Quantitative Techniques — 2021

KNEC Diploma in Supply Chain Management, Module II. Original theory examination questions with explained model answers for revision.


July 2021

Paper codes: 2903/204
Time: Not shown in supplied scan
Instructions: Not shown in supplied scan

Question 1(a) — 12 marks

The following are the revenue (R) function and cost (C) function of PT suppliers.

R=36Q+Q²
C=5Q²−4Q−90

Where Q is the number of units produced and sold. Calculate the: (i) quantity to be sold in order to maximize profit; (ii) price at which profit will be maximized; (iii) maximum profit.

Model answer

Step Working
Profit function π=R−C=40Q−4Q²+90
Maximum condition dπ/dQ=40−8Q=0 ⇒ Q=5 units
Second derivative d²π/dQ²=−8<0
Price R/Q=36+Q=41 monetary units
Maximum profit π(5)=200−100+90=190 monetary units

The supplied negative constant in the cost function is retained as printed.

Question 1(b) — 8 marks

Wema Enterprises buys and sells product W. Its annual demand is 5,000 units. The ordering cost is Ksh 25,000 per order. The cost price per unit is Ksh 400. Annual carrying cost per unit is 10% of the cost price. Calculate the: (i) EOQ; (ii) total relevant costs.

Model answer

Step / item Working Result
Annual holding cost per unit 10% × 400 40
EOQ √(2×5,000×25,000/40) 2,500
Annual ordering cost (5,000/2,500) × 25,000 50,000
Annual holding cost (2,500/2) × 40 50,000
Total relevant annual cost 50,000 + 50,000 100,000

The relevant EOQ cost is ordering plus holding cost; constant annual purchase expenditure is excluded from this decision cost.

Question 2(a) — 10 marks

The following information relates to material MM used by Pendo Manufacturers for the month of May 2016.

Item Value
Maximum consumption 2,100 units per day
Minimum consumption 9,100 units per day
Maximum re-order period 4 days
Minimum re-order period 2 days
Re-order quantity 12,600 units

Calculate the: (i) re-order level; (ii) maximum stock level; (iii) minimum stock level; (iv) average stock level.

Model answer

Source inconsistency: the printed minimum usage of 9,100 exceeds the maximum of 2,100. Normal usage and normal lead time are also not supplied. Consequently there is no valid, unique set of stock levels from these data.

Required level Formula / substitution
Reorder Maximum usage × maximum lead time = 2,100×4 = 8,400 units, formally using the printed maximum.
Maximum stock Reorder + order quantity − minimum usage×minimum lead time = 8,400+12,600−9,100×2 = 2,800 units; this is a formal substitution into inconsistent inputs, not a valid stock policy.
Minimum stock 8,400 − normal daily usage × normal lead time; the normal values are missing.
Average stock Minimum stock + 12,600/2 = minimum stock + 6,300.

Even assuming midpoint values gives normal usage (2,100+9,100)/2=5,600 and lead time 3 days, implying minimum stock 8,400−16,800=−8,400, which confirms the data problem. A corrected minimum usage and a stated normal usage/lead-time convention are needed. No replacement figure is invented.

Question 2(b) — 10 marks

The following table shows the prices and quantities of three commodities for the years 2012 and 2016.

Commodity 2010 price Ksh 2010 quantity kg 2015 price Ksh 2015 quantity kg
Maize 40 315 45 360
Beans 60 200 60 220
Potatoes 36 400 46 390

Calculate the: (i) Laspeyre’s quantity index; (ii) Paasche’s quantity index; (iii) Fisher’s ideal quantity index.

Model answer

The prose says 2012/2016 while the supplied column headings say 2010/2015. The first pair of columns is treated as the base period and the second as the current period; the numerical calculation is unaffected by this label conflict.

Commodity p₀q₀ p₀q₁ p₁q₀ p₁q₁
Maize 12,600 14,400 14,175 16,200
Beans 12,000 13,200 12,000 13,200
Potatoes 14,400 14,040 18,400 17,940
Total 39,000 41,640 44,575 47,340
Step / item Working Result
Laspeyres quantity index 41,640/39,000 × 100 106.77
Paasche quantity index 47,340/44,575 × 100 106.20
Fisher ideal quantity index √(106.769231 × 106.203029) 106.49

Question 3(a) — 10 marks

Fiki Enterprises manufactures product Dx. The following is the quarterly production in units for a period of three years.

