Mineral EconomicsFREE· Reviewed Jun 2026

Mineral Economics & Planning

Discounted cash flow and NPV, annuity present value, break-even cut-off grade and reserve-tonnage estimation — the project-evaluation numerics that decide whether an orebody is worth mining.

Section 1

Recent Trend Analysis (2017–2026)

Mineral economics & planning is worth 3–6 marks and has hardened from definition recall into financial-evaluation NAT.

The decade-long shift in question style:
- 2017–2019 — definitions: resource vs reserve, cut-off grade meaning, simple payback statements. 1-mark theory.
- 2020–2022 — single compute: a present value discounting, a one-line break-even cut-off grade, or a tonnage = volume × density estimate.
- 2023–2026 — full DCF/NPV chains over a project life, annuity present-worth, and cut-off → ore tonnage → contained-metal sequences in one NAT.

Exact recurring themes you must own:
- NPV

= \sum C F_{t} / (1 + i)^{t} - C_{0}

and the annuity short-cut for uniform cash flows.
- Break-even cut-off grade

g_{c} = C / (p y)

— the recovery and price-unit trap.
- Reserve tonnage

T = A t ρ

and contained metal

= T g

.
- Payback vs discounted payback vs IRR (rate where

N P V = 0

).
- Frequent MSQs distinguishing reserve categories (proved/probable) and modifying factors.

Section 2

Master Formula Matrix & Derivations

Time Value of Money & NPV

A future cash flow is worth less today; project worth is the sum of discounted inflows minus the up-front capital.

P V = \frac{F V}{( 1 + i ) ^{n}}, N P V = t = 1 \sum n \frac{C F _{t}}{( 1 + i ) ^{t}} - C_{0}

⚡ Exam shortcut ·

C_{0}

(capital at

t = 0

) is not discounted. Accept the project if

N P V > 0

. Use

i

as a decimal (

10% \to 0.10

Variable Index (SI units)

Symbol	Meaning	SI unit
$N P V$	Net present value	currency
$C F_{t}$	Net cash flow in year $t$	currency
$C_{0}$	Initial capital outlay	currency
$i$	Discount rate (per period)	fraction
$n$	Project life	years

Present Value of an Annuity

A stream of equal annual cash flows collapses to a single closed-form present worth.

P V = A \cdot \frac{1 - ( 1 + i ) ^{- n}}{i}

⚡ Exam shortcut ·

Use this only for uniform

C F

. The bracket is the 'annuity factor' — multiply by the constant annual cash flow

A

. Uneven flows must be discounted year-by-year.

Variable Index (SI units)

Symbol	Meaning	SI unit
$P V$	Present value of the stream	currency
$A$	Uniform annual cash flow	currency/yr
$i$	Discount rate	fraction
$n$	Number of periods	years

Break-even Cut-off Grade

The lowest grade at which the recovered metal value exactly pays the cost of mining and treating one tonne of ore.

g_{c} = \frac{C}{p \cdot y}

⚡ Exam shortcut ·

Keep units consistent:

C

in \

/ t or e,

in \

/t metal ⇒

g_{c}

as a fraction (

\times 100

for %). Higher costs raise

g_{c}

; better recovery

y

or price

p

lowers it.

Variable Index (SI units)

Symbol	Meaning	SI unit
$g_{c}$	Break-even cut-off grade	fraction
$C$	Total cost per tonne of ore	\$/t
$p$	Metal price	\$/t metal
$y$	Mill / metallurgical recovery	fraction

Reserve Tonnage & Contained Metal

Ore tonnage is the orebody volume times bulk density; contained metal scales that by the average grade.

T = A t ρ, M = T \cdot g

⚡ Exam shortcut ·

Volume

V = A t

(area × mean thickness). For cross-sections use the mean-area rule

V = \frac{A _{1} + A _{2}}{2} L

. Metal mass uses grade as a fraction.

Variable Index (SI units)

Symbol	Meaning	SI unit
$T$	Ore tonnage	$t$
$A$	Plan area of orebody	$m^{2}$
$t$	Mean thickness	$m$
$ρ$	Bulk (in-situ) density	$t / m^{3}$
$g$	Average grade (fraction)	dimensionless
$M$	Contained metal	$t$

Payback Period & IRR

Simple payback is the time to recover capital from cash flow; IRR is the discount rate that zeroes the NPV.

n_{p b} = \frac{C _{0}}{C F _{ann u a l}}, N P V (i = I R R) = 0

⚡ Exam shortcut ·

Simple payback ignores the time value of money — discounted payback uses

C F_{t} / (1 + i)^{t}

. Accept a project when

I R R >

the hurdle (discount) rate.

