MGT 403

Operations Management

TU BITM · Semester 7 · BITM curriculum and programme regulation

Requirement
required
Credits
3
Practice sets
5 sets

Practice sets and model questions

5 complete syllabus-aligned practice examinations with model-style questions and step-by-step solutions. Independently prepared by Paper Khoj, not issued by TU.

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Practice Set 1

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PAPER KHOJ

Practice Examination — Set 1

For BITM · Semester 7

Based on the prescribed TU Faculty of Management syllabus and examination structure

Independently prepared by Paper Khoj. This is not an official Tribhuvan University or Dean's Office question paper.

Course: MGT 403 · Operations Management

Level: Bachelor of Information Technology Management (BITM) · Semester 7

Full Marks: 60

Time: 3 hrs.

Candidates are required to give their answers in their own words as far as practicable. The figures in the margin indicate full marks.

Group 'A'

Brief Answer Questions. Attempt all questions.

[5 × 2 = 10]
  1. [2]

    Define Operations Management and state its transformation model.

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    Operations Management:

    The systematic design, direction, and control of business processes that convert organizational resources (inputs) into finished goods and services (outputs).

    • Transformation Model: Inputs→Transformation / Conversion ProcessOutputs→Feedback LoopContinuous Control\text{Inputs} \xrightarrow{\text{Transformation / Conversion Process}} \text{Outputs} \xrightarrow{\text{Feedback Loop}} \text{Continuous Control}. Inputs include 5 Ms (Manpower, Material, Machine, Money, Methods).
  2. [2]

    Write the formulas for Single-Factor Labor Productivity and Multi-Factor Productivity.

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    Productivity Formulas:

    • Single-Factor Labor Productivity:
      Labor Productivity=Total Output (Units or Value)Labor Hours (or Labor Cost)\text{Labor Productivity} = \frac{\text{Total Output (Units or Value)}}{\text{Labor Hours (or Labor Cost)}}
    • Multi-Factor Productivity:
      Multi-Factor Productivity=Total OutputLabor+Materials+Energy+Capital+Overhead\text{Multi-Factor Productivity} = \frac{\text{Total Output}}{\text{Labor} + \text{Materials} + \text{Energy} + \text{Capital} + \text{Overhead}}
  3. [2]

    Define Assembly Line Cycle Time (CC) and write its formula given operating time and desired output.

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    Cycle Time (CC):

    The maximum time that a product is allowed to spend at each individual workstation on an assembly line to achieve the target daily production quota.

    Formula:

    C=Production Time Available per DayDesired Output Units per Day=TDC = \frac{\text{Production Time Available per Day}}{\text{Desired Output Units per Day}} = \frac{T}{D}

  4. [2]

    What does a Six Sigma quality level (3.4 DPMO3.4 \text{ DPMO}) mean?

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    Six Sigma Quality:

    A statistical quality standard indicating that process performance operates at six standard deviations between the mean and nearest specification limit. It translates to no more than 3.4 Defects Per Million Opportunities (DPMO), representing 99.99966%99.99966\% defect-free manufacturing accuracy.

  5. [2]

    Define Kanban in Just-in-Time (JIT) production.

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    Kanban:

    A Japanese visual signaling card or electronic trigger utilized in Lean pull production systems. It authorizes the production or withdrawal of a standardized container of parts from upstream processes only when an immediate downstream process has consumed its existing inventory.

Group 'B'

Short Answer Questions. Attempt any six questions.

[6 × 5 = 30]
  1. [5]

    Nepal Beverage Ltd. produced 100,000 crates of juice last month with 50 workers working 160 hours each at NPR 200/hr. Raw material cost was NPR 3,000,000 and energy cost was NPR 500,000. Each crate sells for NPR 100. Calculate: (a) Labor productivity in units per hour, (b) Labor productivity in value ratio, and (c) Multi-factor productivity ratio.

