Industrial Engineering and Operations Research

Syllabus

• Production planning and control: forecasting models, aggregate production planning, scheduling, materials requirement planning, lean manufacturing.
• Inventory control: deterministic models, safety stock inventory control systems.
• Operations research: linear programming, simplex method, transportation, assignment, network flow models, simple queuing models, PERT and CPM.

Key concepts and formulas

Forecasting. Simple moving average of the last n periods. Exponential smoothing: F_t+1 = αD_t + (1 − α)F_t. Errors: MAD = mean |error|; tracking signal = running sum of errors/MAD.

Inventory.

• Economic order quantity EOQ = √(2DS/H), where D is annual demand, S the ordering cost and H the holding cost per unit per year. At the EOQ, annual ordering cost = annual holding cost.
• Total cost at EOQ = √(2DSH).
• Production (EPQ) model: Q* = √(2DS/(H(1 − d/p))).
• Reorder point = demand during lead time + safety stock.

Scheduling. On one machine, SPT (shortest processing time) minimises the mean flow time; EDD (earliest due date) minimises the maximum lateness. For two machines in series, Johnson's rule minimises the makespan: pick the smallest time; if it is on machine 1, schedule the job first, otherwise last; repeat.

Linear programming. Optimum at a corner point of the feasible region. The simplex method moves between corner points. Transportation problems are balanced when total supply = total demand; a basic feasible solution has m + n − 1 allocations. Assignment problems are solved by the Hungarian method.

Queuing (M/M/1). With arrival rate λ and service rate μ (ρ = λ/μ < 1):

Little's law L = λW holds for any stable queue.

PERT/CPM. Critical path = longest path; total float = LS − ES; PERT t_e = (a + 4m + b)/6, σ = (b − a)/6. Break-even quantity = fixed cost/(price − variable cost per unit).

Worked example

Annual demand is 12000 units, ordering cost ₹300 per order and holding cost ₹20 per unit per year: find the EOQ and the total cost. For an M/M/1 queue with λ = 8 per hour and μ = 10 per hour, find L, L_q, W and W_q. Then sequence five jobs on two machines with Johnson's rule.

import math

D, S, H = 12000, 300, 20
Q = math.sqrt(2 * D * S / H)
print(f"EOQ = {Q:.1f} units, orders per year = {D / Q:.2f}, total cost = {math.sqrt(2 * D * S * H):.1f}")
lam, mu = 8, 10
rho = lam / mu
print(f"L = {rho / (1 - rho):.2f}, Lq = {rho ** 2 / (1 - rho):.2f}, W = {60 / (mu - lam):.0f} min, Wq = {60 * rho / (mu - lam):.0f} min")

jobs = {"A": (5, 2), "B": (1, 6), "C": (9, 7), "D": (3, 8), "E": (10, 4)}
front, back, left = [], [], dict(jobs)
while left:
    j = min(left, key=lambda k: (min(left[k]), k))   # job with the smallest time on either machine
    if left[j][0] <= left[j][1]:
        front.append(j)          # smallest time on machine 1: schedule as early as possible
    else:
        back.insert(0, j)        # smallest time on machine 2: schedule as late as possible
    del left[j]
order = front + back
t1 = t2 = 0
for j in order:
    t1 += jobs[j][0]
    t2 = max(t1, t2) + jobs[j][1]
print("Johnson sequence:", "-".join(order), " makespan =", t2)

EOQ = 600.0 units, orders per year = 20.00, total cost = 12000.0
L = 4.00, Lq = 3.20, W = 30 min, Wq = 24 min
Johnson sequence: B-D-C-E-A  makespan = 30

Practice questions