Quantitative Techniques — 2020
KNEC Diploma in Supply Chain Management, Module II. Original theory examination questions with explained model answers for revision.
July 2020
Paper codes: 2903/204
Time: Not shown in supplied scan
Instructions: Not shown in supplied scan
Question 1(a) — 12 marks
The following information relates to a project to be undertaken by a company.
Activity Preceeding Activity Duration (weeks) A – 1 B A 4 C A 2 D A 3 E B,C 3 F C 6 G D 3 H E,F,G 2 I H 2 (i) Draw a network diagram of the project.
(ii) Determine the: (I) critical path; (II) project duration.
Model answer

| Activity | Predecessors | Duration | ES | EF | LS | LF | Float |
|---|---|---|---|---|---|---|---|
| A | — | 1 | 0 | 1 | 0 | 1 | 0 |
| B | A | 4 | 1 | 5 | 2 | 6 | 1 |
| C | A | 2 | 1 | 3 | 1 | 3 | 0 |
| D | A | 3 | 1 | 4 | 3 | 6 | 2 |
| E | B,C | 3 | 5 | 8 | 6 | 9 | 1 |
| F | C | 6 | 3 | 9 | 3 | 9 | 0 |
| G | D | 3 | 4 | 7 | 6 | 9 | 2 |
| H | E,F,G | 2 | 9 | 11 | 9 | 11 | 0 |
| I | H | 2 | 11 | 13 | 11 | 13 | 0 |
Forward pass: ES = latest predecessor EF; EF = ES + duration. Backward pass: LF = earliest successor LS; LS = LF − duration. Float = LS − ES. Project duration = 13 weeks. Red identifies critical activities and their controlling links.
Critical path: A → C → F → H → I.
| Step / item | Working | Result |
|---|---|---|
| Critical-path duration | 1 + 2 + 6 + 2 + 2 | 13 |
The project takes 13 weeks. Handwritten workings in the supplied scan are not treated as a marking scheme.
Question 1(b) — 8 marks
Explain the meaning of each of the following terms as used in inventory control: (i) re-order level; (ii) maximum stock level; (iii) re-order quantity; (iv) re-order period.
Model answer
- Re-order level is the stock balance that triggers a replenishment order, allowing demand during lead time and any safety-stock requirement to be met.
- Maximum stock level is the upper stock limit the firm intends to hold to avoid excessive storage, tied-up capital and losses through deterioration or obsolescence.
- Re-order quantity is the number of units ordered each time replenishment is initiated; it may be fixed, set by EOQ, or adjusted to an order-up-to level.
- Re-order period is the time associated with replenishment. If used as lead time, it runs from placing an order until receiving usable stock; if used as a review interval, it is the time between successive stock reviews/orders. It is a period of time, unlike the re-order level, which is a stock quantity.
Question 2(a) — 8 marks
Explain the meaning of each of the following types of matrices: (i) square matrix; (ii) scalar matrix; (iii) null matrix; (iv) identify matrix.
Model answer
| Type | Explanation | Example |
|---|---|---|
| Square | Has the same number of rows and columns. | [[1,2],[3,4]] is 2 × 2. |
| Scalar | A square diagonal matrix whose main diagonal entries all equal the same scalar k; all off-diagonal entries are zero. | [[3,0],[0,3]] = 3I. |
| Null | Every entry is zero; it may be rectangular. | [[0,0],[0,0]]. |
| Identity | A square matrix with 1 on every main diagonal entry and 0 elsewhere; IA = AI = A when dimensions permit. | [[1,0],[0,1]]. |
The printed phrase “identify matrix” is reproduced above; the intended standard term is identity matrix.
Question 2(b) — 12 marks
The following are the marks awarded by two judges; judge 1 and judge 2 to seven contestants in a beauty contest.
Contestants Judge 1 Judge 2 A 60 36 B 42 80 C 50 22 D 33 69 E 48 78 F 72 31 G 55 20 (i) calculate the Spearman’s rank correlation coefficient.
(ii) based on the results in (i) above, comment on the correlation between the two judges.
