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Chapter 7

Production and Operations Management

TU BBS past exam questions

Past questions and answers

14 questions set from this chapter. Most repeated first.

  • 2026 · MSC 516 Productions and Operations Management · 15 marks

(a) A steel company has three factories F₁, F₂ and F₃ which supply to three destinations W₁, W₂ and W₃. The unit supply cost (in Rs.) are shown in the following table:
FactoriesDestinationsSupply
W₁W₂W₃
F₁95620
F₂43530
F₃56850
Demand401050100
Find the optimal transportation cost.

Answer

Total supply =20+30+50=100=20+30+50=100 and total demand =40+10+50=100=40+10+50=100.
So the transportation problem is balanced.

Initial solution by VAM

Using Vogel’s Approximation Method:

FactoriesW₁W₂W₃Supply
F₁002020
F₂0102030
F₃4001050
Demand401050100

Transportation cost:

Cost=20(6)+10(3)+20(5)+40(5)+10(8)=120+30+100+200+80=Rs 530\begin{aligned} \text{Cost} &= 20(6)+10(3)+20(5)+40(5)+10(8)\\ &=120+30+100+200+80\\ &=\text{Rs }530 \end{aligned}

Optimality check by MODI method

Basic cells are F1W3, F2W2, F2W3, F3W1, F3W3F_1W_3,\ F_2W_2,\ F_2W_3,\ F_3W_1,\ F_3W_3.

Let u1=0u_1=0. Then from basic cells:

u1+v3=6⇒v3=6u2+v3=5⇒u2=−1u2+v2=3⇒v2=4u3+v3=8⇒u3=2u3+v1=5⇒v1=3\begin{aligned} u_1+v_3&=6 \Rightarrow v_3=6\\ u_2+v_3&=5 \Rightarrow u_2=-1\\ u_2+v_2&=3 \Rightarrow v_2=4\\ u_3+v_3&=8 \Rightarrow u_3=2\\ u_3+v_1&=5 \Rightarrow v_1=3 \end{aligned}

Now compute reduced costs for non-basic cells:

F1W1:9−0−3=6≥0F1W2:5−0−4=1≥0F2W1:4−(−1)−3=2≥0F3W2:6−2−4=0≥0\begin{aligned} F_1W_1 &: 9-0-3=6 \ge 0\\ F_1W_2 &: 5-0-4=1 \ge 0\\ F_2W_1 &: 4-(-1)-3=2 \ge 0\\ F_3W_2 &: 6-2-4=0 \ge 0 \end{aligned}

All reduced costs are non-negative. Hence the solution is optimal.
The zero reduced cost at F3W2F_3W_2 indicates an alternative optimum, but the total cost remains the same.

Optimal transportation cost=Rs 530\boxed{\text{Optimal transportation cost} = \text{Rs }530}

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2026 · MSC 516 Productions and Operations Management · 15 marks

(b) A company has five jobs to be done on five machines. The cost (in Rs) of each jobs on each machine is given in the following table:
MachinesJobs
ABCDE
M₁11910137
M₂89131211
M₃10711810
M₄1214789
M₅10811137
What are the job assignments which will minimize the costs?

Answer

This is a minimisation assignment problem with 5 machines and 5 jobs, solved by the Hungarian method.

Step 1: Row reduction (row minima 7, 8, 7, 7, 7)

ABCDE
M₁42360
M₂01543
M₃30413
M₄57012
M₅31460

Step 2: Column reduction (column minima 0, 0, 0, 1, 0)

ABCDE
M₁42350
M₂01533
M₃30403
M₄57002
M₅31450

Step 3: Test for optimality. M₁ and M₅ have a zero only in column E, so a complete assignment on zeros is impossible. The zeros are covered by 4 lines (columns A and E, rows M₃ and M₄), which is less than 5, so the solution is not yet optimal.

Step 4: Revise. The smallest uncovered element is 1. Subtract 1 from every uncovered element and add 1 at the intersections of the lines.

ABCDE
M₁41240
M₂00423
M₃40404
M₄67003
M₅30340

Step 5: Assign on zeros. M₁ must take E, M₂ takes A, M₅ takes B, M₄ takes C and M₃ takes D.

Optimal assignment

MachineJobCost (Rs)
M₁E7
M₂A8
M₃D8
M₄C7
M₅B8
Total38

The minimum total cost is Rs 38.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2026 · MSC 516 Productions and Operations Management · 15 marks

(a) Explain the types of productivity. Discuss the importance of operations strategy.

Answer

(a) Types of Productivity and Importance of Operations Strategy

1. Meaning of Productivity

Productivity is the ratio of output produced to the inputs used in producing it. It measures how efficiently an organisation converts resources (labour, capital, materials, energy, machines) into goods and services.

