Operations Research and Operations Management

Section 7 of the GATE Production and Industrial Engineering (PI) paper, and its last, is the quantitative core of running a plant. Operations research — linear programming (formulation, the simplex method, duality and sensitivity analysis), transportation and assignment models, integer programming, Markovian queuing models and an introduction to discrete event simulation. Engineering economy and costing — elementary cost accounting and the methods of depreciation, break-even analysis and activity-based costing. Production control — forecasting by causal and time-series models (moving average, exponential smoothing, trend and seasonality), aggregate production planning, master production scheduling, MRP, MRP-II and ERP, routing, scheduling and priority dispatching, push and pull systems with lean and agile manufacturing, logistics, distribution and supply chain management, and inventory — its functions, costs and classifications, the deterministic models, quantity discounts, and perpetual and periodic control systems. Project management — CPM, PERT, the Gantt chart and the time-cost trade-off. This is where the paper sets most of its numerical answer questions; every formula below is one GATE uses directly, and every worked figure was computed twice.

1. Linear programming: formulation, graphical solution, simplex, duality and sensitivity

A linear programme has decision variables, a linear objective to maximise or minimise, linear constraints and non-negativity. Take maximise Z = 3x + 5y subject to x ≤ 4, 2y ≤ 12, 3x + 2y ≤ 18, x, y ≥ 0. The feasible region is a convex polygon and the optimum lies at a corner point: the corners are (0, 0), (4, 0), (4, 3), (2, 6) and (0, 6), with Z = 0, 12, 27, 36 and 30, so the optimum is x = 2, y = 6, Z = 36. If the objective line is parallel to a binding edge there are multiple optima; if the region is unbounded in an improving direction the problem is unbounded; if no point satisfies all constraints it is infeasible.

The simplex method moves from one basic feasible solution (corner) to an adjacent better one. Each ≤ constraint gets a slack variable, each ≥ constraint a surplus and an artificial variable, and each = constraint an artificial variable, handled by the Big-M or two-phase method. For a maximisation the entering variable is the one with the most positive net contribution C_j − Z_j; the leaving variable follows the minimum ratio test (right-hand side divided by the positive entries of the entering column). The tableau is optimal when every C_j − Z_j ≤ 0. A zero C_j − Z_j for a non-basic variable signals alternative optima, a tie in the ratio test signals degeneracy, an entering column with no positive entry signals unboundedness, and an artificial variable left positive in the final tableau signals infeasibility.

Every LP has a dual. For a primal 'maximise c·x subject to Ax ≤ b, x ≥ 0' the dual is 'minimise b·u subject to Aᵀu ≥ c, u ≥ 0': one dual variable per primal constraint, and the roles of objective coefficients and right-hand sides swap. Weak duality: any feasible dual objective bounds the primal. Strong duality: at optimality the two objectives are equal. Complementary slackness: a constraint with slack at the optimum has a zero dual price, and a positive dual price means its constraint is binding. If the primal is unbounded the dual is infeasible. The optimal dual values are the shadow prices — the gain in Z per unit increase in a right-hand side. In the example, x ≤ 4 has slack (x = 2) so its price is 0, and the other two prices are 1.5 and 1: check 12 × 1.5 + 18 × 1 = 36. Sensitivity analysis asks how far a cost coefficient can move before the optimal basis changes (range of optimality) and how far a right-hand side can move with the shadow price still valid (range of feasibility); the reduced cost of a non-basic variable is how much its objective coefficient must improve before it enters.

2. Transportation and assignment models, integer programming, Markovian queues and discrete event simulation

The transportation model ships from m sources to n destinations at least cost. It is balanced when total supply equals total demand; otherwise a dummy row or column with zero cost absorbs the difference. An initial basic feasible solution comes from the north-west corner rule (ignores cost), the least-cost method, or Vogel's approximation (penalties = difference of the two smallest costs in each row and column; allocate where the penalty is largest), which usually starts closest to optimal. A basic solution has m + n − 1 allocations; fewer is degenerate, fixed by placing a tiny ε in an independent empty cell. Optimality is tested by MODI: solve u_i + v_j = c_ij on occupied cells (setting one u = 0), then compute Δ_ij = c_ij − u_i − v_j for empty cells; if all Δ_ij ≥ 0 the solution is optimal, otherwise the most negative cell enters around a closed loop. The assignment model is the special case of n jobs to n agents one-to-one, solved by the Hungarian method: subtract row minima, then column minima, cover all zeros with the fewest lines, and if fewer than n lines are needed subtract the smallest uncovered number from the uncovered cells and add it at intersections; repeat until n lines, then assign on independent zeros. Maximisation problems are converted by subtracting every entry from the largest; unbalanced ones get a dummy row or column.