Model answer
Rank the highest mark as 1 for each judge. There are no tied marks.
| Contestant | Judge 1 mark | Judge 2 mark | Rank 1 | Rank 2 | d | d² |
|---|---|---|---|---|---|---|
| A | 60 | 36 | 2 | 4 | -2 | 4 |
| B | 42 | 80 | 6 | 1 | 5 | 25 |
| C | 50 | 22 | 4 | 6 | -2 | 4 |
| D | 33 | 69 | 7 | 3 | 4 | 16 |
| E | 48 | 78 | 5 | 2 | 3 | 9 |
| F | 72 | 31 | 1 | 5 | -4 | 16 |
| G | 55 | 20 | 3 | 7 | -4 | 16 |
| Total | 90 |
| Step / item | Working | Result |
|---|---|---|
| Number of contestants n | Count the paired ranks | 7 |
| Spearman coefficient | 1 − [6 × 90]/[7(7² − 1)] | -0.61 |
rₛ = -0.607143. The rankings have a moderate negative association: contestants ranked more highly by one judge tend to be ranked less highly by the other. Work with ranks, not differences between the raw marks.
Question 3(a) — 10 marks
The following table shows the prices and quantities of three commodities consumed by a household during the years 2010 and 2016.
Commodity 2010 Price (Ksh) 2010 Quantity (Kgs) 2016 Price (Ksh) 2016 Quantity (Kgs) Rice 40 500 50 480 Sugar 75 120 120 150 Meat 160 96 280 110 (i) Calculate Laspeyre’s price index;
(ii) Paasche’s price index;
(iii) Fisher’s ideal price index.
Model answer
Let 2010 be the base year (p₀,q₀) and 2016 the current year (p₁,q₁). Use price-index formulae, retaining the corresponding quantities as weights.
| Commodity | p₀q₀ | p₁q₀ | p₀q₁ | p₁q₁ |
|---|---|---|---|---|
| Rice | 20,000 | 25,000 | 19,200 | 24,000 |
| Sugar | 9,000 | 14,400 | 11,250 | 18,000 |
| Meat | 15,360 | 26,880 | 17,600 | 30,800 |
| Total | 44,360 | 66,280 | 48,050 | 72,800 |
| Step / item | Working | Result |
|---|---|---|
| Laspeyres price index | 66,280/44,360 × 100 | 149.41 |
| Paasche price index | 72,800/48,050 × 100 | 151.51 |
| Fisher ideal price index | √(149.413886 × 151.508845) | 150.46 |
Question 3(b) — 10 marks
The following are the quarterly sales of Atlantic Suppliers for a period of three years.
Year Quarter 1 (Units) Quarter 2 (Units) Quarter 3 (Units) Quarter 4 (Units) 2013 160 200 210 180 2014 220 250 230 270 2015 290 280 300 320 Calculate:
(i) 3 – period moving average;
(ii) seasonal variations using multiplicative method.
Model answer
(i) Three-period moving averages, as printed. Put each average at the middle quarter of its three observations. The first is (160+200+210)/3 = 190, placed at 2013 Q2.
| Year / quarter | Actual sales Y | Centred 3-period average T | Y/T × 100 |
|---|---|---|---|
| 2013 Q1 | 160 | Not available | Not available |
| 2013 Q2 | 200 | 190.0000 | 105.2632 |
| 2013 Q3 | 210 | 196.6667 | 106.7797 |
| 2013 Q4 | 180 | 203.3333 | 88.5246 |
| 2014 Q1 | 220 | 216.6667 | 101.5385 |
| 2014 Q2 | 250 | 233.3333 | 107.1429 |
| 2014 Q3 | 230 | 250.0000 | 92.0000 |
| 2014 Q4 | 270 | 263.3333 | 102.5316 |
| 2015 Q1 | 290 | 280.0000 | 103.5714 |
| 2015 Q2 | 280 | 290.0000 | 96.5517 |
| 2015 Q3 | 300 | 300.0000 | 100.0000 |
| 2015 Q4 | 320 | Not available | Not available |
(ii) Multiplicative seasonal estimates. For each quarter, average its available Y/T × 100 ratios, then adjust the four averages to sum to 400. This gives mean seasonal factor 1.
| Quarter | Mean available ratio (%) | Adjusted seasonal index (%) | Multiplicative factor |
|---|---|---|---|
| Q1 | 102.554945 | 102.385447 | 1.023854 |
| Q2 | 102.985913 | 102.815703 | 1.028157 |
| Q3 | 99.593220 | 99.428617 | 0.994286 |
| Q4 | 95.528118 | 95.370233 | 0.953702 |
Adjustment multiplier = 400/400.662196 = 0.99834725. An index above 100 means the quarter is above its estimated trend; below 100 means below trend.