Productivity=OutputInput\text{Productivity} = \frac{\text{Output}}{\text{Input}}

Higher productivity means more output from the same inputs, or the same output from fewer inputs — the basis of lower cost, higher profit and national economic growth.


2. Types of Productivity

(i) Partial (Single-Factor) Productivity

Output is related to one input only. It is the most widely used measure because it is simple and easy to compute.

MeasureFormula
Labour productivityOutputLabour hours (or workers)\dfrac{\text{Output}}{\text{Labour hours (or workers)}}
Machine productivityOutputMachine hours\dfrac{\text{Output}}{\text{Machine hours}}
Material productivityOutputQuantity of material used\dfrac{\text{Output}}{\text{Quantity of material used}}
Capital productivityOutputCapital invested\dfrac{\text{Output}}{\text{Capital invested}}
Energy productivityOutputEnergy consumed (kWh)\dfrac{\text{Output}}{\text{Energy consumed (kWh)}}

Example: A firm produces 1,000 units using 250 labour hours → labour productivity = 4 units per labour hour.

Limitation: It ignores other inputs, so a rise in labour productivity may simply be the result of more machinery or better materials.

(ii) Multifactor (Total-Factor) Productivity

Output is related to a combination of two or more inputs, usually measured in money terms:

Multifactor productivity=Output (value)Labour cost+Capital cost+Material cost+Energy cost\text{Multifactor productivity} = \frac{\text{Output (value)}}{\text{Labour cost} + \text{Capital cost} + \text{Material cost} + \text{Energy cost}}

It shows the joint efficiency of the factors combined and is a better indicator of overall performance than a single-factor measure.

(iii) Total Productivity

Output is compared with all inputs taken together:

Total productivity=Total output (goods and services)Total inputs (all resources)\text{Total productivity} = \frac{\text{Total output (goods and services)}}{\text{Total inputs (all resources)}}

It gives the broadest picture of an organisation's or a nation's productive efficiency.

(iv) Other Classifications

  • Physical productivity (units per input) vs value productivity (Rs of output per Rs of input).
  • Static productivity (ratio at a point of time) vs dynamic productivity (change in productivity between two periods, i.e. productivity growth).
  • Direct (measurable at the operation) vs indirect productivity (contribution of support functions).

3. Importance of Operations Strategy

Operations strategy is a long-range plan that specifies how the operations function will deploy its resources — plant, process, technology, people and capacity — to support the business strategy and build competitive advantage. Its importance can be explained as follows:

  1. Links operations with corporate strategy. It translates the mission and business strategy into operational decisions, so that operations works as a competitive weapon rather than a passive support activity.

  2. Defines competitive priorities. It decides the relative emphasis on cost, quality, delivery speed and reliability, flexibility and customization. These become the order winners and order qualifiers in the market.

  3. Guides major operations decisions. It provides direction for process and technology selection, capacity planning, facility location and layout, inventory policy, quality management, supply-chain design and workforce planning.

  4. Improves productivity and reduces cost. By choosing the right process and technology and eliminating waste, it raises output per unit of input, lowering unit cost and strengthening price competitiveness.

  5. Creates competitive advantage. A well-designed operations strategy (e.g. Toyota's lean system, Amazon's distribution network) becomes difficult for rivals to copy and gives sustainable advantage.

  6. Ensures efficient allocation of scarce resources. Capital, skilled manpower and technology are limited; the strategy ensures they are invested where they yield the greatest return.

  7. Provides consistency and direction. It gives a long-term framework that keeps day-to-day decisions consistent, avoiding ad hoc, conflicting choices.

  8. Supports growth and long-term survival. It plans for expansion, new products, new markets and technology up-gradation, helping the firm survive in a changing environment.

  9. Integrates functions. It coordinates operations with marketing (what to make, when, how much), finance (how much to invest) and HR (skills required), so all functions pull in the same direction.

  10. Builds resilience and manages risk. It prepares the firm for supply disruptions, strikes, disasters, pandemics and demand shocks through flexibility, multiple sourcing and buffer capacity.

  11. Promotes sustainability and social responsibility. Green processes, waste reduction and ethical sourcing are increasingly part of operations strategy, meeting customer and regulatory demands.

  12. Relevance to Nepal. For Nepali manufacturing and service firms facing import competition, weak infrastructure and labour migration, a clear operations strategy is essential for quality improvement, import substitution, export competitiveness and efficient delivery of services such as tourism, banking and health care.

Conclusion: Productivity measures how well resources are used, and its various types help managers diagnose performance at the factor, multi-factor and total level. Operations strategy gives the long-term direction that converts these resources into competitive priorities, productivity gains and sustainable competitive advantage.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2026 · MSC 516 Productions and Operations Management · 15 marks

(b) Describe the operational issues in product life cycle. Explain the types of layout with example.