Integer programming requires some or all variables to be integers (0-1 variables for yes/no decisions such as opening a plant). Rounding the LP optimum can be infeasible or far from optimal. Branch and bound solves the LP relaxation, branches on a fractional variable (x ≤ ⌊x⌋ and x ≥ ⌈x⌉), and prunes any branch whose relaxation bound is no better than the best integer solution found; Gomory's cutting-plane method instead adds constraints that cut off the fractional optimum but no integer point. The LP relaxation's optimum is always at least as good as the integer optimum, which is why it serves as the bound.

Markovian queues have Poisson arrivals (exponential inter-arrival times, rate λ) and exponential service (rate μ); Kendall notation (a/b/c):(d/e) names arrival and service distributions, servers, capacity and discipline. For the single-server M/M/1 queue with ρ = λ/μ < 1: P0 = 1 − ρ, P(n) = (1 − ρ)ρⁿ, P(n ≥ k) = ρᵏ, L = ρ/(1 − ρ), L_q = ρ²/(1 − ρ), W = 1/(μ − λ) and W_q = λ/[μ(μ − λ)]. Little's law L = λW (and L_q = λW_q) holds for any stable queue. With λ = 8 per hour and μ = 10 per hour: ρ = 0.8, L = 4, L_q = 3.2, W = 0.5 h = 30 min and W_q = 0.4 h = 24 min. Multi-server M/M/c queues follow the same logic with c servers sharing the load.

Discrete event simulation models a system whose state changes only at instants — arrivals, service completions, breakdowns. A simulation clock jumps from event to event taken from a time-ordered future event list; entities carry attributes; statistics accumulate as the clock advances. Random inputs are generated from uniform random numbers U in (0, 1) by the inverse-transform method: for an exponential variable with rate λ, x = −(1/λ) ln(1 − U), equivalently −(1/λ) ln U. Monte Carlo sampling from an empirical distribution maps random-number ranges to values through the cumulative probabilities. Results are estimates: several independent replications, a warm-up period discarded, and confidence intervals are needed before a conclusion is drawn.

3. Engineering economy and costing: cost accounting, depreciation, break-even analysis and activity-based costing

How the elements of cost build up to the selling price
Build-upConsists of
Prime costDirect material + direct labour + direct expenses
Factory (works) costPrime cost + factory overheads
Cost of production (office cost)Factory cost + administrative overheads
Total cost (cost of sales)Cost of production + selling and distribution overheads
Selling priceTotal cost + profit

Depreciation spreads an asset's cost, less salvage, over its life. With first cost P, salvage S and life n: straight line D = (P − S)/n every year (P = Rs 1 00 000, S = Rs 10 000, n = 9 gives Rs 10 000 a year and a book value of Rs 70 000 after three years); declining balance charges a fixed rate on the opening book value each year, so the charge falls year by year (double-declining uses twice the straight-line rate); sum-of-the-years'-digits charges (remaining life/SYD) × (P − S), SYD = n(n + 1)/2 — for n = 5, SYD = 15 and the first year's charge is (5/15) × 90 000 = Rs 30 000; and the sinking-fund method sets aside equal annual amounts that, with interest, accumulate to P − S. The accelerated methods write off more early, which matches how value is often lost and defers tax.

Break-even analysis separates fixed cost F from variable cost v per unit. At selling price p, profit = Q(p − v) − F, so the break-even quantity is Q_BE = F/(p − v), where p − v is the contribution per unit; the break-even sales value is F divided by the profit-volume ratio (p − v)/p, and the margin of safety is actual sales minus break-even sales. With F = Rs 60 000, p = Rs 50 and v = Rs 30, Q_BE = 60 000/20 = 3000 units. Activity-based costing replaces a single overhead rate (say per direct labour hour) with cost pools for the activities that actually cause overhead — machine set-ups, purchase orders, inspections, material moves — each with its own cost driver and rate: if set-ups cost Rs 50 000 for 100 set-ups, the rate is Rs 500 per set-up, and a product needing 20 set-ups carries Rs 10 000 of set-up overhead whatever its labour hours. It stops low-volume, complex products from being subsidised by high-volume simple ones.

