Advanced Industrial interview questions

125 questions with answers (IND-301 to IND-425). Try each one aloud before reading the answer.

Work Study & Ergonomics

IND-301 · An element's average observed time is 0.80 min, the rating is 120% and PF&D allowance is 15% of normal time. Find standard time and output per 480-minute shift.

NT = 0.80 × 1.20 = 0.96 min. ST = 0.96 × 1.15 = 1.104 min. Output = 480 / 1.104 ≈ 434 units per shift.

IND-302 · How does the result change if the 15% allowance is a percentage of the total workday rather than of normal time?

ST = NT / (1 − 0.15) = 0.96 / 0.85 ≈ 1.129 min, compared with 1.104 min. Allowances based on the workday give a larger standard, so the basis must be stated.

IND-303 · How many machines should one operator tend if load/unload time a = 1 min, operator walk/inspect time b = 0.5 min and automatic machine time t = 4 min?

n′ = (a + t)/(a + b) = 5/1.5 = 3.33. With 3 machines the operator is idle; with 4 the machines wait. Choose by comparing the cost per unit of each option.

IND-304 · In 400 random observations a machine was idle 60 times. Estimate the idle percentage with a 95% confidence interval.

p = 60/400 = 0.15. Half-width = 1.96√(0.15 × 0.85 / 400) ≈ 0.035. The idle time is 15% ± 3.5%, or about 11.5–18.5%.

IND-305 · The first unit takes 100 h on an 80% learning curve. How long does the 8th unit take?

Tₙ = T₁·n^b with b = log 0.8 / log 2 ≈ −0.322. Each doubling multiplies the time by 0.8, so T₈ = 100 × 0.8³ = 51.2 h.

IND-306 · Unit 1 took 50 h and unit 4 took 32 h. What is the learning rate?

Two doublings have occurred, so r² = 32/50 = 0.64 and r = 0.80. This is an 80% learning curve.

IND-307 · How would you design a seated workstation for precision assembly?

Make heights adjustable. Put the work surface slightly above elbow height for precision visual work, and lower for heavier tasks. Place parts within the normal reach of a 5th-percentile user and give a 95th-percentile user leg clearance. Add good task lighting, forearm support and fixtures to free both hands.

IND-308 · Calculate the NIOSH RWL for H = 40 cm, V = 75 cm, D = 25 cm, A = 0°, FM = 0.88, CM = 0.95. Is a 15 kg lift acceptable?

HM = 25/40 = 0.625, VM = 1 − 0.003|75 − 75| = 1, DM = 0.82 + 4.5/25 = 1.0, AM = 1. RWL = 23 × 0.625 × 1 × 1 × 1 × 0.88 × 0.95 ≈ 12.0 kg. LI = 15/12.0 ≈ 1.25 > 1, so the lift carries increased risk and should be redesigned.

IND-309 · How is WBGT calculated for heat stress assessment?

Indoors or outdoors without solar load: WBGT = 0.7·T_nwb + 0.3·T_g. Outdoors with sun: WBGT = 0.7·T_nwb + 0.2·T_g + 0.1·T_db. Here T_nwb is the natural wet-bulb, T_g the globe and T_db the dry-bulb temperature. Limits depend on workload and work–rest regime.

IND-310 · Under OSHA criteria (90 dBA for 8 h, 5 dB exchange rate), a worker spends 4 h at 90 dBA and 2 h at 95 dBA. What is the noise dose?

Permissible times are 8 h at 90 dBA and 4 h at 95 dBA. Dose = 4/8 + 2/4 = 0.5 + 0.5 = 1.0, which is 100% and exactly at the permissible limit.

IND-311 · An operator walks excessively during each cycle. How would you redesign the job?

Draw a spaghetti or string diagram to quantify the walking. Then move parts and tools to point of use within the normal reach envelope, use gravity-feed bins, arrange machines in a U-shaped cell so the start and end are adjacent, and rebalance the elements between operators.

Production Planning & Control

IND-312 · What is the Theory of Constraints and what are its five focusing steps?

TOC holds that system throughput is limited by its constraint (bottleneck). The five steps are: identify the constraint, exploit it (no lost time), subordinate everything else to it, elevate its capacity, then repeat and avoid inertia.

IND-313 · Use Johnson's rule for five jobs on M1→M2: A(5, 2), B(1, 6), C(9, 7), D(3, 8), E(10, 4). What is the sequence and makespan?

The smallest times give B first (1 on M1), A last (2 on M2), D second (3 on M1) and E fourth (4 on M2). The sequence is B–D–C–E–A. M1 finishes at 28 and M2 at 30, so the makespan is 30.

IND-314 · Four jobs with processing times 4, 2, 6 and 3 h wait at one machine. What is the mean flow time under SPT?

The SPT order is 2, 3, 4, 6, giving completion times 2, 5, 9 and 15. Mean flow time = 31/4 = 7.75 h, the minimum possible for this set.

IND-315 · Demand over three months is 800, 1,200 and 1,000 units, with zero opening stock. Compare level and chase plans.

