Industrial Engineering – GATE PI Industrial EngineeringPractice Questions & PYQs

A line balancing problem aims to assign tasks to workstations on an assembly line to equalize the workload, minimizing idle time and bottlenecks for smooth production flow. This problem is inherently solved in the context of a product layout because a product layout (assembly line or flow shop) arranges workstations sequentially for specific product manufacturing steps. Tasks in a product layout are repetitive and standardized, allowing for efficient distribution across workstations to match the desired production rate (takt time). In contrast, a process layout groups similar functions, offering flexibility but lacking a direct sequential flow for line balancing. A fixed position layout involves bringing resources to a stationary product, making line balancing inapplicable, while a production schedule dictates timing rather than the physical arrangement optimized by line balancing.

To find the line efficiency, we use the formula: $\text{Line efficiency} = \frac{T_w}{n \times T_c}$, where $T_w$ is the total work content time, $n$ is the number of workstations, and $T_c$ is the cycle time. Given $n = 6$ and $T_w = 7.2$ minutes. Production of 300 products in 7.5 hours (or 450 minutes) means the cycle time per product is $T_c = \frac{450 \text{ mins}}{300 \text{ products}} = 1.5$ minutes per product. Substituting these values into the formula: $\text{Line efficiency} = \frac{7.2}{6 \times 1.5} \times 100 = 80.00\%$.

This problem asks us to compare the total costs of two manufacturing alternatives to find out when Alternative I is cheaper than Alternative II. We'll set up cost equations for each alternative based on a demand $D$.

For Alternative I (Process P), the total cost ($TC_I$) is its fixed cost plus the variable cost per unit multiplied by the demand $D$:
$TC_I = 100,000 + 75D$
For Alternative II (Process Q), the total cost ($TC_{II}$) is its fixed cost plus the variable cost per unit multiplied by the demand $D$:
$TC_{II} = 190,000 + 60D$
To find when Alternative I is cheaper than Alternative II, we set up the inequality $TC_I < TC_{II}$:
$100,000 + 75D < 190,000 + 60D$
Now, we solve for $D$:
$75D - 60D < 190,000 - 100,000$
$15D < 90,000$
$D < \frac{90,000}{15}$
$D < 6,000$
Thus, Alternative I is cheaper compared to Alternative II when the demand is below 6,000 units.

The assignment problem seeks to pair resources to tasks (e.g., workers to jobs) to optimize an objective, often minimizing total cost. A balanced assignment problem occurs when the number of resources equals the number of tasks. A binary solution means each assignment decision is either 1 (assigned) or 0 (not assigned), represented in a matrix where each row and column has exactly one '1'.

The key reason for the binary nature of the solution to a balanced assignment problem is the uni-modularity property, specifically total uni-modularity, of its constraint matrix. A matrix is totally unimodular if the determinant of every square submatrix is 0, 1, or -1. When a linear programming problem has a totally unimodular constraint matrix and its right-hand side vector consists of integers (as is the case for the assignment problem), the optimal solution obtained from the linear program (even if we initially relax the integer requirement) will automatically be integer-valued. For the assignment problem, this structure specifically forces these integer solutions to be 0s and 1s, leading to a binary solution. This is why algorithms like the Hungarian algorithm guarantee an exact, binary assignment.

Other options are incorrect because:

• Linear formulation alone does not guarantee binary solutions; many linear programs have fractional solutions.
• A non-empty feasible region only means a solution exists, not that it will be binary.
• Approximation algorithms are for problems where exact solutions are too complex; the assignment problem's structure allows for efficient exact, binary solutions.

This problem asks us to find the best location for a new facility to serve four demand points, minimizing the total weighted distance using the rectilinear (Manhattan) distance. The rectilinear distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $d = |x_1 - x_2| + |y_1 - y_2|$. Our objective is to minimize $\sum_{i=1}^{n} w_i (|x - x_i| + |y - y_i|)$, where $(x, y)$ is the facility's location, $(x_i, y_i)$ are demand point coordinates, and $w_i$ are their demands. A crucial property of rectilinear distance is that the optimal $x$ and $y$ coordinates can be found independently.