4. Forecasting; aggregate planning; master production scheduling; MRP, MRP-II and ERP

Causal forecasting relates demand to a driver x by least-squares linear regression y = a + bx, with b = (nΣxy − ΣxΣy)/(nΣx² − (Σx)²) and a = ȳ − b x̄. For (1, 2), (2, 4), (3, 5), (4, 8): n = 4, Σx = 10, Σy = 19, Σxy = 57, Σx² = 30, so b = (228 − 190)/(120 − 100) = 1.9 and a = (19 − 19)/4 = 0, giving a forecast of 9.5 at x = 5. Time-series methods use the demand history alone. A simple moving average of the last N periods smooths noise but lags a trend; a weighted moving average puts more weight on recent data. Exponential smoothing Ft+1 = F_t + α(D_t − F_t) = αD_t + (1 − α)F_t weights past data geometrically: with F_t = 100, D_t = 120 and α = 0.2, Ft+1 = 104. A larger α responds faster but passes on more noise. Trend is handled by double (Holt's) smoothing or by fitting a line to time; seasonality by seasonal indices (the ratio of a season's average to the overall average), deseasonalising, forecasting and then re-seasonalising. Accuracy is measured by MAD, MSE and MAPE, and bias by the tracking signal = running sum of errors/MAD.

Aggregate production planning sets total output, workforce and inventory by period over a medium horizon (typically months) for a product family. Its three pure strategies are a level plan (constant output, inventory absorbs demand swings), a chase plan (output follows demand through hiring, firing or overtime) and a mixed plan; the costs traded are regular and overtime production, hiring and lay-off, inventory holding, backorders and subcontracting. The master production schedule disaggregates the plan into quantities of specific end items by week; its time fences freeze the near term, and the available-to-promise quantity tells sales what can still be committed. MRP explodes the MPS through the bill of materials: for each item, net requirement = gross requirement − on-hand inventory − scheduled receipts (+ safety stock), and planned orders are released one lead time ahead (lead-time offsetting), lot-sized lot-for-lot or otherwise. Gross 500, on hand 120 and scheduled receipts 80 give a net requirement of 300. MRP-II (manufacturing resource planning) closes the loop with capacity requirements planning, shop-floor control and the financial plan; ERP extends the same integrated database to the whole enterprise — sales, purchasing, finance, human resources and the supply chain.

5. Routing, scheduling and priority dispatching; push and pull; lean and agile; logistics and supply chains

Routing fixes the path and sequence of operations a part follows; scheduling fixes when each operation happens on which machine; dispatching releases the work and decides which waiting job goes next. Priority rules on a single machine: SPT (shortest processing time first) minimises mean flow time, mean waiting time and average work-in-process; EDD (earliest due date first) minimises maximum lateness; FCFS is fair but performs poorly; critical ratio = time remaining to due date/processing time remaining, smallest first; Moore's algorithm minimises the number of late jobs. For n jobs through two machines in the same order, Johnson's rule minimises the makespan: find the smallest time anywhere; if it is on machine 1 schedule that job as early as possible, if on machine 2 as late as possible; remove it and repeat. Jobs A (5, 2), B (1, 6), C (9, 7), D (3, 8), E (10, 4) give the order B-D-C-E-A and a makespan of 30.

In a push system production is scheduled in advance from forecasts (MRP is the classic push system) and work is pushed to the next stage whether or not it is ready. In a pull system consumption downstream authorises production upstream, typically through kanban cards, capping work-in-process. Lean manufacturing, from the Toyota Production System, eliminates the seven wastes (muda) — overproduction, waiting, transport, over-processing, inventory, motion and defects — through value-stream mapping, 5S, set-up reduction (SMED), one-piece flow, levelled production (heijunka), mistake-proofing (poka-yoke) and continuous improvement. Agile manufacturing stresses the ability to respond quickly and economically to unpredictable changes in demand and product variety, through reconfigurable resources, modular products and virtual partnerships; lean is strongest with stable demand, agile with volatile demand. Logistics plans and controls the flow and storage of goods and information from origin to consumption; distribution covers warehousing, transport modes, cross-docking and network design; supply chain management coordinates suppliers, manufacturers, distributors and retailers as one system. The bullwhip effect — order variability amplifying as it moves upstream, from order batching, forecast updating, price fluctuation and shortage gaming — is its classic problem, reduced by sharing demand information, smaller batches and vendor-managed inventory.

