Strategic Underground Mine Access Design to Maximise the Net Present Value

Abstract

To date, the scheduling and access design of an underground mine have only been considered as two separate optimisation problems. First, access to the mine is designed and then the scheduling is completed. One drawback of this approach is that the costs of access construction fail to be reflected in the Net Present Value (NPV) calculation. In this paper, designing the access and scheduling its construction are formulated as a single optimisation problem. The underground mine access construction process can be classified according to the number of faces being developed concurrently. An underground mine with a single decline branching at a junction point into two declines is considered. After construction reaches the junction, the two faces of the decline can be constructed sequentially or concurrently. This paper proposes an efficient algorithm for optimally locating a junction point to maximise the NPV where two faces are being developed concurrently. The NPV is defined by taking the locations of ore bodies and their values, decline construction costs, decline development rate and discount rate into account. The variation of the NPV and the optimal locations of the junction point for one and two concurrent development faces for a range of discount rates are discussed and compared. The proposed algorithm is applied in a simulated case study based on hypothetical values for an underground mine.

Appendices

Appendix

In this section, the equations that are used in the discounted junction point algorithm are derived.

To maximise the NPV, differentiate (3) with respect to $x$ and set equal to 0:

$$- (V_{1} + V_{c} )r^{{ - \left( {l_{0} + l_{1} } \right)/D}} \frac{\ln r}{D}\left( {\frac{{\partial l_{0} }}{\partial x} + \frac{{\partial l_{1} }}{\partial x}} \right) - (V_{2} + V_{c} )r^{{ - \left( {l_{0} + l_{2} } \right)/D}} \frac{\ln r}{D}\left( {\frac{{\partial l_{0} }}{\partial x} + \frac{{\partial l_{2} }}{\partial x}} \right) + Cr^{{ - l_{0} /D}} \frac{{\partial l_{0} }}{\partial x} = 0$$

which can be simplified to,

$$\left( {A + B - C} \right)\frac{{\partial l_{0} }}{\partial x} + A\frac{{\partial l_{1} }}{\partial x} + B\frac{{\partial l_{2} }}{\partial x} = 0$$

(4)

where,

$$A = \left( {\frac{{V_{1} \ln r}}{D} + C} \right)r^{{ - l_{1} /D}}$$

(5)

$$B = \left( {\frac{{V_{2} \ln r}}{D} + C} \right)r^{{ - l_{2} /D}}$$

(6)

Similarly, differentiating (3) with respect to $y,z$ and setting equal to 0 yields:

$$\left( {A + B - C} \right)\frac{{\partial l_{0} }}{\partial y} + A\frac{{\partial l_{1} }}{\partial y} + B\frac{{\partial l_{2} }}{\partial y} = 0$$

(7)

$$\left( {A + B - C} \right)\frac{{\partial l_{0} }}{\partial z} + A\frac{{\partial l_{1} }}{\partial z} + B\frac{{\partial l_{2} }}{\partial z} = 0$$

(8)

Equations (4), (7), (8) can be expressed in terms of gradients,

$$\left( {A + B - C} \right)\nabla l_{0} + A\nabla l_{1} + B\nabla l_{2} = 0$$

(9)

If the operating discount rate is zero then the corresponding junction point is mapped to the location which minimises the total length of the network. When $d = 0,r = 1$ and so,

$$A = \mathop {\lim }\nolimits_{ r \to 1} \left( {\frac{{V_{1} \ln r}}{D} + C} \right)r^{{ - l_{1} /D}} = C$$

(10)

$$B = \mathop {\lim }\nolimits_{ r \to 1} \left( {\frac{{V_{2} \ln r}}{D} + C} \right)r^{{ - l_{2} /D}} = C$$

(11)

Therefore, $A = B = C$. By substituting this into (9),

$$\nabla l_{0} + \nabla l_{1} + \nabla l_{2} = 0$$

(12)

From Eq. (12), the optimisation problem is reduced to a length minimisation problem when the discount rate is zero. Therefore by standard Steiner tree theory an angle of $2\uppi/3$ is generated between the junction point and each pair of adjacent points.

