A car servicing company is interested in reducing the waiting time for its customers. They select four customers randomly each…

(a) Control charts for waiting time

Here n = 4, with 25 samples. X̄ = 1000/25 = 40 min and R̄ = 250/25 = 10 min. For n = 4 the tabulated factors are A₂ = 0.729, D₃ = 0, D₄ = 2.282, d₂ = 2.059 and d₃ = 0.880.

(i) X̄ chart: CL = 40, UCL = 40 + 0.729 × 10 = 47.29 min, LCL = 40 − 7.29 = 32.71 min. R chart: CL = 10, UCL = 2.282 × 10 = 22.82 min, LCL = 0.

(ii) σ = R̄/d₂ = 10/2.059 = 4.857 min. For individual waiting times, z = (50 − 40)/4.857 = 2.06, and from the table Φ(2.06) = 0.9803. About 98.0% of customers wait no more than 50 minutes.

(iii) 2σ limits. X̄ chart: 40 ± 2σ/√n = 40 ± 4.857, that is 35.14 and 44.86 min. R chart: σ_R = d₃σ = 0.880 × 4.857 = 4.274, so the limits are 10 ± 8.55, that is 1.45 and 18.55 min.

(b) Lean systems

Functioning: a lean system defines value from the customer's view and removes waste (muda) from the value stream. Production is pulled by downstream demand through kanban, in small batches, at the takt time, rather than pushed by forecasts.

Characteristics: elimination of the seven wastes (overproduction, waiting, transport, over-processing, inventory, motion, defects); just-in-time supply; pull and kanban control; levelled schedules; cellular layout and continuous flow; quick changeovers; multi-skilled, empowered workers; total productive maintenance and built-in quality; close supplier partnerships; visual management; kaizen.

Benefits: lower inventory and WIP, shorter lead times, better quality, lower cost, higher productivity, less floor space, greater flexibility and customer satisfaction.

Risks: with little buffer stock, the system is exposed to supplier failure, transport delay, demand surges and breakdowns; it needs reliable suppliers and disciplined, trained staff; set-up and training costs are high; it may suit highly variable or job-shop demand poorly; and cultural resistance and workforce stress can arise.

(c) Open-die forging of a flat strip (slab method)

Assumptions: plane strain (strip long, no lateral spread); rigid–perfectly plastic, homogeneous material with constant flow stress; flat, rigid, parallel dies; Coulomb friction τ = μp over the interface; stresses uniform across the thickness; negligible inertia and body forces; symmetry about the centre line; sliding (not sticking) friction; no bulging.

Let the strip have thickness h, width L = 2a and unit length. Take a slab of width dx at distance x from the centre. Let p be the die pressure, σₓ the horizontal compressive stress and K the shear yield strength.

Equilibrium in x: σₓh − (σₓ + dσₓ)h − 2τ dx = 0, so dσₓ/dx = −2τ/h.

Plane-strain yield condition: p − σₓ = 2K, so dp = dσₓ. With τ = μp,

dp/dx = −2μp/h.

At the free edge x = a, σₓ = 0, so p = 2K. Integrating from x to a gives

p(x) = 2K exp[2μ(a − x)/h].

The force per unit length is F = 2∫₀ᵃ p dx = (2Kh/μ)[exp(μL/h) − 1].

With Y′ = 2K (the plane-strain flow stress, equal to 2Y/√3 by the von Mises criterion),

F = (Y′h/μ)[exp(μL/h) − 1] per unit length.

The pressure is least at the edges (2K) and highest at the centre.