(a) Obtain the control limits for X̄-chart and R-chart and describe the significance of joint study of these charts. 20…

(a) Control Limits for X̄- and R-Charts; Significance of Joint Study

Let samples of size n have mean X̄ and range R. For an in-control normal process with mean μ and standard deviation σ, the sample mean satisfies X̄ ~ N(μ, σ²/n), so its 3-sigma limits are μ ± 3σ/√n. Since μ and σ are unknown, estimate μ by X̄̄ and σ by R̄/d₂, because E(R)=d₂σ.

Hence for the X̄-chart:

• CL = X̄̄

• UCL = X̄̄ + A₂R̄

• LCL = X̄̄ − A₂R̄

where A₂ = 3/(d₂√n).

For the R-chart, E(R)=d₂σ and SD(R)=d₃σ, so the 3-sigma limits are (d₂ ± 3d₃)σ. Replacing σ by R̄/d₂ gives:

• CL = R̄

• UCL = D₄R̄

• LCL = D₃R̄

where D₄ = 1 + 3d₃/d₂ and D₃ = max(0, 1 − 3d₃/d₂). In standard tables D₃=0 for n≤6; if D₃<0, take LCL=0. Conditions: independent observations, fixed subgroup size, and approximate normality.

Significance of joint study: The X̄-chart detects shifts in the process mean, while the R-chart detects changes in process variability. They must be used together because the X̄-chart limits depend on R̄. If the R-chart is out of control, R̄ is not a reliable estimate of σ, so X̄-chart signals may be misleading. Usually the R-chart is examined first. If R is in control and X̄ is out, the location has shifted. If R is out, variability has changed and may explain apparent mean shifts. Joint patterns help diagnose assignable causes such as sudden shifts, increasing spread, or gradual drift.

(b) Reliability and Hazard Functions of the Weibull Distribution

For a Weibull distribution with scale θ and shape β, the density is

f(t) = (β/θ)(t/θ)^(β−1) exp[−(t/θ)^β], t≥0, θ>0, β>0.

The distribution function is

F(t) = 1 − exp[−(t/θ)^β].

Hence the reliability function is

R(t) = P(T>t) = 1 − F(t) = exp[−(t/θ)^β].

The hazard function is

h(t) = f(t)/R(t) = (β/θ)(t/θ)^(β−1) = β t^(β−1)/θ^β, t≥0,

with unit “per unit time” if t is time.

Interpretation: β<1 gives a decreasing hazard, representing infant mortality. β=1 gives a constant hazard h(t)=1/θ, the exponential memoryless case. β>1 gives an increasing hazard, representing wear-out or aging. The scale θ is the characteristic life because R(θ)=exp(−1).

(c) Simplex Solution: Unbounded LPP

All constraints are of the ≥ type, so subtract surplus variables s₁, s₂, s₃ and add artificial variables a₁, a₂, a₃:

6x₁ + x₂ − s₁ + a₁ = 6; 4x₁ + 3x₂ − s₂ + a₂ = 12; x₁ + 2x₂ − s₃ + a₃ = 4; all variables ≥ 0.

Phase I (minimise w = a₁ + a₂ + a₃):

• Start: basis (a₁, a₂, a₃) = (6, 12, 4), w = 22. The most negative cⱼ − zⱼ is −11 for x₁, so x₁ enters; ratios 6/6 = 1, 12/4 = 3, 4/1 = 4, so a₁ leaves.

• Basis (x₁, a₂, a₃) = (1, 8, 3), w = 11. x₂ enters (cⱼ − zⱼ = −25/6); ratios 6, 24/7 and 18/11, so a₃ leaves.

• Basis (x₁, a₂, x₂) = (8/11, 46/11, 18/11), w = 46/11. s₃ enters (cⱼ − zⱼ = −14/11) and a₂ leaves.

• Basis (x₁, s₃, x₂) = (3/7, 23/7, 24/7), w = 0: all artificials have left, so a basic feasible solution has been found.

Phase II (maximise z = 5x₁ + 2x₂, artificial columns dropped):

• Basis (x₁, s₃, x₂) = (3/7, 23/7, 24/7), z = 9. Both s₁ and s₂ have cⱼ − zⱼ = 1/2 > 0; taking s₁ to enter, s₃ leaves.

• Basis (x₁, s₁, x₂) = (12/5, 46/5, 4/5), z = 68/5. s₂ enters (cⱼ − zⱼ = 8/5) and x₂ leaves.

• Basis (x₁, s₁, s₂) = (4, 18, 4), z = 20. In terms of the non-basic variables, x₁ = 4 − 2x₂ + s₃, s₁ = 18 − 11x₂ + 6s₃, s₂ = 4 − 5x₂ + 4s₃ and z = 20 − 8x₂ + 5s₃.

Now s₃ has cⱼ − zⱼ = +5 > 0, so it should enter, but every entry of the s₃ column is non-positive: increasing s₃ only increases x₁, s₁ and s₂. The ratio test finds no leaving variable, so z = 20 + 5s₃ → ∞.

Conclusion: the LPP has an unbounded solution; z has no finite maximum. This agrees with the geometry: the feasible region is open towards large x₁, and z = 5x₁ + 2x₂ grows without limit along it.