Level: produce 1,000 per month. Ending inventories are 200, 0 and 0, so the cost is 200 unit-months of holding with no workforce changes. Chase: produce 800, 1,200 and 1,000, which needs capacity for +400 then −200 units of change. Choose the plan with lower holding versus hiring/layoff cost.

IND-316 · An item has gross requirement 500 in week 6, on-hand 120, scheduled receipt 80 in week 6 and lead time 2 weeks. What is the planned order release (lot-for-lot)?

Net requirement = 500 − 120 − 80 = 300 units in week 6. Offsetting by the 2-week lead time gives a planned order release of 300 units in week 4.

IND-317 · A serial line has three stations with capacities of 60, 45 and 50 units/h. Demand is 55/h. What do you do?

Throughput is limited to 45/h by station 2, the bottleneck, so 10/h of demand is lost. Exploit station 2 first: no idle time, offload work, quality checks before it. Then elevate it with overtime, a parallel machine or SMED. Station 3 at 50/h becomes the next constraint.

IND-318 · How do you reduce MRP "nervousness"?

Freeze the MPS within a frozen time fence and use firm planned orders. Use lot-sizing rules that damp small changes, apply pegging and exception filters, and add safety stock or safety lead time at the end-item level rather than throughout the BOM.

IND-319 · How does CONWIP compare with kanban?

CONWIP holds the total WIP in a line constant with a single card loop from the end of the line back to the start. Kanban caps WIP between each pair of stations. CONWIP is simpler for high-mix lines and lets WIP collect naturally at the bottleneck.

IND-320 · Which algorithm minimizes the number of tardy jobs on a single machine?

The Moore–Hodgson algorithm. Sequence jobs by EDD. When a job becomes tardy, remove the longest job scheduled so far and put it at the end. Repeat until no remaining job in the sequence is tardy.

IND-321 · Annual demand is 200,000 units at 3 min/unit, each machine has 2,000 h/yr and efficiency is 85%. How many machines are needed?

Required hours = 200,000 × 3/60 = 10,000 h. Effective hours per machine = 2,000 × 0.85 = 1,700 h. Machines = 10,000/1,700 ≈ 5.88, so 6 machines.

IND-322 · How do you choose between make-to-stock, assemble-to-order, make-to-order and engineer-to-order?

Compare the customer's tolerable lead time with the total production lead time, and consider product variety. MTS suits standard, forecastable items. ATO keeps modules in stock for configurable products. MTO suits high variety. ETO suits unique designs. The choice sets the order decoupling point.

Inventory Management

IND-323 · D = 4,000 units/yr, S = $40 and H = 20% of price. The price is $10 for Q < 500 and $9.50 for Q ≥ 500. What should you order?

At $10: EOQ = √(2×4000×40/2) = 400, which is feasible. TC = 40,000 + 400 + 400 = $40,800. At $9.50: EOQ ≈ 410 < 500, so use Q = 500. TC = 38,000 + 320 + 475 = $38,795. Order 500 units.

IND-324 · Daily demand averages 50 with σ = 10, lead time is 9 days and the target is 95% cycle service. Find safety stock and ROP.

SS = 1.645 × 10 × √9 = 49.4 ≈ 50 units. ROP = 50 × 9 + 50 = 500 units.

IND-325 · How is safety stock calculated when both demand and lead time vary?

σ_LTD = √(L̄·σ_d² + d̄²·σ_L²) and SS = z × σ_LTD. Lead-time variability is often the larger term, so reducing supplier variability can cut safety stock sharply.

IND-326 · A seasonal item sells at $12, costs $6 and salvages at $2. Demand ~ N(1000, 200). How many should you order?

Cu = 12 − 6 = 6 and Co = 6 − 2 = 4. The critical ratio = 6/10 = 0.6, so z ≈ 0.253. Q* = 1000 + 0.253 × 200 ≈ 1,051 units.

IND-327 · Find the EPQ for D = 12,000/yr, p = 48,000/yr, S = $100 per setup and H = $2/unit/yr.

EPQ = √(2 × 12,000 × 100 / (2 × (1 − 0.25))) = √1,600,000 ≈ 1,265 units. Maximum inventory = 1,265 × 0.75 ≈ 949 units.

IND-328 · What is echelon stock and why does it matter in multi-echelon inventory?

Echelon stock at a stage is all inventory at that stage plus everything downstream of it, including stock in transit. Policies based on echelon stock coordinate stages and avoid double-buffering, unlike independent installation-stock policies.

IND-329 · How would you design inventory policies for A, B and C items?

A: continuous review, accurate forecasts, frequent small orders, tight safety stock analysis, monthly cycle counts and supplier collaboration. B: periodic review with moderate control. C: simple rules such as two-bin systems or large infrequent orders, with yearly counts.

IND-330 · How do you calculate the fill rate of a (Q, R) policy?

Expected units short per cycle = σ_L × G(z), where G(z) is the standard normal loss function. Fill rate = 1 − σ_L·G(z)/Q. A larger Q raises the fill rate even at the same safety stock.