First, let's find the optimal x-coordinate. We need to find the weighted median of the x-coordinates.
The demand points and their demands are:
Point 1: $(1, 2)$, $w_1 = 700$
Point 2: $(2, 3)$, $w_2 = 100$
Point 3: $(3, 5)$, $w_3 = 300$
Point 4: $(4, 1)$, $w_4 = 500$

The total demand is $W = 700 + 100 + 300 + 500 = 1600$. Half of the total demand is $W/2 = 1600 / 2 = 800$.
Now, we sort the x-coordinates and calculate the cumulative demand:
| X-coordinate ($x_i$) | Demand ($w_i$) | Cumulative Demand |
|:----------------------:|:---------------:|:-----------------:|
| 1 | 700 | 700 |
| 2 | 100 | $700 + 100 = 800$ |
| 3 | 300 | $800 + 300 = 1100$|
| 4 | 500 | $1100 + 500 = 1600$|
The cumulative demand first reaches or exceeds $W/2 = 800$ at $x=2$. Since the cumulative demand exactly equals 800 at $x=2$, the optimal x-coordinate can be any value between $x=2$ and the next x-coordinate, $x=3$. So, the optimal x-coordinate is in the interval $[2, 3]$.

Next, we find the optimal y-coordinate by finding the weighted median of the y-coordinates.
The total demand is still $W = 1600$, and $W/2 = 800$.
Sort the y-coordinates and calculate the cumulative demand:
| Y-coordinate ($y_i$) | Demand ($w_i$) | Cumulative Demand |
|:----------------------:|:---------------:|:-----------------:|
| 1 | 500 | 500 |
| 2 | 700 | $500 + 700 = 1200$|
| 3 | 100 | $1200 + 100 = 1300$|
| 5 | 300 | $1300 + 300 = 1600$|
The cumulative demand first reaches or exceeds $W/2 = 800$ at $y=2$ (where the cumulative demand is 1200). Thus, the optimal y-coordinate is $y=2$.

Combining these results, the optimal facility location $(x, y)$ must satisfy $x \in [2, 3]$ and $y = 2$. This describes the line segment connecting the points $(2, 2)$ and $(3, 2)$. This matches option A.

The Work Content represents the total time required to complete a job, including all necessary allowances. It's calculated by first finding the Normal Time (or Basic Work Content) and then adding various allowances.

Here's how we calculate it:

1. Basic Work Content (Normal Time): This is the time an average worker would take if working at a normal pace.
   $\text{Normal Time} = \text{Observed time} \times \text{Rating factor}$
   $\text{Normal Time} = 15 \text{ minutes} \times 0.9 = 13.5 \text{ minutes}$

2. Total Work Content: This includes the Normal Time plus allowances for personal needs and fatigue. Allowances for contingencies (like delays due to management issues) are generally considered separately as they relate to excess work content due to external factors, not directly tied to the operator's effort or basic process time.

   The allowance factor is applied to account for these additional times. The formula used for applying allowances in this context is:
   $\text{Allowance Factor} = \frac{100}{100 - \text{allowance \in percentage}}$

   Personal need allowance ($4\%$) and fatigue allowance ($3\%$) are applied sequentially as they relate to the manufacturing method. Contingency allowances for work ($3\%$) and delay ($2\%$) are related to shortcomings of management and are not included in this specific calculation of Total Work Content based on the provided methodology.

   $\text{Total Work Content} = \text{Normal Time} \times \frac{100}{100 - \text{Personal need allowance}} \times \frac{100}{100 - \text{Fatigue allowance}}$
   $\text{Total Work Content} = 13.5 \times \frac{100}{100 - 4} \times \frac{100}{100 - 3}$
   $\text{Total Work Content} = 13.5 \times \frac{100}{96} \times \frac{100}{97}$
   $\text{Total Work Content} = 14.497 \approx 14.5 \text{ minutes}$

This problem asks us to determine the Lower Control Limit (LCL) and Upper Control Limit (UCL) for a p-chart, a tool used in Statistical Process Control (SPC) to monitor the proportion of defective items. The control limits are set at 3 standard deviations from the process average.