Year Q1 units Q2 units Q3 units Q4 units
2014 320 300 380 480
2015 500 450 480 550
2016 580 570 600 630

Calculate: (i) 3-period moving averages; (ii) seasonal variations using additive method.

Model answer

Take the mean of each consecutive three-quarter window and centre it on the middle quarter. Additive residual = actual output − moving-average trend.

Centred quarter Actual output 3-quarter total 3-period MA Actual − MA
2014 Q2 300 1,000 333.33 -33.33
2014 Q3 380 1,160 386.67 -6.67
2014 Q4 480 1,360 453.33 26.67
2015 Q1 500 1,430 476.67 23.33
2015 Q2 450 1,430 476.67 -26.67
2015 Q3 480 1,480 493.33 -13.33
2015 Q4 550 1,610 536.67 13.33
2016 Q1 580 1,700 566.67 13.33
2016 Q2 570 1,750 583.33 -13.33
2016 Q3 600 1,800 600 0
Quarter Sum of residuals Number Mean residual Adjusted seasonal variation
Q1 36.67 2 18.33 16.53
Q2 -73.33 3 -24.44 -26.25
Q3 -20 3 -6.67 -8.47
Q4 40.00 2 20.00 18.19

The four mean residuals sum to 7.222222; subtract their average, 1.805556, from each to make the additive seasonal effects sum to zero. The first and last quarters have no centred three-period average. A four-quarter seasonal cycle is usually smoothed using centred four-quarter averages, but the calculation above follows the question’s explicit three-period requirement.

Question 3(b) — 10 marks

The following information relates to a project to be undertaken by a firm.

Activity Preceding activities Duration days
A — 2
B — 4
C A 5
D A 7
E B,D 1
F E 3
G C,F 4
H G 6

(i) Draw a network diagram of the project.

(ii) Determine the: (I) critical path; (II) project duration.

Model answer

2021 July Q3(b): project network — days
2021 July Q3(b): project network — days
Activity Predecessors Duration ES EF LS LF Float
A — 2 0 2 0 2 0
B — 4 0 4 5 9 5
C A 5 2 7 8 13 6
D A 7 2 9 2 9 0
E B,D 1 9 10 9 10 0
F E 3 10 13 10 13 0
G C,F 4 13 17 13 17 0
H G 6 17 23 17 23 0

Forward pass: ES = latest predecessor EF; EF = ES + duration. Backward pass: LF = earliest successor LS; LS = LF − duration. Float = LS − ES. Project duration = 23 days. Red identifies critical activities and their controlling links.

Critical path A → D → E → F → G → H takes 2+7+1+3+4+6=23 days.

Question 4(a) — 12 marks

The following information shows the advertising costs and corresponding sales of Moja Enterprises for eight consecutive periods.

Advertising costs x Ksh 000 Sales y Ksh 000
32 250
40 300
42 400
46 420
48 420
50 450
56 500
60 580

(i) Determine the linear regression equation y on x.

(ii) Using the regression equation in (i) above, estimate the sales when advertising costs are Ksh 52,000.

Model answer

x y x² xy
32 250 1,024 8,000
40 300 1,600 12,000
42 400 1,764 16,800
46 420 2,116 19,320
48 420 2,304 20,160
50 450 2,500 22,500
56 500 3,136 28,000
60 580 3,600 34,800
374 3,320 18,044 161,580

b = [nΣxy−ΣxΣy]/[nΣx²−(Σx)²] = (8×161,580.00−374.00×3,320.00)/(8×18,044.00−374.00²) = 11.385165.
a = ȳ−bx̄ = -117.256479.

Regression equation: ŷ=-117.256479+11.385165x, with both variables in Ksh thousands. At x=52, ŷ=474.772118 thousand, so estimated sales = Ksh 474,772.12.

Question 4(b) — 8 marks

Explain the meaning of each of the following terms as used in network analysis: (i) activity; (ii) project; (iii) event; (iv) dummy activity.

Model answer

  1. An activity is a defined task that takes time and may consume resources; it is represented by an arrow or node depending on the network convention.
  2. A project is a set of interrelated activities undertaken to achieve a specified objective within defined time and resource limits.
  3. An event is a milestone marking the start or completion of one or more activities; it consumes no time or resources itself.
  4. A dummy activity is a zero-duration, zero-resource arrow used in an activity-on-arrow network to preserve precedence relationships or distinguish activities.

Question 5(a) — 10 marks

Explain five roles of quantitative techniques in business decision making.