IND-331 · How do you compute EOQs for several items under a storage or budget limit?

If the unconstrained EOQs violate the limit, add a Lagrange multiplier λ: Qᵢ = √(2DᵢSᵢ / (Hᵢ + 2λwᵢ)), where wᵢ is the space or cost per unit. Increase λ until the constraint is just met.

IND-332 · What is the EOQ when planned backorders are allowed?

Q* = √(2DS/H) × √((H + B)/B), where B is the backorder cost per unit per year. The maximum backorder is Q*·H/(H + B). As B → ∞ the model reduces to the basic EOQ.

IND-333 · Management wants a 30% inventory cut without hurting service. Where do you look?

Split inventory into cycle, safety, pipeline and obsolete stock. Cycle stock: cut setup or ordering cost, since EOQ ∝ √S, using SMED and e-ordering. Safety stock: reduce lead time and its variability and improve forecasts. Pipeline stock: shorten lead times. Also dispose of obsolete stock.

Operations Research

IND-334 · What is linear programming and how does the simplex method work?

LP optimizes a linear objective under linear constraints. The optimum lies at a vertex of the convex feasible region. Simplex starts at one basic feasible vertex and moves to adjacent vertices that improve the objective until no improvement is possible.

IND-335 · Products A and B earn $40 and $30. Machine: 2A + B ≤ 100 h; labour: A + B ≤ 80 h. Find the optimal mix.

The corners are (0,0), (50,0), (20,60) and (0,80), with Z = 0, 2000, 2600 and 2400. The optimum is A = 20, B = 60, Z = $2,600, and both constraints are binding.

IND-336 · S1 supplies 30 and S2 supplies 20. D1 needs 25 and D2 needs 25. Costs: S1–D1 4, S1–D2 6, S2–D1 5, S2–D2 3. Find the optimal plan.

Ship S1→D1 25, S1→D2 5 and S2→D2 20. Cost = 100 + 30 + 60 = 190. MODI: u₁ = 0, v₁ = 4, v₂ = 6, u₂ = −3, so Δ₂₁ = 5 − 1 = 4 ≥ 0 and the plan is optimal.

IND-337 · Assign three workers to three jobs with costs W1 [9, 2, 7], W2 [6, 4, 3], W3 [5, 8, 1].

The optimum is W1→J2 (2), W2→J1 (6) and W3→J3 (1), for a total cost of 9. The Hungarian method reaches this after row and column reduction and one adjustment.

IND-338 · What does sensitivity analysis provide for an LP solution?

It gives the range of optimality for each objective coefficient, over which the current basis stays optimal, and the range of feasibility for each RHS, over which shadow prices stay valid. It also gives reduced costs, which show how much a non-basic variable's coefficient must improve before it enters the solution.

IND-339 · How is an integer programming problem solved?

Branch and bound: solve the LP relaxation, then branch on a fractional variable (xⱼ ≤ ⌊v⌋ or xⱼ ≥ ⌈v⌉) and prune by bound or infeasibility. Cutting planes (Gomory cuts) and branch-and-cut are also used. Rounding the LP optimum may be infeasible or suboptimal.

IND-340 · What is goal programming?

A multi-objective method in which each goal becomes a constraint with deviation variables d⁻ (under-achievement) and d⁺ (over-achievement). The objective minimizes the unwanted deviations, either weighted or in pre-emptive priority order.

IND-341 · What is dynamic programming?

A method that splits a problem into stages with states and solves it recursively using Bellman's principle of optimality: an optimal policy's remaining decisions must be optimal from any intermediate state. Applications include shortest routes, resource allocation and Wagner–Whitin lot sizing.

IND-342 · What is a transshipment problem and how is it solved?

A transportation problem in which goods may pass through intermediate nodes, such as warehouses, or between sources and destinations. Add a buffer of total supply to each transshipment node's supply and demand, then solve it as a standard transportation problem or as a minimum-cost network flow.

IND-343 · What is a saddle point in a two-person zero-sum game?

A payoff that is both the minimum of its row and the maximum of its column, so maximin = minimax. Both players then use pure strategies, and the saddle-point value is the value of the game. Without a saddle point, mixed strategies are found by LP or by graphical methods.

IND-344 · Machine hours have a shadow price of $15/h with an allowable increase of 20 h. Should you rent 30 extra hours at $10/h?

Rent only 20 hours. Within the allowable increase each hour adds $15 − $10 = $5 net, or $100 for 20 h. Beyond 20 h the shadow price changes, possibly to zero, so the LP must be re-solved before buying more.

Queueing Theory & Simulation

IND-345 · What are the key M/M/1 formulas?

ρ = λ/μ, L = ρ/(1 − ρ) = λ/(μ − λ), Lq = ρ²/(1 − ρ), W = 1/(μ − λ) and Wq = λ/(μ(μ − λ)), valid for ρ < 1.

