Quantitative Techniques — Undated papers
KNEC Diploma in Supply Chain Management, Module II. Original theory examination questions with explained model answers for revision.
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Question 1(a) — 10 marks
Explain five areas where quantitative techniques are applicable in business decision making.
Model answer
- Production planning: allocate labour, materials and machine capacity among competing products using optimisation models to meet an objective such as contribution maximisation.
- Inventory control: estimate economic order quantities, reorder points and safety stock so stock availability is balanced against inventory costs.
- Marketing and sales: use forecasting, sampling and regression to estimate demand and assess relationships between sales and factors such as advertising or price.
- Finance and investment: use cash-flow, interest, risk and capital-budgeting calculations to compare funding or investment alternatives on an appropriate basis.
- Logistics and project management: determine transport routes, resource schedules and critical activities so deliveries and projects can meet cost and time targets.
Question 1(b) — 10 marks
The following information shows the ranking of 10 employees based on their performance by two different agencies.
Employees Rank 1 Rank 2 A 8 7 B 6 5 C 4 2 D 1 1 E 3 4 F 5 8 G 7 6 H 2 3 I 10 9 J 9 10 (i) determine Spearman’s rank correlation coefficient.
(ii) interpret the results in (i) above.
Model answer
The observations are already ranks, so do not rank them again. There are no ties.
| Employee | Rank 1 | Rank 2 | d | d² |
|---|---|---|---|---|
| A | 8 | 7 | 1 | 1 |
| B | 6 | 5 | 1 | 1 |
| C | 4 | 2 | 2 | 4 |
| D | 1 | 1 | 0 | 0 |
| E | 3 | 4 | -1 | 1 |
| F | 5 | 8 | -3 | 9 |
| G | 7 | 6 | 1 | 1 |
| H | 2 | 3 | -1 | 1 |
| I | 10 | 9 | 1 | 1 |
| J | 9 | 10 | -1 | 1 |
| Total | 20 |
| Step / item | Working | Result |
|---|---|---|
| Spearman coefficient | 1 − (6 × 20)/[10(10² − 1)] | 0.88 |
rₛ = 0.878788, indicating strong positive agreement between the two agencies’ rankings. High-ranked employees for one agency generally rank highly for the other.
Question 2(a) — 10 marks
Explain five types of matrices, giving an example in each case.
Model answer
| Type | Explanation | Example |
|---|---|---|
| Row matrix | Contains exactly one row. | [2 4 6], order 1 × 3. |
| Column matrix | Contains exactly one column. | [[2],[4],[6]], order 3 × 1. |
| Square matrix | Has equal numbers of rows and columns. | [[1,2],[3,4]], order 2 × 2. |
| Null / zero matrix | All entries are zero. | [[0,0,0],[0,0,0]], order 2 × 3. |
| Identity / unit matrix | A square matrix with main diagonal entries 1 and all other entries 0. | I₂ = [[1,0],[0,1]]. |
Question 2(b) — 10 marks
The total revenue (TR) function of a company is given as TR = 25q − 0.5q², where q is the level of output sold. Determine the:
(i) marginal revenue (MR) function;
(ii) level of output when MR = 0;
(iii) maximum total revenue.
Model answer
| Step | Working | Result |
|---|---|---|
| Marginal revenue | MR = dTR/dq = d(25q − 0.5q²)/dq | MR = 25 − q |
| Set MR to zero | 25 − q = 0 | q = 25 units |
| Maximum check | d²TR/dq² = −1 < 0 | Revenue is maximised. |
| Maximum revenue | TR(25) = 25 × 25 − 0.5 × 25² | 312.5 monetary units |
No currency unit is supplied for this revenue function.
Question 3(a) — 10 marks
Explain five requirements of a linear programming problem.
Model answer
- Defined decision variables: identify the quantities to choose, with clear units, such as units of two products to manufacture.
- A numerical objective: state what is to be maximised or minimised, such as total contribution or total cost.
- Linear relationships: the objective and constraints must be linear combinations of the decision variables, satisfying proportionality and additivity.
- Explicit resource or requirement constraints: represent the limited capacity, minimum demand or other restrictions using equations or inequalities with specified coefficients.
- A defined feasible domain: impose non-negativity and other necessary variable bounds; feasible choices must satisfy every constraint. Ordinary continuous LP allows fractional variable values, while indivisible decisions require an integer model.
Question 3(b) — 10 marks
The following data shows the ages and weights of 10 employees randomly selected in an organization:
Ages in years (x) Weight in kilogrammes (y) 25 55 40 70 24 50 55 68 45 62 52 80 27 60 42 90 60 85 35 70 (i) determine the regression equation y on x;
(ii) using the equation in (i) above, estimate the weight of an employee who is 50 years old.