Model answer

  1. Forecasting demand and sales from data helps determine future production, purchasing and staffing requirements.
  2. Optimisation allocates scarce labour, materials and machine time to the activities that best meet a stated objective.
  3. Inventory models balance ordering, holding and shortage costs to guide replenishment decisions.
  4. Project scheduling identifies activity dependencies, completion times and critical tasks so managers can control delivery.
  5. Probability and statistical analysis measure uncertainty, compare alternatives and test claims, supporting more consistent risk-informed decisions.

Question 5(b) — 10 marks

Simon supplied 12 calculators and 30 geometric sets to United College valued at a total of Ksh 21,600. He also supplied 18 calculators and 28 geometric sets to New Life College valued at a total of Ksh 22,200. The items are sold on the same terms to both colleges.

(i) Formulate simultaneous equations from the information above.

(ii) Solve the equations in (i) above using matrices to determine the price of: (I) one calculator; (II) one geometric set.

Model answer

Let x be calculator price and y be geometric-set price. 12x+30y=21,600; 18x+28y=22,200. A[x,y]ᵀ=b with A=

12 30
18 28

det A=12×28−30×18=−204. A⁻¹=(−1/204)

28 -30
-18 12

Multiply A⁻¹ by [21,600,22,200]ᵀ:

Step / item Working Result
Calculator x (28×21,600−30×22,200)/(−204) 300
Geometric set y (−18×21,600+12×22,200)/(−204) 600

Check: 12×300+30×600=21,600; 18×300+28×600=22,200.

Question 6(a) — 10 marks

Alpha Manufacturers produces two products P and Q whose contribution per unit is Ksh 60 and Ksh 40 respectively. The resource requirements for each product are as follows:

Resource P Q Total available
Skilled labour hours 1 2 6,000
Unskilled labour hours 2 3 9,000
Material kg 4 1 2,000

(i) Formulate a linear programming problem.

(ii) Using the graphical method, determine the optimal product mix.

Model answer

Let x and y be units of P and Q. Maximise Z=60x+40y subject to x+2y≤6,000, 2x+3y≤9,000, 4x+y≤2,000, x,y≥0.

2021 July Q6(a): maximise contribution (x=P, y=Q); shaded area is feasible
2021 July Q6(a): maximise contribution (x=P, y=Q); shaded area is feasible
Feasible corner x y Objective cost
0 0 0
-0 2,000 80,000
500 0 30,000

Optimal mix: P=0, Q=2,000; maximum contribution Ksh 80,000. This uses all 2,000 kg of material, 4,000 skilled hours and 6,000 unskilled hours. Material is the binding resource; skilled and unskilled capacity remain available.

Question 6(b) — 10 marks

Kajum invested a total of Ksh 1,200,000 in two investment funds A and B. The returns from each fund was 4% and 3% respectively. The total returns from the funds was Ksh 40,000. Calculate the amount of money invested in: (i) fund A; (ii) fund B.

Model answer

Step Working
Equations A+B=1,200,000; 0.04A+0.03B=40,000
Eliminate B Subtract 0.03(A+B)=36,000: 0.01A=4,000
Fund A A=Ksh 400,000
Fund B B=1,200,000−400,000=Ksh 800,000
Check 0.04×400,000+0.03×800,000=16,000+24,000=40,000

Question 7(a) — 8 marks

Explain the meaning of each of the following terms as used in probability theory: (i) elementary event; (ii) mutually exclusive events; (iii) equiprobable events; (iv) complementary events.

Model answer

  1. An elementary event contains exactly one outcome in the sample space, such as rolling a 3 on one die.
  2. Mutually exclusive events cannot occur together in the same trial: their intersection is empty and P(A∩B)=0.
  3. Equiprobable events have equal probabilities, such as the individual faces of a fair die.
  4. Complementary events are an event A and its non-occurrence Aᶜ; together they exhaust the sample space and P(Aᶜ)=1−P(A).

Question 7(b) — 12 marks

A supervisor in a supermarket claims that 20% of the shoppers leave without making a purchase. In a random sample of 80 shoppers selected, it was found that 25 shoppers left without making a purchase. Test the supervisor’s claim at 5% significance level.

Model answer

Step Working / conclusion
Hypotheses H₀:p=0.20; H₁:p≠0.20 (two-sided).
Sample proportion p̂=25/80=0.3125
Normal approximation check np₀=16 and n(1−p₀)=64, both sufficiently large. Assume an independent random sample.
Standard error under H₀ √(0.20×0.80/80)=0.04472136
Test statistic z=(0.3125−0.20)/0.04472136=2.515576
Critical region Reject H₀ if z1.96.
Decision 2.5156>1.96, so reject H₀.
Conclusion The sample provides evidence that the true non-purchase proportion differs from 20%; the observed direction is higher. The two-sided p-value is approximately 0.0119.