IND-346 · Customers arrive at 8/h and one server serves 10/h (M/M/1). Find ρ, L, Lq, W and Wq.

ρ = 0.8, L = 0.8/0.2 = 4, Lq = 0.64/0.2 = 3.2, W = 1/(10 − 8) = 0.5 h (30 min) and Wq = 8/(10 × 2) = 0.4 h (24 min).

IND-347 · With λ = 8/h and μ = 5/h per server, compare one pooled M/M/2 queue with two separate M/M/1 queues of 4/h each.

M/M/2: a = 1.6, ρ = 0.8, P₀ = 1/(1 + 1.6 + 6.4) = 0.111, Lq ≈ 2.84 and Wq ≈ 0.36 h. Separate queues: Wq = 4/(5 × 1) = 0.8 h each. Pooling more than halves the waiting time.

IND-348 · How do you choose the optimal service rate economically?

Minimize the total cost per hour, TC = C_s·μ + C_w·L, where C_s is the cost per unit of service rate and C_w the waiting cost per customer per hour. For M/M/1, L = λ/(μ − λ), and setting dTC/dμ = 0 gives μ* = λ + √(C_w·λ/C_s).

IND-349 · How do you analyse an M/M/1/K queue with finite capacity?

With ρ ≠ 1: P₀ = (1 − ρ)/(1 − ρᴷ⁺¹) and Pₙ = P₀ρⁿ for n ≤ K. Arrivals finding the system full are lost, so the effective arrival rate is λ_eff = λ(1 − P_K), and Little's Law uses λ_eff. The system is stable even when ρ ≥ 1.

IND-350 · What is Kingman's (VUT) approximation?

For a G/G/1 queue: Wq ≈ [(c_a² + c_s²)/2] × [ρ/(1 − ρ)] × t_s, which is Variability × Utilization × Time. Here c_a and c_s are the coefficients of variation of interarrival and service times. Waiting time falls by reducing either variability or utilization.

IND-351 · How many simulation replications are needed for a given precision?

Make a pilot run of n₀ replications to estimate the standard deviation s. Then n ≈ (t_(α/2) × s / e)², where e is the desired half-width of the confidence interval. Replications use independent random-number streams.

IND-352 · How would you staff a call centre to answer 80% of calls within 20 seconds?

Compute the offered load in Erlangs, a = λ/μ. Use Erlang C to find the probability of waiting for c agents and P(Wq ≤ 20 s) = 1 − P_wait·e^(−(cμ − λ)t). Increase c until the target is met, then add shrinkage for breaks and absenteeism.

IND-353 · What variance reduction techniques are used in simulation?

Common random numbers, which use the same streams when comparing alternatives. Antithetic variates, which pair U with 1 − U. Control variates, which correct with a correlated variable whose mean is known. Importance sampling and stratified sampling are also used.

IND-354 · What is a Jackson network?

An open network of M/M/cᵢ nodes with Poisson external arrivals and probabilistic routing. Solve the traffic equations λᵢ = γᵢ + Σⱼλⱼpⱼᵢ for each node's total arrival rate. Each node then behaves as an independent M/M/cᵢ queue (product-form solution).

IND-355 · What is the machine-repair (finite-source) queueing model?

A population of N machines, each failing at rate λ while running, is served by c repairers. The arrival rate falls as more machines are down: λₙ = (N − n)λ. It is solved with birth–death balance equations and is used to size maintenance crews.

Statistical Quality Control

IND-356 · USL = 10.6 mm, LSL = 9.4 mm, μ = 10.1 mm, σ = 0.15 mm. Find Cp and Cpk and interpret them.

Cp = 1.2/0.9 = 1.33. Cpk = min(0.5, 0.7)/0.45 = 1.11. The spread is acceptable, but the process is off-centre towards the USL. Re-centring to 10.0 mm would raise Cpk to 1.33.

IND-357 · Twenty subgroups of n = 5 give X̿ = 50.0 and R̄ = 4.0. Compute the X̄ and R chart limits.

X̄ chart: 50 ± 0.577 × 4 = 47.69 to 52.31. R chart: UCL = 2.114 × 4 = 8.46 and LCL = 0 × 4 = 0.

IND-358 · Estimate process σ from R̄ = 4.0 with n = 5.

σ̂ = R̄/d₂ = 4.0/2.326 ≈ 1.72. This short-term (within-subgroup) estimate is the one used for Cp and Cpk.

IND-359 · Find p-chart limits for p̄ = 0.04 and n = 200.

σ_p = √(0.04 × 0.96 / 200) ≈ 0.0139. UCL = 0.04 + 3 × 0.0139 ≈ 0.082. The LCL is negative, so it is set to 0.

IND-360 · When should CUSUM or EWMA charts be used instead of Shewhart charts?

When small, sustained shifts of about 0.5–1.5σ must be detected quickly. CUSUM and EWMA accumulate information from past points, so their ARL for small shifts is much shorter. Shewhart charts remain better for large, sudden shifts.

IND-361 · What is the average run length (ARL)?