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    Step 1: Compute Inputs & Outputs

    • Total Output in Units = 100,000 crates
    • Total Output Value = 100,000×100=NPR 10,000,000100,000 \times 100 = \text{NPR } 10,000,000
    • Total Labor Hours = 50×160=8,000 hours50 \times 160 = 8,000 \text{ hours}
    • Total Labor Cost = 8,000×200=NPR 1,600,0008,000 \times 200 = \text{NPR } 1,600,000
    • Material Cost = NPR 3,000,000
    • Energy Cost = NPR 500,000
    • Total Multi-Factor Inputs = 1,600,000+3,000,000+500,000=NPR 5,100,0001,600,000 + 3,000,000 + 500,000 = \text{NPR } 5,100,000

    Step 2: Compute Productivity Metrics

    (a) Labor Productivity (Physical Units):

    Productivity=100,000 crates8,000 hours=12.5 crates per labor hour\text{Productivity} = \frac{100,000 \text{ crates}}{8,000 \text{ hours}} = \mathbf{12.5 \text{ crates per labor hour}}

    (b) Labor Productivity (Value Ratio):

    Productivity=NPR 10,000,000NPR 1,600,000=6.25\text{Productivity} = \frac{\text{NPR } 10,000,000}{\text{NPR } 1,600,000} = \mathbf{6.25}
    (Each rupee spent on labor generates NPR 6.25 in revenue).

    (c) Multi-Factor Productivity:

    MFP=NPR 10,000,000NPR 5,100,000=1.96\text{MFP} = \frac{\text{NPR } 10,000,000}{\text{NPR } 5,100,000} = \mathbf{1.96}
    (Every NPR 1.00 of combined operational input yields NPR 1.96 of revenue).

  2. [5]

    An assembly line operates 8 hours (480 minutes) per day with a desired daily output of 240 units. The sum of all task times is 8 minutes. (a) Calculate the required cycle time. (b) Find the theoretical minimum number of workstations. (c) If 5 workstations are actually used, calculate the line balancing efficiency and balance delay.

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    (a) Cycle Time (CC):

    C=Operating Time per DayDesired Output=480 minutes240 units=2.0 minutes per unitC = \frac{\text{Operating Time per Day}}{\text{Desired Output}} = \frac{480 \text{ minutes}}{240 \text{ units}} = \mathbf{2.0 \text{ minutes per unit}}

    (b) Theoretical Minimum Workstations (NminN_{min}):

    Nmin=∑tC=8.0 minutes2.0 minutes=4 workstationsN_{min} = \frac{\sum t}{C} = \frac{8.0 \text{ minutes}}{2.0 \text{ minutes}} = \mathbf{4 \text{ workstations}}

    (c) Line Efficiency & Balance Delay (Nactual=5N_{actual} = 5):

    • Line Efficiency:
      Efficiency=∑tNactual×C×100%=8.05×2.0×100%=8.010.0×100%=80.0%\text{Efficiency} = \frac{\sum t}{N_{actual} \times C} \times 100\% = \frac{8.0}{5 \times 2.0} \times 100\% = \frac{8.0}{10.0} \times 100\% = \mathbf{80.0\%}
    • Balance Delay (Idle Time Percentage):
      Balance Delay=100%−Efficiency=100%−80%=20.0%\text{Balance Delay} = 100\% - \text{Efficiency} = 100\% - 80\% = \mathbf{20.0\%}
  3. [5]

    Himalayan Herbal Ltd. has an annual demand of 10,000 bottles of essential oil. Ordering cost is NPR 500 per order, and annual holding cost is NPR 4 per bottle. (a) Calculate the Economic Order Quantity (EOQ). (b) Determine the optimal number of orders per year. (c) Compute total annual inventory cost (ordering + holding) at EOQ.

