Supplemental File S1. Analysis of the effect of pathway length on the predicted flux
distribution in a small metabolic network.
Using 1000 different sets of randomly chosen weighting factors and averaging the flux
solutions, cost-weighted FBA of the small metabolic network shown in Fig. S1 predicted
fluxes of approximately one third through R1, R2 and R3, five sixths through R4 and one
sixth through R5 and R6.
R4
R1
W_in
W
R2
X
Z
R5
R3
Z_out
R6
Y
Figure S1. Model for investigating the utility of the cost-weighted flux minimisation objective
function. There are three equivalent pathways between W and X, and two inequivalent
pathways between X and Z – one involving a single step and the other requiring two steps.
In conventional FBA, flux minimization would predict that R4 would be the optimum route
between X and Z. In contrast, cost-weighted FBA considers the possibility that the cost of
providing the two pathways between X and Z may not be directly proportional to the number
of steps. Assigning random values to the unknown costs and averaging the results of multiple
simulations highlights the possibility that the network could use the longer pathway (R5/R6)
in preference to the shorter pathway (R4). The likelihood of this outcome is equal to the
probability that the sum of two random numbers is less than a third random number. The two
proofs below show that this is 1/6 in agreement with the simulations.
Proof 1
Let X, Y and Z be the random variables and assume that they each have a uniform probability
density in the interval [0,1]. Each allowed combination lies within a cube of side 1 (Fig. S2):
Figure S2. The geometric proof: the volume within the bold lines is proportional to the
probability that the sum of two random numbers is less than a third random number.
1
If the sum of the values of X and Y is less than the value of Z, then the values must lie within
the volume defined by the bold lines. For example if z = 1, then x and y must lie with the
triangle defined by y = 1 – x and the x and y axes. More generally, for any value of z, x and y
must lie within the triangle defined by y = z – x and the x and y axes.
The area of this triangle is ½z2, because the maximum values of x and y are both z, and the
volume of the enclosed space equals:
1
1
1
 2 z dz  6
2
0
Thus the probability that reactions R5 and R6 will be used in preference to R4 in costweighted FBA is 1/6.
Proof 2
The probability density function - fZ(z) - of the sum (Z) of two continuous random variables
X and Y, with probability density functions fX(x) and fY(y) respectively, is given by:
f Z ( z )=
¥
ò f ( z - y) f ( y) dy
X
Y
-¥
This is a standard result which states that the probability density function of the sum is the
convolution of the probability density functions of X and Y; for a proof of this result, see
http://www.statisticalengineering.com/sums_of_random_variables.htm
Step 1 Calculating the density function for the sum of two random variables
Let X and Y be the random variables and assume that they each have a uniform probability
density in the interval [0,1], i.e.
f X ( x )= fY ( y ) = 1
if 0 ≤ x ≤ 1, or 0 outside these limits.
The density function for the sum is given by:
f Z ( z )=
¥
ò f ( z - y) f ( y) dy
X
Y
-¥
but since fY(y) equals 0 unless 0 ≤ y ≤ 1 the convolution simplifies to:
2
1
fZ ( z )= ò f X ( z - y ) dy
0
The integrand disappears unless 0 ≤ (z – y) ≤ 1 when fX(z - y) = 1.
The allowed range of z is between 0 and 2 (the minimum sum of the two random numbers is
0 and the maximum is 2) and the inequality for (z – y) can be satisfied as follows.
If 0 ≤ z ≤ 1 then y can only take values between 0 and z, and the density function reduces to:
z
fZ ( z )= ò dy = z
0
If 1 ≤ z ≤ 2 then y can only take values between z-1 and 1, and the density function reduces
to:
f Z ( z )=
1
ò dy = 2 - z
z-1
fZ(z) is zero for all other values of z and the sum of the distributions can be represented as
below:
Step 2 Calculating the density function corresponding to the subtraction of a random number
from the sum of two random numbers
This can be visualized as follows:
3
where the shape of the probability density function for T is unknown.
The density function for the sum is given by:
fT ( t )=
¥
ò f (t - s ) f ( s ) ds
R
S
-¥
but since fS(s) equals 0 unless -1 ≤ s ≤ 0 the convolution simplifies to:
0
fT ( t )= ò fR ( t - s ) ds
-1
From step 1, fR(t-s) = t-s if 0 ≤ t-s ≤ 1 and fT(t) can be evaluated by working out the allowed
values of s.
If -1 ≤ t ≤ 0 then s can only take values between -1 and t, and the density function reduces to:
t
t
é
s2 ù
fT ( t )= ò ( t - s ) ds = êts - ú = 0.5t 2 + t + 0.5
2 û1
ë
-1
If 0 ≤ t ≤ 1 then s can only take values between t-1 and 0, and the density function reduces to:
0
é
s2 ù
fT ( t )= ò ( t - s ) ds = êts - ú = -0.5t 2 + 0.5
2 ût-1
ë
t-1
0
If 1 ≤ t ≤ 2 then there are no allowed values of s and the contribution to the density function
is zero.
Also from step 1, fR(t-s) = 2-(t-s) if 1 ≤ t-s ≤ 2 and the contributions to fT(t) in this range can
be evaluated in the same way as before by working out the allowed values of s.
4
If -1 ≤ t ≤ 0 then there are no allowed values of s and the contribution to the density function
is zero.
If 0 ≤ t ≤ 1 then s can only take values between -1 and t-1, and the density function reduces
to:
t-1
t-1
é
s2 ù
fT ( t )= ò ( 2 - t + s ) ds = ê(2 - t)s + ú = -0.5t 2 + t
2 û-1
ë
-1
If 1 ≤ t ≤ 2 then s can only take values between t-2 and 0, and the density function reduces to:
0
é
s2 ù
fT ( t )= ò ( 2 - t + s ) ds = ê(2 - t)s + ú = 0.5t 2 - 2t + 2
2 ût-2
ë
t-2
0
Note that the two contributions to the density function for the range 0 ≤ t ≤ 1 have to be
summed, so the complete the density function corresponding to the subtraction of a random
number from the sum of two random numbers is:
fT ( t )= 0.5t 2 + t + 0.5
if -1 ≤ t ≤ 0
fT ( t )= -t 2 + t + 0.5
if 0 ≤ t ≤ 1
fT ( t )= 0.5t 2 - 2t + 2
if 1 ≤ t ≤ 2
The probability density function is zero for all other values of t.
Step 3 Calculating the probability of the sum of two random numbers being less than a third
The probability that the sum of two random numbers is less than a third is the fraction of the
probability distribution function for the sum calculated in Step 2 that corresponds to the range
-1 ≤ t ≤ 0. This can be solved by integration:
0
é t3 t2
ù
1
(0.5t
+
t
+
0.5)dt
=
ê 0.5 3 + 2 + 0.5t ú = 6
ò-1
ë
û -1
0
2
This value is the probability that cost-weighted FBA will use reactions R5 and R6 in
preference to R4.
5