Answer

(b) Operational Issues in Product Life Cycle and Types of Layout

1. Product Life Cycle (PLC) — Concept

The product life cycle is the sequence of stages through which a product passes from its introduction to its withdrawal from the market. Operationally, it is the time span over which a product generates sales and profit. The four standard stages are:

  1. Introduction – product is launched; sales are low and grow slowly.
  2. Growth – sales rise rapidly; competitors enter.
  3. Maturity – sales peak and stabilize; competition is intense.
  4. Decline – sales fall as the product becomes obsolete.

2. Operational Issues in the Product Life Cycle

Operations management must respond differently at each stage, because volume, variety, cost, quality and flexibility requirements change over time.

A. Introduction Stage

  • Low volume, high variety: demand is uncertain, so operations must be flexible rather than efficient.
  • High unit cost: fixed costs (design, tooling, plant set-up) are spread over few units.
  • Design instability: frequent design changes require flexible processes and skilled labour.
  • Quality risk: defects damage reputation; strict quality control is essential.
  • Supply chain uncertainty: suppliers must be developed and reliable delivery ensured.
  • Capacity under-utilization: excess capacity is normal; the issue is whether to build capacity ahead of demand.

B. Growth Stage

  • Rapid demand increase: capacity must be expanded quickly without hurting quality.
  • Standardization begins: product design stabilizes, allowing more efficient processes.
  • Cost pressure: competitors force price reduction, so operations must lower unit cost.
  • Forecasting difficulty: demand may outrun supply, causing stock-outs and lost sales.
  • Distribution expansion: wider markets require new channels and logistics planning.
  • Workforce training: new workers must be trained fast to maintain quality.

C. Maturity Stage

  • High volume, low variety: operations focus on efficiency, cost reduction and productivity.
  • Intense competition: cost, quality, delivery speed and reliability become order winners.
  • Process automation: repetitive operations justify specialized machinery and standardization.
  • Inventory management: large volumes tie up working capital; inventory control is critical.
  • Product proliferation: minor variants increase variety and complicate scheduling.
  • Maintenance and reliability: unplanned downtime is very costly at high volume.

D. Decline Stage

  • Shrinking demand: capacity must be reduced or redeployed; downsizing issues arise.
  • Narrow product line: operations focus on a few profitable items only.
  • Cost of obsolescence: obsolete stock, spare parts and tooling must be written off.
  • Resource reallocation: labour and machines shifted to new or growing products.
  • Service and spare parts: after-sales support must continue for existing customers.
  • Exit decisions: deciding when to withdraw the product is a major operational and strategic issue.

Summary of operational issues across PLC:

IssueIntroductionGrowthMaturityDecline
VolumeLowRisingHighFalling
VarietyHighDecreasingLowLow
ProcessFlexible/jobBatchRepetitive/flowReduced
Cost focusDesign costScale economiesCost controlCost cutting
Key concernDesign & qualityCapacity & deliveryEfficiency & reliabilityExit & reallocation

3. Types of Layout

Layout is the physical arrangement of machines, equipment, workers and materials within a facility so that production can be carried out efficiently. The main types are:

1. Process Layout (Functional Layout)

  • Similar machines and functions are grouped together — e.g., all lathes in one section, all drilling machines in another.
  • Suitable for low volume, high variety (job shop) production.
  • Example: A general engineering workshop, a hospital (separate departments for X-ray, surgery, pharmacy), a university campus.

2. Product Layout (Line Layout)

  • Machines and workers are arranged in the sequence of operations required to make the product.
  • Suitable for high volume, low variety (mass production).
  • Example: Automobile assembly line, bottling plant, television assembly line.

3. Fixed Position Layout

  • The product remains stationary; men, machines and materials are brought to it.
  • Used when the product is too heavy, bulky or fragile to move.
  • Example: Shipbuilding, construction of a bridge or dam, aircraft manufacturing.

4. Group Layout (Cellular Layout)

  • Machines are grouped into cells to produce a family of similar parts (group technology).
  • Combines the flexibility of process layout with the efficiency of product layout.
  • Example: A cell producing all types of shafts or gears in a machine shop.

5. Hybrid Layout

  • A combination of two or more basic layouts in one facility.
  • Example: A hospital using process layout for OPD and fixed position layout for operation theatres; a factory using product layout for assembly and process layout for fabrication.

Comparison of Layout Types

BasisProcessProductFixed PositionCellular
VolumeLowHighVery lowMedium
VarietyHighLowVery highMedium
FlowIntermittentContinuousProduct stationaryGrouped flow
FlexibilityHighLowHighModerate
ExampleHospital, job shopAssembly lineShipbuildingGear cell

Conclusion: The product life cycle forces operations managers to shift from flexible, design-oriented processes at introduction to efficient, cost-oriented processes at maturity, and finally to capacity reduction and resource reallocation at decline. The choice of layout must match this changing volume–variety position: process layout suits the early stages, product layout suits maturity, fixed position suits unique large products, and cellular layout offers a balanced compromise.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2026 · MSC 516 Productions and Operations Management · 15 marks

(a) A manufacturing company produces iron rods. The following table shows the length of iron rods (in centimeter).
Sample numberLength of iron rods
1410424408414412
2414399402409402
3407394399404413
4407409414398399
5416444424404414
Construct mean control chart and range control chart. Also, comment the results. (For n = 5, A₂ = 0.577, D₃ = 0 & D₄ = 2.115)

Answer

There are k=5k = 5 samples, each of size n=5n = 5.