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    (a) Economic Order Quantity (EOQ):

    Given:

    • Annual Demand (DD) = 10,000 units
    • Ordering Cost (SS) = NPR 500
    • Holding Cost (HH) = NPR 4 per unit/year
    EOQ=2DSH=2×10,000×5004=10,000,0004=2,500,000=1,581.14≈1,581 bottles\text{EOQ} = \sqrt{\frac{2DS}{H}} = \sqrt{\frac{2 \times 10,000 \times 500}{4}} = \sqrt{\frac{10,000,000}{4}} = \sqrt{2,500,000} = \mathbf{1,581.14} \approx \mathbf{1,581 \text{ bottles}}

    (b) Optimal Number of Orders per Year (NN):

    N=DEOQ=10,0001,581.14≈6.32 orders per yearN = \frac{D}{\text{EOQ}} = \frac{10,000}{1,581.14} \approx \mathbf{6.32 \text{ orders per year}}

    (c) Total Annual Inventory Cost (TIC):

    TIC=(DEOQ×S)+(EOQ2×H)\text{TIC} = \left( \frac{D}{\text{EOQ}} \times S \right) + \left( \frac{\text{EOQ}}{2} \times H \right)
    Annual Ordering Cost=6.3245×500=NPR 3,162.28\text{Annual Ordering Cost} = 6.3245 \times 500 = \text{NPR } 3,162.28
    Annual Holding Cost=1,581.142×4=NPR 3,162.28\text{Annual Holding Cost} = \frac{1,581.14}{2} \times 4 = \text{NPR } 3,162.28
    Total Annual Cost=3,162.28+3,162.28=NPR 6,324.56\text{Total Annual Cost} = 3,162.28 + 3,162.28 = \mathbf{\text{NPR } 6,324.56}

    (Note: At EOQ, Annual Ordering Cost exactly equals Annual Holding Cost).

  4. [5]

    Explain the Center-of-Gravity method for facility location. A logistics firm wants to locate a central distribution warehouse to serve four retail hubs with coordinates and shipments:

    • Hub 1: (10, 20), Shipments = 100
    • Hub 2: (30, 40), Shipments = 300
    • Hub 3: (50, 10), Shipments = 200
    • Hub 4: (70, 50), Shipments = 400 Find the optimal (X,Y)(X, Y) coordinates for the distribution warehouse.
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    Center-of-Gravity Method:

    Finds the location that minimizes total weighted transportation costs based on Euclidean coordinates (xi,yi)(x_i, y_i) and shipment volumes (Wi)(W_i):

    X∗=∑(xi⋅Wi)∑Wi,Y∗=∑(yi⋅Wi)∑WiX^* = \frac{\sum (x_i \cdot W_i)}{\sum W_i}, \quad Y^* = \frac{\sum (y_i \cdot W_i)}{\sum W_i}

    Calculations:

    Hub xix_i yiy_i Volume (WiW_i) xi⋅Wix_i \cdot W_i yi⋅Wiy_i \cdot W_i
    1 10 20 100 1,000 2,000
    2 30 40 300 9,000 12,000
    3 50 10 200 10,000 2,000
    4 70 50 400 28,000 20,000
    Total 1,000 48,000 36,000
    X∗=48,0001,000=48.0X^* = \frac{48,000}{1,000} = \mathbf{48.0}
    Y∗=36,0001,000=36.0Y^* = \frac{36,000}{1,000} = \mathbf{36.0}

    The optimal location coordinate for the central warehouse is (48.0, 36.0).

  5. [5]

    Explain the DMAIC methodology of Six Sigma. How does it improve service and manufacturing processes?

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    The DMAIC Methodology:

    A closed-loop, data-driven 5-phase quality improvement framework:

    1. Define: Explicitly define customer Critical-to-Quality (CTQ) parameters, project boundaries, and business goals using a Project Charter.
    2. Measure: Quantify current baseline performance by gathering data on defect frequencies and process cycle times.
    3. Analyze: Identify root causes of variation and defects using Ishikawa (Fishbone) diagrams, 5-Whys, and Pareto charts.
    4. Improve: Develop, pilot, and deploy targeted solutions to eliminate root causes; optimize settings using Design of Experiments (DOE).
    5. Control: Institute Statistical Process Control (SPC) charts, standard operating procedures, and mistake-proofing (Poka-Yoke) to ensure process gains are permanently sustained.
  6. [5]

    Distinguish between Chase Strategy and Level Strategy in Aggregate Operations Planning. When should a manufacturer adopt each?