6. Inventory: functions, costs and classifications; deterministic models; quantity discounts; perpetual and periodic systems

Inventory functions: cycle stock (ordering in lots), safety stock (buffer against uncertainty), anticipation stock (seasonal build-up), pipeline stock (in transit) and decoupling stock (letting stages run independently). Costs: ordering or set-up cost per order (S), holding or carrying cost per unit per year (H — capital, storage, insurance, obsolescence, often a percentage of the unit price), shortage or stock-out cost, and the purchase cost itself. Classifications: ABC by annual consumption value (a few A items take most of the value and get tight control); VED by criticality (vital, essential, desirable); FSN by movement (fast, slow, non-moving); HML by unit price; SDE by availability (scarce, difficult, easy).

Deterministic inventory models (D annual demand, S cost per order, H holding cost per unit per year)
ModelOptimal quantityNotes
Basic EOQ (instant replenishment, no shortages)Q* = √(2DS/H)At Q*, annual ordering cost = annual holding cost; total = √(2DSH)
Production (EPQ), gradual replenishment at rate p > dQ* = √[2DS/(H(1 − d/p))]Maximum inventory Q(1 − d/p), not Q
Planned shortages (backorder cost B per unit per year)Q* = √(2DS/H) × √[(H + B)/B]Larger lots than EOQ; tends to EOQ as B grows

With D = 10 000 units a year, S = Rs 200 and H = Rs 4, Q* = √(2 × 10 000 × 200/4) = 1000 units, ordering and holding cost Rs 2000 each, Rs 4000 together. Quantity discounts: compute the EOQ at each price (H being a percentage of price); if an EOQ is feasible in its price range keep it, otherwise move it to the nearest boundary of that range; then compare the total cost including purchase cost at every candidate and take the least. With H = 20 % of price, Rs 20 below 2000 units and Rs 19 at 2000 or more: at Rs 20 the EOQ of 1000 is valid, total Rs 2 04 000; at Rs 19 the EOQ of 1026 is below 2000, so the candidate is 2000, total 1 90 000 + 1000 + 3800 = Rs 1 94 800 — order 2000. The reorder point in a continuous system is ROP = dL + safety stock (d demand per day, L lead time in days).

Perpetual (continuous-review, Q) system: stock is tracked continuously and a fixed quantity Q, usually the EOQ, is ordered whenever stock falls to the reorder point; the order interval varies, and safety stock covers only the lead time — fewer units of safety stock, more record-keeping (the two-bin system is its manual form). Periodic (fixed-interval, P) system: stock is reviewed every T periods and an order brings it up to a target level M, so the order quantity varies; safety stock must cover the review period plus the lead time, so more is needed, but orders for many items from one supplier can be combined.

⚠️ Quantity discounts are decided on total cost, purchase price included
Comparing only ordering plus holding cost would reject the discount in the example above (Rs 4000 at 1000 units against Rs 4800 at 2000 units) and throw away Rs 10 000 of price saving. The purchase cost D × price is the largest term, and it is the one that changes between price breaks.

7. Project management: CPM, PERT, the Gantt chart and the time-cost trade-off

A project network shows activities on arrows (AOA, with dummy activities where needed to show dependence) or on nodes (AON). The forward pass gives each activity's earliest start ES and earliest finish EF = ES + t (ES = the largest EF of its predecessors); the backward pass from the project end gives the latest finish LF and latest start LS = LF − t (LF = the smallest LS of its successors). Total float = LS − ES = LF − EF is how far an activity can slip without delaying the project; free float is how far it can slip without delaying any successor's earliest start. The critical path is the longest path, made of zero-float activities, and its length is the project duration; there can be more than one. Example: A(3) and B(4) start; C(2) and D(5) follow A; E(3) follows B and C; F(2) follows D and E. Forward: A 0-3, B 0-4, C 3-5, D 3-8, E 5-8, F 8-10, so the project takes 10; backward, B has LS 1 and ES 0, a total float of 1, and both A-D-F and A-C-E-F are critical.