Model answer
Fit the least-squares line ŷ = a + bx, predicting weight from age.
| x | y | x² | xy |
|---|---|---|---|
| 25 | 55 | 625 | 1,375 |
| 40 | 70 | 1,600 | 2,800 |
| 24 | 50 | 576 | 1,200 |
| 55 | 68 | 3,025 | 3,740 |
| 45 | 62 | 2,025 | 2,790 |
| 52 | 80 | 2,704 | 4,160 |
| 27 | 60 | 729 | 1,620 |
| 42 | 90 | 1,764 | 3,780 |
| 60 | 85 | 3,600 | 5,100 |
| 35 | 70 | 1,225 | 2,450 |
| 405 | 690 | 17,873 | 29,015 |
b = [nΣxy−ΣxΣy]/[nΣx²−(Σx)²] = (10×29,015.00−405.00×690.00)/(10×17,873.00−405.00²) = 0.727644.
a = ȳ−bx̄ = 39.530432.
Regression equation: ŷ = 39.530432 + 0.727644x.
| Step / item | Working | Result |
|---|---|---|
| Predicted weight at age 50 | 39.530432 + 0.727644 × 50 | 75.91 |
The estimated weight is 75.91 kg. This is a fitted average association for these observations, not a guaranteed weight for an individual.
Question 4(a) — 10 marks
Explain five types of costs associated with holding of inventory in a firm.
Model answer
- Capital cost: funds tied up in stock incur borrowing interest or forgo the return that those funds could earn in another use.
- Storage-space cost: inventory occupies rented or owned warehouse space, with related utilities, shelving and premises costs.
- Insurance and stock-related taxes: premiums and any applicable inventory levies arise from carrying the stock and may vary with its value.
- Inventory handling and protection cost: receiving, moving, counting, preserving and securing stored goods consume labour, equipment and supplies.
- Risk-of-loss cost: damage, deterioration, expiry, obsolescence, shrinkage and theft reduce the amount recoverable from stored stock and may require write-offs.
Question 4(b) — 10 marks
The following data shows the prices and quantities of five commodities in a market for the years 2012 and 2016.
Commodities 2012 Prices (Ksh) 2012 Quantities (Units) 2016 Prices (Ksh) 2016 Quantities (Units) 1 50 200 40 180 2 120 150 100 200 3 60 180 100 150 4 50 100 80 80 5 40 50 50 40 Determine the:
(i) Paasche quantity index;
(ii) Laspéyre’s quantity index;
(iii) Fisher’s ideal quantity index.
Model answer
2012 is the base year (p₀,q₀); 2016 is the current year (p₁,q₁). For quantity indices, quantities change while prices provide the weights. Lq = Σp₀q₁/Σp₀q₀ × 100; Pq = Σp₁q₁/Σp₁q₀ × 100; Fq = √(LqPq).
| Commodity | p₀q₀ | p₀q₁ | p₁q₀ | p₁q₁ |
|---|---|---|---|---|
| 1 | 10,000 | 9,000 | 8,000 | 7,200 |
| 2 | 18,000 | 24,000 | 15,000 | 20,000 |
| 3 | 10,800 | 9,000 | 18,000 | 15,000 |
| 4 | 5,000 | 4,000 | 8,000 | 6,400 |
| 5 | 2,000 | 1,600 | 2,500 | 2,000 |
| Total | 45,800 | 47,600 | 51,500 | 50,600 |
| Step / item | Working | Result |
|---|---|---|
| Laspeyres quantity index | 47,600/45,800 × 100 | 103.93 |
| Paasche quantity index | 50,600/51,500 × 100 | 98.25 |
| Fisher ideal quantity index | √(103.930131 × 98.252427) | 101.05 |
Fisher’s quantity index is 101.0514 with 2012 = 100. It indicates a weighted aggregate quantity change of 1.0514% relative to 2012.
Question 5(a) — 12 marks
Distinguish between each of the following terms as used in probability theory: (i) mutually exclusive events and exhaustive events; (ii) independent events and dependent events; (iii) combinations and permutations.
Model answer
| Pair | First term | Second term |
|---|---|---|
| Mutually exclusive / exhaustive | Mutually exclusive events cannot occur together: A ∩ B = ∅. Example: an odd and an even result in a single die roll. | Exhaustive events collectively cover every possible outcome: their union is the sample space. For example, results 1–6 on one die. A set may be exhaustive without being mutually exclusive. |
| Independent / dependent | Independent events do not change one another’s probabilities: P(A ∩ B) = P(A)P(B). | For dependent events, occurrence of one changes the probability of the other: P(A ∩ B) = P(A)P(B|A). |
| Combinations / permutations | A combination is an unordered selection: nCᵣ = n!/[r!(n−r)!]. Selecting a committee is an example. | A permutation is an ordered selection: nPᵣ = n!/(n−r)!. Assigning distinct chair, secretary and treasurer roles is an example. |
Question 5(b) — 8 marks
Kiweto Enterprises is considering two investment options, A and B.
Option A – deposit Ksh 500,000 in a bank account earning compound interest at the reate of 9% per annum for a period of 5 years.
Option B – deposit Ksh 150,000 per year in a bank account earning interest at the rate of 8% compounded annually.
(i) For each investment option, determine the: (I) amount in the bank account at the end of the investment period. (II) interest earned.
(ii) Advise the management on the investment option to select.