First, we calculate the average proportion of defectives ($\bar{p}$).
Total items inspected = Number of samples $\times$ Sample size = $24 \times 50 = 1200$ items.
$\bar{p} = \frac{\text{Total defective items}}{\text{Total items inspected}} = \frac{52}{1200} = \frac{13}{300} \approx 0.04333$

Next, we calculate the standard deviation ($\sigma_p$) for the p-chart using the formula:
$\sigma_p = \sqrt{\frac{\bar{p}(1-\bar{p})}{n}}$
Plugging in the values:
$\sigma_p = \sqrt{\frac{0.04333 (1 - 0.04333)}{50}} = \sqrt{\frac{0.04333 \times 0.95667}{50}} = \sqrt{\frac{0.041445}{50}} = \sqrt{0.0008289} \approx 0.02879$

Now, we calculate the LCL and UCL.
$\text{LCL} = \bar{p} - 3\sigma_p = 0.04333 - 3 \times 0.02879 = 0.04333 - 0.08637 = -0.04304$
Since the proportion of defectives cannot be negative, the practical LCL is set to $0$.

$\text{UCL} = \bar{p} + 3\sigma_p = 0.04333 + 3 \times 0.02879 = 0.04333 + 0.08637 = 0.12970$
Rounding to two decimal places, UCL $\approx 0.13$.

Thus, the lower and upper control limits are $(0, 0.13)$.

According to Taylor's differential piece-rate wage incentive plan, if a worker's actual output is less than the standard production, the total wage is calculated using the standard piece rate.

Here, the actual output ($18$ units) is less than the standard production ($20$ units).
So, the total wage paid to the worker is:
$\text{Total wage} = \text{Actual no. of units produced} \times \text{Price per unit}$
$\text{Total wage} = 18 \times 500 = 9000$
Therefore, the wage paid to the worker as per Taylor's differential piece-rate wage incentive plan is Rs. $9000$.

Computer-Aided Process Planning (CAPP) uses computer software to automate manufacturing process planning. The question asks about determining a process sequence for any part design without predefined standard plans. Let's look at the different CAPP approaches:

• Retrieval Type Process Planning: This method searches a database for a similar, already-planned part and retrieves its existing plan, which might then be modified. It relies on a library of existing plans.
• Variant Type Process Planning: Similar to retrieval, this type also uses existing plans but systematically modifies them for a new part that is sufficiently similar. It often uses coding and classification to group parts.
• Generative Type Process Planning: This is the key. It does not rely on predefined or existing standard plans. Instead, it automatically creates a process plan directly from the part's design data (like CAD models), manufacturing rules, and available machinery using algorithms and expert knowledge. This directly matches the scenario of determining a process sequence without predefined standard plans.
• Group Technology (GT) Based Process Planning: GT is a classification method for grouping similar parts, often used to support variant or retrieval CAPP systems by finding similar plans. It's a classification technique, not a plan generation method itself.

Therefore, determining the process sequence for any part design without predefined standard plans is known as Generative type process planning.

Material Requirements Planning (MRP) manages inventory and plans production by determining what materials are needed, how much, and when.

• Bill of Materials (BOM): This is a fundamental input, listing all raw materials, components, and quantities required for a product. MRP uses BOM to calculate material needs.
• Inventory Level: Crucial for MRP, this indicates current stock. MRP uses it to calculate net requirements, determining what more needs to be ordered or produced.
• Production Schedule: MRP integrates with this schedule (often from a Master Production Schedule), which outlines products, quantities, and deadlines, driving the procurement and production planning of components.
• Material Price: While material price impacts purchasing and cost, it is not a direct input or calculation component within the core logic of MRP. MRP focuses on material needs ('what', 'how much', 'when'), while pricing is typically handled by purchasing/finance, which uses MRP's output.

Therefore, material price is not included in Material Requirements Planning.

In Value Engineering, the 'Value' of a product or service is defined as the ratio of its 'Function' to its 'Cost'. The function represents the benefits derived, while the cost is what is paid for it.

$\text{Value} = \frac{\text{Function}}{\text{Cost}}$

Here, the 'Function' of air travel is the sum of its monetized comfort and the monetary value of time saved:
Monetary Function = Monetized comfort + Monetary time saved
Monetary Function = Rs. 3000 + Rs. 3000 = Rs. 6000

The 'Cost' of air travel is the ticket price, which is Rs. 4000.

