Systems Engineering for Mines: Reliability, Maintainability and Availability, Linear Programming, Transportation and Assignment, Network Analysis, Inventory and Queuing
1. Reliability, maintainability and availability
Reliability R(t) is the probability that an item performs its function for a time t under stated conditions. With a constant failure (hazard) rate λ — the flat middle of the bathtub curve, after early (infant-mortality) failures have been weeded out and before wear-out begins — R(t) = e−λt and the mean time between failures MTBF = 1/λ. A pump with λ = 0.001 per hour survives 200 h with probability e−0.2 = 0.8187.
| Arrangement | System reliability | Example |
|---|---|---|
| Series — all must work | Rs = R₁ R₂ … Rₙ | 0.9 × 0.95 × 0.98 = 0.8379 |
| Parallel (active redundancy) — one suffices | Rp = 1 − (1 − R₁)(1 − R₂) … (1 − Rₙ) | two pumps of 0.8: 1 − 0.2² = 0.96 |
| Mixed | reduce parallel groups first, then multiply in series | 0.95 in series with the pair: 0.95 × 0.96 = 0.912 |
| k-out-of-n (identical, R each) | Σ from i = k to n of C(n, i) Rⁱ(1 − R)ⁿ⁻ⁱ | two of three fans needed |
Maintainability is the probability that a failed item is restored within a given time, summarised by the mean time to repair (MTTR); with a constant repair rate μ, MTTR = 1/μ. Inherent availability is the fraction of time an item is able to work: A = MTBF/(MTBF + MTTR) — 90 h between failures and 10 h to repair give 0.9. In mine equipment reporting, mechanical availability counts only maintenance downtime, and utilisation the fraction of available time the machine is actually used; their product with performance governs output. Preventive maintenance replaces parts before wear-out raises the failure rate; condition monitoring — vibration, oil analysis, thermography — times it by need.
2. Linear programming, transportation and assignment
A linear programme maximises or minimises a linear objective subject to linear constraints and non-negativity. With two variables it is solved graphically: the constraints bound a convex feasible region, and the optimum lies at a corner point. For max z = 3x + 5y subject to x ≤ 4, 2y ≤ 12, 3x + 2y ≤ 18, the corners are (0, 0), (4, 0), (4, 3), (2, 6) and (0, 6), with z = 0, 12, 27, 36 and 30, so z* = 36 at (2, 6). Larger problems use the simplex method, moving from corner to corner along edges that improve z, with slack variables turning inequalities into equations. Every LP has a dual, whose optimal value equals the primal’s, and whose variables are the shadow prices of the primal constraints — the value of one more unit of each resource. Mines use LP for ore blending (meeting grade and impurity specifications at least cost), production planning and equipment allocation.
The transportation problem ships from m sources with supplies to n destinations with demands at least cost. When total supply equals total demand (add a dummy row or column otherwise), a basic feasible solution has m + n − 1 allocations; fewer means it is degenerate and an ε must be placed. An initial solution comes from the north-west corner rule, the least-cost method or Vogel’s approximation (usually the best start), and it is tested and improved by the MODI (u–v) method: find uᵢ + vⱼ = cᵢⱼ for occupied cells, and if any empty cell has cᵢⱼ − uᵢ − vⱼ < 0, bring it in round a closed loop. The assignment problem — n jobs to n machines or operators, one each — is a special transportation problem solved by the Hungarian method: subtract row and column minima, cover all zeros with the fewest lines, and adjust by the smallest uncovered element until n lines are needed.
3. Network analysis: CPM and PERT
A project is drawn as a network of activities with precedence. The critical path method (CPM) makes a forward pass for the earliest start and finish times (ES, EF = ES + duration, each activity starting at the latest EF of its predecessors) and a backward pass for the latest times (LF, LS = LF − duration). An activity’s total float LS − ES is the delay it can absorb without delaying the project; the critical path is the chain of zero-float activities, and its length is the project duration. For A (3) and B (6) at the start, C (2) and D (4) after A, E (3) after B and C, and F (2) after D and E: the paths are A–C–E–F = 10, A–D–F = 9 and B–E–F = 11, so B–E–F is critical, the project takes 11, and D has a float of 2. Crashing shortens the project by buying time on critical activities in order of lowest cost slope.
PERT handles uncertain durations with three estimates — optimistic a, most likely m, pessimistic b — assuming a beta distribution: expected time te = (a + 4m + b)/6 and variance σ² = ((b − a)/6)². For a = 4, m = 6, b = 14, te = 42/6 = 7 and σ² = (10/6)² = 2.78. The project duration is taken as normally distributed with mean the sum of te and variance the sum of σ² along the critical path, so the probability of finishing by a target T is Φ((T − ΣTe)/√Σσ²).
