Model paper

Dean's Office Official Model Question Paper

MGT 205 · Operations Management

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Programme
BBM
Academic year
Semester 5
Paper type
Official Model Question
Sitting
Dean's Office Blueprint
Full marks
60
Duration
180 minutes

Tribhuvan University

Faculty of Management

Office of the Dean

Official Model Question Paper / Dean's Office Blueprint

Course: MGT 205 · Operations Management

Level: Bachelor of Business Management (BBM) · Semester 5

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. Define Operations Management and illustrate the Input-Transformation-Output model.

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    Answer: Operations Management: The systematic direction and control of the processes that transform inputs (raw materials, labor, capital, information) into finished goods and services.

    Inputs (5 Ms: Men, Machine, Material, Money, Methods)    Transformation Process (Value Addition)    Outputs (Goods & Services)\text{Inputs (5 Ms: Men, Machine, Material, Money, Methods)} \implies \text{Transformation Process (Value Addition)} \implies \text{Outputs (Goods \& Services)}
  2. Distinguish between Manufacturing Operations and Service Operations on any two criteria.

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    Answer:

    Basis Manufacturing Operations Service Operations
    Tangibility & Storage Produces tangible, physical goods that can be stored in inventory. Produces intangible experiences that cannot be stored; perishable.
    Customer Contact Low customer contact during actual factory production. High customer interaction; customer is often a co-producer of service.
  3. What is Total Quality Management (TQM)? Name two prominent TQM pioneers.

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    Answer: Total Quality Management (TQM): An organization-wide management philosophy committed to the continuous improvement of products, processes, and culture to satisfy or exceed customer expectations. Two Pioneers:

    1. W. Edwards Deming (Deming’s 14 Points, PDCA Cycle)
    2. Joseph M. Juran (Juran’s Quality Trilogy: Planning, Control, Improvement)
  4. Differentiate between Design Capacity and Effective Capacity.

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    Answer:

    • Design Capacity: The theoretical maximum output rate that a facility can achieve under ideal, continuous operating conditions with zero breakdowns or interruptions.
    • Effective Capacity: The maximum realistic output that a facility can sustain under normal operating conditions, accounting for work breaks, preventive maintenance, setup times, and scheduling delays:
      Efficiency=Actual OutputEffective Capacity×100%\text{Efficiency} = \frac{\text{Actual Output}}{\text{Effective Capacity}} \times 100\%
  5. Define the Just-In-Time (JIT) manufacturing philosophy and state its ultimate goal.

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    Answer: Just-In-Time (JIT): An operations philosophy originated by Toyota aiming to produce and deliver finished goods exactly when needed, in the exact quantities demanded, with zero waste. Ultimate Goal: Total elimination of non-value-adding waste (Muda), achieving zero inventory, zero defects, and zero machine downtime.

Group B

Descriptive Answer Questions. Attempt any THREE questions.

[3 × 10 = 30]
  1. A cement manufacturing company is evaluating three candidate locations in Nepal for establishing a new clinker grinding unit: Hetauda, Bhairahawa, and Birgunj. The management has established six location factors, their respective importance weights, and ratings (on a 100-point scale) for each site:

    Location Factor Weight Hetauda Bhairahawa Birgunj
    Proximity to Raw Materials (Limestone) 0.25 80 65 60
    Transportation & Railway Logistics 0.20 70 85 90
    Availability of Stable Grid Power 0.15 75 80 70
    Proximity to Core Markets (Kathmandu/Pokhara) 0.20 90 70 75
    Skilled Labor Availability & Costs 0.10 65 75 80
    Environmental Regulations & Land Cost 0.10 70 60 65

    a) Compute the Weighted Location Score for all three candidate sites using the Factor Rating Method. b) Which location should management select based on the quantitative analysis? c) Discuss the qualitative factors that could override purely quantitative location decisions in Nepal.