Let ${\mathbf{u}}_{0} ,{\mathbf{u}}_{1} ,{\mathbf{u}}_{2}$ be the unit vectors directed from the fixed points $p_{0} ,p_{1} ,p_{2}$ towards the junction point as shown in Fig. 7. Let $\theta_{0} ,\theta_{1} ,\theta_{2}$ be the angles between ${\mathbf{u}}_{0}$ and ${\mathbf{u}}_{1} ,{\mathbf{u}}_{1}$ and ${\mathbf{u}}_{2} ,{\mathbf{u}}_{2}$ and ${\mathbf{u}}_{0}$ respectively. The unit vectors are expressed using the corresponding gradients ${\mathbf{u}}_{0} = \nabla l_{0} ,{\mathbf{u}}_{1} = \nabla l_{1} ,{\mathbf{u}}_{2} = \nabla l_{2} .$ Hence, Eq. (9) becomes,

Fig. 7

Vector representation of the problem

$$\left( {A + B - C} \right){\mathbf{u}}_{0} + A{\mathbf{u}}_{1} + B{\mathbf{u}}_{2} = 0$$

(13)

Equation (13) can be rewritten,

$$A\left( {{\mathbf{u}}_{0} + {\mathbf{u}}_{1} } \right) + B\left( {{\mathbf{u}}_{0} + {\mathbf{u}}_{2} } \right) - C{\mathbf{u}}_{0} = 0$$

(14)

Also note that ${\mathbf{u}}_{0} \cdot {\mathbf{u}}_{1} = \cos \theta_{0} ,{\mathbf{u}}_{1} \cdot {\mathbf{u}}_{2} = \cos \theta_{1} ,{\mathbf{u}}_{2} \cdot {\mathbf{u}}_{0} = \cos \theta_{2}$ and since this is a planar problem, $\theta_{0} + \theta_{1} + \theta_{2} = 2\pi$ and so $\theta_{0} = 2\pi - \left( {\theta_{1} + \theta_{2} } \right).$

By taking the dot product of (13) with ${\mathbf{u}}_{0}$,

$$\left( {{\text{A}} + {\text{B}} - {\text{C}}} \right){\mathbf{u}}_{0} .{\mathbf{u}}_{0} + {\text{A}}{\mathbf{u}}_{1} .{\mathbf{u}}_{0} + {\text{B}}{\mathbf{u}}_{2} .{\mathbf{u}}_{0} = 0$$

$$\cos \left( {\uptheta_{1} +\uptheta_{2} } \right) = \frac{{{\text{C}} - {\text{A}} - {\text{B}} - {\text{B}}\cos\uptheta_{2} }}{\text{A}}$$

(15)

By taking the dot product of (14) with ${\mathbf{u}}_{0} - {\mathbf{u}}_{1}$,

$$A\left( {{\mathbf{u}}_{0} + {\mathbf{u}}_{1} } \right) \cdot \left( {{\mathbf{u}}_{0} - {\mathbf{u}}_{1} } \right) + B\left( {{\mathbf{u}}_{0} + {\mathbf{u}}_{2} } \right) \cdot \left( {{\mathbf{u}}_{0} - {\mathbf{u}}_{1} } \right) - C{\mathbf{u}}_{0} \cdot \left( {{\mathbf{u}}_{0} - {\mathbf{u}}_{1} } \right) = 0$$

(16)

Note that $\left( {{\mathbf{u}}_{0} + {\mathbf{u}}_{1} } \right) \cdot \left( {{\mathbf{u}}_{0} - {\mathbf{u}}_{1} } \right) = \left| {{\mathbf{u}}_{0} } \right|^{2} - \left| {{\mathbf{u}}_{1} } \right|^{2} = 1 - 1 = 0$.

