(a) An HSS drill during its life can drill 150 through holes in a 10 mm thick brass plate at a drill speed of 400 rpm. Another…

(a) Let the drill speed be N rpm, feed be f mm/rev, plate thickness be L = 10 mm, and tool life be T minutes. The cutting speed V = πDN/1000, so for the same drill diameter D, V ∝ N. Taylor’s tool-life equation is V Tⁿ = C, hence N Tⁿ = C₁.

Time required for one through hole = L/(f N). Therefore, number of holes in tool life T is H = T/(L/(f N)) = T f N/L. So T = H L/(f N).

Substitute T in Taylor’s equation: N (H L/(f N))ⁿ = C₂. Since L and f are constant, N^(1−n) Hⁿ = C₃.

Using the two given data sets: 400^(1−n) × 150ⁿ = 600^(1−n) × 75ⁿ. Taking logarithms: (1−n) ln(400/600) + n ln(150/75) = 0 (1−n) ln(2/3) + n ln 2 = 0 (1−n)(−0.405465) + n(0.693147) = 0 −0.405465 + 0.405465n + 0.693147n = 0 1.098612n = 0.405465 n = 0.3691.

Now for N₃ = 800 rpm: 400^(1−n) × 150ⁿ = 800^(1−n) × H₃ⁿ H₃ = 150 × (400/800)^((1−n)/n) H₃ = 150 × (1/2)^((1−0.3691)/0.3691) H₃ = 150 × (1/2)^1.7095 H₃ = 150 × 0.3058 = 45.87.

Therefore, the drill will produce about 46 holes theoretically. Strictly, the limiting life is reached after about 45.9 holes, so 45 complete through holes are certain before the drill needs attention.

(b) Functions of coating on a Shielded Metal Arc Welding electrode:

• Arc stabilisation: Coating contains potassium, sodium, calcium and titanium compounds which ionise the arc gap. This makes arc striking easier and keeps the arc stable, reducing arc blow and spatter.

• Atmospheric shielding: The coating decomposes under arc heat and generates shielding gases such as CO₂, CO and water vapour. These gases form an envelope around the molten weld pool and prevent contact with atmospheric oxygen and nitrogen.

• Slag formation: The flux forms a molten slag that floats on the weld pool. It protects the solidifying weld from oxidation, slows cooling, refines grain structure and supports the weld bead in vertical and overhead positions.

• Deoxidation and purification: Elements such as manganese, silicon, titanium and aluminium in the coating deoxidise the molten metal and combine with sulphur, phosphorus and oxides, which then enter the slag and are removed.

• Alloying additions: Ferroalloys in the coating can add chromium, nickel, molybdenum, manganese, vanadium etc. to the weld deposit. This improves strength, hardness, toughness, creep resistance or corrosion resistance.

• Control of bead shape and penetration: The coating controls slag viscosity and surface tension, influencing weld bead profile, penetration, ripples and appearance. It also helps control spatter and metal transfer.

• Electrical insulation and arc direction: The coating insulates the core wire from accidental contact and confines the arc to the electrode tip, giving directional control to the welder.

• Mechanical protection and binding: Silicate binders give mechanical strength to the coating, prevent it from chipping during handling, protect the core wire from corrosion and help maintain the electrode shape.

• Deposition improvement: Coatings containing iron powder increase deposition rate, improve arc efficiency and reduce welding time in suitable positions.

(c) Capacity timing and sizing decisions determine when and how much production capacity is to be added. Two common strategies are Expansionist and Wait-and-See.

Expansionist strategy: This is an aggressive, lead-capacity strategy. Capacity is added before demand actually appears, often in large increments. The firm anticipates growth and creates capacity ahead of the market. Its advantages are: it captures market share, reduces the risk of shortage, discourages competitors, exploits economies of scale, lowers unit cost and improves service level. It is suitable when demand is expected to grow steadily, capital is available, and delay would mean permanent loss of market to competitors. The main risks are overcapacity, idle facilities, high fixed cost and financial loss if demand does not materialise.

Wait-and-See strategy: This is a conservative, lag-capacity strategy. Capacity is added only after demand has actually increased and is confirmed. Additions are usually smaller and more frequent. Its advantages are lower risk of overbuilding, better utilisation of existing capacity, reduced capital exposure and greater flexibility under uncertain demand. It is suitable when demand is volatile, forecasts are unreliable, technology changes rapidly or capital is scarce. Its disadvantages are lost sales, poor customer service, inability to meet sudden demand and loss of market share to an expansionist competitor. Later expansion may also become costlier due to rising prices or scarce resources.

In short, Expansionist policy favours leading demand with large capacity chunks, while Wait-and-See policy favours following demand with small, cautious increments. Modern practice often uses a mixed or average-capacity strategy, supported by overtime, subcontracting, inventories and flexible facilities.

(d) The Kanban formula is: N = (D × L + S)/C where N = number of Kanbans or containers, D = daily demand, L = production lead time in days, S = safety stock in units, and C = container capacity.

Given: D = 500 units/day L = 3 days Safety stock = one day demand = 500 units C = 400 units/container

Substitute: N = (500 × 3 + 500)/400 N = (1500 + 500)/400 N = 2000/400 N = 5.

Therefore, the number of Kanbans required is 5 containers. If the result had not been exact, it would have been rounded up to the next whole container.

(e) Let demand per period be a random variable with mean μ and standard deviation σ. Let lead time L be constant. Demand during lead time is the sum of L independent demand periods: X = d₁ + d₂ + ... + d_L.

By linearity of expectation: E(X) = μ + μ + ... + μ = L μ.

Since the demand periods are independent, variances add: Var(X) = σ² + σ² + ... + σ² = L σ². Hence the standard deviation of demand during lead time is: σ_X = σ√L.

For a desired service level, say probability p of not stocking out during lead time, let z be the standard normal variate such that P(Z ≤ z) = p. Then: P(X ≤ ROP) = p P(Z ≤ (ROP − L μ)/(σ√L)) = p.

Therefore: z = (ROP − L μ)/(σ√L) ROP = L μ + z σ√L.

The term z σ√L is the safety stock: SS = z σ√L.

Given: μ = 18 units/week σ = 5 units/week L = 2 weeks Service level = 95%, so from the standard normal table, z = 1.645.

Mean demand during lead time: L μ = 2 × 18 = 36 units.

Standard deviation during lead time: σ_X = σ√L = 5√2 = 7.071 units.

Safety stock: SS = z σ_X = 1.645 × 7.071 = 11.632 units.

Reorder point: ROP = L μ + SS = 36 + 11.632 = 47.632 units.

Therefore, the safety stock is about 11.63 units and the reorder point is about 47.63 units, i.e. practically 48 units. This holds when lead time is constant, successive demands are independent, and demand during lead time is approximated by a normal distribution.