PERT treats activity times as uncertain, with optimistic a, most likely m and pessimistic b estimates from a beta distribution: expected time t_e = (a + 4m + b)/6 and standard deviation σ = (b − a)/6 — for 4, 6 and 14 days, t_e = 7 and σ = 1.67. The project time is taken as normal with mean T_e = Σt_e and variance Σσ² along the critical path (the variances add, not the standard deviations), so the probability of finishing by T is Φ(Z) with Z = (T − T_e)/σ_project. A Gantt chart draws each activity as a horizontal bar against a time scale — clear for tracking progress, but it does not show the dependencies that a network does. Time-cost trade-off (crashing): each activity has a normal time and cost and a shorter crash time at a higher crash cost; its cost slope = (crash cost − normal cost)/(normal time − crash time). To shorten the project, crash the critical activity with the least cost slope, one unit at a time, re-checking the critical path after each step; when several paths are critical, all must be shortened together. Total cost = direct cost (rising as the project is crashed) + indirect cost (falling as it shortens), and the optimum duration minimises their sum.

⚠️ Add variances along the critical path, never standard deviations
Three critical activities with σ = 1, 1 and 1 day give a project σ of √3 = 1.73 days, not 3. Summing standard deviations overstates the spread, understates Z and so understates the probability of finishing on time.

Key takeaways

  • LP optima sit at corners; simplex is optimal when all C_j − Z_j ≤ 0; the dual of a max is a min with one variable per constraint, equal objective at the optimum, and zero shadow price on any slack constraint.
  • Transportation needs m + n − 1 allocations and is tested by MODI; assignment is solved by the Hungarian method; M/M/1: L = ρ/(1 − ρ), W = 1/(μ − λ), W_q = λ/[μ(μ − λ)], L = λW.
  • Break-even Q = F/(p − v); SYD first-year charge n/[n(n + 1)/2] × (P − S); ABC costing assigns overhead by cost drivers; exponential smoothing Ft+1 = F_t + α(D_t − F_t); regression b = (nΣxy − ΣxΣy)/(nΣx² − (Σx)²).
  • SPT minimises mean flow time, EDD maximum lateness, Johnson's rule the two-machine makespan; MRP is push, kanban is pull; net requirement = gross − on hand − scheduled receipts.
  • EOQ √(2DS/H) with equal ordering and holding costs; judge discounts on total cost including price; PERT t_e = (a + 4m + b)/6 and σ = (b − a)/6 with variances added on the critical path; crash the cheapest critical activity first.

Practice questions (24)

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.

  1. Maximise Z = 3x + 5y subject to x ≤ 4, 2y ≤ 12, 3x + 2y ≤ 18 and x, y ≥ 0. What is the optimal value of Z?

    Numerical answer — type the value.

    Show answer

    Answer: 36

    The corner points are (0, 0), (4, 0), (4, 3), (2, 6) and (0, 6), giving Z = 0, 12, 27, 36 and 30. The optimum is at (2, 6), where 2y ≤ 12 and 3x + 2y ≤ 18 both bind: Z = 6 + 30 = 36. Stopping at (4, 3), where x is at its largest, gives 27, and (0, 6) gives 30 — neither is the best corner.
  2. For the LP maximise Z = 3x + 5y subject to x ≤ 4, 2y ≤ 12, 3x + 2y ≤ 18, x, y ≥ 0, what is the shadow price (optimal dual value) of the constraint 3x + 2y ≤ 18?

    1. 0
    2. 1
    3. 1.5
    4. 3
    Show answer

    Answer: B — 1

    At the optimum (2, 6) the first constraint is slack, so u1 = 0. The dual constraints for x and y then read 3u3 = 3 and 2u2 + 2u3 = 5, so u3 = 1 and u2 = 1.5; check 12 × 1.5 + 18 × 1 = 36 = Z. Raising 18 to 19 moves the corner to (7/3, 6) and Z to 37, a gain of exactly 1. 1.5 belongs to the constraint 2y ≤ 12, and 0 to the slack constraint x ≤ 4.
  3. Which of the following statements about linear programming and duality are correct?

    1. A ≤ constraint is converted to an equation by adding a slack variable
    2. When both the primal and dual have optimal solutions, their optimal objective values are equal
    3. A constraint that has positive slack at the optimum has a positive shadow price
    4. If the primal is unbounded, the dual is infeasible
    Show answer

    Answer: A — A ≤ constraint is converted to an equation by adding a slack variable; B — When both the primal and dual have optimal solutions, their optimal objective values are equal; D — If the primal is unbounded, the dual is infeasible

    Slack variables convert ≤ constraints, strong duality equates the optimal objectives, and an unbounded primal leaves the dual with no feasible point (weak duality would otherwise bound it). By complementary slackness, a constraint with slack at the optimum has a ZERO shadow price — more of a resource that is not used up is worth nothing — so the third statement is false.
  4. A balanced transportation problem has 3 sources and 4 destinations. How many allocations does a non-degenerate basic feasible solution have?