The average number of points plotted before an out-of-control signal. In control, ARL₀ = 1/α ≈ 1/0.0027 ≈ 370 for 3σ limits. ARL₁ = 1/(1 − β) after a shift should be small.

IND-362 · Find the probability of acceptance for a plan with n = 50, c = 1 and lot fraction defective p = 0.02.

Using the Poisson approximation with np = 1: Pa = P(d ≤ 1) = e⁻¹(1 + 1) ≈ 0.736. About 74% of such lots would be accepted.

IND-363 · What is the difference between Cpk and Ppk?

Cpk uses the short-term, within-subgroup σ (R̄/d₂ or s̄/c₄) and shows potential capability. Ppk uses the overall long-term sample standard deviation and shows actual performance. A large gap means there are between-subgroup shifts or drift.

IND-364 · A process is in statistical control but Cpk = 0.7. What do you do?

The process is stable but not capable, so control-chart actions will not help. Reduce common-cause variation by changing the process, materials, machines or methods (for example through DOE), or re-centre the mean, or negotiate the tolerances. Meanwhile, contain the product with screening.

IND-365 · How do you apply SPC in short-run, high-mix production?

Use deviation-from-nominal (DNOM) charts, which plot X − target for several part numbers on one chart. When variances differ, use standardized (Z or Q) charts. This allows charting by process rather than by part number.

Six Sigma

IND-366 · Outline a DMAIC project to reduce invoice errors from 6% to 1%.

Define: charter, SIPOC, CTQ = error-free invoice. Measure: validate data entry, baseline DPMO. Analyze: Pareto chart of error types, fishbone diagram, hypothesis tests on shift and source. Improve: mistake-proof the entry fields, standard work, pilot. Control: p-chart, control plan, process owner handover.

IND-367 · A 2² experiment gives (−,−) = 20, (+,−) = 30, (−,+) = 25 and (+,+) = 45. Find the effects of A, B and AB.

Effect A = (30 + 45)/2 − (20 + 25)/2 = 15. Effect B = (25 + 45)/2 − (20 + 30)/2 = 10. Interaction AB = [(20 + 45) − (30 + 25)]/2 = 5.

IND-368 · A process has 99.38% long-term yield. What is its sigma level?

The defect rate is 0.62%, or 6,200 DPMO. The long-term z ≈ 2.5, and adding the 1.5σ shift gives about 4σ.

IND-369 · A Gage R&R shows 35% of total variation. The team wants to start DOE. What do you advise?

Stop and fix the measurement system first: at more than 30% it is unacceptable, so it may hide real effects or produce false ones. Investigate repeatability (gauge, fixture, resolution) and reproducibility (method, training). Then re-run the MSA.

IND-370 · How should regression results be judged in the Analyze phase?

Check the p-values of the coefficients, R² and adjusted R², and the residual plots for normality, constant variance and independence. Check for multicollinearity (VIF) and influential points. A significant correlation is not proof of causation without process knowledge or experiments.

IND-371 · Why can RTY be much lower than final yield?

Final yield counts only units that pass final inspection, including those that were reworked. RTY counts first-pass success at every step. The difference is the "hidden factory" of rework that adds cost and lead time without being seen.

IND-372 · What is the Taguchi quality loss function?

L(y) = k(y − m)², where m is the target and k = A/Δ², with A the loss at tolerance limit Δ. Loss rises quadratically with any deviation from target, even inside the specification.

IND-373 · What are Taguchi's signal-to-noise ratios?

Smaller-the-better: S/N = −10 log(Σy²/n). Larger-the-better: S/N = −10 log(Σ(1/y²)/n). Nominal-the-best: S/N = 10 log(ȳ²/s²). Maximizing S/N gives designs that are robust to noise factors.

IND-374 · How is QFD used in DFSS?

Quality Function Deployment's House of Quality maps customer needs (WHATs), with importance weights, to technical characteristics (HOWs) through a relationship matrix. It ranks engineering targets and shows trade-offs in the correlation roof. Later houses cascade down to part and process characteristics.

IND-375 · How would you select and prioritize Six Sigma projects?

Link them to strategic goals and the cost of poor quality. Score candidates on financial benefit, customer impact, data availability, scope (completion in 3–6 months) and likelihood of success. Use a Pareto analysis or prioritization matrix, and avoid "boil the ocean" scopes.

Lean Manufacturing & TPM

IND-376 · Daily demand is 400 units, lead time 0.5 day, safety factor 20% and container size 25. How many kanbans are needed?

N = 400 × 0.5 × 1.2 / 25 = 9.6, so 10 kanbans. The maximum WIP is 10 × 25 = 250 units.

IND-377 · A value stream map shows 12 days of lead time and 45 minutes of value-added time. What does this mean?

Process cycle efficiency = VA time / lead time ≈ 45 / (12 × 480 min, assuming one 8 h shift) ≈ 0.8%. Most of the lead time is queue and inventory, so reduce WIP and batch sizes and create flow and pull rather than speeding up processes.