Now, substituting these values into the formula:
$\text{Value} = \frac{\text{Rs. 6000}}{\text{Rs. 4000}} = 1.5$

The Ishikawa diagram, also known as a cause-and-effect or fishbone diagram, is a visualization tool designed to identify and categorize the potential causes leading to a specific problem or "effect." It systematically maps out these causes, showing their relationship to the problem.

• Option A: different types of quality defects - This diagram doesn't list defect types; it focuses on the causes of one specific defect.
• Option B: quantitative relation between the extent of defect and a process parameter - This describes statistical analysis tools, not the qualitative cause-and-effect identification of an Ishikawa diagram.
• Option C: relation between defects and their causes - This accurately captures the diagram's purpose: to visually link a problem (defect) to its contributing causes.
• Option D: prioritized quality defects - Prioritization is typically done by tools like Pareto charts, not by the Ishikawa diagram itself.

Therefore, the Ishikawa diagram represents the relation between defects and their causes.

To find the total tardiness for the given sequence, we need to calculate the completion time and tardiness for each product in the specified order: R-P-S-T-Q. Tardiness for a job ($T_i$) is the delay past its due date, calculated as $T_i = \max(0, C_i - D_i)$, where $C_i$ is the completion time and $D_i$ is the due date. The total tardiness is the sum of individual job tardiness values.

Let's break down the calculation for each product:

1. Job R:

   • $R_R = 0$, $P_R = 2$, $D_R = 15$.
   • Start Time ($S_R$) = $\max(\text{Release Time}, \text{Machine Available Time}) = \max(0, 0) = 0$.
   • Completion Time ($C_R$) = $S_R + P_R = 0 + 2 = 2$.
   • Tardiness ($T_R$) = $\max(0, C_R - D_R) = \max(0, 2 - 15) = 0$.
   • Machine available at time 2.

2. Job P:

   • $R_P = 3$, $P_P = 10$, $D_P = 9$.
   • Start Time ($S_P$) = $\max(\text{Release Time}, \text{Machine Available Time}) = \max(3, 2) = 3$.
   • Completion Time ($C_P$) = $S_P + P_P = 3 + 10 = 13$.
   • Tardiness ($T_P$) = $\max(0, C_P - D_P) = \max(0, 13 - 9) = 4$.
   • Machine available at time 13.

3. Job S:

   • $R_S = 1$, $P_S = 5$, $D_S = 11$.
   • Start Time ($S_S$) = $\max(\text{Release Time}, \text{Machine Available Time}) = \max(1, 13) = 13$.
   • Completion Time ($C_S$) = $S_S + P_S = 13 + 5 = 18$.
   • Tardiness ($T_S$) = $\max(0, C_S - D_S) = \max(0, 18 - 11) = 7$.
   • Machine available at time 18.

4. Job T:

   • $R_T = 11$, $P_T = 1$, $D_T = 13$.
   • Start Time ($S_T$) = $\max(\text{Release Time}, \text{Machine Available Time}) = \max(11, 18) = 18$.
   • Completion Time ($C_T$) = $S_T + P_T = 18 + 1 = 19$.
   • Tardiness ($T_T$) = $\max(0, C_T - D_T) = \max(0, 19 - 13) = 6$.
   • Machine available at time 19.

5. Job Q:

   • $R_Q = 2$, $P_Q = 4$, $D_Q = 9$.
   • Start Time ($S_Q$) = $\max(\text{Release Time}, \text{Machine Available Time}) = \max(2, 19) = 19$.
   • Completion Time ($C_Q$) = $S_Q + P_Q = 19 + 4 = 23$.
   • Tardiness ($T_Q$) = $\max(0, C_Q - D_Q) = \max(0, 23 - 9) = 14$.
   • Machine available at time 23.

The individual tardiness values are $T_R = 0$, $T_P = 4$, $T_S = 7$, $T_T = 6$, and $T_Q = 14$.

The total tardiness is the sum of these values:
$\text{Total Tardiness} = T_R + T_P + T_S + T_T + T_Q = 0 + 4 + 7 + 6 + 14 = 31$

The Product Life Cycle (PLC) has four stages: Introduction, Growth, Maturity, and Decline. Initially, product variety is low in the Introduction stage. It increases during the Growth and Maturity stages, peaking in Maturity as companies offer a wide range of variations to compete. However, as the product passes through ALL stages, the question focuses on the overall trend. In the Decline stage, demand drops, leading companies to streamline operations and discontinue unpopular or unprofitable variations. This results in a significant decrease in product variety. Therefore, considering the entire progression, product variety ultimately decreases, especially in the Decline stage due to product line rationalization.

