(a) Following data were observed while machining two different work materials X and Y using a centre lathe machine: If work…

(a) The data are X: (V,T) = (125 m/min, 20 min) and (150 m/min, 15 min); Y: (250 m/min, 45 min) and (300 m/min, 25 min).

• Method: Taylor’s tool-life equation, V Tⁿ = C. It is valid for the same tool, feed, depth of cut and cooling; the 50 min value is an extrapolation from the tested range. Natural logarithms are used; the base is immaterial because it cancels in the ratio.
• For X: 125×20ⁿ = 150×15ⁿ. Hence (20/15)ⁿ = 150/125 = 6/5, so nₓ = ln(6/5)/ln(4/3) ≈ 0.6338. Then Cₓ = 125×20ⁿₓ ≈ 834 m/min, with T in min. The second point gives the same Cₓ. At T = 50 min, Vₓ = Cₓ/50ⁿₓ = 125(20/50)ⁿₓ = 125(2/5)ⁿₓ ≈ 69.9 m/min.
• For Y: 250×45ⁿ = 300×25ⁿ. Hence (45/25)ⁿ = 6/5, so nᵧ = ln(6/5)/ln(9/5) ≈ 0.3102. Then Cᵧ = 250×45ⁿᵧ ≈ 814 m/min. The second point gives the same Cᵧ. At T = 50 min, Vᵧ = 250(45/50)ⁿᵧ = 250(9/10)ⁿᵧ ≈ 242.0 m/min.
• Relative machinability with X as standard at a common tool life of 50 min is M = (V of material / Vₓ)×100. The ratio is dimensionless. Substituting the logarithmic n values in the speed ratio gives 345.8, rounded to 346. Final: Mₓ = 100; Mᵧ ≈ 346, i.e. Y is about 3.46 times as machinable as X at 50 min tool life.

(b) All jobs are available at time 0. All times are in days. For each sequence, completion time C is cumulative processing time; flow time F = C; waiting time W = C - processing time; tardiness = max(0, C - due date). Tardiness is used rather than lateness because negative lateness is not penalised. The final completion time is 31 days for all four sequences because total processing time is 31 days.

• (i) First Come First Serve (FCFS): order A-B-C-D-E. C = 5, 15, 17, 25, 31 days. Flow sum = 93 days, average = 18.6 days. W = 0, 5, 15, 17, 25 days; sum = 62 days, average = 12.4 days. Tardiness = 0, 0, 12, 13, 23 days; total = 48 days, average = 9.6 days, maximum = 23 days, late jobs = 3.
• (ii) Earliest Due Date (EDD): sort by due date: C(5), E(8), A(10), D(12), B(15). C = 2, 8, 13, 21, 31 days. Flow sum = 75 days, average = 15.0 days. W = 0, 2, 8, 13, 21 days; sum = 44 days, average = 8.8 days. Tardiness = 0, 0, 3, 9, 16 days; total = 28 days, average = 5.6 days, maximum = 16 days, late jobs = 3.
• (iii) Minimum Slack (SLACK): slack = due date - processing time: A 5, B 5, C 3, D 4, E 2. Minimum-slack order E-C-D-A-B, with the A/B tie taken as A before B. C = 6, 8, 16, 21, 31 days. Flow sum = 82 days, average = 16.4 days. W = 0, 6, 8, 16, 21 days; sum = 51 days, average = 10.2 days. Tardiness = 0, 3, 4, 11, 16 days; total = 34 days, average = 6.8 days, maximum = 16 days, late jobs = 4.
• (iv) Shortest Processing Time (SPT): sort by processing time: C(2), A(5), E(6), D(8), B(10). C = 2, 7, 13, 21, 31 days. Flow sum = 74 days, average = 14.8 days. W = 0, 2, 7, 13, 21 days; sum = 43 days, average = 8.6 days. Tardiness = 0, 0, 5, 9, 16 days; total = 30 days, average = 6.0 days, maximum = 16 days, late jobs = 3.
• SPT gives the lowest average flow and waiting times. EDD gives the lowest total and average tardiness, and its maximum tardiness is tied for best. Since due dates are explicitly supplied, due-date performance should dominate. Recommended rule: EDD; if the only objective were minimum flow time, SPT would be recommended.

(c)

• Variable overhead per unit = 50% × (1000 + 500) = 750 ₹/unit.
• Variable cost per unit v = 1000 + 500 + 750 = 2250 ₹/unit. Fixed cost F = 10,00,000 ₹, and D is the number of units sold. The break-even condition is total revenue = F + vD, assuming fixed cost is constant, variable cost is linear with D, all units produced are sold, and the revenue function is valid over the break-even range.
• In break-even notation S is normally total sales. Taking the given S = 4D² as total revenue, 4D² = 10,00,000 + 2250D. 4D² - 2250D - 1,000,000 = 0. D = [2250 + √(2250² + 4×4×1,000,000)]/(2×4) = [2250 + √21,062,500]/8 = 125(9 + √337)/4. The positive root is selected because demand must be positive. Check whole units: f(854) = 4×854² - 2250×854 - 1,000,000 = -4,236 ₹, and f(855) = 350 ₹. Hence the minimum whole number of units is 855.
• If S is read literally as unit selling price, total revenue is D×S = 4D³. Then 4D³ = 10,00,000 + 2250D. f(65) = -47,750 ₹ and f(66) = 1,484 ₹, so the minimum whole number would be 66 units. Final: 855 units under the usual total-sales reading; 66 units under the literal unit-price reading.