November 2021

Paper codes: 2903/204
Time: Not shown in supplied scan
Instructions: Not shown in supplied scan

Question 1(a) — 8 marks

The probability that a train arrives on time at terminus A is 5/6 while the probability that it arrives on time at terminus B is 3/4. Determine the probability that the train arrives: (i) on time at both terminus A and B; (ii) on time in one of the terminals; (iii) late at both terminals.

Model answer

Assume arrival-time events at the two terminals are independent. This is required for multiplying the marginal probabilities; it is not explicitly stated in the question.

Part Working Probability
Both on time (5/6)(3/4) 15/24 = 5/8 = 0.625
Exactly one on time (5/6)(1/4)+(1/6)(3/4) 8/24 = 1/3 ≈ 0.333333
Both late (1/6)(1/4) 1/24 ≈ 0.041667

“One of the terminals” is interpreted as exactly one. If it means at least one, the answer is 1−1/24=23/24. Without independence or a joint probability, the three events cannot be determined uniquely from the marginals alone.

Question 1(b) — 12 marks

Mwamba Limited intends to invest in a project. The following are the activities, preceding activities and durations of the project.

Activity Preceding activity Duration weeks
A — 7
B — 10
C A 4
D A 30
E A 7
F B,C 12
G B,C 15
H E,F 11
I E,F 25
J E,F 6
K D,H 19
L G,J 25
M K,I,L 2

(i) Draw a network diagram of the project.

(ii) Determine: I. critical path; II. project duration.

Model answer

2021 November Q1(b): project network
2021 November Q1(b): project network
Activity Predecessors Duration ES EF LS LF Float
A — 7 0 7 0 7 0
B — 10 0 10 3 13 3
C A 4 7 11 9 13 2
D A 30 7 37 7 37 0
E A 7 7 14 18 25 11
F B,C 12 11 23 13 25 2
G B,C 15 11 26 16 31 5
H E,F 11 23 34 26 37 3
I E,F 25 23 48 31 56 8
J E,F 6 23 29 25 31 2
K D,H 19 37 56 37 56 0
L G,J 25 29 54 31 56 2
M K,I,L 2 56 58 56 58 0

Forward pass: ES = latest predecessor EF; EF = ES + duration. Backward pass: LF = earliest successor LS; LS = LF − duration. Float = LS − ES. Project duration = 58 weeks. Red identifies critical activities and their controlling links.

Critical path A → D → K → M, taking 7+30+19+2=58 weeks.

Question 2(a) — 8 marks

Explain the meaning of each of the following components of a time series: (i) trend; (ii) seasonal variation; (iii) cyclical variation; (iv) random variation.

Model answer

  1. Trend is the persistent long-term upward, downward or stable movement in a series after shorter fluctuations are smoothed.
  2. Seasonal variation is a pattern that repeats within a year, such as quarterly or monthly sales changes associated with weather or holidays.
  3. Cyclical variation is a longer-term rise and fall around the trend, often associated with business cycles and without a fixed within-year period.
  4. Random variation is irregular, unpredictable movement caused by unusual events or chance influences after systematic components are accounted for.

Question 2(b) — 12 marks

The following information shows the advertising costs and the respective sales revenue of Mika Enterprises for six consecutive periods.

Advertising costs x Ksh Sales revenue y Ksh
2,100 13,600
2,800 15,800
2,200 14,500
3,000 16,200
2,600 14,900
2,500 15,000

(i) Calculate the: I. Pearson’s coefficient of correlation; II. coefficient of determination.

(ii) Interpret the results obtained in (II) above.

Model answer

x y x² y² xy
2,100 13,600 4,410,000 184,960,000 28,560,000
2,800 15,800 7,840,000 249,640,000 44,240,000
2,200 14,500 4,840,000 210,250,000 31,900,000
3,000 16,200 9,000,000 262,440,000 48,600,000
2,600 14,900 6,760,000 222,010,000 38,740,000
2,500 15,000 6,250,000 225,000,000 37,500,000
15,200 90,000 39,100,000 1,354,300,000 229,540,000

r = [nΣxy − ΣxΣy]/√{[nΣx² − (Σx)²][nΣy² − (Σy)²]} = 9,240,000.0000/9,583,736.2234 = 0.964133.

