Network Analysis - Critical Path Method

Step 1: Given Data

Step 2: Forward Pass (ES, EF)

Using $EF = ES + t$ and $ES = \max(EF \text{ of predecessors})$:

Project completion time = 54 weeks

Step 3: Backward Pass (LF, LS)

Using $LS = LF - t$ and $LF = \min(LS \text{ of successors})$. Project LF = 54.

Successor mapping:

• H → end; G → H; C, E, F → G; D → E; A → C, E; B → D, F

Step 4: Slack and Critical Activities

$\text{Slack} = LS - ES$

Step 5: Critical Path

Critical activities (zero slack): B, F, G, H

$$\text{Critical Path: } B \rightarrow F \rightarrow G \rightarrow H$$

Verification of length: $$12 + 6 + 14 + 22 = 54 \text{ weeks}$$

Network Diagram

        A(5) ── C(6) ─┐
       /              │
Start                 │
       \              ├── G(14) ── H(22) ── End
        B(12) ─ D(3) ─ E(2) ┐        
             \              │
              F(6) ─────────┘

Answer Summary

Check the slack before calling an activity critical: activity A has slack = 7 (LS = 7, ES = 0), so A is not critical. With a project duration of 54 weeks and critical path B -> F -> G -> H, the critical activities are only B, F, G, H.

Hungarian Method - Assignment Problem (Maximization)

STEP 1 - EXTRACT: Given Data

Profit matrix (bids in thousands Rs.), one bid accepted per individual:

Objective: maximize total revenue via one-to-one assignment.

STEP 2 - SOLVE

Convert Maximization to Minimization

Maximum element = 115. Subtract each element from 115 to get the cost matrix.

Row Reduction

Row minimums: A = 5, B = 5, C = 10, D = 0.

Column Reduction

Column minimums: Honda = 0, Hero = 10, Bajaj = 0, Yamaha = 0.

Cover Zeros with Minimum Lines

Zeros locations:

• A: Bajaj
• B: Honda, Hero
• C: Honda, Hero, Yamaha
• D: Honda

Minimum lines to cover all zeros:

• Line 1: Column Honda (covers B, C, D zeros in Honda)
• Line 2: Column Bajaj (covers A)
• Line 3: Row C (covers Hero, Yamaha zeros of C)
• Line 4: Row B (covers Hero zero of B)

That is 4 lines. Since number of lines (4) = order of matrix (4), an optimal assignment exists.

Make Optimal Assignment

Assign row by row using unique zeros:

• Row A: only zero at Bajaj → A → Bajaj. Strike column Bajaj.
• Row D: only zero at Honda → D → Honda. Strike column Honda.
• Row B: zeros at Honda (gone), Hero. → B → Hero. Strike column Hero.
• Row C: zeros at Honda (gone), Hero (gone), Yamaha. → C → Yamaha.

Assignment:

• A → Bajaj
• B → Hero
• C → Yamaha
• D → Honda

Compute Maximum Revenue

Using original bids:

• A → Bajaj = 110
• B → Hero = 100
• C → Yamaha = 105
• D → Honda = 115

$$\text{Total} = 110 + 100 + 105 + 115 = 430$$

Maximum Revenue = Rs. 430,000 (Rs. 430 thousand).

Final Answer

Maximum Revenue = Rs. 430,000

Queue and Single Channel Queuing Model

What is a Queue?

A queue is a linear data structure that follows the FIFO (First-In-First-Out) principle, where elements are inserted at the rear end and removed from the front end. In the context of queuing theory, a queue refers to a waiting line of customers or entities awaiting service.

Operating Characteristics of Single Channel Queuing Model

The single channel queuing model (also called M/M/1 queue) has the following key operating characteristics:

1. Single Server/Channel

• Only one service facility or server is available
• All arriving customers must wait in a single queue if the server is busy
• Customers are served one at a time in the order they arrive

2. FIFO Service Discipline

• Customers are served in the order of their arrival (First-Come-First-Served)
• No priority or preemption is allowed
• Each customer waits until all previous customers are served

3. Arrival Pattern

• Arrivals follow a Poisson distribution with arrival rate λ (lambda)
• Arrivals are random and independent
• The inter-arrival times are exponentially distributed

4. Service Time Distribution

• Service times follow an exponential distribution with service rate μ (mu)
• Service times are independent and identically distributed
• The server works continuously when customers are present

5. Queue Capacity

• The queue has unlimited capacity (infinite waiting space)
• Customers do not balk or renege (they always join and wait)

6. Steady State Condition

• The system operates under the condition λ < μ (arrival rate less than service rate)
• This ensures the queue does not grow indefinitely

7. Key Performance Metrics

• Utilization factor (ρ) = λ/μ
• Average number of customers in system (L) = λ/(μ - λ)
• Average waiting time in queue (Wq) = λ/[μ(μ - λ)]