4. Inventory models and queuing theory
Mines hold large inventories of spares, explosives, tyres, fuel and consumables. In the basic economic order quantity model — constant annual demand D, ordering cost S per order, holding cost H per unit per year, instant replenishment, no shortages — the total annual cost DS/Q + HQ/2 is least at Q* = √(2DS/H), where the ordering and holding costs are equal, and the minimum total is √(2DSH). With D = 24 000 units, S = ₹300 and H = ₹40, Q* = √360 000 = 600 units and the minimum cost is ₹24 000 a year. The reorder point is demand during lead time plus safety stock against variable demand; ABC analysis puts control effort on the few items that carry most of the value.
A queue forms whenever arrivals meet a server of limited capacity — trucks at a shovel or a crusher, cars at a loading point, broken machines at a workshop. In the M/M/1 model (Poisson arrivals at rate λ, exponential service at rate μ, one server) the utilisation ρ = λ/μ must be below 1 for the queue to be stable, and: L = ρ/(1 − ρ) units in the system, Lq = ρ²/(1 − ρ) in the queue, W = 1/(μ − λ) time in the system and Wq = λ/(μ(μ − λ)) waiting. Little’s law, L = λW, holds for any stable queue. With λ = 10 trucks/h and μ = 12/h, ρ = 0.833, L = 5, Lq = 4.17, W = 0.5 h and Wq = 25 min — a shovel that is busy 83% of the time already keeps four trucks waiting.
Key takeaways
- R(t) = e−λt, MTBF = 1/λ; series multiply reliabilities, parallel multiply unreliabilities; A = MTBF/(MTBF + MTTR).
- An LP optimum lies at a corner of the feasible region; the dual’s variables are shadow prices; blending is the classic mine LP.
- A transportation basic solution has m + n − 1 allocations; MODI tests optimality; the Hungarian method solves assignment.
- CPM: the critical path has zero float and sets the duration; PERT te = (a + 4m + b)/6, σ² = ((b − a)/6)².
- EOQ = √(2DS/H) with minimum cost √(2DSH); M/M/1: ρ = λ/μ < 1, L = ρ/(1 − ρ), Wq = λ/(μ(μ − λ)), and L = λW.
Practice questions (16)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
A conveyor line has three drives in series with reliabilities 0.90, 0.95 and 0.98 over a shift. The reliability of the line, to four decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.8379
Series: Rs = 0.90 × 0.95 × 0.98 = 0.8379, lower than the weakest drive. Adding the unreliabilities (1 − 0.17 = 0.83) is only an approximation for small failure probabilities, and the parallel formula would give 0.9999.A dewatering system has a main valve of reliability 0.95 in series with two identical pumps in parallel, each of reliability 0.8 (either pump alone is sufficient). The system reliability, to three decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.912
Parallel pumps: 1 − (1 − 0.8)² = 1 − 0.04 = 0.96; in series with the valve: 0.95 × 0.96 = 0.912. Treating all three in series gives 0.608, and 0.95 × 0.8 = 0.76 ignores the standby pump altogether.A pump has a constant failure rate of 0.001 per hour. The probability that it runs 200 hours without failure, to four decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.8187
R(t) = e−λt = e−0.001 × 200 = e−0.2 = 0.8187. The linear approximation 1 − λt = 0.8 is close only for small λt, and the MTBF of 1000 h is the mean life, not a guaranteed one — only 36.8% of pumps reach it.A dragline has a mean time between failures of 90 h and a mean time to repair of 10 h. Its inherent availability, as a fraction, is ____.
Numerical answer — type the value.
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Answer: 0.9
A = MTBF/(MTBF + MTTR) = 90/(90 + 10) = 0.9. Dividing MTTR by MTBF (0.11) gives the ratio of down to up time, not availability, and 1 − 10/90 = 0.889 uses the wrong denominator.During the useful-life period of the bathtub curve, the failure rate of equipment is:
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Answer: B — approximately constant
Early failures from defects make the rate fall at first, wear-out makes it rise at the end, and in between random failures give a roughly constant rate — the region where R = e−λt and MTBF = 1/λ apply. The rate is never zero while the machine runs.Maximise z = 3x + 5y subject to x ≤ 4, 2y ≤ 12, 3x + 2y ≤ 18, x ≥ 0, y ≥ 0. The optimal value of z is ____.
Numerical answer — type the value.