Coefficient of determination R²=r²=0.929553, or 92.96%. In the fitted simple linear relationship, about 92.96% of the sample variation in sales revenue is associated with variation in advertising expenditure. This is explanatory association within the model, not proof that advertising alone causes that share of sales.

Question 3(a) — 12 marks

Mjengo Limited manufactures two products, P and Q using machines A, B and C. Product P requires 2 minutes of machine A, 3 minutes of machine B and 1 minute of machine C. Product Q requires 3 minutes of machine A, 2 minutes of machine B and 1 minute of machine C.

The capacity available per week for each machine is 1,500, 1,500 and 600 for A, B and C respectively. The profit per unit of products P and Q is Ksh 10 and Ksh 12 respectively.

(i) Formulate a linear programming model from the information above.

(ii) Using graphical method, determine the weekly production that maximizes profit.

Model answer

Let x=P units and y=Q units. Maximise Z=10x+12y subject to 2x+3y≤1,500, 3x+2y≤1,500, x+y≤600, x,y≥0. Capacities are interpreted in minutes, consistently with the unit processing times.

2021 November Q3(a): maximise weekly profit (x=P, y=Q); shaded area is feasible
2021 November Q3(a): maximise weekly profit (x=P, y=Q); shaded area is feasible
Feasible corner x y Objective cost
0 0 0
-0 500 6,000
300 300 6,600
500 0 5,000

Solve 2x+3y=1,500 and 3x+2y=1,500: x=y=300; x+y=600. Produce 300 P and 300 Q per week for maximum profit Ksh 6,600.

Question 3(b) — 8 marks

Explain four roles of quantitative techniques in business decision making.

Model answer

  1. Forecasting demand and sales from data helps determine future production, purchasing and staffing requirements.
  2. Optimisation allocates scarce labour, materials and machine time to the activities that best meet a stated objective.
  3. Inventory models balance ordering, holding and shortage costs to guide replenishment decisions.
  4. Project scheduling identifies activity dependencies, completion times and critical tasks so managers can control delivery.

Question 4(a) — 8 marks

Moses bought 2 vests and 3 shirts from a shop and paid a total of Ksh 3,100. Mohammed bought 3 similar vests and one similar shirt from the same shop and paid a total of Ksh 2,200. Using matrices, determine the price of: (i) one vest; (ii) one shirt.

Model answer

Let x=vest price and y=shirt price. Equations: 2x+3y=3,100 and 3x+y=2,200. A=

2 3
3 1

det A=2−9=−7; A⁻¹=(−1/7)

1 -3
-3 2

Multiply by b=[3,100,2,200]ᵀ:

Step / item Working Result
Vest x (3,100−3×2,200)/(−7) 500
Shirt y (−3×3,100+2×2,200)/(−7) 700

Check: 2×500+3×700=3,100 and 3×500+700=2,200.

Question 4(b) — 12 marks

Distinguish between each of the following terms as used in probability theory: (i) dependent events and independent events; (ii) permutations and combinations; (iii) compound events and mutually exclusive events.

Model answer

Pair First term Second term
Dependent / independent Dependent: occurrence of one event changes the probability of the other; P(A∩B)=P(A)P(B|A). Independent: occurrence of one does not change the probability of the other; P(A∩B)=P(A)P(B).
Permutations / combinations Permutations count ordered selections: nPᵣ=n!/(n−r)!. Combinations count unordered selections: nCᵣ=n!/[r!(n−r)!].
Compound / mutually exclusive A compound event contains more than one elementary outcome, e.g. rolling an even number on a die. Mutually exclusive describes two or more events that cannot occur together, e.g. an odd and an even result in one roll. This describes a relationship between events, not their number of outcomes.

Question 5(a) — 10 marks

Panda Limited produces and sells product kx. The average price (P) and total cost (TC) functions are as follows:

P=200−8Q
TC=Q²−16Q

Where Q is the number of units produced and sold. Determine the: (i) total revenue (TR) function; (ii) number of units that maximizes profit; (iii) number of units that maximizes revenue; (iv) maximum profit.

Model answer

Part Working / result
TR function TR=PQ=200Q−8Q²
Profit function π=TR−TC=216Q−9Q²
Profit-maximising quantity dπ/dQ=216−18Q=0 ⇒ Q=12; d²π/dQ²=−18<0
Revenue-maximising quantity dTR/dQ=200−16Q=0 ⇒ Q=12.5; second derivative −16<0
Maximum profit π(12)=216×12−9×144=1,296 monetary units

If whole units are mandatory, revenue is the same at Q=12 and Q=13 (1,248 monetary units); the continuous maximum at 12.5 is 1,250. The question does not impose integrality.

