which can only be achieved if at least one syndicate is at the minimum cap.

(* :Note that this absolute value is not strictly achievable in practice as starting from the situation where 5 syndicates are at maximum and 1 is at minimum and trying to increase the latter while spreading losses, some positive standing will be lost in the first standing acquisitions as 5 of the syndicates are already capped.)

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