(a) What are the various computer languages used for NC machines? Explain the merits and limitations of each language. (10…

(a)

NC part programs are written either manually in word-address code or with a computer-assisted part-programming language, whose output is post-processed into the controller's code. The main languages are:

• Manual G- and M-code (ISO 6983 / EIA RS-274 word-address format): preparatory (G), miscellaneous (M), feed (F), speed (S) and tool (T) words executed directly by the controller. Merits: no computer or compiler is needed, it runs directly on the machine, and it is ideal for simple 2-D parts and shop-floor edits. Limitations: every coordinate must be calculated by hand, it is error-prone and slow for complex contours, and codes vary between controller makes.
• APT (Automatically Programmed Tools, MIT/US Air Force, 1956): the most comprehensive language; it defines geometry (POINT, LINE, CIRCLE, PLANE, surfaces), tool motion (GOFWD, GOLFT) and post-processor statements, with up to 5-axis capability. Merits: handles complex 3-D sculptured surfaces, is machine-independent through post-processors, and is the base of later languages. Limitations: a large vocabulary and compiler, a mainframe origin and long learning time; it is uneconomical for simple parts.
• ADAPT (ADaptation of APT, IBM): a subset of APT for smaller computers. Merits: needs less memory and is easier to learn. Limitations: limited to about 2½-axis contouring and simpler geometry.
• EXAPT (EXtended subset of APT, Germany): EXAPT 1 for point-to-point drilling, EXAPT 2 for turning and EXAPT 3 for milling. Merits: adds technological (process-planning) capability, automatically computing feeds, speeds and cut sequences from material and tool data. Limitations: needs a large technological database and is less widely supported.
• AUTOSPOT (AUTOmatic System for POsitioning Tools, IBM): a point-to-point language for drilling and boring. Merits: simple and fast for hole patterns. Limitations: not suited to contouring.
• COMPACT II (Manufacturing Data Systems Inc.): a conversational language widely used via time-sharing. Merits: English-like statements, easy to learn and interactive. Limitations: proprietary, dependent on the vendor's system, and weaker for complex surfaces.
• SPLIT and AUTOPROMPT: SPLIT (Sundstrand) has post-processing built in and supports up to 5 axes; AUTOPROMPT (IBM) was for 3-D milling. Merits: suited to specific machine families. Limitations: machine-specific and now largely superseded.

Today most programming is done through CAD/CAM systems (graphical tool-path generation followed by post-processing to G-code), which combine the geometric power of APT with interactive verification; conversational (MDI) programming on the controller remains common for simple jobs.

(b)

Given:

• Target weight: 60 g
• Sample size (n): 5
• Constant A₂ for n=5: 0.577
• Weights for the first 5 days are taken from the table given in the question.

Step 1: Calculate Daily Means (X̄) and Ranges (R) The mean is the sum of the 5 samples divided by 5. The range is the difference between the maximum and minimum values in the sample.

• 1: Samples (g) — 59.8, 60.2, 60.0, 60.1, 59.9; Mean (X̄) — 60.00; Range (R) — 0.4
• 2: Samples (g) — 60.1, 60.3, 60.2, 60.0, 59.8; Mean (X̄) — 60.08; Range (R) — 0.5
• 3: Samples (g) — 59.7, 60.0, 60.1, 59.9, 60.2; Mean (X̄) — 59.98; Range (R) — 0.5
• 4: Samples (g) — 60.0, 59.8, 60.2, 60.1, 59.9; Mean (X̄) — 60.00; Range (R) — 0.4
• 5: Samples (g) — 60.2, 60.3, 60.1, 60.0, 59.9; Mean (X̄) — 60.10; Range (R) — 0.4

Step 2: Calculate Grand Mean (X̿) and Mean Range (R̄) X̿ = (60.00 + 60.08 + 59.98 + 60.00 + 60.10)/5 = 300.16/5 = 60.032 g R̄ = (0.4 + 0.5 + 0.5 + 0.4 + 0.4)/5 = 2.2/5 = 0.44 g

Step 3: Calculate Control Limits The formula for the control limits of an X-bar chart is: UCL = X̿ + (A₂ × R̄) LCL = X̿ − (A₂ × R̄)

Substituting the values: UCL = 60.032 + (0.577 × 0.44) = 60.032 + 0.25388 = 60.286 g LCL = 60.032 - (0.577 × 0.44) = 60.032 - 0.25388 = 59.778 g

Interpretation: The process is in statistical control if the daily means fall between 59.778 g and 60.286 g.

(c)

Given:

• Order Quantity (Q): 5,00,000 units
• Welding: Fixed Cost (FC_w) = ₹15,000; Variable Cost (VC_w) = ₹5/unit
• Forging: Fixed Cost (FC_f) = ₹94,000; Variable Cost (VC_f) = ₹4.25/unit

Step 1: Determine the Break-Even Point (BEP) The BEP is the quantity where the total cost of both methods is equal. Total Cost = Fixed Cost + (Variable Cost × Q) 15,000 + 5Q = 94,000 + 4.25Q 5Q - 4.25Q = 94,000 - 15,000 0.75Q = 79,000 Q_BEP = (79,000)/0.75 = 1,05,333 units

Step 2: Compare Costs for 5,00,000 Units Since the order quantity (5,00,000) is significantly higher than the BEP (1,05,333), the method with the lower variable cost (Forging) will be more economical.

• Total Cost (Welding): 15,000 + (5 × 5,00,000) = 15,000 + 25,00,000 = ₹ 25,15,000
• Total Cost (Forging): 94,000 + (4.25 × 5,00,000) = 94,000 + 21,25,000 = ₹ 22,19,000

Indicative Graph for BEP: The graph plots Total Cost (Y-axis) against Quantity (X-axis).

• Welding Line: Starts at ₹15,000 on the Y-axis with a steeper slope (5).
• Forging Line: Starts at ₹94,000 on the Y-axis with a shallower slope (4.25).
• Intersection: The lines cross at Q = 1,05,333. To the right of this point, the Forging line is lower (cheaper).

Step 3: Calculate the Loss of a Wrong Choice If the factory chooses Welding (the wrong choice for this volume), the loss is the difference in total costs: Loss = Cost_Welding - Cost_Forging Loss = 25,15,000 - 22,19,000 = ₹ 2,96,000

Result: Forging is the most economical method. If the factory incorrectly chooses Welding, it will incur a financial loss of ₹ 2,96,000 compared to the optimal choice.