Model Answer: TV Repairman Queuing Problem (M/M/1)

STEP 1 - Given Data

• Service time: Exponential, mean = 30 minutes
• Arrival: Poisson, average rate = 10 sets per 8-hour day
• Working day = 8 hours = 480 minutes

STEP 2 - Solution

Rates (converted to consistent units, per hour)

Service rate: mean service time 30 min = 0.5 hour, so $$\mu = \frac{1}{0.5} = 2 \text{ sets/hour}$$

Arrival rate: $$\lambda = \frac{10}{8} = 1.25 \text{ sets/hour}$$

Traffic intensity

$$\rho = \frac{\lambda}{\mu} = \frac{1.25}{2} = 0.625$$

Since $\rho < 1$, the system is stable.

Expected idle time each day

Probability system is empty (repairman idle): $$P_0 = 1 - \rho = 1 - 0.625 = 0.375$$

Expected idle time per 8-hour day: $$= P_0 \times 8 = 0.375 \times 8 = 3 \text{ hours}$$

Expected number of jobs ahead of the set just brought in

The number of jobs "ahead" of an arriving set is the expected number in the system $L$ (all sets already present that must be dealt with before/including the one in service):

$$L = \frac{\lambda}{\mu - \lambda} = \frac{1.25}{2 - 1.25} = \frac{1.25}{0.75} = \frac{5}{3} \approx 1.67 \text{ jobs}$$

Final Answers

• Expected idle time each day: $3$ hours (180 minutes)
• Expected number of jobs ahead of the set just brought in: $L = \dfrac{5}{3} \approx 1.67$ jobs

Note on interpretation

The standard textbook interpretation of "how many jobs are ahead of the set just brought in" is the expected number in the system $L = 5/3 \approx 1.67$ (the newly arrived job sees, on average, $L$ jobs ahead of it, by PASTA). The alternative reading, the number waiting in the queue rather than in the system, gives $L_q \approx 1.04$. The conventional answer to this classic problem is $L = 5/3 \approx 1.67$.

Critical Path Analysis - Verified Solution

Given Data

Step 1: Forward Pass (ES, EF)

$EF = ES + \text{Duration}$, and $ES = \max(EF \text{ of predecessors})$.

Project duration = 42 weeks.

Step 2: Backward Pass (LF, LS)

$LS = LF - \text{Duration}$, and $LF = \min(LS \text{ of successors})$.

Successors:

• J → (end)
• I → J
• H → I
• G → I
• F → G, H
• E → G
• D → H
• C → F
• B → C, D
• A → C, E

Step 3: Slack (LS − ES)

Step 4: Critical Path

Zero-slack activities: B → C → F → H → I → J

Check: $8 + 4 + 7 + 6 + 8 + 9 = 42$ weeks.

Result

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

Model Answer: Marginal Analysis & Arithmetic Method in Game Theory

a) Marginal Analysis Approach in Decision Making (2.5 marks)

Definition: Marginal analysis is a decision-making technique that examines the incremental changes in costs and benefits resulting from a specific decision or action, rather than analyzing total values.

Key Concepts:

• Marginal Cost (MC): The additional cost incurred by producing or doing one more unit of output.
• Marginal Benefit (MB): The additional benefit or revenue gained from one more unit of output.

Decision Rule:

• Continue an activity as long as Marginal Benefit > Marginal Cost
• Stop when Marginal Benefit = Marginal Cost (optimal point)
• Do not proceed when Marginal Benefit < Marginal Cost

Application in Decision Making:

• Used in production decisions: whether to increase output
• Used in resource allocation: determining optimal level of investment
• Used in cost-benefit analysis: evaluating project feasibility
• Helps identify the point of diminishing returns

Advantage: Focuses on relevant incremental changes rather than total figures, making it practical for tactical business decisions.

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b) Arithmetic Method in Game Theory (2.5 marks)

Definition: The arithmetic method is a technique used to solve 2×2 games (two players, two strategies each) to find the optimal mixed strategy for each player when no pure strategy equilibrium exists.

Procedure:

1. Identify the payoff matrix for both players

2. Check for saddle point (pure strategy equilibrium) - if exists, that is the solution

3. If no saddle point exists, calculate probability of each strategy:

   For Player A (Row player):

   • Probability of Strategy 1: p = (d - c)/(a - b - c + d)
   • Probability of Strategy 2: q = 1 - p

   For Player B (Column player):

   • Probability of Strategy 1: r = (d - b)/(a - b - c + d)
   • Probability of Strategy 2: s = 1 - r

   Where a, b, c, d are payoffs in the 2×2 matrix

4. Calculate expected payoff using the probabilities

Advantage: Provides a systematic algebraic approach to find optimal mixed strategies when players should randomize between strategies.

Limitation: Applicable only to 2×2 games; larger games require linear programming or other methods.