By substituting the result above into (16),

$$B\left( {{\mathbf{u}}_{0} + {\mathbf{u}}_{2} } \right) \cdot \left( {{\mathbf{u}}_{0} - {\mathbf{u}}_{1} } \right) - C{\mathbf{u}}_{0} \cdot \left( {{\mathbf{u}}_{0} - {\mathbf{u}}_{1} } \right) = 0$$

$$\left( {C - B} \right)\cos \left( {\theta_{1} + \theta_{2} } \right) + B\,\cos \theta_{2} - B\,\cos \theta_{1} = C - B$$

(17)

By substituting the value of $\cos \left( {\theta_{1} + \theta_{2} } \right)$ into (17),

$$B\,\cos \theta_{2} \left( {B - C + A} \right) - AB\,\cos \theta_{1} = \left( {C - B} \right)\left( {2A + B - C} \right)$$

(18)

Similarly, by taking the dot product of (14) with ${\mathbf{u}}_{0} - {\mathbf{u}}_{2}$ and following the same steps as above,

$$A\,\cos \left( {\theta_{1} + \theta_{2} } \right) - A\,\cos \theta_{1} + \left( {C - A} \right)\cos \theta_{2} = C - A$$

(19)

Then, by substituting the value of $\cos \left( {\theta_{1} + \theta_{2} } \right)$ into (19),

$$\cos \theta_{2} \left( {C - A - B} \right) - A\,\cos \theta_{1} = B$$

(20)

Equation (20) is multiplied by $B$ and subtracted from (18),

$$\cos \theta_{2} = - 1 + \frac{{C\left( {C - 2A} \right)}}{{2B\left( {C - A - B} \right)}}$$

(21)

By substituting the value of $\cos \theta_{2}$ into (21),

$$\cos \theta_{1} = 1 + \frac{{C\left( {C - 2A - 2B} \right)}}{2AB}$$

(22)

Equations (21), (22) can be verified for zero discount rate as follows. For the zero discount rate $\theta_{1} ,\theta_{2}$ should be $2\pi /3$.

If $d = 0,$ Eqs. (10), (11) imply $A = B = C$ and so,

$$\begin{array}{*{20}c} {\cos \theta_{2} = - 1 + \frac{{A\left( {A - 2A} \right)}}{{ 2A\left( {A - A - A} \right)}} = - 0.5} & {\theta_{2} = \frac{2\pi }{3}} \\ {\cos \theta_{1} = 1 + \frac{{A\left( {A - 2A - 2A} \right)}}{2AA} = - 0.5} & {\theta_{1} = \frac{2\pi }{3}} \\ \end{array}$$

Hence, Eqs. (21), (22) are correct in this case. Equations (21), (22) are called the discount equations since $\theta_{1} ,\theta_{2}$ depend on $V_{1} ,V_{2}$ and $d$. These two discount equations are used in the iterative process in the DJPA.

Let $\gamma_{0} = \angle sp_{0} p_{2} ,\gamma_{1} = \angle sp_{1} p_{2} ,\nu = \angle p_{0} p_{1} p_{2}$ as shown in Fig. 8. The distances $d_{0} ,d_{1}$ are from $p_{0}$ to $p_{2}$ and from $p_{1}$ to $p_{2} .$ By applying the Sine rule to triangles $p_{0} sp_{2} ,p_{1} sp_{2}$,

Fig. 8

The geometric parameters

$$\begin{array}{*{20}c} {\gamma_{0} = \sin^{ - 1} \left( {\frac{{l_{2} \sin \theta_{2} }}{{d_{0} }}} \right)} & {\gamma_{1} = \sin^{ - 1} \left( {\frac{{l_{2} \sin \theta_{1} }}{{d_{1} }}} \right)} \\ \end{array}$$

Also, by applying the Sine rule to the triangle $p_{1} sp_{2}$,

$$l_{1} = \frac{{d_{1} \sin \left( {\theta_{1} + \gamma_{1} } \right)}}{{\sin \theta_{1} }}$$

(23)

