(a) An orthogonal cutting of metal is performed using cutting speed of 120 m/min, tool rake angle 5° and width of cut 5 mm…

(a) Given: cutting speed v = 120 m/min, tool rake angle α = 5°, width of cut w = 5 mm, chip thickness t₂ = 0.4 mm, uncut chip thickness t₁ = 0.2 mm, main cutting force F_c = 600 N and thrust force F_t = 300 N.

By definition of chip thickness ratio in orthogonal cutting, r = t₁/t₂ = 0.2/0.4 = 0.5.

Using Merchant’s force-circle resolution at the chip-tool interface: F = F_c sin α + F_t cos α F = 600 sin 5° + 300 cos 5° F = 600(0.08716) + 300(0.99619) F = 52.30 + 298.86 = 351.16 N.

Also, normal force N = F_c cos α − F_t sin α = 600(0.99619) − 300(0.08716) = 597.72 − 26.15 = 571.57 N, so friction coefficient μ = F/N = 351.16/571.57 = 0.614.

For percentage of total cutting energy used to overcome chip-tool friction, chip velocity is V_c = r v = 0.5 × 120 = 60 m/min. Total cutting power = F_c v = 600 × 120 = 72000 N·m/min = 1200 W. Friction power at chip-tool interface = F V_c = 351.16 × 60 = 21069.6 N·m/min = 351.16 W. Percentage = (F V_c)/(F_c v) × 100 = (351.16 × 0.5/600) × 100 = 29.26%.

This assumes orthogonal cutting, continuous chip, steady cutting forces and no built-up-edge effect.

(b) Let x₁ = tons/day of exterior paint and x₂ = tons/day of interior paint. Using the stated objective Z = 480000x₁ + 320000x₂, the linear programming problem is:

Maximize Z = 480000x₁ + 320000x₂ subject to 6x₁ + 4x₂ ≤ 24 … raw material M1 x₁ + 2x₂ ≤ 6 … raw material M2 x₂ − x₁ ≤ 1 … interior demand constraint x₂ ≤ 2 … maximum daily interior demand x₁, x₂ ≥ 0.

Using the graphical method, convert M1 to 3x₁ + 2x₂ ≤ 12. The feasible region vertices are: (0,0), (0,1), (1,2), (2,2), (3,3/2), (4,0).

Evaluate Z: Z(0,0) = 0 Z(0,1) = 320000 Z(1,2) = 480000 + 640000 = 1120000 Z(2,2) = 960000 + 640000 = 1600000 Z(3,3/2) = 480000×3 + 320000×3/2 = 1440000 + 480000 = 1920000 Z(4,0) = 480000×4 = 1920000.

Thus the objective has multiple optimal solutions. The objective vector (480000, 320000) is proportional to the M1 constraint normal (6,4), since 480000/6 = 80000 and 320000/4 = 80000. Hence the objective line is parallel to the M1 constraint. Every point on the M1 edge between (3,3/2) and (4,0) is optimal.

So the optimal product mix is x₂ = 6 − 3x₁/2, with 3 ≤ x₁ ≤ 4, and the maximum daily profit is ₹19,20,000.

If the table profit ₹4,00,000 for exterior paint is used instead, the unique optimum occurs at x₁ = 3 tons/day, x₂ = 1.5 tons/day, giving Z = ₹16,80,000/day.

(c) A quality costing system classifies all costs incurred to ensure good quality and all costs caused by poor quality. Its main elements are:

• Prevention costs: quality planning, training, supplier evaluation, process design, preventive maintenance.
• Appraisal costs: inspection, testing, measurement, audits and quality checks.
• Internal failure costs: scrap, rework, re-inspection, downtime and corrective action before delivery.
• External failure costs: warranty claims, returns, complaints, recalls, field service and liability.
• Supporting elements: cost data collection, classification, reporting, trend analysis and corrective-action feedback.

Organizations benefit because the system quantifies quality problems in money terms, identifies the most costly failure areas, and justifies investment in prevention and appraisal. It reduces rework, scrap, warranty costs and customer dissatisfaction, improves product reliability and delivery performance, supports continuous improvement and helps prioritise quality projects. Ultimately, it lowers total cost of quality, improves profitability and strengthens competitive advantage.