Variable Index (SI units)

Symbol	Meaning	SI unit
$n_{p b}$	Payback period	years
$C_{0}$	Initial capital	currency
$C F_{ann u a l}$	Annual net cash flow	currency/yr
$I R R$	Internal rate of return	fraction

Section 3

The "IIT Trap" Warning System

Discount rate as a decimal. $(1 + i)^{t}$ needs $i = 0.10$ , not $10$ . Plugging the percentage value blows the denominator apart.
**Don't discount $C_{0}$ .** The initial capital sits at $t = 0$ ; only the future inflows $C F_{t}$ get the $(1 + i)^{t}$ divisor.
Annuity formula is for uniform flows only. If cash flows vary year-to-year, discount each one separately — the closed form silently gives a wrong answer.
Cut-off grade: recovery is in the denominator. $g_{c} = C / (p y)$ . Forgetting $y$ (or putting it on top) mis-states the break-even grade.
Grade as fraction vs percent. A $1.2%$ grade is $0.012$ in $M = T g$ . Also $1% = 10000 pp m$ — watch ppm-quoted grades.
Tonnage vs contained metal. $T = A tρ$ is ore; multiply by grade for metal. Reporting tonnage as metal (or vice-versa) is a classic slip.
Bulk density units. $ρ$ in $t / m^{3}$ gives tonnes directly; a $k g / m^{3}$ value must be divided by 1000.
Reserve vs resource. A *reserve* is the economically/legally mineable part of a *resource*; cut-off grade and modifying factors convert one to the other.
Payback ≠ profitability. A short payback can still have a negative NPV; use discounted measures (NPV, IRR) for the accept/reject decision.

Section 4

High-Fidelity Core Examples

Example 12-mark complexity

A mining project needs an initial capital of

C_{0} = ₹ 50

crore and is expected to generate a uniform net cash flow of

C F = ₹ 15

crore per year for

5

years. At a discount rate of

i = 10%

, evaluate the project NPV and state whether it is viable.

Given Parameters Matrix (clean SI)

Initial capital ( $C_{0}$ )	$₹ 50$ crore
Annual cash flow ( $A$ )	$₹ 15$ crore/yr
Project life ( $n$ )	$5$ years
Discount rate ( $i$ )	$10% = 0.10$

Algebraic Derivation Track

Step 1 — Annuity factor (uniform cash flows):

\frac{1 - ( 1 + i ) ^{- n}}{i} = \frac{1 - ( 1.10 ) ^{- 5}}{0.10}

(1.10)^{5} = 1.61051 \Rightarrow (1.10)^{- 5} = 0.62092

= \frac{1 - 0.62092}{0.10} = \frac{0.37908}{0.10} = 3.7908

Step 2 — Present value of inflows:

P V = A \times 3.7908 = 15 \times 3.7908 = 56.86 crore

Step 3 — NPV:

N P V = P V - C_{0} = 56.86 - 50 = 6.86 crore

🎯 Final target & accepted range ·

N P V \approx ₹ 6.86 crore

(accept

6.8 - 6.9

). Since

N P V > 0

, the project is viable at a

10%

hurdle rate.

Example 22-mark complexity

For a copper deposit the operating cost is

C=\

p er t o nn eo f or e, t h eco pp er p r i ce i s

p=\

6000

per tonne of metal and the mill recovery is

y = 0.85

. The tabular orebody has plan area

A = 250000 m^{2}

, mean thickness

t = 8 m

, bulk density

ρ = 2.8 t / m^{3}

and average grade

g = 1.2%

Cu. Find (a) the break-even cut-off grade and (b) the ore tonnage and contained copper.

Given Parameters Matrix (clean SI)

Operating cost ( $C$ )	$\$ 40/t$ ore
Copper price ( $p$ )	$\$ 6000/t$ metal
Recovery ( $y$ )	$0.85$
Area × thickness ( $A, t$ )	$250000 m^{2} \times 8 m$
Bulk density ( $ρ$ )	$2.8 t / m^{3}$
Average grade ( $g$ )	$1.2% = 0.012$

Algebraic Derivation Track

Step 1 — Break-even cut-off grade:

g_{c} = \frac{C}{p y} = \frac{40}{6000 \times 0.85} = \frac{40}{5100} = 0.00784

g_{c} = 0.00784 \times 100 = 0.78% Cu

Step 2 — Ore tonnage (

T = A t ρ

T = 250000 \times 8 \times 2.8 = 2000000 \times 2.8 = 5.6 \times 1 0^{6} t

Step 3 — Contained copper (

M = T g

M = 5.6 \times 1 0^{6} \times 0.012 = 6.72 \times 1 0^{4} t

🎯 Final target & accepted range ·

Cut-off grade

g_{c} \approx 0.78% Cu

; ore tonnage

T = 5.6 Mt

; contained copper

M \approx 67200 t

(

67.2 k t

). The

1.2%

average comfortably exceeds the

0.78%

cut-off, so the block is ore.