    1. 12
    2. 7
    3. 6
    4. 4
    Show answer

    Answer: C — 6

    A basic solution has m + n − 1 = 3 + 4 − 1 = 6 allocations, because one of the m + n supply and demand equations is redundant once the problem is balanced. Seven (m + n) forgets that redundancy, and fewer than six means the solution is degenerate and MODI cannot be applied until an ε allocation is added.
  5. Three jobs are to be assigned one each to three machines. The cost matrix (rows jobs 1-3, columns machines 1-3) is: job 1: 9, 2, 7; job 2: 6, 4, 3; job 3: 5, 8, 1. What is the minimum total cost?

    Numerical answer — type the value.

    Show answer

    Answer: 9

    Row reduction gives (7, 0, 5), (3, 1, 0), (4, 7, 0); column reduction subtracts 3 from column 1, giving (4, 0, 5), (0, 1, 0), (1, 7, 0). Independent zeros: job 1 → machine 2, job 2 → machine 1, job 3 → machine 3, cost 2 + 6 + 1 = 9. Taking each job's cheapest machine greedily (2, 3, 1) is infeasible because jobs 2 and 3 would both want machine 3 — the reason the Hungarian method exists.
  6. Which method solves integer programming problems by repeatedly solving LP relaxations, splitting on fractional variables and pruning branches by their bounds?

    1. Vogel's approximation method
    2. Branch and bound
    3. The Hungarian method
    4. The MODI method
    Show answer

    Answer: B — Branch and bound

    Branch and bound branches on a fractional variable (x ≤ ⌊x⌋ or x ≥ ⌈x⌉) and prunes any branch whose LP bound cannot beat the best integer solution so far. Vogel and MODI are transportation-model methods, and the Hungarian method solves the assignment problem.
  7. Customers arrive at a single-server counter in a Poisson stream at 8 per hour and service times are exponential with a mean rate of 10 per hour. What is the average time a customer waits in the queue before service, in minutes?

    Numerical answer — type the value.

    Show answer

    Answer: 24

    W_q = λ/[μ(μ − λ)] = 8/(10 × 2) = 0.4 h = 24 min. The time in the system, W = 1/(μ − λ) = 0.5 h = 30 min, includes the 6-minute mean service and is the tempting wrong answer; the two differ by exactly 1/μ.
  8. In an M/M/1 queue with utilisation ρ = 0.8, what is the probability that there are 3 or more customers in the system?

    1. 0.512
    2. 0.102
    3. 0.640
    4. 0.200
    Show answer

    Answer: A — 0.512

    P(n) = (1 − ρ)ρⁿ, so P(n ≥ k) = ρᵏ = 0.8³ = 0.512. 0.102 is P(n = 3) alone, (0.2)(0.512), which misses 4, 5 and more; 0.640 is ρ², the probability of 2 or more.
  9. In a discrete event simulation, exponentially distributed inter-arrival times with rate λ are generated from uniform random numbers U in (0, 1) by

    1. x = λU
    2. x = −(1/λ) ln U
    3. x = e−λU
    4. x = U/λ
    Show answer

    Answer: B — x = −(1/λ) ln U

    Inverse transform: setting U = F(x) = 1 − e−λx and solving gives x = −(1/λ) ln(1 − U); since 1 − U is itself uniform, −(1/λ) ln U is equivalent. U/λ and λU produce uniform, not exponential, times, and e−λU lies between 0 and 1 and is not a time at all.
  10. A product sells for Rs 50 per unit, its variable cost is Rs 30 per unit and fixed costs are Rs 60 000 per year. What is the break-even quantity in units per year?

    Numerical answer — type the value.

    Show answer

    Answer: 3000

    Each unit contributes p − v = Rs 20 towards fixed costs, so Q_BE = F/(p − v) = 60 000/20 = 3000 units. Dividing by the price alone gives 1200 units, which ignores that the variable cost of each unit must be covered first.
  11. A machine costs Rs 1 00 000, has a salvage value of Rs 10 000 and a life of 5 years. What is the depreciation charge in year 1 by the sum-of-the-years'-digits method?