Project crashing aims to reduce the project duration by shortening critical activities, which usually increases direct costs but can reduce indirect (overhead) costs.

First, let's identify the critical path and normal project duration.
Given the activities:

• 1-2: Normal Time = 6 days
• 1-3: Normal Time = 7 days
• 1-4: Normal Time = 10 days
• 2-4: Normal Time = 4 days
• 3-4: Normal Time = 5 days

Paths and their normal durations:

• Path 1-2-4: $6 + 4 = 10$ days
• Path 1-3-4: $7 + 5 = 12$ days
• Path 1-4: $10$ days

The critical path is 1-3-4 with a normal duration of 12 days.
The initial total project cost (considering only overhead) is:
Normal Duration $\times$ Overhead Cost per Day $= 12 \text{ days} \times \text{Rs. } 200/\text{day} = \text{Rs. } 2400$.

Now, let's calculate the project duration and crashing costs after crashing activities as per the table:

• Activity 1-2: Normal = 6, Crashed by 1 day $\implies$ New Time = 5 days. Crashing Cost = $1 \times \text{Rs. } 100 = \text{Rs. } 100$.
• Activity 1-3: Normal = 7, Crashed by 2 days $\implies$ New Time = 5 days. Crashing Cost = $2 \times \text{Rs. } 100 = \text{Rs. } 200$.
• Activity 1-4: Normal = 10, Crashed by 1 day $\implies$ New Time = 9 days. Crashing Cost = $1 \times \text{Rs. } 100 = \text{Rs. } 100$.
• Activity 2-4: Normal = 4, Not Crashed $\implies$ New Time = 4 days. Crashing Cost = $0 \times \text{Rs. } 200 = \text{Rs. } 0$.
• Activity 3-4: Normal = 5, Crashed by 1 day $\implies$ New Time = 4 days. Crashing Cost = $1 \times \text{Rs. } 200 = \text{Rs. } 200$.

With the crashed times:

• Path 1-2-4: $5 + 4 = 9$ days
• Path 1-3-4: $5 + 4 = 9$ days
• Path 1-4: $9$ days

After crashing, all paths become critical paths, and the new project duration is 9 days.
The total crashing cost is: $100 + 0 + 200 + 100 + 200 = \text{Rs. } 600$.
The overhead cost for the crashed project duration is: $9 \text{ days} \times \text{Rs. } 200/\text{day} = \text{Rs. } 1800$.
The total project cost after crashing is: Overhead Cost + Total Crashing Cost $= \text{Rs. } 1800 + \text{Rs. } 600 = \text{Rs. } 2400$.
The net change in project cost is: Cost after crashing $-$ Cost before crashing $= \text{Rs. } 2400 - \text{Rs. } 2400 = \text{Rs. } 0$.

The core concept here is the Productivity Index, which measures how much productivity has changed relative to a base period. It's calculated as the ratio of current productivity to base period productivity:

$\text{Productivity Index} = \frac{\text{Present Productivity}}{\text{Base Period Productivity}}$

First, let's calculate the Present Productivity: This combines the revenue generated and the total labor cost for the current year.
$\text{Present Productivity} = 200,000 + (1000 \times 22) = 222,000$

Next, we calculate the Base Year Productivity (from the previous year's data):
$\text{Base Year Productivity} = 220,000 + (1200 \times 18) = 241,600$

Finally, we calculate the Productivity Index:
$\text{Productivity Index} = \frac{222,000}{241,600} \approx 0.9$
Thus, the labor productivity index for this plant is $0.9$.

The core concept here is that the total time spent producing items (Net Production Time) is equal to the time it takes to produce one item (Cycle Time) multiplied by the total number of units produced.