Question 5(b) — 10 marks

Matemu Limited used material V20 in its production process. The annual demand is 2,000 units. The following details relate to the material.

Particulars Ksh
Cost per unit Ksh 5
Ordering cost per order Ksh 500
Annual carrying cost per unit Ksh 200
Working days per year 300

Determine the: (i) EOQ; (ii) number of orders to be placed in one year; (iii) time interval between orders; (iv) total inventory cost.

Model answer

Step / item Working Result
EOQ √(2×2,000×500/200) 100
Orders per year 2,000/100 20
Time between orders 300/20 working days 15
Annual ordering cost 20×500 10,000
Annual holding cost 100/2×200 10,000
Relevant ordering + holding cost 10,000+10,000 20,000
Annual purchase cost 2,000×5 10,000
Total inventory cost including purchase 20,000+10,000 30,000

The unusually high holding cost is used exactly as supplied. Basic EOQ assumptions include steady demand, no shortages and no safety stock.

Question 6(a) — 10 marks

Tom deposited Ksh 150,000 in an account paying compound interest at the rate of 10% per annum for four years.

(i) Calculate the: I. amount in his account at the end of the fourth year; II. interest earned at the end of the fourth year.

(ii) Peter wanted to have the same amount as in (I) above by depositing a certain sum of money in an account for a period of 3 years at a compound interest rate of 10% per annum. Calculate the amount to be deposited at the beginning of the three year period.

Model answer

Step / item Working Result
Tom’s future amount 150,000(1.1)⁴ 219,615.00
Tom’s interest 219,615−150,000 69,615.00
Peter’s required present deposit 219,615/(1.1)³ 165,000

Check: 165,000(1.1)³=219,615.

Question 6(b) — 10 marks

The manager of Pona Pharmaceuticals Limited claims that the average content of medicine in a bottle is 0.5 litres. A random sample of 100 bottles is selected and the mean content is found to be 0.45 litres with a standard deviation of 0.10 litres. Test the manager’s claim at 5% level of significance.

Model answer

Step Working / decision
Hypotheses H₀:μ=0.50 litre; H₁:μ≠0.50 litre.
Method Large-sample two-sided mean test using s to estimate σ; assume independent random sampling.
Standard error 0.10/√100=0.01 litre
Statistic z=(0.45−0.50)/0.01=−5.00
Critical values ±1.96 at α=0.05
Decision |−5|>1.96: reject H₀.
Conclusion The data provide strong evidence that the population mean content differs from 0.5 litre; the sample indicates a lower mean. A one-sample t-test with df=99 also rejects at 5% (critical value about ±1.984).

Question 7(a) — 8 marks

Explain the meaning of each of the following terms as used in network analysis: (i) dangling activity; (ii) loop; (iii) slack; (iv) dummy activity.

Model answer

  1. A dangling activity is an activity or path left improperly disconnected from the project’s intended start/finish flow, creating an incomplete logical network.
  2. A loop is a directed cycle that returns to an earlier event/activity, creating circular precedence; it is not permitted in a conventional project network.
  3. Event slack is latest permissible event time minus earliest event time; it measures how long that event may be delayed without delaying project completion.
  4. A dummy activity is a zero-duration, zero-resource arrow used in an activity-on-arrow network to preserve precedence logic or uniquely identify activities.

Question 7(b) — 12 marks

The following table shows the prices and quantities of three commodities for the years 2010 and 2016.

Commodity 2010 price p₀ Ksh 2010 quantity q₀ 2016 price p₁ Ksh 2016 quantity q₁
Bread 65 20 135 30
Sugar 95 8 160 7
Rice 150 5 320 8

(i) Calculate: I. Laspeyre’s price index; II. Paasche’s price index; III. Fisher’s ideal price index.

(ii) Interpret the result in (III) above.

Model answer

Commodity p₀q₀ p₁q₀ p₀q₁ p₁q₁
Bread 1,300 2,700 1,950 4,050
Sugar 760 1,280 665 1,120
Rice 750 1,600 1,200 2,560
Total 2,810 5,580 3,815 7,730
Step / item Working Result
Laspeyres price index 5,580/2,810 × 100 198.58
Paasche price index 7,730/3,815 × 100 202.62
Fisher ideal price index √(198.576512 × 202.621232) 200.59

With 2010=100, Fisher’s ideal price index is 200.59, indicating an aggregate weighted price rise of approximately 100.59% by 2016.