    1. Rs 18 000
    2. Rs 30 000
    3. Rs 33 333
    4. Rs 6 000
    Show answer

    Answer: B — Rs 30 000

    SYD = 5 × 6/2 = 15, and the year-1 charge is (5/15) × (1 00 000 − 10 000) = Rs 30 000. Rs 18 000 is the straight-line charge, Rs 33 333 applies 5/15 to the first cost without removing the salvage value, and Rs 6000 is the year-5 charge, (1/15) × 90 000.
  12. In activity-based costing, machine set-up costs of Rs 50 000 are incurred for 100 set-ups in a period. A product requiring 20 set-ups is charged how much set-up overhead?

    1. Rs 2 500
    2. Rs 10 000
    3. Rs 20 000
    4. An amount proportional to its direct labour hours
    Show answer

    Answer: B — Rs 10 000

    The cost driver is the number of set-ups: rate = 50 000/100 = Rs 500 per set-up, so 20 set-ups carry Rs 10 000. Allocating by direct labour hours is the traditional method ABC replaces, because it charges a simple high-volume product for set-ups it does not cause.
  13. The forecast for a month was 100 units and actual demand was 120 units. Using simple exponential smoothing with α = 0.2, what is the forecast for the next month?

    Numerical answer — type the value.

    Show answer

    Answer: 104

    Ft+1 = F_t + α(D_t − F_t) = 100 + 0.2 × 20 = 104. Weighting the old forecast by α instead, 0.2 × 100 + 0.8 × 120 = 116, swaps the weights — α goes on the newest demand.
  14. Sales y (thousand units) against advertising spend x (lakh rupees) over four periods were (1, 2), (2, 4), (3, 5) and (4, 8). Fitting y = a + bx by least squares, what sales are forecast for x = 5? (Answer to one decimal place.)

    Numerical answer — type the value.

    Show answer

    Answer: 9.5

    n = 4, Σx = 10, Σy = 19, Σxy = 2 + 8 + 15 + 32 = 57, Σx² = 30. b = (4 × 57 − 10 × 19)/(4 × 30 − 10²) = 38/20 = 1.9 and a = (19 − 1.9 × 10)/4 = 0, so y(5) = 9.5. Extrapolating the last step (from 5 to 8, a rise of 3) gives 11, which is not a least-squares fit.
  15. For a component in a given week, the gross requirement is 500, projected on-hand inventory is 120 and a scheduled receipt of 80 is due. With no safety stock, what is the net requirement?

    1. 500
    2. 380
    3. 300
    4. 420
    Show answer

    Answer: C — 300

    Net requirement = gross − on hand − scheduled receipts = 500 − 120 − 80 = 300, and a planned order for 300 is then released one lead time earlier. 380 forgets the scheduled receipt and 420 forgets the stock on hand.
  16. Five jobs pass through machine M1 then machine M2. Processing times (M1, M2) in hours are A (5, 2), B (1, 6), C (9, 7), D (3, 8), E (10, 4). Using Johnson's rule, what is the minimum makespan in hours?

    Numerical answer — type the value.

    Show answer

    Answer: 30

    Smallest times in turn: B's 1 on M1 (first), A's 2 on M2 (last), D's 3 on M1 (second), E's 4 on M2 (fourth), leaving C third: B-D-C-E-A. M1 finishes jobs at 1, 4, 13, 23, 28; M2 runs B 1-7, D 7-15, C 15-22, E 23-27 (waiting for M1), A 28-30. Makespan = 30 h. Adding all M1 times plus the last M2 time (28 + 2) happens to give 30 here only because M2 never makes M1 wait at the end.
  17. Which of the following statements about production control and supply chains are correct?

    1. On a single machine, the shortest-processing-time rule minimises the mean flow time
    2. On a single machine, the earliest-due-date rule minimises the maximum lateness
    3. A kanban system is a push system driven by the forecast
    4. The bullwhip effect is the amplification of order variability as orders move upstream in a supply chain
    Show answer

    Answer: A — On a single machine, the shortest-processing-time rule minimises the mean flow time; B — On a single machine, the earliest-due-date rule minimises the maximum lateness; D — The bullwhip effect is the amplification of order variability as orders move upstream in a supply chain

    SPT and EDD are the two classical single-machine optimality results, and the bullwhip effect is exactly the upstream amplification of variability. Kanban is the standard example of a PULL system: a card released by consumption downstream authorises production upstream, whereas MRP driven by forecasts is push — so the third statement is false.
  18. Annual demand is 10 000 units, the ordering cost is Rs 200 per order and the holding cost is Rs 4 per unit per year. What is the minimum total annual ordering-plus-holding cost, in rupees?

    Numerical answer — type the value.