Method note: the scan explicitly asks for a 3-period average, so that instruction is followed. A quarterly seasonal cycle normally calls for a 4-quarter moving average followed by centring. A 3-period average does not fully remove quarterly seasonality, so the indices above are the literal requested-method estimates rather than the conventional 4-quarter estimates.
Question 4(a) — 8 marks
Outline four rules that need to be followed when constructing a network diagram.
Model answer
- Preserve precedence: an activity may start only after every required predecessor is completed; show all logical dependencies accurately.
- Use a clear overall start and finish, and connect every activity into a valid start-to-finish path; avoid dangling activities and disconnected sections.
- Do not introduce directed loops or circular dependencies. Time and event numbering should move consistently forward through the network.
- In an activity-on-arrow diagram, identify activities uniquely by their start/end events and use zero-duration dummy arrows where necessary to preserve precedence without duplicating event pairs. Arrow length does not represent duration.
Question 4(b) — 12 marks
Vertex processors uses material VX in its production. The average monthly demand is 4,000 units. The cost price per unit is Ksh 300 while ordering cost per order is Ksh 4,800. The annual carrying cost per unit is Ksh 80.
Calculate the:
(i) Economic Order Quantity (EOQ);
(ii) total costs associated with the inventory.
Model answer
Use annual demand because the holding cost is annual. Assume steady demand, immediate replenishment, no shortages, no quantity discounts and no safety stock.
| Step / item | Working | Result |
|---|---|---|
| Annual demand D | 4,000 × 12 | 48,000 |
| EOQ Q* | √(2DS/H) = √(2 × 48,000 × 4,800/80) | 2,400 |
| Orders per year | 48,000/2,400 | 20 |
| Annual ordering cost | (48,000/2,400) × 4,800 | 96,000 |
| Annual holding cost | (2,400/2) × 80 | 96,000 |
| Relevant annual inventory cost | 96,000 + 96,000 | 192,000 |
| Annual purchase cost | 48,000 × 300 | 14,400,000 |
| Total annual cost including purchases | 14,400,000 + 192,000 | 14,592,000 |
EOQ is 2,400 units. Ordering plus holding cost is Ksh 192,000 per year; including the unchanged purchase cost gives Ksh 14,592,000 per year.
Question 5(a) — 12 marks
The following are the Revenue (R) and Cost (C) functions of Turbo manufacturers in millions of shillings:
Function Expression Revenue (R) function Q² + 4Q Cost (C) function 1.5Q² + 2Q Where Q is the number of units produced and sold. Calculate the:
(i) number of units to be produced in order to maximize profits;
(ii) price that maximizes profits;
(iii) maximum profit.
Model answer
| Step | Working | Result |
|---|---|---|
| Profit function | π = R − C = Q² + 4Q − (1.5Q² + 2Q) | π = −0.5Q² + 2Q |
| Stationary point | dπ/dQ = −Q + 2 = 0 | Q = 2 units |
| Maximum check | d²π/dQ² = −1 < 0 | Maximum profit |
| Price (average revenue) | P = R/Q = Q + 4; P(2) = 2 + 4 | 6 million shillings per unit |
| Revenue at Q = 2 | 2² + 4 × 2 | 12 million shillings |
| Cost at Q = 2 | 1.5 × 2² + 2 × 2 | 10 million shillings |
| Maximum profit | 12 − 10 | 2 million shillings |
The functions are used exactly as printed; revenue is not assumed to have a negative quadratic term.
Question 5(b) — 8 marks
Outline four areas where quantitative techniques are applicable in a business organization.
Model answer
- Production planning: use linear programming to allocate labour, machine time and materials among products subject to capacity limits.
- Inventory management: calculate order quantities, reorder points and safety stocks to balance availability against ordering and holding costs.
- Marketing and sales forecasting: analyse sales trends and associations with price or advertising to plan demand and budgets.