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Answer: 36
The corners of the feasible region are (0, 0), (4, 0), (4, 3), (2, 6) and (0, 6), giving z = 0, 12, 27, 36 and 30; the best is 36 at (2, 6), where y = 6 and 3x + 2y = 18 meet. Taking (4, 6) gives 42 but violates 3x + 2y ≤ 18.A balanced transportation problem ships coal from 3 mines to 4 washeries. A non-degenerate basic feasible solution has how many occupied cells?
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Answer: B — 6
m + n − 1 = 3 + 4 − 1 = 6: there are m + n supply and demand equations, but one is redundant because total supply equals total demand. Seven would contain a loop, 12 is every cell, and fewer than six means the solution is degenerate and needs an ε allocation.The Hungarian method is used to solve:
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Answer: B — the assignment problem
The Hungarian method assigns n operators to n machines at least total cost by reducing rows and columns and covering zeros; it is the special structure of the assignment problem, every supply and demand equal to 1, that makes it work. Critical paths use forward and backward passes, and EOQ is a closed-form formula.A PERT activity has an optimistic time of 4 days, a most likely time of 6 days and a pessimistic time of 14 days. Its expected time, in days, is ____.
Numerical answer — type the value.
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Answer: 7
te = (a + 4m + b)/6 = (4 + 24 + 14)/6 = 42/6 = 7 days. The simple average (4 + 6 + 14)/3 = 8 gives the most likely time too little weight, and the most likely time itself, 6, ignores the long pessimistic tail.For the same PERT activity (a = 4, m = 6, b = 14 days), the variance of its duration, in days², to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 2.78
σ² = ((b − a)/6)² = (10/6)² = 1.6667² = 2.78 days². The standard deviation is 1.67 days; forgetting to square gives that figure instead of the variance, which is what adds along a critical path.A project has activities A (3 days) and B (6 days) at the start; C (2 days) and D (4 days) follow A; E (3 days) follows both B and C; and F (2 days) follows both D and E. The project duration, in days, is ____.
Numerical answer — type the value.
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Answer: 11
Forward pass: A ends at 3, B at 6; C ends at 5, D at 7; E starts at max(6, 5) = 6 and ends at 9; F starts at max(7, 9) = 9 and ends at 11. The critical path is B–E–F; A–C–E–F (10) and A–D–F (9) have float. Summing every activity (20) treats the network as one chain.A mine uses 24 000 drill bits a year. Each order costs ₹300 to place, and holding one bit in stock costs ₹40 a year. The economic order quantity, in bits, is ____.
Numerical answer — type the value.
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Answer: 600
Q* = √(2DS/H) = √(2 × 24 000 × 300/40) = √360 000 = 600 bits, i.e. 40 orders a year, with total ordering plus holding cost √(2DSH) = ₹24 000. Omitting the 2 gives 424, and inverting S and H gives 80.Trucks arrive at a shovel at random at an average of 10 per hour, and the shovel loads them one at a time at an average rate of 12 per hour (exponential service). The average time a truck waits in the queue before loading begins, in minutes, is ____.
Numerical answer — type the value.
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Answer: 25
Wq = λ/(μ(μ − λ)) = 10/(12 × 2) = 0.4167 h = 25 min. The time in the system, W = 1/(μ − λ) = 0.5 h = 30 min, includes the 5 min of loading itself; using W instead of Wq is the usual slip.For the same shovel (λ = 10 per hour, μ = 12 per hour, one server), the average number of trucks in the system (waiting and being loaded) is ____.
Numerical answer — type the value.
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Answer: 5
ρ = 10/12 = 0.8333, so L = ρ/(1 − ρ) = 0.8333/0.1667 = 5. The number waiting is Lq = ρ²/(1 − ρ) = 4.17, and Little’s law checks it: L = λW = 10 × 0.5 = 5.Which of the following statements are correct? (Select all that apply.)
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Answer: B — An M/M/1 queue is stable only if the arrival rate is less than the service rate; C — Little’s law L = λW holds for any stable queue; D — At the economic order quantity, the annual ordering cost equals the annual holding cost
ρ = λ/μ must be below 1 or the queue grows without limit; Little’s law needs only steady state; and at Q* the two cost terms DS/Q and HQ/2 are equal. Since Q* = √(2DS/H), a higher holding cost makes it smaller, not larger — dear stock is ordered in smaller lots.Which of the following statements about linear programming are correct? (Select all that apply.)
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Answer: A — If an optimal solution exists, one lies at a corner point of the feasible region; B — The optimal values of the primal and its dual are equal; D — A shadow price is the change in the optimal objective per unit increase in a constraint’s right-hand side
A linear objective over a convex polygon reaches its optimum at a vertex; strong duality makes the primal and dual optima equal; and a shadow price is the marginal value of a resource. The intersection of half-planes is always convex, so a non-convex feasible region cannot arise in an LP.