Given that the plant operates for $2$ shifts of $8$ hours each, the Net Production Time in seconds is calculated as $2 \times 8 \times 3600 = 57600$ seconds. The total Number of Units Produced is $30000$.
Using the formula:
$Cycle\; time = \frac{Net \;Production\; time}{Number\;of\;Units\;Produced}$
We substitute the values:
$Cycle\; time = \frac{57600}{30000} = 1.92$
Therefore, the Cycle time is $1.92$ seconds, which can be rounded to $1.9$ to $2$ seconds.

To set up a control chart for the number of non-conformities per unit (a c-chart), we first need to calculate the average number of non-conformities ($\bar{c}$). This chart is used when we're counting defects or non-conformities on a single item.

We are given:
Total non-conformities = 16
Total units inspected = 30 (which acts as the number of subgroups, $k$)

First, calculate the central line, which is the average non-conformities per unit ($\bar{c}$):
$\bar{c} = \frac{\text{Total non-conformities}}{\text{Number of subgroups}} = \frac{16}{30} \approx 0.5333$

Next, we use the standard formulas for the Upper Control Limit (UCL) and Lower Control Limit (LCL) for a c-chart. These limits are set at 3 standard deviations from the average:
UCL $= \bar{c} + 3\sqrt{\bar{c}}$
LCL $= \bar{c} - 3\sqrt{\bar{c}}$

Let's calculate the values:
$\sqrt{\bar{c}} = \sqrt{\frac{16}{30}} \approx \sqrt{0.5333} \approx 0.7303$
For UCL:
$UCL \approx 0.5333 + (3 \times 0.7303)$
$UCL \approx 0.5333 + 2.1909$
$UCL \approx 2.7242$
For LCL:
$LCL \approx 0.5333 - (3 \times 0.7303)$
$LCL \approx 0.5333 - 2.1909$
$LCL \approx -1.6576$

Since the number of non-conformities cannot be negative, any calculated LCL less than zero must be set to 0. Therefore:
$LCL = 0.00$

Comparing our calculated values (UCL $\approx 2.72$, LCL $= 0.00$) with the options, Option B (UCL = 2.71, LCL = 0.00) is the closest match.

A Just-In-Time (JIT) manufacturing system focuses on minimizing waste and improving efficiency by receiving materials and producing goods only when needed. Let's analyze the given options in the context of JIT characteristics:

• A. Reduction of lot sizes: JIT systems strive to produce in smaller batches, ideally single units, to reduce inventory holding, shorten lead times, and increase flexibility. This is a core JIT principle.
• C. Small but frequent deliveries: To support minimal inventory, JIT relies on frequent, small deliveries from suppliers that precisely match production requirements. This aligns with JIT.
• D. Higher productivity: By eliminating waste (e.g., excess inventory, waiting time, defects) and streamlining processes, JIT systems inherently lead to increased overall productivity. This is a direct outcome of JIT implementation.
• B. Efficient use of buffer inventory: The fundamental principle of JIT is to minimize or eliminate all forms of inventory, including buffer inventory (extra stock held to cover variations or disruptions). JIT aims for a "zero inventory" state, relying on strong supplier relationships, high quality control, flexible workforce, and preventive maintenance to achieve smooth production flow without buffers. Therefore, "efficient use of buffer inventory" contradicts the JIT philosophy, which seeks to reduce, not just efficiently manage, buffer inventory.

We need to find the number of feasible corner points and check if the optimal solution is degenerate for the given Linear Programming Problem (LPP):
Maximize: $z = 2x_1 + 3x_2$
Subject to:
$c_1: x_1 + 2x_2 \le 60$
$c_2: 2x_1 + x_2 \le 30$
$c_3: x_1 - x_2 \ge -10 \quad (\text{or } -x_1 + x_2 \le 10)$
$c_4: x_1 \ge 0$
$c_5: x_2 \ge 0$

Step 1: Identify all potential corner points by finding intersections of boundary lines.
The boundary lines are:

1. $x_1 + 2x_2 = 60$
2. $2x_1 + x_2 = 30$
3. $x_1 - x_2 = -10$
4. $x_1 = 0$
5. $x_2 = 0$

Let's find the intersection points and check their feasibility:

• P1: Intersection of $x_1=0$ and $x_2=0$
  Point: $(0, 0)$
  Feasibility Check: $0+2(0)\le 60$ (True), $2(0)+0\le 30$ (True), $0-0\ge -10$ (True).
  Result: Feasible.

