(a) A cutting tool is used to machine alloy steel at cutting speed of 40 m/min. Considering the values of constants c and n for…

(a) Use Taylor’s tool-life equation, V Tⁿ = C , where V is cutting speed in m/min, T is tool life in min, C = 300 and n = 0·5 for the given tool material.

At initial cutting speed V₁ = 40 m/min:

40 × T₁⁰·⁵ = 300

T₁⁰·⁵ = 300 / 40 = 7·5

T₁ = 7·5² = 56·25 min

So the initial tool life is 56·25 min.

If cutting speed is increased by 80%, the new speed is:

V₂ = 40 + 0·80 × 40 = 40 × 1·80 = 72 m/min

Again using Taylor’s equation:

72 × T₂⁰·⁵ = 300

T₂⁰·⁵ = 300 / 72 = 25 / 6 = 4·1667

T₂ = (25 / 6)² = 625 / 36 = 17·3611 min

Percentage change in tool life:

= [(T₂ − T₁) / T₁] × 100

= [(17·3611 − 56·25) / 56·25] × 100

= [−38·8889 / 56·25] × 100

= −69·1358%

Therefore, tool life decreases by 69·14%. Final answers: T₁ = 56·25 min, T₂ = 17·36 min, percentage change = −69·14%.

(b) The stepper motor has sensitivity 3°/pulse. Therefore, the number of pulses required for one complete revolution of the motor shaft is:

Pulses per revolution = 360° / 3° per pulse = 120 pulses/rev

The lead screw pitch is 2·4 mm/rev. Assuming the motor is directly coupled to the lead screw and there is no backlash or gear reduction, the linear movement per pulse is:

Linear resolution = pitch / pulses per revolution

= 2·4 mm/rev / 120 pulses/rev

= 0·02 mm/pulse

Thus, one pulse moves the slide by 0·02 mm. Hence the expected positioning accuracy, in the ideal direct-coupled case, is:

Positioning accuracy = 0·02 mm per pulse If expressed as a tolerance about the nearest achievable step, this corresponds to ±0·01 mm under ideal no-backlash conditions.

(c) Available time per day:

Total shifts = 2 shifts/day Each shift = 8 hours

Available time = 2 × 8 × 60 × 60 = 57,600 seconds/day

Assuming 28,000 tube lights are produced per day, the actual cycle time of the plant operation is:

Cycle time = available time / production quantity

= 57,600 s/day / 28,000 units/day

= 2·057142857 s/unit

≈ 2·057 s/unit

The total work content is the sum of all task times:

Σ task times = 10 seconds

The theoretical minimum number of workstations required to maintain this production level is:

N_min = ceil(Σ task times / cycle time)

= ceil(10 / 2·057142857)

= ceil(4·861111)

= 5 workstations

Therefore, the actual cycle time is approximately 2·057 s/unit and the minimum number of workstations required is 5, assuming tasks can be combined feasibly and no single task time exceeds the cycle time.

(d) This is a continuous-review deterministic inventory problem. Use the Economic Order Quantity, EOQ, model.

Given:

Annual demand, D = 2500 boxes/year Unit cost, C = ₹1250/box Ordering cost, S = ₹1200/order Carrying cost rate = 20% per unit per annum Working days = 250 days/year

Carrying cost per box per year:

H = 0·20 × 1250 = ₹250/box/year

EOQ formula:

Q* = √(2DS / H)

Q* = √(2 × 2500 × 1200 / 250)

Q* = √(6,000,000 / 250)

Q* = √24,000

Q* = 20√60 ≈ 154·92 boxes

Rounding to a practical order size:

EOQ ≈ 155 boxes per order

Number of orders per year:

N = D / Q* = 2500 / 154·92 ≈ 16·14 orders/year

Time between orders in working days:

T = (Q* / D) × working days

= (154·92 / 2500) × 250

= 15·49 working days

≈ 15·5 working days

Average inventory:

Q* / 2 = 154·92 / 2 = 77·46 boxes

Daily demand:

d = D / 250 = 2500 / 250 = 10 boxes/day

Annual ordering cost and carrying cost at EOQ:

Total variable inventory cost = √(2DSH)

= √(2 × 2500 × 1200 × 250)

= √1,500,000,000

≈ ₹38,729·83/year

Purchase cost:

= D × C = 2500 × 1250 = ₹31,25,000/year

Total annual cost including purchase:

= ₹31,25,000 + ₹38,729·83

= ₹31,63,729·83/year

Thus, the company should follow a continuous-review EOQ policy: order about 155 boxes every 15·5 working days. The relevant inventory cost is ₹38,729·83/year, and the total annual cost including purchase is ₹31,63,729·83/year. Since lead time is not given, the reorder point cannot be fixed; if lead time is L days and no safety stock is kept, reorder point = 10L boxes.

(e) Let X = number of new marriages in the area and Y = demand for toys. The market research concludes that demand is a simple linear function of X. Therefore, fit a least-squares linear regression model:

Y = a + bX

Using the given data:

n = 8

ΣX = 200 + 235 + 210 + 195 + 225 + 240 + 217 + 225 = 1747

ΣY = 165 + 184 + 180 + 145 + 190 + 169 + 180 + 170 = 1383

ΣX² = 200² + 235² + 210² + 195² + 225² + 240² + 217² + 225²

= 40,000 + 55,225 + 44,100 + 38,025 + 50,625 + 57,600 + 47,089 + 50,625

= 383,289

ΣXY = 200×165 + 235×184 + 210×180 + 195×145 + 225×190 + 240×169 + 217×180 + 225×170

= 33,000 + 43,240 + 37,800 + 28,275 + 42,750 + 40,560 + 39,060 + 38,250

= 302,935

Mean values:

X̄ = ΣX / n = 1747 / 8 = 218·375

Ȳ = ΣY / n = 1383 / 8 = 172·875

The slope b is:

b = [nΣXY − ΣXΣY] / [nΣX² − (ΣX)²]

= [8×302,935 − 1747×1383] / [8×383,289 − 1747²]

= [2,423,480 − 2,416,101] / [3,066,312 − 3,052,009]

= 7,379 / 14,303

= 0·515906

The intercept a is:

a = Ȳ − bX̄

= 172·875 − 0·515906 × 218·375

= 172·875 − 112·6609

= 60·2141

Hence the forecasting model is:

Y = 60·2141 + 0·515906X

where X is the number of new marriages and Y is the predicted demand for toys.

Interpretation: each additional newlywed couple increases toy demand by about 0·516 toys, and the model has a mathematical base demand of about 60·214 toys at X = 0, though this value should not be interpreted outside the observed range. The model is valid approximately for X between 195 and 240 new marriages. For example, if X = 220, predicted demand is:

Y = 60·2141 + 0·515906 × 220

= 60·2141 + 113·4993

= 173·71 toys

The correlation coefficient is r ≈ 0.59 (r² ≈ 0.34), so the number of newlyweds explains only about a third of the variation in demand; other factors should also be considered in forecasting.