Sample means and ranges

SampleMean Xˉ\bar{X}Range RR
1413.6424 − 408 = 16
2405.2414 − 399 = 15
3403.4413 − 394 = 19
4405.4414 − 398 = 16
5420.4444 − 404 = 40

Xˉˉ=413.6+405.2+403.4+405.4+420.45=409.6\bar{\bar{X}} = \frac{413.6+405.2+403.4+405.4+420.4}{5} = 409.6

Rˉ=16+15+19+16+405=21.2\bar{R} = \frac{16+15+19+16+40}{5} = 21.2

Mean chart (Xˉ\bar{X} chart)

CL=Xˉˉ=409.6CL = \bar{\bar{X}} = 409.6

UCL=Xˉˉ+A2Rˉ=409.6+0.577×21.2=421.83UCL = \bar{\bar{X}} + A_2\bar{R} = 409.6 + 0.577 \times 21.2 = 421.83

LCL=Xˉˉ−A2Rˉ=409.6−12.23=397.37LCL = \bar{\bar{X}} - A_2\bar{R} = 409.6 - 12.23 = 397.37

Range chart (RR chart)

CL=Rˉ=21.2CL = \bar{R} = 21.2

UCL=D4Rˉ=2.115×21.2=44.84UCL = D_4\bar{R} = 2.115 \times 21.2 = 44.84

LCL=D3Rˉ=0×21.2=0LCL = D_3\bar{R} = 0 \times 21.2 = 0

Comment

  • Every sample mean (403.4 to 420.4) lies between 397.37 and 421.83, so the mean chart shows no point out of control.
  • Every sample range (15 to 40) lies between 0 and 44.84, so the range chart is also in control.
  • Therefore the process is statistically in control and the variation is due to chance causes only. Sample 5 is close to the upper limits (mean 420.4 and range 40), so it should be watched, but no corrective action is needed now.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 · MSC 516 Production and Operations Management

(b) Perform ABC inventory chart using following data:
ItemAnnual DemandItem cost per unit (Rs)
A3,00050
B4,00012
C1,50015
D6,00010
E1,00020
F500500
G3001,500
H60020
I1,75010
J2,5005

Answer

Step 1: Compute Annual Usage Value of each item

Annual Usage Value = Annual Demand × Item cost per unit

ItemAnnual DemandCost per unit (Rs)Annual Usage Value (Rs)
A3,000501,50,000
B4,0001248,000
C1,5001522,500
D6,0001060,000
E1,0002020,000
F5005002,50,000
G3001,5004,50,000
H6002012,000
I1,7501017,500
J2,500512,500

Total Annual Usage Value = Rs 10,42,500

Step 2: Rank items in descending order of usage value

RankItemAnnual Usage Value (Rs)% of Total ValueCumulative % of ValueCumulative % of ItemsClass
1G4,50,00043.1743.1710A
2F2,50,00023.9867.1520A
3A1,50,00014.3981.5430A
4D60,0005.7687.3040B
5B48,0004.6091.9050B
6C22,5002.1694.0660B
7E20,0001.9295.9870B
8I17,5001.6897.6680C
9J12,5001.2098.8690C
10H12,0001.15100.00100C
Total10,42,500100

Step 3: Classification (ABC Chart)

ClassItems% of ItemsAnnual Usage Value (Rs)% of Value
AG, F, A30% (3 items)8,50,00081.54%
BD, B, C, E40% (4 items)1,50,50014.44%
CI, J, H30% (3 items)42,0004.03%
Total100%10,42,500100%

ABC Chart

Plotting cumulative % of items (x-axis) against cumulative % of usage value (y-axis) gives a curve that rises steeply for the first three items (to about 82% of value), flattens through the next four (to about 96%) and then creeps to 100% for the last three.

Interpretation of the chart:

  • Class A (G, F, A): Only 30% of the items but 81.5% of the total annual usage value. These are high-value items and need tight inventory control — frequent review, low safety stock, accurate records, and close monitoring.
  • Class B (D, B, C, E): 40% of the items contributing about 14% of the value. These need moderate control — periodic review and normal record-keeping.
  • Class C (I, J, H): 30% of the items contributing only about 4% of the value. These are low-value items; simple control with large order quantities and less frequent review is sufficient.

