Study Case 4.1 Lane’s Algorithm

Variable definition:
Fixed costs … f
Selling price …s
(Recovery … y is 100% and gets ignored for the following calculations)
Mine: Quantity Qm, Capacity M, unit costs m
Plant: Quantity Qp, Capacity P, unit costs p
Beneficiation: Quantity Qb, Capacity B, unit costs b

1)
LOM is 10 years. We mine 65kt/a.
With gc=0.4 → (260+195+32.5)/10 → From 65kt mined (Qm), 48.75 go to the plant (Qp), 16.25 are sent to the dump.
Average grade of the material for the plant is: (26*0.5+19.5*0.7+3.25*0.9)/48.75 = 0.61 kg/t
So, 48.75kt/a*0.61kg/t [of ore] = 29.74 t/a [of Antimony] sent to Beneficiation (Qb)
Profit = Revenues – Costs
Revenues = s * Qb (we sell everything that was refined).
Costs = m*Qm + p*Qp + b*Qb + f(*T)
Profit per year: (s-b)Qb-mQm-pQp-f
= (28-2) €/kg * 29.74 t/a – 2€/t * 65kt/a – 4€/t * 48.75kt/a – 250k€/a
= 26 k€/t * 29.74 t/a – 130 k€/a – 195 k€/a – 250 k€/a = 198.24 k€/a

2)
Economic cut-off grade:
P=sQb-(mQm+pQp+bQb+fT) = (s-b)Qb-mQm-pQp-fT
a) Mining is constraint
T = Qm/M
P=(s-b)Qb-mQm-pQp-f(Qm/M)
P=(s-b)Qb-pQp-(m+f/M)Qm
To find maximum of a function → derive function (after grade) and set zero
First derivation:

The Quantity mined is not linked to the grade → dQm/dg = 0
0=(s-b) dQb/dg – p dQp/dg [1]
Further function needed – link Qb and Qp (like at the beginning):
Qb = average grade (gav) * Qp
dQb/dg = gav dQp/dg [2]
[1] and [2] → gav = p/(s-b)
Inserting values from above: gav = 4€/t/(28€/kg-2€/kg) = 0.15 kg/t → optimal average cut-off grade, when mining is the constraint, so we call it gm.
gm=0.15
b) Processing is the constraint
T=Qp/P, using same process we get:

gav=gp=(p+f/P)/(s-b) and with our values gav= (4+250/50)/(26) = 0.35 kg/t
gp=0.35
c) Beneficiation is the constraint
T=Qb/B
gav=gb=p/((s-b-f/B)) = 4/(28-2-250/35)=0.21 kg/t
gb=0.21

3)
if we plot the Profit-function for all values we can easily find g-values when balancing two operation- steps (graphically) → gmp, gmb, gpb

According to Lane’s algorithm (Lane 1964) we build 3 groups and select a value according to the selection criteria:

Group1 (Mining, Processing): gm=0.15, gp=0.35, gmp=0.37; Gmp=gp=0.35
Group 2 (Processing and Beneficiation): gp=0.35, gb=0.21, gpb=0.56; Gbp=gp=0.35
Group 3 (Mining, Beneficiation): gm=0.15, gb=0.21, gmb=0; Gmb=gm=0.15
To get the median we sort the values and take the middle value: 0.15, 0.35, 0.35 → g=0.35

4)
Now, what is the profit for g=0.35? (calculate gav with gc=0.35 → 0.59)
Our total quantities would be: Qm=650kt; Qp= 520kt; Qb=520*0.59=306.8t
With the given capacity constraints we get the time T:
650kt/65kt/a → 10a; 520kt/50kt/a → 10.4a; 306.8t/35t/a →8.8a
So, we need to look at 10.4 years:
P=(s-b)Qb-pQp-mQb-fT=(28-2)306.8-4*520-2*650-250*10.4=1996.8 k€
Makes per year 1996.8/10.4 → 192 k€