ORS255 · TU past paper
Operations Research 2080 question paper
The complete TU 2080 exam paper for Operations Research (ORS255), all 12 questions with solved model answers written to the mark scheme.
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- 110 marksNumericalPERT network diagramsHideAnswer
PERT Network Analysis Problem
A project consists of nine activities whose time estimates (in weeks) and other characteristics are given below:
Activities A B C D E F G H I Preceding activities - - - A A B, D B, D C, F E Optimistic time 2 6 6 2 11 8 3 9 4 Most likely time 4 6 12 5 14 10 6 15 10 Pessimistic time 6 6 24 8 23 12 9 27 16 (a) Show the PERT network diagram for the project.
(b) Identify the critical activities.
(c) What is the expected project completing time and its variance?
[10]
Activity Predecessors a (opt) m (likely) b (pess) ------------------------------------------------------ A - 2 4 6 B - 6 6 6 C - 6 12 24 D A 2 5 8 E A 11 14 23 F B, D 8 10 12 G B, D 3 6 9 H C, F 9 15 27 I E 4 10 16 Note: Activities G and...
- 210 marksNumericalSimplex method for solving LPPHideAnswer
Solve the given linear programming problem (LPP) by using simplex method and interpret the findings.
Maximize Z = $4A+3B+6C$
Subject to Constraints:
$2A+3B+2C \leq 440$
$4A+3C \leq 470$
$2A+5B \leq 430$
$A,B,C \geq 0$
[10]
Objective: Maximize $Z = 4A + 3B + 6C$ Constraints: - $2A + 3B + 2C \le 440$ - $4A + 0B + 3C \le 470$ - $2A + 5B + 0C \le 430$ - $A, B, C \ge 0$ Add slack variables $s1, s2, s3$: - $2A + 3B + 2C + s1 = 440$ - $4A + 0B + 3C + s2 = 470$ - ...
- 310 marksNumericalExpected Monetary ValueHideAnswer
Milk Salesman Decision Problem
A milk salesman estimates the probability of the demand for a litre of milk as follows: He purchases a litre of milk @ Rs. 60 and sells it @ Rs. 70. Assuming that the unsold milk has no scrap value, find:
(a) Find optimum quantity that would obtain Max. EMV. (b) Find the minimum value of EOL. (c) What is the value of expected profit with perfect information (EPPI)?
Demand 11 12 13 14 15 Probability 0.10 0.15 0.30 0.25 0.20 [10]
- Purchase cost = Rs. 60 per litre - Selling price = Rs. 70 per litre - Profit per litre sold = $70 - 60 = $ Rs. 10 - Loss per litre unsold (no scrap) = Rs. 60 - Demand distribution: Demand 11 12 13 14 15 --------------------------------...
- 45 marksNumericalCritical path identificationHideAnswer
Network Diagram and Critical Path Analysis
Network Diagram and Critical Path Analysis
Step 1 - Given Data
Activity A B C D E F G Predecessors - A A A B,C C,D E,F Time (days) 7 4 5 6 7 5 6 Step 2 - Solve
Network Diagram
B(4) ┌────────► ③ │ \ ①──A(7)──►②──C(5)──►④───E(7)──►⑤──G(6)──►⑥ │ \ / │ \/ └─D(6)──►(F merges C,D)──F(5)──►⑤Logical structure of dependencies:
- A starts the project.
- B, C, D all follow A.
- E follows B and C.
- F follows C and D.
- G follows E and F (final activity).
Forward Pass (Earliest Times)
$$ES = \max(EF \text{ of all predecessors}), \quad EF = ES + t$$
- A: ES = 0, EF = 0 + 7 = 7
- B: ES = 7, EF = 7 + 4 = 11
- C: ES = 7, EF = 7 + 5 = 12
- D: ES = 7, EF = 7 + 6 = 13
- E: ES = max(EF$_B$, EF$_C$) = max(11, 12) = 12, EF = 12 + 7 = 19
- F: ES = max(EF$_C$, EF$_D$) = max(12, 13) = 13, EF = 13 + 5 = 18
- G: ES = max(EF$_E$, EF$_F$) = max(19, 18) = 19, EF = 19 + 6 = 25
Project Duration = 25 days
Backward Pass (Latest Times)
$$LF = \min(LS \text{ of all successors}), \quad LS = LF - t$$
- G: LF = 25, LS = 25 - 6 = 19
- F: LF = 19, LS = 19 - 5 = 14
- E: LF = 19, LS = 19 - 7 = 12
- D: LF = LS$_F$ = 14, LS = 14 - 6 = 8
- C: LF = min(LS$_E$, LS$_F$) = min(12, 14) = 12, LS = 12 - 5 = 7
- B: LF = LS$_E$ = 12, LS = 12 - 4 = 8
- A: LF = min(LS$_B$, LS$_C$, LS$_D$) = min(8, 7, 8) = 7, LS = 7 - 7 = 0
Slack Table
Activity ES EF LS LF Slack (LS-ES) Critical? A 0 7 0 7 0 ✓ B 7 11 8 12 1 ✗ C 7 12 7 12 0 ✓ D 7 13 8 14 1 ✗ E 12 19 12 19 0 ✓ F 13 18 14 19 1 ✗ G 19 25 19 25 0 ✓ Critical Path
Activities with zero slack: A → C → E → G
$$\text{Duration} = 7 + 5 + 7 + 6 = 25 \text{ days}$$
Bottleneck Activities
Bottleneck (critical) activities with zero slack that control the project completion:
$$\boxed{A,\ C,\ E,\ G}$$
Any delay in these directly delays the whole project.