Setting the sum of the angles in the quadrilateral $p_{0} sp_{1} p_{2}$ equal to $2\pi$ and substituting the values for $\gamma_{0} ,\gamma_{1}$ the distance $l_{2}$ can be shown to be,

$$l_{2} = \frac{{d_{0} d_{1} \left| {\sin \left( {\theta_{1} + \theta_{2} + \nu } \right)} \right|}}{{\sqrt {d_{0}^{2} { \sin }^{2} \theta_{1} + d_{1}^{2} { \sin }^{2} \theta_{2} + 2\cos \left( {\theta_{1} + \theta_{2} + \nu } \right)\sin \theta_{1} \sin \theta_{2} d_{0} d_{1} } }}$$

(24)

Equations (23), (24) are called the geometric equations since $l_{1} ,l_{2}$ depend on the constants $d_{0} ,d_{1} ,\nu$ that define the geometry of the network. These two geometric equations are also used in the iterative process in the DJPA.

The Junction Point at One of the Vertices

The critical angle is the minimum angle for each vertex for which the junction point coincides with the breakout point $p_{0}$ or draw points $p_{1} ,p_{2}$. The critical angles associated with the degenerate point are calculated by using (21), (22), (23), (24). The angles $\nu ,\mu ,\lambda$ are calculated by applying the cosine rule to the triangle $p_{0} p_{1} p_{2}$. The critical angles are given by $\psi ,\omega ,\varphi$ when the junction point is at the points $p_{2} ,p_{1} ,p_{0}$ repectively.

The Junction Point at the Breakout Point or Surface Portal p₀

The conditions that need to be satisfied so that the junction point is at the breakout point $p_{0}$ are:

$l_{1} = d_{2} ,l_{2} = d_{0} ,\theta_{1} = \varphi .$ By substituting these values into (5), (6), (22)

$$A = \left( {\frac{{V_{1} \ln r}}{D} + C} \right)r^{{ - d_{2} /D}}$$

(25)

$$B = \left( {\frac{{V_{2} \ln r}}{D} + C} \right)r^{{ - d_{0} /D}}$$

(26)

$$\cos \varphi = 1 + \frac{{C\left( {C - 2A - 2B} \right)}}{2AB}$$

By substituting the values of $A,B$ into the equation above, we get:

$$\varphi = \cos^{ - 1} \left( {1 + \frac{{CD\left( {CD(1 - 2 r^{{ - d_{2} /D}} - 2 r^{{ - d_{0} /D}} - 2\left( {V_{1} \ln r + V_{2} \ln r} \right)} \right)}}{{2\left( {V_{1} \ln r + CD} \right)\left( {V_{2} \ln r + CD} \right)r^{{ - \left( {d_{0} + d_{2} } \right)/D}} }}} \right)$$

(27)

The maximum NPV at this point is $NPV_{{s = p_{0} }}$ where,

$$NPV_{{s = p_{0} }} = (V_{1} + V_{c} )r^{{ - d_{2} /D}} + (V_{2} + V_{c} )r^{{ - d_{0} /D}} - V_{c} + NPV_{fixed}$$

(28)

The Junction Point at the Drawpoint p₁

The conditions that need to be satisfied so that the junction point is at the draw point $p_{1}$ are:

$l_{1} = 0,l_{2} = d_{1} ,\theta_{2} = \omega .$ By substituting these values into (5), (6), (23)

$$A = \left( {\frac{{V_{1} \ln r}}{D} + C} \right)$$

(29)

$$B = \left( {\frac{{V_{2} \ln r}}{D} + C} \right)r^{{ - d_{1} /D}}$$

(30)

$$\cos \omega = - 1 + \frac{{C\left( {C - 2A} \right)}}{{2B\left( {C - A - B} \right)}}$$

(31)

By substituting the values of $A,B$ into (32),

$$\omega = \cos^{ - 1} \left( { - 1 + \frac{{CD\left( {CD + 2V_{1} \ln r} \right)}}{{2\left( {V_{2} \ln r + CD} \right)r^{{ - d_{1} /D}} \left( {V_{1} \ln r + \left( {V_{2} \ln r + CD} \right)r^{{ - d_{1} /D}} } \right)}}} \right)$$

