A car servicing company is interested in reducing the waiting time for its customers. They select four customers randomly each day and find the waiting time for each customer while his/her car is serviced. From these observations, the sample average

Question

A car servicing company is interested in reducing the waiting time for its customers. They select four customers randomly each day and find the waiting time for each customer while his/her car is serviced. From these observations, the sample average and range are found. This process is repeated for 25 days. The summary data for these observations are as under:

Σᵢ₌₁²⁵ X̄ᵢ = 1000, Σᵢ₌₁²⁵ Rᵢ = 250

(i) Find out X̄ and R chart control limits.

(ii) Assuming that the process is in control and the distribution of waiting time is normal, find the percentage of customers who will not have to wait for more than 50 minutes.

(iii) Find the 2σ control limits.

(Factors for Computing Centerline and Three-Sigma Control Limits and Cumulative Standard Normal Distribution table are appended in the question paper)

(b) How do lean systems function? What are the characteristics of lean systems? Also, discuss the benefits and risks of lean systems.

(c) In a case of open die forging, derive the expression for determining forging force per unit length for forging a flat strip between two parallel dies. Also, state the assumptions made while deriving the above mentioned expression.

UPSC Answer Check · Accepted Answer

(i)
For n = 4, m = 25:
X̄_g = Σ X̄ᵢ / 25 = 1000/25 = 40 min
R̄ = Σ Rᵢ / 25 = 250/25 = 10 min

From the appended table, for n = 4:
A₂ = 0.729, D₃ = 0, D₄ = 2.282.

X̄ chart control limits:
CL = X̄_g = 40 min
UCL = X̄_g + A₂ R̄ = 40 + 0.729 × 10 = 47.29 min
LCL = X̄_g − A₂ R̄ = 40 − 0.729 × 10 = 32.71 min

R chart control limits:
CL = R̄ = 10 min
UCL = D₄ R̄ = 2.282 × 10 = 22.82 min
LCL = D₃ R̄ = 0 × 10 = 0 min

(ii)
The process standard deviation is estimated by:
σ = R̄ / d₂ = 10 / 2.059 = 4.857 min.

For a normal distribution:
z = (50 − 40) / 4.857 = 2.059.

From the standard normal table:
Φ(2.059) ≈ 0.9803.

Therefore, 98.03% of customers will not have to wait for more than 50 minutes.

(iii)
2σ control limits:

X̄ chart:
σ_X̄ = σ / √n = 4.857 / 2 = 2.428 min
2σ_X̄ = 4.857 min
Limits = 40 ± 4.86 = 35.14 min and 44.86 min

R chart:
σ_R = d₃ σ = 0.880 × 4.857 = 4.274 min
2σ_R = 8.548 min
Limits = 10 ± 8.548 = 1.45 min and 18.55 min

(b)
Lean systems function by identifying value from the customer’s viewpoint and eliminating waste, called muda, from the value stream. Production is pulled by downstream demand rather than pushed by forecasts. Kanban, takt time, continuous flow, small lots, quick changeovers, levelled schedules, standard work, visual control, total productive maintenance, total quality management and supplier partnerships keep material and information flowing smoothly. Problems are exposed and removed through kaizen and employee involvement.

Characteristics: waste elimination—overproduction, waiting, transport, overprocessing, inventory, motion and defects; JIT delivery; pull/Kanban; one-piece or small-batch flow; cellular layout; multi-skilled workers; continuous improvement; respect for people; stable and levelled schedules; close supplier links; visual management.

Benefits: lower inventory and WIP, shorter lead times, improved quality, lower cost, higher productivity, reduced space, greater flexibility, better customer satisfaction and higher employee morale.

Risks: little buffer stock makes the system vulnerable to supplier failure, transport delay, demand spikes and machine breakdown; it demands reliable suppliers, disciplined workers and strong management; implementation and training costs are high; cultural resistance can occur; it may not suit highly variable demand or job-shop production; continuous pressure can cause worker burnout.

(c)
Let a flat strip of thickness h and total width L = 2a be forged between parallel dies; depth is taken as unity. Let p be die pressure, σₓ horizontal compressive stress, τ friction shear stress, μ coefficient of friction and K shear yield strength. For plane strain, the yield condition is p − σₓ = 2K. At the free edge x = a, σₓ = 0, so p(a) = 2K.

Using the slab method, consider a slab of width dx at distance x from the centre in the right half. Equilibrium in the x-direction gives:
σₓ h − (σₓ + dσₓ)h − 2τ dx = 0
⇒ dσₓ/dx = −2τ/h.

Taking Coulomb friction τ = μp and using p = σₓ + 2K:
dp/dx = −2μp/h.

Integrating from x to a:
∫ dp/p (from p(x) to p(a)) = −(2μ/h) ∫ dx (from x to a)
⇒ ln[p(a)/p(x)] = −(2μ/h)(a − x)
⇒ p(x) = 2K exp[2μ(a − x)/h].

Total forging force per unit depth on one die:
F = ∫ p(x) dx (from −a to a) = 2∫ p(x) dx (from 0 to a)
= 4K ∫ exp[2μ(a − x)/h] dx (from 0 to a)
= (2K h/μ)[exp(2μa/h) − 1].

With L = 2a and Y′ = 2K, this becomes:
F = (Y′ h/μ)[exp(μL/h) − 1] per unit length.

For von Mises plane strain, Y′ = 2Y/√3, so:
F = (2Y h/(√3 μ))[exp(μL/h) − 1] per unit length.

Assumptions: plane strain; homogeneous, isotropic, rigid-plastic material with constant flow stress; rigid, parallel, flat dies; Coulomb sliding friction τ = μp over the whole interface; stresses uniform over any vertical section; inertia and body forces neglected; symmetry about the centreline; no bulging and no strain hardening; quasi-static deformation.

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