Conclusion: The firm should concentrate its inventory management effort on items G, F and A (Class A), since they account for nearly four-fifths of the total annual usage value.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 · MSC 516 Production and Operations Management

(a) A truck company has 5 trucks for sending material to 5 destinations. The cost of sending material from destinations by different trucks will be different as given by the cost (Rs) matrix below:
DestinationsTrucks
ABCDE
12016223312
21418301424
3111214813
43122161731
52618162118
Find the minimizing cost by assignment problem.

Answer

Cost matrix (Rs):

ABCDE120162233122141830142431112148134312216173152618162118\begin{matrix} & A & B & C & D & E\\ 1 & 20 & 16 & 22 & 33 & 12\\ 2 & 14 & 18 & 30 & 14 & 24\\ 3 & 11 & 12 & 14 & 8 & 13\\ 4 & 31 & 22 & 16 & 17 & 31\\ 5 & 26 & 18 & 16 & 21 & 18 \end{matrix}

Step 1: Row reduction
Subtract each row minimum.
Row minima are (12,14,8,16,16).

84102100416010346051560115102052\begin{matrix} 8 & 4 & 10 & 21 & 0\\ 0 & 4 & 16 & 0 & 10\\ 3 & 4 & 6 & 0 & 5\\ 15 & 6 & 0 & 1 & 15\\ 10 & 2 & 0 & 5 & 2 \end{matrix}

Step 2: Column reduction
Column minima are (0,2,0,0,0). Subtract (2) from column B:

82102100216010326051540115100052\begin{matrix} 8 & 2 & 10 & 21 & 0\\ 0 & 2 & 16 & 0 & 10\\ 3 & 2 & 6 & 0 & 5\\ 15 & 4 & 0 & 1 & 15\\ 10 & 0 & 0 & 5 & 2 \end{matrix}

Step 3: Assignment of zeros

We can assign:

  • Destination 1 → Truck E
  • Destination 2 → Truck A
  • Destination 3 → Truck D
  • Destination 4 → Truck C
  • Destination 5 → Truck B

All these are zero positions in the reduced matrix. Since the reduced cost of this assignment is zero, it is optimal.

Minimum cost:

12+14+8+16+18=Rs 6812+14+8+16+18 = \text{Rs }68

Answer:
The minimum assignment cost is Rs 68, with assignment
1–E, 2–A, 3–D, 4–C, 5–B.

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  • 2025 · MSC 516 Production and Operations Management

(b) Construct mean and range charts for the following data in which samples of 5 being taken every hour. Comment on whether the process seems to be under control or not. (For n = 5, A2 = 0.577, D3 = 0 & D4 = 2.115)
Sample NumbersHours
1st2nd3rd4th5th6th
1444143616369
2674656918075
3778270939683
479837910010197
5788177989995

Answer

Given:

  • Sample size, n=5n = 5
  • Number of samples, k=6k = 6 (one sample taken every hour for 6 hours)
  • Factors: A2=0.577A_2 = 0.577, D3=0D_3 = 0, D4=2.115D_4 = 2.115

Note: The table lists 6 columns (1st–6th hour) and 5 rows (observations within each sample). Thus each column is a sample of size 5.

Step 1: Compute mean (Xˉ\bar{X}) and range (RR) of each sample

Sample No.ObservationsTotalMean Xˉ\bar{X}Range RR
144, 67, 77, 79, 7834569.0035
241, 46, 82, 83, 8133366.6042
343, 56, 70, 79, 7732565.0036
461, 91, 93, 100, 9844388.6039
563, 80, 96, 101, 9943987.8038
669, 75, 83, 97, 9541983.8028

Step 2: Compute grand mean and average range

Xˉˉ=69.00+66.60+65.00+88.60+87.80+83.806=460.806=76.80\bar{\bar{X}} = \frac{69.00 + 66.60 + 65.00 + 88.60 + 87.80 + 83.80}{6} = \frac{460.80}{6} = 76.80

Rˉ=35+42+36+39+38+286=2186=36.33\bar{R} = \frac{35 + 42 + 36 + 39 + 38 + 28}{6} = \frac{218}{6} = 36.33

Step 3: Control limits for Xˉ\bar{X}-chart

UCLXˉ=Xˉˉ+A2Rˉ=76.80+0.577(36.33)=76.80+20.96=97.76UCL_{\bar{X}} = \bar{\bar{X}} + A_2\bar{R} = 76.80 + 0.577(36.33) = 76.80 + 20.96 = 97.76

LCLXˉ=Xˉˉ−A2Rˉ=76.80−20.96=55.84LCL_{\bar{X}} = \bar{\bar{X}} - A_2\bar{R} = 76.80 - 20.96 = 55.84

Step 4: Control limits for RR-chart

UCLR=D4Rˉ=2.115(36.33)=76.84UCL_R = D_4\bar{R} = 2.115(36.33) = 76.84

LCLR=D3Rˉ=0(36.33)=0LCL_R = D_3\bar{R} = 0(36.33) = 0

Step 5: Interpretation

Sample No.Xˉ\bar{X}RRXˉ\bar{X} within limits?RR within limits?
169.0035YesYes
266.6042YesYes
365.0036YesYes
488.6039YesYes
587.8038YesYes
683.8028YesYes

Comment:

  • In the Xˉ\bar{X}-chart, all sample means lie within the control limits (55.84 to 97.76).
  • In the RR-chart, all sample ranges lie within the control limits (0 to 76.84).
  • Since no point falls outside the control limits in either chart, the process is under control. No assignable cause of variation is indicated, and production may continue.