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    Solution: Facility Location Evaluation via Factor Rating Method


    Part (a): Weighted Location Score Computations

    Weighted Score=(Factor Weight×Rating)\text{Weighted Score} = \sum (\text{Factor Weight} \times \text{Rating})
    Factor Weight Hetauda Score Bhairahawa Score Birgunj Score
    Limestone Proximity 0.25 0.25×80=20.000.25 \times 80 = 20.00 0.25×65=16.250.25 \times 65 = 16.25 0.25×60=15.000.25 \times 60 = 15.00
    Logistics & Rail 0.20 0.20×70=14.000.20 \times 70 = 14.00 0.20×85=17.000.20 \times 85 = 17.00 0.20×90=18.000.20 \times 90 = 18.00
    Power Grid 0.15 0.15×75=11.250.15 \times 75 = 11.25 0.15×80=12.000.15 \times 80 = 12.00 0.15×70=10.500.15 \times 70 = 10.50
    Market Proximity 0.20 0.20×90=18.000.20 \times 90 = 18.00 0.20×70=14.000.20 \times 70 = 14.00 0.20×75=15.000.20 \times 75 = 15.00
    Skilled Labor 0.10 0.10×65=6.500.10 \times 65 = 6.50 0.10×75=7.500.10 \times 75 = 7.50 0.10×80=8.000.10 \times 80 = 8.00
    Land & Regulations 0.10 0.10×70=7.000.10 \times 70 = 7.00 0.10×60=6.000.10 \times 60 = 6.00 0.10×65=6.500.10 \times 65 = 6.50
    Total Composite Score 1.00 76.75 72.75 73.00

    Part (b): Recommendation

    • Hetauda Composite Score: 76.75
    • Birgunj Composite Score: 73.00
    • Bhairahawa Composite Score: 72.75

    Recommendation: Management should select Hetauda. It achieves the highest composite weighted score (76.75), driven primarily by its immediate proximity to Makwanpur limestone deposits and lower freight distance to the dominant consumer construction markets of the Kathmandu Valley.


    Part (c): Qualitative Overriding Factors in Nepal

    1. Political & Labor Union Climate: Frequent regional strikes (bandhs) or aggressive local trade union interference in specific industrial corridors can paralyze operations regardless of physical advantages.
    2. Community & Environmental Clearances: Stringent opposition from local community settlements regarding cement dust pollution and EIA delays.
    3. Disaster Vulnerability: Highway landslides along the Tribhuvan Highway and Narayangadh-Mugling corridor during monsoon seasons that sever transit access.
  2. Compare and contrast Product Layout (Line Layout) with Process Layout (Functional Layout). What is Cellular Manufacturing (Group Technology) and what advantages does it offer?

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    1. Comparative Matrix: Product Layout vs. Process Layout

    Product (Line) Layout (Continuous Flow)
    [Raw Materials] -> [Operation 1] -> [Operation 2] -> [Operation 3] -> [Finished Good]
    
    Process (Functional) Layout (Intermittent Flow)
    [Lathe Dept]    [Milling Dept]
         ^                |
         +--> [Welding Dept] <---+ [Assembly Dept]
    
    Dimension Product Layout (Line Layout) Process Layout (Functional Layout)
    Layout Logic Equipment arranged sequentially according to product assembly sequence. Similar machines and functions grouped together in departments.
    Product Variety & Volume Standardized products; very high production volume. Customized products; low to moderate volume.
    Material Handling Cost Very low; automated conveyors and direct flow. High; materials travel back and forth between functional departments.
    Machine Utilization Dedicated, specialized machines; high vulnerability to single-point line stoppages. General-purpose machines; work can be rerouted if one machine breaks down.
    Examples Automobile assembly lines, beverage bottling plants. Machine job shops, multi-specialty hospitals, university faculties.

    2. Cellular Manufacturing (Group Technology)

    Cellular Manufacturing is an innovative hybrid layout that groups dissimilar machines into dedicated autonomous work cells designed to produce families of parts with similar geometric shapes or processing sequences.

