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

July 2020 Q1(a): project network
July 2020 Q1(a): project network
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

  1. 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.
  2. 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.
  3. 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.
  4. 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

  1. Preserve precedence: an activity may start only after every required predecessor is completed; show all logical dependencies accurately.
  2. Use a clear overall start and finish, and connect every activity into a valid start-to-finish path; avoid dangling activities and disconnected sections.
  3. Do not introduce directed loops or circular dependencies. Time and event numbering should move consistently forward through the network.
  4. 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

  1. Production planning: use linear programming to allocate labour, machine time and materials among products subject to capacity limits.
  2. Inventory management: calculate order quantities, reorder points and safety stocks to balance availability against ordering and holding costs.
  3. Marketing and sales forecasting: analyse sales trends and associations with price or advertising to plan demand and budgets.
  4. 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

  1. Linearity: proportionality and additivity may not fit quantity discounts, economies of scale, learning effects or interactions between activities.
  2. Certainty: ordinary linear programming treats objective and constraint coefficients as known constants, although demand, prices and processing times may change.
  3. Divisibility: a continuous model may recommend fractions of workers, vehicles or other indivisible items; integer programming is needed where whole units are essential.
  4. 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.