    Show answer

    Answer: 4000

    EOQ = √(2 × 10 000 × 200/4) = 1000 units. Ordering cost = (10 000/1000) × 200 = Rs 2000 and holding cost = (1000/2) × 4 = Rs 2000, total Rs 4000 = √(2DSH). Using Q × H instead of (Q/2) × H for holding doubles that term to Rs 4000 and gives Rs 6000 in all — average inventory is half the lot.
  19. Annual demand is 10 000 units, ordering cost Rs 200 per order and holding cost 20 % of the unit price per year. The price is Rs 20 per unit for orders below 2000 units and Rs 19 for orders of 2000 or more. What order quantity minimises the total annual cost?

    1. 1000 units
    2. 1026 units
    3. 2000 units
    4. 10 000 units
    Show answer

    Answer: C — 2000 units

    At Rs 20, H = 4 and EOQ = 1000 (valid): total = 2 00 000 + 2000 + 2000 = Rs 2 04 000. At Rs 19, H = 3.8 and EOQ = 1026, below the 2000 break, so evaluate Q = 2000: total = 1 90 000 + 1000 + 3800 = Rs 1 94 800. The discount wins by Rs 9200. 1026 is infeasible at that price, and 1000 is optimal only if purchase cost is left out of the comparison.
  20. Compared with a perpetual (continuous-review, fixed-quantity) system, a periodic (fixed-interval) inventory system

    1. orders a fixed quantity at varying intervals
    2. needs more safety stock, because it must cover the review period plus the lead time
    3. needs continuous records of every withdrawal
    4. needs no safety stock at all
    Show answer

    Answer: B — needs more safety stock, because it must cover the review period plus the lead time

    In a periodic system stock is checked only every T periods, so a demand surge just after a review is not seen until the next one; safety stock must protect T + L, not L alone. Fixed quantity at varying intervals and continuous records describe the perpetual (Q) system instead.
  21. The critical path of a project has an expected duration of 22 days and a standard deviation of 2 days. Assuming a normal distribution, what is the probability of completing the project within 25 days? (Φ(1.5) = 0.9332)

    1. 0.0668
    2. 0.8413
    3. 0.9332
    4. 0.9987
    Show answer

    Answer: C — 0.9332

    Z = (T − T_e)/σ = (25 − 22)/2 = 1.5, and P(T ≤ 25) = Φ(1.5) = 0.9332. 0.0668 is 1 − Φ(1.5), the probability of taking longer; 0.9987 would be Φ(3), from forgetting to divide the 3 days by σ.
  22. In a project, activities A (3 days) and B (4 days) start at time zero; C (2 days) and D (5 days) follow A; E (3 days) follows both B and C; F (2 days) follows both D and E. What is the total float of activity B, in days?

    Numerical answer — type the value.

    Show answer

    Answer: 1

    Forward pass: A 0-3, B 0-4, C 3-5, D 3-8, E starts at max(4, 5) = 5 and ends at 8, F 8-10. Backward: F LS 8, so E LF 8 and LS 5, which makes B's LF 5 and LS 1. Total float of B = LS − ES = 1 − 0 = 1 day. Both A-D-F and A-C-E-F are critical, so B is the only activity with any float.
  23. A critical activity has a normal time of 10 days at Rs 5000 and a crash time of 7 days at Rs 6500. The project's indirect cost is Rs 700 per day. Crashing this activity by one day, with the critical path unchanged, changes the total project cost by

    1. an increase of Rs 500
    2. a decrease of Rs 200
    3. an increase of Rs 1500
    4. a decrease of Rs 700
    Show answer

    Answer: B — a decrease of Rs 200

    Cost slope = (6500 − 5000)/(10 − 7) = Rs 500 per day of direct cost added, while one day less saves Rs 700 of indirect cost: net change −Rs 200, so crashing pays. An increase of Rs 500 counts only the direct cost, and Rs 1500 is the cost of crashing all three days at once.
  24. Which statement best describes a Gantt chart?

    1. A network showing the precedence relationships between activities
    2. Horizontal bars showing each activity's start, duration and progress against a time scale
    3. A plot of project cost against project duration
    4. A probability distribution of project completion time
    Show answer

    Answer: B — Horizontal bars showing each activity's start, duration and progress against a time scale

    A Gantt chart draws each activity as a bar on a calendar, which makes planned versus actual progress easy to see. Its weakness is that it does not show dependencies, which is what the CPM/PERT network adds; the cost-duration plot belongs to the time-cost trade-off, and the completion-time distribution to PERT.