Minimum Transportation Cost - Verified Solution

Step 1: Extract Given Data

I must read the matrix carefully. The LaTeX has two conflicting layouts. The trailing (cleaner) array is:

$$ \begin{array}{|c|ccc|c|}\hline \text{Destination/Source} & D1 & D2 & D3 & \text{Units Available} \ \hline S1 & 12 & 15 & 11 & 200 \ S2 & 16 & 12 & 12 & 150 \ S3 & 10 & 10 & 12 & 100 \ \hline \text{Units Demanded} & 200 & 200 & 50 & 450 \ \hline \end{array} $$

So sources are S1, S2, S3 (supplies 200, 150, 100) and destinations are D1, D2, D3 (demands 200, 200, 50).

Cost matrix $c_{ij}$ (row = source, column = destination):

Note: the same table is sometimes laid out with sources as columns; the numeric cells are then the transpose of this one, and the allocations are unchanged.

Step 2: Check Balance

• Supply $= 200 + 150 + 100 = 450$
• Demand $= 200 + 200 + 50 = 450$

Balanced. Proceed with VAM.

Step 3: Vogel's Approximation Method

Iteration 1

Row penalties (2nd smallest − smallest):

• S1: rows $12,15,11 \Rightarrow 12-11=1$
• S2: $16,12,12 \Rightarrow 12-12=0$
• S3: $10,10,12 \Rightarrow 10-10=0$

Column penalties:

• D1: $12,16,10 \Rightarrow 12-10=2$
• D2: $15,12,10 \Rightarrow 12-10=2$
• D3: $11,12,12 \Rightarrow 12-11=1$

Max penalty $= 2$ (D1 or D2). Take D1. Min cost in D1 = 10 (S3). Allocate $\min(100,200)=100$ to S3-D1. S3 exhausted; D1 remaining = 100.

Iteration 2 (S3 removed)

Column penalties: D1 $=16-12=4$, D2 $=15-12=3$, D3 $=12-11=1$. Max $=4$ (D1). Min cost in D1 = 12 (S1). Allocate $\min(200,100)=100$ to S1-D1. D1 satisfied; S1 remaining = 100.

Iteration 3 (D1 removed)

Column penalties: D2 $=15-12=3$, D3 $=12-11=1$. Max $=4$ (S1). Min cost in S1 = 11 (D3). Allocate $\min(100,50)=50$ to S1-D3. D3 satisfied; S1 remaining = 50.

Iteration 4

Remaining cells: S1-D2 (supply 50), S2-D2 (supply 150), demand D2 = 200. Allocate $50$ to S1-D2 and $150$ to S2-D2.

Step 4: Final Allocation Table

Number of allocations = 5 = $m+n-1 = 3+3-1$. Non-degenerate. ✓

Step 5: Total Cost

$$Z = 1000 + 1200 + 550 + 750 + 1800 = \mathbf{Rs.\ 5300}$$

Common Mistakes to Avoid

Allocating 50 to S3-D2 when S3 has already supplied 100 to S3-D1 makes S3's total 150, which exceeds S3's supply of 100 and violates the row constraint, so such an allocation is infeasible. The correct VAM solution sends the leftover 50 units of S1 to D2 (cost 15), giving $Rs.\ 5300$, not $Rs.\ 5050$.

Final Answer

Minimum Transportation Cost = Rs. 5300

Critical Path Analysis

Step 1: Given Data

Step 2: Forward Pass (Earliest Times)

$ES$ = max of predecessors' $EF$; $EF = ES + \text{duration}$.

Project completion time = 42 weeks

Step 3: Backward Pass (Latest Times)

Project finish $LF = 42$. $LF$ = min of successors' $LS$; $LS = LF - \text{duration}$; Slack $= LS - ES$.

Successor mapping:

• J → none; I → J; G, H → I; F → G, H; E → G; D → H; C → F; A → C, E; B → C, D

Step 4: Identify Critical Path

Activities with Slack = 0: B, C, F, H, I, J.

$$\textbf{Critical Path: } B \to C \to F \to H \to I \to J$$

Duration check: $$8 + 4 + 7 + 6 + 8 + 9 = 42 \text{ weeks}$$

Final Answer

• Critical Path: $B \to C \to F \to H \to I \to J$
• Project Duration: $42$ weeks

STEP 1 - EXTRACT: Given Data

Products: Tables ($x_1$), Chairs ($x_2$), Desks ($x_3$)

Department time limits per week:

• Cutting: 300 hours
• Assembly: 450 hours
• Painting: 200 hours

Resource requirements and profit:

Balance constraint: Total production $\leq 150$ units per week.

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STEP 2 - SOLVE: LPP Formulation

Decision Variables

Let:

• $x_1$ = number of Tables produced per week
• $x_2$ = number of Chairs produced per week
• $x_3$ = number of Desks produced per week

Objective Function (Maximize Profit)

$$\text{Maximize } Z = 1000x_1 + 800x_2 + 1200x_3 \quad \text{(Rs. per week)}$$

Constraints

Cutting department (300 hours): $$4x_1 + 3x_2 + 2x_3 \leq 300$$

Assembly department (450 hours): $$5x_1 + 4x_2 + 3x_3 \leq 450$$

Painting department (200 hours): $$1x_1 + 1x_2 + 2x_3 \leq 200$$

Production balance (max 150 total units): $$x_1 + x_2 + x_3 \leq 150$$

Non-negativity: $$x_1, x_2, x_3 \geq 0$$

Complete Model

$$\text{Max } Z = 1000x_1 + 800x_2 + 1200x_3$$

Subject to: $$ \begin{aligned} 4x_1 + 3x_2 + 2x_3 &\leq 300 \ 5x_1 + 4x_2 + 3x_3 &\leq 450 \ x_1 + x_2 + 2x_3 &\leq 200 \ x_1 + x_2 + x_3 &\leq 150 \ x_1, x_2, x_3 &\geq 0 \end{aligned} $$

All coefficients match the given data exactly. The formulation is correct and complete.

Applications of Operations Research in Different Fields

Operations Research (OR) is a quantitative discipline that applies mathematical and analytical methods to solve complex decision-making problems. Here are its major applications across different fields:

1. Manufacturing and Production

• Production scheduling and planning
• Inventory management and control
• Quality control and process optimization
• Resource allocation and capacity planning
• Minimizing production costs while maintaining quality standards

2. Transportation and Logistics

• Vehicle routing and fleet management
• Route optimization to minimize fuel and time
• Warehouse location and distribution network design
• Supply chain management
• Port and airport operations planning

3. Finance and Banking

• Portfolio optimization and investment decisions
• Risk management and analysis
• Credit allocation and loan management
• Capital budgeting
• Financial forecasting and planning

4. Healthcare

• Hospital resource allocation (beds, staff, equipment)
• Patient scheduling and appointment systems
• Ambulance routing and emergency response optimization
• Drug inventory management
• Treatment planning and resource optimization

5. Telecommunications

• Network design and optimization
• Bandwidth allocation
• Call routing and switching
• Infrastructure planning

6. Agriculture

• Crop planning and resource allocation
• Irrigation scheduling
• Pest management optimization
• Farm equipment utilization

7. Government and Public Sector

• Urban planning and development
• Traffic management
• Public resource allocation
• Policy analysis and decision-making

8. Retail and Commerce

• Inventory management
• Store location decisions
• Pricing strategies
• Demand forecasting

Key Benefit: OR helps organizations make optimal decisions under constraints, leading to cost reduction, efficiency improvement, and better resource utilization across all sectors.