- Project and logistics planning: use network analysis, scheduling and transport models to allocate resources and identify completion times or lower-cost distribution plans.
Question 6(a) — 8 marks
Explain four limitations of a linear programming model.
Model answer
- Linearity: proportionality and additivity may not fit quantity discounts, economies of scale, learning effects or interactions between activities.
- Certainty: ordinary linear programming treats objective and constraint coefficients as known constants, although demand, prices and processing times may change.
- Divisibility: a continuous model may recommend fractions of workers, vehicles or other indivisible items; integer programming is needed where whole units are essential.
- Model scope: a single numerical objective and measurable constraints can omit quality, employee relations, risk or other qualitative and competing goals, so managerial judgement remains necessary.
Question 6(b) — 12 marks
Mwaka has deposited Ksh 150,000 in a fixed account for a period of 4 years. The rate of interest is 6% per annum for the first 3 years and 8% per annum for the fourth year compounded annually. Thara intends to deposit a certain amount in an account earning simple interest at the rate of 7.5% per annum for a period of 4 years.
Calculate the:
(i) interest earned by Mwaka;
(ii) amount to be deposited by Thara in order to earn the same interest as Mwaka.
Model answer
| Step / item | Working | Result |
|---|---|---|
| Mwaka: balance after year 3 | 150,000(1.06)³ | 178,652.40 |
| Balance after year 4 | 150,000(1.06)³(1.08) | 192,944.59 |
| Mwaka: total interest | 192,944.592 − 150,000 | 42,944.59 |
| Thara: simple-interest equation | I = Prt = P × 0.075 × 4; solve P = I/0.30 | 143,148.64 |
Mwaka earns Ksh 42,944.59. Thara must deposit Ksh 143,148.64. Use unrounded intermediate amounts before rounding the final currency answers.
Question 7(a) — 8 marks
Emmax Limited manufactures and sells two products A and B. The probability that product A will be in the market in 5 years time is 0.8 while that of product B is 0.75.
Determine the probability that in 5 years time:
(i) both products will not be in the market;
(ii) only one product will be in the market;
(iii) at least one product will be in the market.
Model answer
Assume the two market-survival events are independent; the question gives marginal probabilities only. P(not A) = 0.2 and P(not B) = 0.25.
| Part | Working | Probability |
|---|---|---|
| Neither survives | 0.2 × 0.25 | 0.05 |
| Exactly one survives | 0.8 × 0.25 + 0.2 × 0.75 | 0.35 |
| At least one survives | 1 − 0.05 | 0.95 |
If independence is not assumed, the probabilities cannot be determined uniquely without joint-event information.
Question 7(b) — 12 marks
The mean lifetime of lighting bulbs manufactured by a company is 1200 hours with a standard deviation of 60 hours. The company has adopted a new technology in the production of the bulbs. The production manager claims that the mean lifetime of the bulbs will be prolonged. To test this claim, a random sample of 36 bulbs is selected from the new bulbs and found to have a mean lifetime of 1220 hours.
(i) Test the manager’s claim at 5% significance level. (8 marks)
(ii) Differentiate between one trailed test and two tailed test as used in hypothesis testing. (4 marks)
Model answer
(i) Right-tailed mean test. Treat the stated 60-hour standard deviation as the population value and assume independent random sampling.
| Step | Working / decision |
|---|---|
| Hypotheses | H₀: μ ≤ 1,200 hours (boundary μ = 1,200); H₁: μ > 1,200 hours. |
| Standard error | σ/√n = 60/√36 = 10 hours |
| Test statistic | z = (1,220 − 1,200)/10 = 2.00 |
| Critical region | Reject H₀ if z > 1.645 at α = 0.05. |
| Decision | 2.00 > 1.645; reject H₀. The one-sided p-value is about 0.0228. |
| Conclusion | At 5%, the sample supports the claim that the new population mean lifetime exceeds 1,200 hours. |
(ii) One-tailed versus two-tailed. A one-tailed test specifies a direction (greater than or less than), allocating all α to that tail. A two-tailed test asks whether a parameter differs in either direction, splitting α equally between both tails. The direction must be chosen from the research claim before inspecting the result. The printed “one trailed” refers to a one-tailed test.