(32)

The maximum NPV at this point is $NPV_{{s = p_{1} }}$ where,

$$NPV_{{s = p_{1} }} = (V_{1} + V_{c} ) r^{{ - d_{2} /D}} + (V_{2} + V_{c} )r^{{ - \left( {d_{0} + d_{2} } \right)/D}} - V_{c} \left( {r^{{ - d_{2} /D}} + 1} \right) + NPV_{fixed}$$

(33)

The Junction Point at the Drawpoint p₂

The conditions that need to be satisfied so that the junction point is at the draw point $p_{2}$ are:

$l_{2} = 0,l_{1} = d_{1} ,2\pi - (\theta_{1} + \theta_{2} ) = \psi .$ By substituting these values into (5), (6)

$$A = \left( {\frac{{V_{1} \ln r}}{D} + C} \right)r^{{ - d_{1} /D}}$$

(34)

$$B = \left( {\frac{{V_{2} \ln r}}{D} + C} \right)$$

(35)

First, the value of $\cos \left( {\theta_{1} + \theta_{2} } \right)$ is calculated in terms of $A,B,C$. By substituting the values of $\cos \theta_{1} ,\cos \theta_{2} ,$ into (19),

$$\psi = \cos^{ - 1} \left( {\frac{{\left( {C - A - B} \right)\left( {4B\left( {C - A} \right) + C\left( {C - 2A} \right)} \right) + C\left( {C - A} \right)\left( {C - 2A} \right)}}{{2AB\left( {C - A - B} \right)}}} \right)$$

(36)

where ${\text{A}},{\text{B}}$ are given in (35), (36) respectively.

The maximum NPV at this point is $NPV_{{s = p_{2} }}$ where,

$$NPV_{{s = p_{2} }} = (V_{1} + V_{c} )r^{{ - \left( {d_{0} + d_{1} } \right)/D}} + (V_{2} + V_{c} )r^{{ - d_{0} /D}} - V_{c} \left( {r^{{ - \left( {d_{0} + d_{1} } \right)/D}} + 1} \right) + NPV_{fixed}$$

(37)

Let $l_{1}^{*} ,l_{2}^{*} ,\theta_{1}^{*} ,\theta_{2}^{*}$ be the optimal values obtained from the DJPA. The distance $l_{0}^{*}$ is calculated by applying the Sine rule to the triangle $p_{0} sp_{2}$.

$$l_{0}^{*} = \frac{{d_{0} \sin \left( {\theta_{2} + \gamma_{0} } \right)}}{{\sin \theta_{2} }}$$

(38)

Therefore, the maximum NPV is calculated using (3) and is given by $NPV^{ *}$ where,

$$NPV^{*} = (V_{1} + V_{c} )r^{{ - \left( { l_{0}^{*} + l_{1}^{*} } \right)/D}} + (V_{2} + V_{c} )r^{{ - \left( {l_{0}^{*} + l_{2}^{*} } \right)/D}} - V_{c} \left( {r^{{ - l_{0}^{*} /D}} - 1} \right) + NPV_{fixed}$$

(39)

Since $l_{0}^{ *} ,l_{1}^{ *} ,l_{2}^{ *}$ are known, the junction point coordinates $x,y,z$ can be calculated by solving three quadratic simultaneous Eqs. (40), (41), (42).

$$l_{0}^{*2} = \left( {x_{0} - x} \right)^{2} + \left( {y_{0} - y} \right)^{2} + \left( {z_{0} - z} \right)^{2}$$

(40)

$$l_{1}^{*2} = \left( {x - x_{1} } \right)^{2} + \left( {y - y_{1} } \right)^{2} + \left( {z - z_{1} } \right)^{2}$$

(41)

$$l_{2}^{*2} = \left( {x - x_{2} } \right)^{2} + \left( {y - y_{2} } \right)^{2} + \left( {z - z_{2} } \right)^{2}$$

(42)