Keep the node table consistent with the forward pass: values such as E at ES = 11, EF = 18, or G at ES = 13, EF = 19, contradict the forward pass computed above. The critical path A -> C -> E -> G and the 25-day duration are unaffected, but an internally inconsistent table loses marks.
- 55 marksNumericalHungarian Assignment MethodHideAnswer
The ABC Company Job Assignment Problem
The ABC company has three jobs to be done on three machines. Each job must be done on one and only one machine. The cost of each job on each machine is given in the following table. By using Hungarian method, find the job assignments which will minimize the machine cost.
$$\begin{array}{|c|c|c|c|}\hline \text{Jobs\backslash Machine} & X & Y & Z \ \hline A & 4 & 6 & 8 \ B & 2 & 3 & 4 \ C & 4 & 8 & 5 \ \hline \end{array}$$
[5]
Cost matrix (Jobs A, B, C on Machines X, Y, Z): $$\begin{array}{cccc}\hline \text{Job/Machine} & X & Y & Z \ \hline A & 4 & 6 & 8 \ B & 2 & 3 & 4 \ C & 4 & 8 & 5 \ \hline \end{array}$$ Objective: minimize total assignment cost, each ...
- 65 marksNumericalFormulation of linear programming problemsHideAnswer
Food X contains 6 units of vitamin A and 7 units of vitamin B and costs Rs. 5 per gram. Food Y contains 8 units of vitamin A and 12 units of vitamin B and costs Rs. 18 per gram. The daily minimum requirement of vitamins A and B are respectively 100 units and 138 units respectively. Formulate the problem as a LPP with the objective function minimizing the cost. [5]
Item Vitamin A (units/gram) Vitamin B (units/gram) Cost (Rs./gram) ------------------------------------------------------------------- Food X 6 7 5 Food Y 8 12 18 Minimum daily requirement 100 138 - Objective: minimize cost. Let: - $x$ =...
- 75 marksNumericalAverage number of customers in system and HideAnswer
Customers arrive at a bank having single counter at the rate of 25 customers per hour. Time required to serve a customer has exponential distribution and average number of customers served per hour is 30. Find the average number of customers in queue and in system as well. [5]
- Arrival rate: $\lambda = 25$ customers/hour - Service rate: $\mu = 30$ customers/hour - Single counter with exponential service time → M/M/1 model $$\rho = \frac{\lambda}{\mu} = \frac{25}{30} = \frac{5}{6} \approx 0.833$$ Since
- 85 marksNumericalVogel's Approximation MethodHideAnswer
Find the initial solution by using Vogel's Approximation Method (VAM).
$$\begin{array}{|c|c|c|c|c|c|}\hline \text{From} & P & Q & R & S & \text{Supply} \ \hline A & 19 & 30 & 50 & 10 & 7 \ B & 70 & 30 & 40 & 60 & 9 \ C & 40 & 8 & 70 & 20 & 18 \ \hline \text{Demand} & 5 & 8 & 7 & 14 & 34 \ \hline \end{array}$$
[5]
Cost matrix: From P Q R S Supply ------------------ A 19 30 50 10 7 B 70 30 40 60 9 C 40 8 70 20 18 Demand 5 8 7 14 Total supply = 7 + 9 + 18 = 34; total demand = 5 + 8 + 7 + 14 = 34. Balanced. Penalty = difference between two smallest c...
- 95 marksNumericalOptimal strategies for playersHideAnswer
The following is the payoff matrix of a game being played by A and B. Determine the optimal strategies for the players and the value of the game.
$$\begin{array}{|c|c|c|c|}\hline \text{A's strategies} & B_1 & B_2 & B_3 \ \hline A_1 & 120 & -80 & -20 \ A_2 & 60 & 70 & 30 \ A_3 & -100 & 70 & 20 \ \hline \end{array}$$
[5]
Payoff matrix (A's gains): A \ B $B1$ $B2$ $B3$ ------------ $A1$ 120 -80 -20 $A2$ 60 70 30 $A3$ -100 70 20 Row minima (A's guaranteed minimum for each strategy): - $A1$: $\min(120, -80, -20) = -80$ - $A2$: $\min(60, 70, 30) = 30$ - $A3$...
- 105 marksDominance rule methodHideAnswer
Describe dominance rule method for solving a problem of game theory. [5]
The dominance rule method is a technique used to simplify and solve game theory problems by eliminating strategies that are clearly inferior (dominated) for a player, regardless of what the opponent does. A strategy is dominated if there...
- 115 marksQueuing disciplines and service orderHideAnswer
Describe different types of queuing disciplines used in serving a customer in a queue. [5]
A queuing discipline (or queue discipline) is the rule that determines the order in which customers waiting in a queue are selected for service. - Also called FCFS (First Come First Served) - Customers are served in the exact order they ...
- 125 marksMarginal analysis approachHideAnswer
Write short notes on: (a) Marginal analysis approach in decision making. Write short notes on: (b) Modified distribution (MODI) method. [2.5+2.5]
Definition: Marginal analysis is a decision-making technique that examines the additional (incremental) benefits and costs resulting from a small change in an activity or decision variable. Key Principles: 1. Marginal Concept: Focuses on...