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 · MSC 516 Production and Operations Management · 5 marks

A firm requires 1500 units of items per month. Ordering cost is estimated to be Rs 240 per order, unit cost of item is Rs 25 and annual inventory holding cost is charged at 20%. Find (a) the economic lot size to be ordered and (b) the total minimum cost.

Answer

Given:

  • Monthly requirement = 1,500 units
  • Annual demand, (D = 1,500 \times 12 = 18{,}000) units/year
  • Ordering cost, (S = \text{Rs } 240) per order
  • Unit cost, (C = \text{Rs } 25)
  • Holding cost rate = 20% of unit cost
  • Annual holding cost per unit, (H = 0.20 \times 25 = \text{Rs } 5)

(a) Economic lot size (EOQ)

[ EOQ=\sqrt{\frac{2DS}{H}} ]

[ EOQ=\sqrt{\frac{2 \times 18{,}000 \times 240}{5}} ]

[ EOQ=\sqrt{1{,}728{,}000}=1{,}314.53 \text{ units} ]

[ \boxed{EOQ \approx 1{,}315 \text{ units per order}} ]


(b) Total minimum cost

At EOQ, ordering cost = holding cost.

[ \text{Ordering cost}=\frac{D}{Q}S ]

[ =\frac{18{,}000}{1{,}314.53}\times 240 ]

[ = \text{Rs } 3{,}286.34 ]

[ \text{Holding cost}=\frac{Q}{2}H ]

[ =\frac{1{,}314.53}{2}\times 5 ]

[ = \text{Rs } 3{,}286.34 ]

[ \text{Minimum inventory cost}=3{,}286.34+3{,}286.34 ]

[ \boxed{\text{Minimum inventory cost} = \text{Rs } 6{,}572.68} ]

If purchase cost is also included:

[ \text{Purchase cost}=D \times C=18{,}000 \times 25=\text{Rs } 4{,}50{,}000 ]

[ \text{Total annual cost}=4{,}50{,}000+6{,}572.68 ]

[ \boxed{\text{Total annual cost including purchase} = \text{Rs } 4{,}56{,}572.68} ]


Answer:

[ \boxed{EOQ = 1{,}315 \text{ units approximately}} ]

[ \boxed{\text{Minimum inventory cost} = \text{Rs } 6{,}572.68} ]

[ \boxed{\text{Total annual cost including purchase} = \text{Rs } 4{,}56{,}572.68} ]

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2025 · MSC 516 Production and Operations Management · 5 marks

Product A consists of five units of B and three units of C. Product B is made of three units of D and two units of E. Product C includes four units of D and three units of F. Product E is prepared from three units of F and five units of H. (a) Construct product structure tree (Bill of materials file) (b) If 1,000 units of A are required, how many units of each component are needed?

Answer

(a) Product structure tree (BOM file)

A
├── B (5)
│   ├── D (3)
│   └── E (2)
│       ├── F (3)
│       └── H (5)
└── C (3)
    ├── D (4)
    └── F (3)

BOM file table

LevelItemParentQuantity per parent
0A—1
1BA5
1CA3
2DB3
2EB2
2DC4
2FC3
3FE3
3HE5

(b) Component requirements for 1,000 units of A

ComponentCalculationRequired units
B(5 \times 1{,}000)5,000
C(3 \times 1{,}000)3,000
D((3 \times 5{,}000) + (4 \times 3{,}000))27,000
E(2 \times 5{,}000)10,000
F((3 \times 3{,}000) + (3 \times 10{,}000))39,000
H(5 \times 10{,}000)50,000

Therefore, to produce 1,000 units of A, the required components are:

  • B = 5,000 units
  • C = 3,000 units
  • D = 27,000 units
  • E = 10,000 units
  • F = 39,000 units
  • H = 50,000 units

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2025 · MSC 516 Production and Operations Management · 5 marks

Convert the following primal problem into dual problem.
Max. Z = 4x1 + 3x2 + 5x3
Subject to constraints:
3x1 + 4x2 + 5x3 ≥ 4 4x1 + x2 + 5x3 ≤ 9 2x1 + 3x2 + x3 = 7
x1, x2, x3 ≥ 0