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    Chase Strategy vs. Level Strategy:

    Feature Chase Strategy Level Strategy
    Production Rate Fluctuates strictly to match forecasted demand in each period. Kept completely constant; smooth production rate.
    Workforce Management Hires and lays off workers dynamically as demand shifts. Maintains a stable, constant workforce size.
    Inventory Levels Kept to near-zero; minimal finished goods holding cost. Fluctuates heavily; absorbs demand surges via buffer inventory.
    Major Costs High hiring, firing, severance, and worker training costs. High inventory holding, obsolescence, and back-order costs.
    Ideal Application Highly perishable products, customized services, agricultural packing. High capital-investment automated plants, high-skilled labor environments.

Group 'C'

Comprehensive Answer Questions. Attempt all questions.

[2 × 10 = 20]
  1. [10]

    An IT infrastructure project involves the following activities, durations (in weeks), and immediate predecessors:

    Activity Description Predecessor Duration (Weeks)
    A Requirements Gathering - 3
    B Architecture Design A 4
    C Hardware Procurement A 2
    D Core Software Development B 6
    E Database Configuration B 3
    F System Integration C, D 4
    G User Acceptance Testing (UAT) E, F 2

    (a) Draw the CPM project network diagram. (b) Calculate Earliest Start (ES), Earliest Finish (EF), Latest Start (LS), and Latest Finish (LF) for each activity. (c) Calculate Total Float (Slack) for each activity and determine the Critical Path and total project duration.

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    Step 1: Forward Pass (ES & EF)

    EF=ES+Duration,ES=max⁡(EFpredecessors)EF = ES + \text{Duration}, \quad ES = \max(EF_{\text{predecessors}})
    • A (Dur=3): ES=0,  EF=0+3=3ES = 0, \; EF = 0 + 3 = 3
    • B (Dur=4): ES=3,  EF=3+4=7ES = 3, \; EF = 3 + 4 = 7
    • C (Dur=2): ES=3,  EF=3+2=5ES = 3, \; EF = 3 + 2 = 5
    • D (Dur=6): ES=7,  EF=7+6=13ES = 7, \; EF = 7 + 6 = 13
    • E (Dur=3): ES=7,  EF=7+3=10ES = 7, \; EF = 7 + 3 = 10
    • F (Dur=4, Pred: C, D): ES=max⁡(5,13)=13,  EF=13+4=17ES = \max(5, 13) = 13, \; EF = 13 + 4 = 17
    • G (Dur=2, Pred: E, F): ES=max⁡(10,17)=17,  EF=17+2=19ES = \max(10, 17) = 17, \; EF = 17 + 2 = 19

    Total Project Duration = 19 weeks.


    Step 2: Backward Pass (LF & LS)

    LS=LF−Duration,LF=min⁡(LSsuccessors)LS = LF - \text{Duration}, \quad LF = \min(LS_{\text{successors}})
    • G (Dur=2): LF=19,  LS=19−2=17LF = 19, \; LS = 19 - 2 = 17
    • F (Dur=4, Succ: G): LF=17,  LS=17−4=13LF = 17, \; LS = 17 - 4 = 13
    • E (Dur=3, Succ: G): LF=17,  LS=17−3=14LF = 17, \; LS = 17 - 3 = 14
    • D (Dur=6, Succ: F): LF=13,  LS=13−6=7LF = 13, \; LS = 13 - 6 = 7
    • C (Dur=2, Succ: F): LF=13,  LS=13−2=11LF = 13, \; LS = 13 - 2 = 11
    • B (Dur=4, Succ: D, E): LF=min⁡(7,14)=7,  LS=7−4=3LF = \min(7, 14) = 7, \; LS = 7 - 4 = 3
    • A (Dur=3, Succ: B, C): LF=min⁡(3,11)=3,  LS=3−3=0LF = \min(3, 11) = 3, \; LS = 3 - 3 = 0