IND-378 · Shift = 480 min, planned breaks 30 min, breakdowns 45 min, ideal cycle 1 min, output 350 units with 14 rejects. Find OEE.

Availability = (450 − 45)/450 = 0.90. Performance = 350 × 1 / 405 = 0.864. Quality = 336/350 = 0.96. OEE = 0.90 × 0.864 × 0.96 ≈ 74.7%, which equals 336 good units / 450 min.

IND-379 · A press changeover takes 90 minutes. How would you apply SMED?

Video the changeover and list each element. Classify elements as internal or external. Move preparation to external time: pre-staged dies, preheating, checklists, tool carts. Standardize die heights, and use quick clamps, locating pins and parallel operators. Eliminate adjustments, then standardize and train.

IND-380 · How do you implement pull in a high-mix, low-volume environment?

Replenishment kanbans for every part are impractical. Use supermarkets only for common components, sequenced pull or CONWIP for make-to-order items, a heijunka box to level the mix, and FIFO lanes with capped WIP between processes.

IND-381 · What are the seven steps of autonomous maintenance?

• Initial cleaning and inspection. 2. Eliminate sources of contamination and hard-to-reach areas. 3. Set cleaning and lubrication standards. 4. General inspection training. 5. Autonomous inspection. 6. Standardization and workplace organization. 7. Full autonomous management.

IND-382 · What is EPEI and how is it used?

"Every Part Every Interval" is the time needed to cycle through all part numbers on a shared resource. It depends on the time available for changeovers after production time. Reducing setup time with SMED shortens EPEI, which permits smaller batches and lower inventory.

IND-383 · A line has 600 units of WIP and completes 200 units per day. What is the lead time, and how can it be cut?

By Little's Law, lead time = WIP / throughput = 600/200 = 3 days. Halving WIP with CONWIP or kanban, at the same throughput, halves lead time to 1.5 days.

IND-384 · Total work content is 18 min/unit and takt time is 2.5 min. How many operators does the cell need?

Operators = 18 / 2.5 = 7.2, so 8, or 7 with kaizen to remove 0.5 min of waste. Balance the elements so each operator's load is just under takt.

IND-385 · How do you decide between preventive, predictive and run-to-failure maintenance for a machine?

Use criticality and failure-pattern analysis (RCM). Run-to-failure suits cheap, non-critical items with redundancy. Time-based preventive maintenance suits age-related wear. Condition-based or predictive maintenance (vibration, thermography, oil analysis) suits critical assets with detectable degradation. TPM operator checks support all of them.

Supply Chain & Logistics

IND-386 · What is the bullwhip effect and how is it reduced?

Demand variability is amplified as orders move upstream. Causes are demand-signal processing, order batching, price promotions and shortage gaming. Reduce it by sharing POS data (VMI, CPFR), cutting lead times, ordering in smaller and more frequent batches, using everyday low pricing and allocating by past sales.

IND-387 · Four warehouses each have independent demand σ = 100 units over lead time, with z = 1.65. How much safety stock is saved by centralizing?

Decentralized: 4 × 1.65 × 100 = 660 units. Centralized: σ = √4 × 100 = 200, so SS = 330 units. Safety stock falls by 50%, but outbound transport cost and delivery time may rise.

IND-388 · How is a distribution network design problem formulated?

As a mixed-integer program. Binary variables yⱼ decide whether facility j opens, and continuous xᵢⱼ are flows. The objective is to minimize Σ fixed cost·yⱼ + Σ transport and handling cost·xᵢⱼ, subject to demand met, capacity xᵢⱼ ≤ Capⱼ·yⱼ and service-distance limits.

IND-389 · How does transit time affect transport mode selection?

Total cost = freight + pipeline inventory + safety stock at the destination. Annual in-transit holding = D × unit value × h × transit days / 365. For high-value goods, a faster but costlier mode can lower total cost.

IND-390 · Supplier X scores Quality 8, Cost 6, Delivery 9; Supplier Y scores 7, 9, 6. Weights are 0.5, 0.3 and 0.2. Which supplier is chosen?

X = 0.5×8 + 0.3×6 + 0.2×9 = 4.0 + 1.8 + 1.8 = 7.6. Y = 3.5 + 2.7 + 1.2 = 7.4. X is preferred, but a sensitivity check on the weights is advisable.

IND-391 · What is postponement and when is it useful?

Delaying product differentiation, such as final assembly, localization, packaging or colour, until actual demand is known. Inventory is held in generic form. It suits high-variety products with common platforms and uncertain demand for each variant, because it pools risk across the variants.

IND-392 · What does the Beer Distribution Game teach?

Even with stable end demand, local decisions, delays and lack of information cause order oscillation and amplification (bullwhip) across retailer, wholesaler, distributor and factory. Players also tend to ignore inventory already in the pipeline. The structure of the system drives its behaviour.

IND-393 · Find the centre-of-gravity location for demand points (2,3) with 100 units, (6,1) with 300 and (4,7) with 200.