    Key Advantages:

    1. Compressed Setup Times: Dedicated tooling within the cell eliminates lengthy machine re-setups.
    2. Drastic Reduction in Work-In-Process (WIP): Parts move directly from machine to machine within the cell without waiting in transit buffers.
    3. Enhanced Worker Motivation & Teamwork: Cross-trained small work teams take collective ownership of complete component production, enhancing quality control.
  3. The project activities, precedence relationships, and estimated durations (in days) for establishing an automated dairy processing plant are given below:

    Activity Immediate Predecessors Duration (Days)
    A 5
    B 7
    C A 6
    D A 4
    E B, C 8
    F D 5
    G E, F 6

    a) Draw the Activity-on-Node (AON) Network Diagram. b) Determine the Earliest Start (ES), Earliest Finish (EF), Latest Start (LS), and Latest Finish (LF) times for each activity. c) Identify the Critical Path and the Total Project Completion Duration. d) Calculate the Total Float (Slack) for activities D and F.

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    Solution: Project Network Analysis (CPM)


    Part (a) & (b): Forward Pass and Backward Pass Computations

    • Forward Pass: EF=ES+DurationEF = ES + \text{Duration}; ES=max(EF of all predecessors)ES = \max(\text{EF of all predecessors}).
    • Backward Pass: LS=LFDurationLS = LF - \text{Duration}; LF=min(LS of all successors)LF = \min(\text{LS of all successors}).
    Activity Duration Predecessors ES EF LF LS Slack (LSESLS - ES) Critical?
    A 5 0 5 5 0 0 YES
    B 7 0 7 11 4 4 No
    C 6 A 5 11 11 5 0 YES
    D 4 A 5 9 14 10 5 No
    E 8 B, C max(7,11)=11\max(7, 11) = 11 19 19 11 0 YES
    F 5 D 9 14 19 14 5 No
    G 6 E, F max(19,14)=19\max(19, 14) = 19 25 25 19 0 YES

    Part (c): Critical Path and Total Project Duration

    • Possible Paths:
      1. ACEG=5+6+8+6=25 DaysA \to C \to E \to G = 5 + 6 + 8 + 6 = \mathbf{25 \text{ Days}}
      2. BEG=7+8+6=21 DaysB \to E \to G = 7 + 8 + 6 = \mathbf{21 \text{ Days}}
      3. ADFG=5+4+5+6=20 DaysA \to D \to F \to G = 5 + 4 + 5 + 6 = \mathbf{20 \text{ Days}}Critical Path=ACEG\mathbf{\text{Critical Path}} = \mathbf{A \to C \to E \to G}$
        Total Project Completion Duration=25 Days\mathbf{\text{Total Project Completion Duration}} = \mathbf{25 \text{ Days}}

    Part (d): Total Float (Slack) for Activities D and F

    Total Float=LSES=LFEF\text{Total Float} = LS - ES = LF - EF
    • Activity D: FloatD=105=5 Days\text{Float}_D = 10 - 5 = \mathbf{5 \text{ Days}}
    • Activity F: FloatF=149=5 Days\text{Float}_F = 14 - 9 = \mathbf{5 \text{ Days}}

    Interpretation: Activities D and F have 5 days of allowable delay before they impact the overall 25-day project deadline.

  4. Explain the concept of Statistical Process Control (SPC). Differentiate between Chance (Common) Causes and Assignable (Special) Causes of process variation. How do Xˉ\bar{X} and RR charts help maintain manufacturing quality?

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    1. Concept of Statistical Process Control (SPC)

    Statistical Process Control (SPC) is the application of statistical techniques to measure, monitor, and control a production process to ensure it operates at its full potential, producing conforming output with minimal defect rates.