Answer

Primal (maximisation)

Max Z=4x1+3x2+5x3\text{Max } Z = 4x_1 + 3x_2 + 5x_3

subject to

3x1+4x2+5x3≥43x_1 + 4x_2 + 5x_3 \ge 4 4x1+x2+5x3≤94x_1 + x_2 + 5x_3 \le 9 2x1+3x2+x3=72x_1 + 3x_2 + x_3 = 7 x1,x2,x3≥0x_1, x_2, x_3 \ge 0

Rules used (primal maximisation to dual minimisation)

  • One dual variable is created for each primal constraint: y1,y2,y3y_1, y_2, y_3.
  • The primal right-hand sides (4, 9, 7) become the dual objective coefficients, and the primal objective coefficients (4, 3, 5) become the dual right-hand sides.
  • A ≥\ge constraint gives y1≤0y_1 \le 0, a ≤\le constraint gives y2≥0y_2 \ge 0, and an == constraint gives y3y_3 unrestricted in sign.
  • Since x1,x2,x3≥0x_1, x_2, x_3 \ge 0, each dual constraint is of the ≥\ge type.

Dual

Min W=4y1+9y2+7y3\text{Min } W = 4y_1 + 9y_2 + 7y_3

subject to

3y1+4y2+2y3≥43y_1 + 4y_2 + 2y_3 \ge 4 4y1+y2+3y3≥34y_1 + y_2 + 3y_3 \ge 3 5y1+5y2+y3≥55y_1 + 5y_2 + y_3 \ge 5 y1≤0,y2≥0,y3 unrestricted in signy_1 \le 0,\quad y_2 \ge 0,\quad y_3 \text{ unrestricted in sign}

Each dual constraint is built from one column of the primal constraint matrix: column x1x_1 is (3, 4, 2), column x2x_2 is (4, 1, 3) and column x3x_3 is (5, 5, 1).

Answer checked by Claude Sonnet · Oct 21, 2026

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  • 2025 · MSC 516 Production and Operations Management · 5 marks

Describe about value engineering.

Answer

Value Engineering

Meaning: Value engineering (VE) is a systematic, organized, and creative approach to analyzing the function of a product, service, or project with the aim of achieving the required function at the lowest possible total cost, without compromising quality, reliability, or performance. It seeks to improve the "value" of an item, where value is the ratio of function to cost:

Value=Function (performance, quality, reliability)Cost\text{Value} = \frac{\text{Function (performance, quality, reliability)}}{\text{Cost}}

Value can therefore be increased by improving function at the same cost, keeping function the same at lower cost, or improving function more than the increase in cost.

Origin and Concept: Value engineering was developed by Lawrence D. Miles at General Electric (USA) during the Second World War, when shortages of materials and skilled labour forced the search for substitute materials and cheaper methods without losing function. It was originally called "value analysis" (VA); when applied to new products at the design stage it is called value engineering, and when applied to existing products it is called value analysis. The two terms are often used interchangeably.

Objectives:

  1. To reduce cost without reducing quality or performance.
  2. To eliminate unnecessary costs arising from over-design, unnecessary features, and inefficient processes.
  3. To improve the functional performance of the product or project.
  4. To identify and use better alternative materials, methods, and designs.
  5. To increase profitability and competitiveness of the organization.

Basic Steps / Job Plan of Value Engineering:

  1. Information phase: Collect all facts about the product — its functions, costs, quantities, specifications, and customer requirements.
  2. Function analysis phase: Identify the basic function (the essential purpose) and secondary functions, and express them in two words (verb + noun), e.g., "transmit torque", "hold liquid".
  3. Creative phase: Generate alternative ideas and means of performing each function through brainstorming, without judging them at this stage.
  4. Evaluation phase: Evaluate and rank the alternatives on the basis of cost, feasibility, quality, and time.
  5. Development phase: Develop the selected alternatives into workable proposals with cost estimates and technical details.
  6. Presentation and implementation phase: Present recommendations to management, obtain approval, implement the changes, and monitor the results.

Techniques Used: Function analysis system technique (FAST) diagram, brainstorming, brainwriting, checklists, life-cycle costing, cost–benefit analysis, and value tests.

Applications: VE is widely applied in construction projects, manufacturing industries, product design, procurement, and service organizations. In construction, for example, it may replace an expensive imported material with a locally available material of equal function, or simplify a structural design, thereby saving cost and time.

Advantages:

  1. Reduces unnecessary cost and improves profit.
  2. Improves quality, reliability, and performance of the product.
  3. Encourages innovation and creative thinking among employees.
  4. Helps in better utilization of scarce resources and materials.
  5. Improves the competitive position of the firm through lower prices.

Limitations:

  1. It requires trained and experienced personnel, which may be costly.
  2. It is time-consuming and may delay the project if not properly managed.
  3. It may face resistance from employees and designers who fear change.
  4. Wrong application may reduce quality and harm the reputation of the firm.