Cx = (2×100 + 6×300 + 4×200)/600 = 2800/600 ≈ 4.67. Cy = (3×100 + 1×300 + 7×200)/600 = 2000/600 ≈ 3.33. The starting site is (4.67, 3.33).

IND-394 · A critical part comes from a single overseas supplier. How do you reduce disruption risk?

Map tier-n suppliers and assess the probability and impact of failure. Options are qualifying a second or nearer source (dual sourcing), holding strategic safety stock, buying capacity reservation contracts, designing for interchangeable parts and monitoring supplier health. Weigh the extra cost against the expected loss.

IND-395 · How does the Clarke–Wright savings algorithm work?

Start with a separate depot-to-customer round trip for each customer. Compute savings Sᵢⱼ = d₀ᵢ + d₀ⱼ − dᵢⱼ for joining customers i and j on one route. Merge routes in descending order of savings while vehicle capacity and route-length limits are met.

Facility Layout & Material Handling

IND-396 · What is assembly line balancing and what is its objective?

Assigning tasks to workstations in an order that respects precedence, so that no station's load exceeds the cycle time (takt). The objective is to minimize the number of stations or the idle time, which maximizes line efficiency.

IND-397 · Six tasks total 2.8 min. 400 min are available and demand is 200 units per shift. Find the cycle time, minimum stations and efficiency.

C = 400/200 = 2.0 min. N_min = ⌈2.8/2.0⌉ = 2. If precedence allows two stations, efficiency = 2.8/(2 × 2.0) = 70% and balance delay = 30%.

IND-398 · What is the ranked positional weight (RPW) method?

A line-balancing heuristic (Helgeson–Birnie). Each task's positional weight is its own time plus the times of all its successors. Tasks are ranked in descending weight and assigned in that order to the earliest station where precedence and cycle time allow.

IND-399 · How does CRAFT evaluate a pairwise exchange?

Layout cost = Σᵢ Σⱼ fᵢⱼ·cᵢⱼ·dᵢⱼ, with rectilinear distances between department centroids. For each candidate swap of adjacent or equal-area departments, estimate the new cost, make the swap with the largest reduction, and repeat until no swap improves. The result is only a local optimum.

IND-400 · Why is the facility layout problem hard to solve optimally?

Assigning n departments to n locations to minimize Σ flow × distance is the quadratic assignment problem, which is NP-hard. Exact methods are practical only for small n, so heuristics such as CRAFT, simulated annealing and genetic algorithms are used.

IND-401 · How is the cycle time of a unit-load AS/RS estimated?

The S/R machine moves horizontally and vertically at the same time, so travel time to a location = max(t_h, t_v). A single command (one store or retrieve) is about 2 × average travel + 2 × pick/deposit time. A dual command (store + retrieve in one trip) raises throughput per trip.

IND-402 · When would you choose AGVs over a fixed conveyor?

AGVs suit variable routes, layout changes, low-to-medium flow rates and shared aisles. Conveyors suit high-volume, fixed-path, continuous flow at a low cost per move. Compare flow rate, flexibility needs, capital cost, floor obstruction and safety.

IND-403 · What is the cube-per-order index (COI) rule?

COI = space required by an item ÷ number of picks per period. Items with the lowest COI, which are small and fast-moving, get the locations closest to the I/O point. This minimizes expected travel in dedicated storage.

IND-404 · Rank order clustering leaves some parts needing machines in two cells. How do you handle these exceptional elements?

Options: duplicate the bottleneck machine if utilization and cost justify it, reroute the part to an alternative process, subcontract the operation, redesign the part, or accept inter-cell moves. Choose by comparing the cost of each with the cost of the extra handling.

IND-405 · Departments A–B have 50 trips/day and A–C have 10. Layout 1 puts A–B 30 m apart and A–C 10 m. Layout 2 puts A–B 10 m and A–C 30 m. Which is better?

Layout 1: 50×30 + 10×10 = 1,600 trip-m/day. Layout 2: 50×10 + 10×30 = 800 trip-m/day. Layout 2 halves material movement because the high-flow pair is adjacent.

Project Management

IND-406 · Activities: A(3), B(4, after A), C(2, after A), D(5, after B), E(3, after C), F(2, after D, E). Find the critical path and the floats.

Paths: A-B-D-F = 14 and A-C-E-F = 10. The critical path is A-B-D-F, with a 14-day duration. C and E each have total float 4. E has free float 4, and C has free float 0.

IND-407 · A critical path has Tₑ = 30 days and variance 16 days². What is the probability of finishing within 34 days?

σ = 4 days. Z = (34 − 30)/4 = 1.0, so P ≈ 84.1%.

IND-408 · In IND-406, B can be crashed 1 day at $300 and D 2 days at $500/day. The other path is 10 days. How do you shorten the project to 12 days?

Crash B by 1 day ($300) to reach 13 days, then D by 1 day ($500) to reach 12 days. The total extra direct cost is $800. Path A-C-E-F (10 days) stays non-critical.