    2. Sources of Process Variation

                          Sources of Process Variation
                                       |
         +-----------------------------+-----------------------------+
         |                                                           |
    Chance (Common) Causes                                   Assignable (Special) Causes
    - Inherent, random background noise                      - Specific, identifiable disruptive events
    - Natural machine vibration, temperature shifts          - Operator error, tool breakage, bad raw material
    - Cannot be eliminated without major system redesign     - Can and must be identified and eliminated
    - Process remains in statistical control                 - Process is out of statistical control
    

    3. Role of Xˉ\bar{X}-Chart and RR-Chart in Quality Maintenance

    Control charts are graphical tracking tools displaying a Central Line (CL), Upper Control Limit (UCL), and Lower Control Limit (LCL):

    1. Mean Chart (Xˉ\bar{X}-Chart):
      • Monitors the central tendency of the process over time.
      • Detects process shifts, tool wear, or gradual calibration drift.
    2. Range Chart (RR-Chart):
      • Monitors the dispersion / variability of the process.
      • Detects increased erratic scattering or machine bearing looseness.

    Operational Value: By plotting sample statistics in real-time, operators receive early warnings of assignable causes before defective goods are manufactured, shifting quality management from reactive post-inspection sorting to proactive defect prevention.

Group C

Comprehensive Answer / Case Analysis Question.

[1 × 20 = 20]
  1. Read the operations case scenario and answer all questions:

    Case Scenario: Jagadamba Steel Products Ltd. (JSPL) Jagadamba Steel Products Ltd. (JSPL) operates a structural steel rolling mill in Simara, producing TMT rebar for domestic construction projects. The manufacturing process consists of four sequential production workstations:

    1. Billet Reheating Furnace: Design capacity = 60 tons per hour.
    2. Continuous Rolling Mill: Design capacity = 50 tons per hour.
    3. Thermax Quenching & Cooling Bed: Design capacity = 45 tons per hour.
    4. Shearing, Bundling & Quality Inspection: Design capacity = 55 tons per hour.

    The factory operates two 8-hour shifts per day (16 operating hours daily), 300 days a year.

    During the previous quarter, actual operating records revealed:

    • The plant operated 16 hours per day but lost an average of 2 hours per day to unscheduled mechanical breakdowns and electricity grid tripping.
    • Total actual output produced during the quarter (75 working days) was 43,200 tons of finished TMT rebar.
    • Inventory records show that scrap metal and billets require an annual carrying cost of Rs. 400 per ton. Placing an order for imported billets from India costs Rs. 8,000 per order. The factory consumes 160,000 tons of billets annually.

    Required: (a) Identify the operational Process Bottleneck workstation and determine the System (Design) Capacity of the plant in tons per hour and tons per day. (5 Marks) (b) Calculate the Plant Capacity Utilization and Plant Operating Efficiency for the previous quarter. (5 Marks) (c) Compute the Economic Order Quantity (EOQ) for raw billet procurement and determine the annual total inventory carrying and ordering costs. (5 Marks) (d) As Vice President of Operations, propose an actionable Lean Manufacturing & Total Productive Maintenance (TPM) strategy to eliminate the 2 hours of daily unplanned downtime and elevate bottleneck throughput. (5 Marks)

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    Case Solution: Jagadamba Steel Products Ltd. (JSPL)


    Part (a): Identification of Bottleneck & System Capacity (5 Marks)

    The bottleneck is the operation with the lowest throughput capacity, which constrains the maximum output rate of the entire sequential system:

    • Reheating Furnace: 60 tons/hr
    • Continuous Rolling Mill: 50 tons/hr
    • Quenching & Cooling Bed: 45 tons/hr (LOWEST)
    • Shearing & Bundling: 55 tons/hr
    Bottleneck Workstation=Thermax Quenching & Cooling Bed\mathbf{\text{Bottleneck Workstation}} = \mathbf{\text{Thermax Quenching \& Cooling Bed}}
    System Design Capacity=45 tons per hour\mathbf{\text{System Design Capacity}} = \mathbf{45 \text{ tons per hour}}
    System Capacity per Day (16 hrs)=45×16=720 tons per day\text{System Capacity per Day (16 hrs)} = 45 \times 16 = \mathbf{720 \text{ tons per day}}