    Step 3: Slack (Total Float) & Critical Path Table

    Activity Duration ES EF LS LF Slack (LS−ESLS - ES) On Critical Path?
    A 3 0 3 0 3 0 YES
    B 4 3 7 3 7 0 YES
    C 2 3 5 11 13 8 No
    D 6 7 13 7 13 0 YES
    E 3 7 10 14 17 7 No
    F 4 13 17 13 17 0 YES
    G 2 17 19 17 19 0 YES

    Conclusions:

    • Critical Path: A →\rightarrow B →\rightarrow D →\rightarrow F →\rightarrow G
    • Total Project Completion Time: 19 weeks
    • Non-critical activities C and E possess 8 weeks and 7 weeks of slack respectively, allowing resource leveling without delaying the completion date.
  2. [10]

    Explain Total Quality Management (TQM) and the 7 Basic Quality Control (QC) Tools. Discuss how Statistical Process Control (SPC) uses Xˉ\bar{X} and RR charts to separate common cause variation from assignable cause variation in production.

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    1. Total Quality Management (TQM):

    An organization-wide management philosophy aimed at continuously improving the quality of products, processes, and corporate culture to exceed customer expectations.

    • Core pillars: Customer focus, Total employee involvement, Process-centered thinking, Integrated system, Continuous improvement (Kaizen), and Data-based decision making.

    2. The Seven Basic Quality Control (QC) Tools (Ishikawa’s Tools):

    1. Check Sheet: Structured form for real-time manual defect logging.
    2. Histogram: Visual graphic showing frequency distributions and central tendencies of measurements.
    3. Pareto Chart: Bar chart illustrating the 80/20 rule (vital few causes produce majority of defects).
    4. Cause-and-Effect (Fishbone / Ishikawa) Diagram: Maps categorized root causes (Man, Machine, Material, Method, Measurement, Milieu).
    5. Scatter Diagram: Graphical plot testing mathematical correlation between two variables.
    6. Flowchart / Process Map: Visual sequence of process steps highlighting bottlenecks.
    7. Control Chart (Shewhart Chart): Time-series plot with statistical Upper and Lower Control Limits (UCL,LCLUCL, LCL) to monitor process stability.

    3. Statistical Process Control (SPC) & Control Charts:

    SPC uses statistical sampling to monitor a process over time and distinguish between two types of variation:

    • Common Cause (Random) Variation: Natural, inherent background noise of a stable system (e.g. slight temperature fluctuations). Cannot be eliminated without fundamentally redesigning the process.
    • Assignable (Special Cause) Variation: Specific, identifiable anomalies that shift the process out of control (e.g. worn machine tool, operator fatigue, bad raw material batch).

    Xˉ\bar{X} and RR Charts:

    • Xˉ\bar{X} Chart (Mean Chart): Monitors changes in the central process average over time.
      UCLXˉ=Xˉˉ+A2Rˉ,LCLXˉ=Xˉˉ−A2RˉUCL_{\bar{X}} = \bar{\bar{X}} + A_2 \bar{R}, \quad LCL_{\bar{X}} = \bar{\bar{X}} - A_2 \bar{R}
    • RR Chart (Range Chart): Monitors process dispersion and variability.
      UCLR=D4Rˉ,LCLR=D3RˉUCL_R = D_4 \bar{R}, \quad LCL_R = D_3 \bar{R}
      If sample points fall within bounds without non-random trends, the process is deemed in statistical control; any point crossing UCL/LCLUCL/LCL triggers immediate root-cause intervention before defective output reaches customers.