IND-409 · How is the optimal (minimum-cost) project duration found?

Total cost = direct cost (rises as the project is crashed) + indirect cost (falls with shorter duration, at a fixed rate per day). Crash critical activities in order of cost slope while each slope is less than the daily indirect cost. Stop when the next slope exceeds it.

IND-410 · BAC = $100k, PV = $50k, EV = $40k and AC = $45k. Assess the project.

CPI = 40/45 = 0.89 (over budget) and SPI = 40/50 = 0.80 (behind schedule). EAC = BAC/CPI ≈ $112.5k, so the forecast overrun is about $12.5k.

IND-411 · How are resource-constrained schedules built?

Use priority-rule heuristics in serial or parallel schedule-generation schemes. Rules include minimum slack, longest processing time, most successors and greatest resource demand. They schedule activities as resources allow and may extend the project beyond the CPM duration. Exact solutions use integer programming for small networks.

IND-412 · What is merge bias in PERT?

Where several near-critical paths converge, the project finishes only when the slowest path finishes. Considering only the critical path therefore underestimates the expected duration and overstates the probability of on-time completion. Monte Carlo simulation captures this effect.

IND-413 · How is Monte Carlo simulation used for schedule risk analysis?

Assign duration distributions (triangular, beta or PERT) to activities and sample them thousands of times, computing the project finish each time. The output is a completion-date distribution (P50, P80), criticality indices and sensitivity rankings for setting contingency.

IND-414 · A project is 3 weeks behind with a fixed deadline. What options do you evaluate?

Re-check the critical and near-critical paths. Then consider fast-tracking overlapping activities (rework risk), crashing the lowest cost-slope critical activities, reducing or deferring scope with stakeholder agreement, adding shifts, or re-sequencing resources. Quantify the cost and risk of each and choose the cheapest feasible mix.

IND-415 · How are buffers sized in critical chain project management?

Task estimates are cut to aggressive values, often about 50% of the safe estimate. A common rule sizes the project buffer at about half the removed safety on the critical chain, or the root-sum-square of individual safeties. Progress is managed by buffer consumption against chain completion.

Engineering Economics & Forecasting

IND-416 · An investment of $10,000 returns $4,000 a year for 3 years. At 10%, should it be accepted?

(P/A, 10%, 3) = 2.4869. PW = 4000 × 2.4869 = $9,948, so NPV = −$52. Reject it, though it is marginal and the decision is sensitive to the estimates.

IND-417 · Estimate the IRR of the project in IND-416.

At 9%, P/A = 2.5313, giving NPV = +$125. At 10%, NPV = −$52. Interpolating: IRR ≈ 9% + 125/177 × 1% ≈ 9.7%, which is below a 10% MARR.

IND-418 · How is the economic life of an asset determined?

For each possible retention period n, compute the equivalent uniform annual cost: capital recovery (P − S_n)(A/P, i, n) + S_n·i, plus the annualized O&M cost. The n with the minimum EUAC is the economic life, which is the best replacement interval.

IND-419 · How do you compare mutually exclusive alternatives with unequal lives?

Use annual worth, which implicitly assumes repeatability, or present worth over the least common multiple of the lives, or a stated study period with estimated salvage values. Do not compare PWs over different lives directly.

IND-420 · A machine costs $50,000, lasts 5 years with zero salvage and has O&M of $5,000/yr. Find its equivalent annual cost at 10%.

(A/P, 10%, 5) = 0.2638. Capital recovery = 50,000 × 0.2638 = $13,190. EAC = 13,190 + 5,000 = $18,190/yr.

IND-421 · How is benefit–cost analysis applied to public projects?

B/C = PW(benefits − disbenefits) / PW(costs). Accept if B/C ≥ 1. For mutually exclusive alternatives, use incremental B/C between successive alternatives ranked by cost, not the highest ratio.

IND-422 · How do Holt's and Holt–Winters' methods extend exponential smoothing?

Holt adds a trend term: level Lₜ = αAₜ + (1 − α)(Lₜ₋₁ + Tₜ₋₁), trend Tₜ = β(Lₜ − Lₜ₋₁) + (1 − β)Tₜ₋₁, and forecast Fₜ₊ₖ = Lₜ + kTₜ. Holt–Winters adds a seasonal index with a third smoothing constant γ, in additive or multiplicative form.

IND-423 · The last forecast was 100, actual demand was 120 and α = 0.3. What is the next forecast?

F = 0.3 × 120 + 0.7 × 100 = 36 + 70 = 106.

IND-424 · How are seasonal indices computed and used?

Average the demand for each season over several cycles and divide by the overall average per season. The indices average 1.0. Deseasonalize the data by dividing by the index, forecast the trend, then reseasonalize by multiplying by the index.

IND-425 · A part can be bought for $12 or made at $8 variable cost plus $20,000/yr fixed cost. When should you make it?

Make when 20,000 + 8Q < 12Q, that is, Q > 20,000/4 = 5,000 units/yr. Also consider capacity, quality, strategic core competence and supplier risk.

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