    Part (b): Capacity Utilization and Plant Efficiency (5 Marks)

    • Quarter working days = 75 days
    • Design Capacity for Quarter: 75 days×720 tons/day=54,000 tons75 \text{ days} \times 720 \text{ tons/day} = \mathbf{54,000 \text{ tons}}
    • Actual Output for Quarter = 43,200 tons\mathbf{43,200 \text{ tons}}
    • Operating hours planned = 75×16=1,200 hours75 \times 16 = 1,200 \text{ hours}
    • Lost hours (2 hrs/day) = 75×2=150 hours75 \times 2 = 150 \text{ hours}
    • Effective operating hours = 1,200150=1,050 hours1,200 - 150 = 1,050 \text{ hours}
    • Effective Capacity for Quarter: 1,050 hours×45 tons/hr=47,250 tons1,050 \text{ hours} \times 45 \text{ tons/hr} = \mathbf{47,250 \text{ tons}}

    1. Capacity Utilization:

    Utilization=Actual OutputDesign Capacity×100%=43,20054,000×100%=80.00%\mathbf{\text{Utilization}} = \frac{\text{Actual Output}}{\text{Design Capacity}} \times 100\% = \frac{43,200}{54,000} \times 100\% = \mathbf{80.00\%}

    2. Operating Efficiency:

    Efficiency=Actual OutputEffective Capacity×100%=43,20047,250×100%=91.43%\mathbf{\text{Efficiency}} = \frac{\text{Actual Output}}{\text{Effective Capacity}} \times 100\% = \frac{43,200}{47,250} \times 100\% = \mathbf{91.43\%}

    Part (c): Economic Order Quantity (EOQ) (5 Marks)

    • Annual Demand (DD): 160,000 tons
    • Ordering Cost per order (SS): Rs. 8,000
    • Annual Carrying Cost per ton (HH): Rs. 400
    EOQ=2DSH=2×160,000×8,000400=2,560,000,000400=6,400,000=2,529.82    2,530 tons\mathbf{\text{EOQ}} = \sqrt{\frac{2 D S}{H}} = \sqrt{\frac{2 \times 160,000 \times 8,000}{400}} = \sqrt{\frac{2,560,000,000}{400}} = \sqrt{6,400,000} = \mathbf{2,529.82 \implies 2,530 \text{ tons}}

    Total Annual Inventory Costs at EOQ:

    Annual Ordering Cost=DEOQ×S=160,0002,529.82×8,000=Rs. 505,964\text{Annual Ordering Cost} = \frac{D}{\text{EOQ}} \times S = \frac{160,000}{2,529.82} \times 8,000 = \text{Rs. } 505,964
    Annual Carrying Cost=EOQ2×H=2,529.822×400=Rs. 505,964\text{Annual Carrying Cost} = \frac{\text{EOQ}}{2} \times H = \frac{2,529.82}{2} \times 400 = \text{Rs. } 505,964
    Total Inventory Cost (TIC)=Rs. 505,964+505,964=Rs. 1,011,928\mathbf{\text{Total Inventory Cost (TIC)}} = \text{Rs. } 505,964 + 505,964 = \mathbf{\text{Rs. } 1,011,928}

    Part (d): Lean & TPM Turnaround Strategy (5 Marks)

    1. Total Productive Maintenance (TPM) - 5S & Autonomous Maintenance: Empower machine operators to perform daily cleaning, lubrication, and visual tightening (Jishu Hozen), preventing unexpected mechanical failures.
    2. Dedicated Backup Generator / Industrial Dedicated Feeder: Install automated high-voltage industrial UPS/generator changeovers to prevent rolling mill thermal freezing during NEA national grid tripping.
    3. Debottlenecking the Cooling Bed: Invest in auxiliary forced-air cooling blowers and widen the transfer run-out tables, elevating the quenching bed throughput from 45 tons/hr to 55 tons/hr, aligning it with the continuous rolling mill and unlocking an immediate 22% expansion in plant capacity.