• P2: Intersection of $x_2=0$ and $2x_1+x_2=30$
  Substituting $x_2=0$ into $2x_1+x_2=30 \implies 2x_1 = 30 \implies x_1=15$.
  Point: $(15, 0)$
  Feasibility Check: $15+2(0)\le 60$ (True), $2(15)+0\le 30$ (True), $15-0\ge -10$ (True).
  Result: Feasible.

• P3: Intersection of $x_1=0$ and $x_1-x_2=-10$
  Substituting $x_1=0$ into $x_1-x_2=-10 \implies -x_2 = -10 \implies x_2=10$.
  Point: $(0, 10)$
  Feasibility Check: $0+2(10)\le 60$ (True), $2(0)+10\le 30$ (True), $0-10\ge -10$ (True).
  Result: Feasible.

• P4: Intersection of $2x_1+x_2=30$ and $x_1-x_2=-10$
  Adding the two equations: $(2x_1+x_2) + (x_1-x_2) = 30 - 10 \implies 3x_1 = 20 \implies x_1 = 20/3$.
  Substitute $x_1=20/3$ into $2x_1+x_2=30 \implies 2(20/3)+x_2=30 \implies 40/3+x_2=90/3 \implies x_2=50/3$.
  Point: $(20/3, 50/3)$
  Feasibility Check: $x_1\ge 0, x_2\ge 0$ (True). $20/3+2(50/3) = 120/3 = 40 \le 60$ (True). $2(20/3)+50/3 = 90/3 = 30 \le 30$ (True). $20/3-50/3 = -30/3 = -10 \ge -10$ (True).
  Result: Feasible.

• Intersection of $x_1+2x_2=60$ and $2x_1+x_2=30$
  Multiply the second equation by 2: $4x_1+2x_2=60$.
  Subtract the first equation ($x_1+2x_2=60$) from this: $(4x_1+2x_2) - (x_1+2x_2) = 60-60 \implies 3x_1=0 \implies x_1=0$.
  Substitute $x_1=0$ into $2x_1+x_2=30 \implies x_2=30$.
  Point: $(0, 30)$
  Feasibility Check: $0-30 = -30 \not\ge -10$.
  Result: Not Feasible.

• Intersection of $x_1+2x_2=60$ and $x_1-x_2=-10$
  From $x_1-x_2=-10 \implies x_1=x_2-10$.
  Substitute into $x_1+2x_2=60 \implies (x_2-10)+2x_2=60 \implies 3x_2=70 \implies x_2=70/3$.
  Then $x_1 = 70/3 - 10 = 40/3$.
  Point: $(40/3, 70/3)$
  Feasibility Check: $2(40/3)+70/3 = 80/3+70/3 = 150/3 = 50 \not\le 30$.
  Result: Not Feasible.

We have found 4 feasible corner points: $(0, 0)$, $(15, 0)$, $(0, 10)$, and $(20/3, 50/3)$.

Step 2: Determine if the optimal solution is degenerate.
An optimal solution is degenerate if, at that corner point, more constraints than the number of variables are binding (active as equalities). Here, we have $n=2$ variables ($x_1, x_2$). A corner point is degenerate if more than 2 constraints are binding.

Let's check the binding constraints for each feasible corner point:

• P1 $(0, 0)$: Constraints $x_1=0$ and $x_2=0$ are binding. (2 constraints = $n$).
• P2 $(15, 0)$: Constraints $x_2=0$ and $2x_1+x_2=30$ are binding. (2 constraints = $n$).
• P3 $(0, 10)$: Constraints $x_1=0$ and $x_1-x_2=-10$ are binding. (2 constraints = $n$).
• P4 $(20/3, 50/3)$: Constraints $2x_1+x_2=30$ and $x_1-x_2=-10$ are binding. (2 constraints = $n$).

Since exactly two constraints (equal to the number of variables) are binding at each feasible corner point, none of the feasible corner points are degenerate. This implies the optimal solution, which must occur at one of these non-degenerate corner points, is also not degenerate. (The objective function value at P4 is $z = 2(20/3) + 3(50/3) = 40/3 + 150/3 = 190/3$, which is the optimal value.)

Thus, there are 4 feasible corner point solutions, and the optimal solution is not degenerate.