Conclusion: Value engineering is thus a systematic, function-oriented, and cost-reduction technique that aims to provide the required function at the lowest total cost while maintaining quality. It is an important tool of project and production management for improving value, profitability, and customer satisfaction.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2025 · MSC 516 Production and Operations Management · 5 marks

What is transformation model? Discuss.

Answer

Transformation Model

The transformation model is a model of organizational behaviour and management that views an organization as a system which takes inputs (resources) and converts them into outputs (goods/services) through a transformation process, while interacting with the external environment. It is also called the input–process–output (IPO) model or the open-system model of the organization.

Components of the transformation model:

  1. Inputs — The resources the organization draws from its environment: human resources (people, skills), physical resources (materials, machinery, technology), financial resources (capital), and information resources (data, knowledge).

  2. Transformation process — The core activity that adds value to inputs. It may be:

    • Physical transformation — manufacturing, assembling (e.g., raw cotton into cloth).
    • Locational transformation — transportation, delivery (e.g., courier service).
    • Exchange transformation — retailing, banking (e.g., a shop selling goods).
    • Informational transformation — processing data into useful reports (e.g., accounting, consulting).
    • Physiological/psychological transformation — health care, education, entertainment.
  3. Outputs — The results of transformation: products, services, profits, customer satisfaction, employee satisfaction, and social contribution. Outputs are returned to the environment as goods/services and also fed back into the system.

  4. Feedback — Information about outputs that is returned to the system to correct and improve the transformation process.

  5. Environment — The external forces (economic, political, social, technological, legal) that supply inputs and absorb outputs, and to which the organization must adapt.

Features:

  • It treats the organization as an open system — continuously exchanging resources with its environment.
  • It emphasizes value addition at each stage of conversion.
  • It highlights the interdependence of subsystems (production, marketing, finance, HR).
  • It focuses on goal attainment — converting inputs into valued outputs efficiently and effectively.

Importance:

  • Helps managers understand the organization as a whole and coordinate its parts.
  • Provides a framework for measuring efficiency (input–output ratio) and effectiveness (goal achievement).
  • Explains how environmental changes affect organizational performance and the need for adaptation.
  • Useful in designing organizational structure, processes, and control systems.

Conclusion: The transformation model is a systematic framework that explains how organizations create value by converting inputs into outputs through a transformation process, while remaining responsive to the external environment. It is a foundational concept in understanding organizational behaviour and management.

Answer checked by Claude Haiku · Oct 21, 2026

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  • 2025 · MSC 516 Production and Operations Management · 5 marks

Define the term quality in your words. Explain the total quality management.

Answer

Quality refers to the degree to which a product or service meets or exceeds the expectations of customers by consistently satisfying their stated and implied needs. In simple words, quality means "fitness for use" — doing the right thing, in the right way, at the first time and every time, so that the customer remains satisfied. It is not merely the absence of defects; it is the total sum of features and characteristics of a product or service that bear on its ability to satisfy a given need. Quality is judged by the customer, not by the producer, and it covers the whole organisation — its people, processes, products and services.

Total Quality Management (TQM) is a management philosophy and a set of guiding principles that represent the foundation of a continuously improving organisation. It is an organisation-wide, customer-focused, continuous improvement approach that integrates all functions, all levels and all people of an enterprise to meet customer requirements at the lowest cost. TQM treats quality as everybody's responsibility, from top management to the lowest worker, and seeks to build quality into the process rather than inspecting it into the product at the end.

The main elements of TQM are:

  1. Top management commitment — quality must be led from the top, with a clear vision, mission and quality policy.
  2. Customer focus — internal and external customers define quality; their needs drive all activities.
  3. Employee involvement and empowerment — every employee participates through teams, suggestions and quality circles.
  4. Continuous improvement — an ongoing effort (often through the PDCA cycle) to improve processes, products and people.
  5. Process-centred approach — quality is built into the process; prevention is preferred over detection.
  6. Fact-based decision making — decisions rest on data, measurement and statistical tools, not on guesswork.
  7. Strategic and systematic approach — quality is made part of the organisation's strategy and planning.
  8. Communication — free flow of information at all levels to keep everyone aligned with quality goals.
  9. Training and education — continuous development of skills and quality awareness.
  10. Supplier partnership — long-term, mutually beneficial relationships with suppliers to ensure quality inputs.
  11. Benchmarking — comparing with the best practices of leading organisations to learn and improve.

Benefits: higher customer satisfaction and loyalty, reduced rework, scrap and warranty costs, improved productivity and morale, better market image and competitive advantage.

Conclusion: TQM is not a one-time programme but a continuous journey of cultural change in which everyone in the organisation works together to satisfy the customer and to improve quality at every stage.

Answer checked by Claude Haiku · Oct 21, 2026

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