Model answer
Assumptions for B: use the same five-year horizon as A, with five deposits at each year-end (an ordinary annuity). The question does not state B’s duration or deposit timing explicitly.
| Step / item | Working | Result |
|---|---|---|
| A: future amount | 500,000(1.09)⁵ | 769,311.98 |
| A: total interest | 769311.97745 − 500,000 | 269,311.98 |
| B: total deposits | 150,000 × 5 | 750,000 |
| B: future amount | 150,000[(1.08)⁵ − 1]/0.08 | 879,990.14 |
| B: total interest | 879990.14400 − 750,000 | 129,990.14 |
Advice: B produces the higher five-year balance (Ksh 879,990.14) under these assumptions, but it also requires Ksh 750,000 of total contributions, compared with Ksh 500,000 for A. A pays the higher annual rate (9% versus 8%) and earns Ksh 269,311.98 interest on the initial lump sum. Choose A if Ksh 500,000 is available now and return per invested shilling is the priority; choose B for the specified regular-saving plan and higher target balance if those annual payments are affordable. Higher final balance alone does not establish a superior return because the cash flows differ.
If B’s deposits are instead at the beginning of each year, its annuity-due value is 879,990.14400 × 1.08 = Ksh 950,389.36; interest is Ksh 200,389.36. A different duration for B requires a different answer.
Question 6(a) — 8 marks
Explain the four components of a time series.
Model answer
- Trend is the persistent long-term upward, downward or stable movement in a series after shorter fluctuations are smoothed.
- Seasonal variation is a pattern that repeats within a year, such as quarterly or monthly sales changes associated with weather or holidays.
- Cyclical variation is a longer-term rise and fall around the trend, often associated with business cycles and without a fixed within-year period.
- Random variation is irregular, unpredictable movement caused by unusual events or chance influences after systematic components are accounted for.
Question 6(b) — 12 marks
The following information relates to a project:
Activities Preceding activities Duration (week) A – 5 B A 10 C A 4 D A 2 E B 6 F C 8 G C 4 H E,F 10 I D,G 12 J I,H 6 (i) Draw a network diagram for the project;
(ii) Determine the: (I) critical path; (II) project duration.
Model answer
| Activity | Predecessors | Duration | ES | EF | LS | LF | Float |
|---|---|---|---|---|---|---|---|
| A | — | 5 | 0 | 5 | 0 | 5 | 0 |
| B | A | 10 | 5 | 15 | 5 | 15 | 0 |
| C | A | 4 | 5 | 9 | 9 | 13 | 4 |
| D | A | 2 | 5 | 7 | 17 | 19 | 12 |
| E | B | 6 | 15 | 21 | 15 | 21 | 0 |
| F | C | 8 | 9 | 17 | 13 | 21 | 4 |
| G | C | 4 | 9 | 13 | 15 | 19 | 6 |
| H | E,F | 10 | 21 | 31 | 21 | 31 | 0 |
| I | D,G | 12 | 13 | 25 | 19 | 31 | 6 |
| J | I,H | 6 | 31 | 37 | 31 | 37 | 0 |
Forward pass: ES = latest predecessor EF; EF = ES + duration. Backward pass: LF = earliest successor LS; LS = LF − duration. Float = LS − ES. Project duration = 37 weeks. Red identifies critical activities and their controlling links.
Critical path: A → B → E → H → J.
| Step / item | Working | Result |
|---|---|---|
| Project duration | 5 + 10 + 6 + 10 + 6 | 37 |
The project therefore takes 37 weeks.
Question 7(a) — 10 marks
Explain five problems that may be encountered when constructing index numbers.
Model answer
- Choosing a representative basket: the selected commodities and outlets may fail to reflect the population’s actual consumption or purchasing behaviour.
- Selecting a base period: an unusually high- or low-price year distorts comparisons, while an old base may no longer represent current conditions.
- Obtaining comparable data: missing observations, different units, unreliable quotations and changes in outlet coverage can make prices or quantities inconsistent over time.
- Allowing for quality and product change: new products, discontinued goods, different pack sizes and improvements in quality make like-for-like comparison difficult.
- Choosing weights and a formula: spending patterns change, substitutions occur and alternative weighting systems or formulae produce different results; collecting updated weights may also be costly.
Question 7(b) — 10 marks
The general manager of Sutox Limited claims that the average age of employees in the company is 35 years. A sample of 36 employees in the company was taken and found to have a mean age of 42 years with a standard deviation of 8 years. Test the manager’s claim at 5% level of significance.
Model answer
| Step | Working / decision |
|---|---|
| Hypotheses | H₀: μ = 35 years; H₁: μ ≠ 35 years (two-tailed). |
| Standard error estimate | s/√n = 8/√36 = 1.333333 years |
| Large-sample statistic | z ≈ (42 − 35)/1.333333 = 5.25 |
| Critical values | −1.96 and +1.96 at α = 0.05 |
| Decision | |5.25| > 1.96: reject the manager’s null hypothesis. |
| Conclusion | The sample gives significant evidence that the population mean age differs from 35 years; the observed direction is higher. |
Because 8 years is the sample standard deviation, the formal normal-population test is a one-sample t-test: t = 5.25 with 35 degrees of freedom. Its two-tailed 5% critical value is about ±2.030, giving the same rejection. Assume the sample is random and independent.