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Compiled by: Samuel Maina
Cooperative managers in Kenya rely on index numbers to analyze changes in prices, quantities, or values over time. Index numbers help cooperatives monitor inflation, cost of inputs, and market trends critical for decision-making and financial planning. This chapter focuses on the mathematical computation of index numbers, providing formulas and step-by-step calculations for commonly used methods in cooperative management.
Index numbers summarize relative changes in economic variables such as prices or quantities between two periods, often called the base period and the current period. The key formulae for index numbers involve weighted averages where weights can be quantities or prices depending on the method. This section covers the main formulae used in cooperative contexts: Laspeyre’s, Paasche’s, Fisher’s ideal, and Marshall’s index numbers.
Laspeyre’s index number uses base period quantities as weights to measure price changes. It reflects how prices have changed if the quantities had remained constant at the base period level. The formula is:
$$ I_L = \frac{\sum (P_1 \times Q_0)}{\sum (P_0 \times Q_0)} \times 100 $$
where \(P_1\) = price in current period, \(P_0\) = price in base period, \(Q_0\) = quantity in base period.
Example 1: A cooperative sells maize. Base period price \(P_0\) = Ksh 50/kg, current price \(P_1\) = Ksh 60/kg, base period quantity \(Q_0\) = 1000 kg. Compute Laspeyre’s price index.
Given: \(P_0 = 50\), \(P_1 = 60\), \(Q_0 = 1000\)
$$ I_L = \frac{(60 \times 1000)}{(50 \times 1000)} \times 100 $$
$$ I_L = \frac{60000}{50000} \times 100 $$
$$ I_L = 1.2 \times 100 = 120 $$
Answer: 120 (Price increased by 20%)
Example 2: A dairy cooperative has three products with base prices \(P_0 = [40, 60, 80]\), current prices \(P_1 = [50, 70, 90]\), and base quantities \(Q_0 = [200, 150, 100]\). Calculate Laspeyre’s price index.
Given: \(P_0 = 40, 60, 80\); \(P_1 = 50, 70, 90\); \(Q_0 = 200, 150, 100\)
$$ I_L = \frac{(50 \times 200) + (70 \times 150) + (90 \times 100)}{(40 \times 200) + (60 \times 150) + (80 \times 100)} \times 100 $$
$$ I_L = \frac{10000 + 10500 + 9000}{8000 + 9000 + 8000} \times 100 $$
$$ I_L = \frac{29500}{25000} \times 100 $$
$$ I_L = 1.18 \times 100 = 118 $$
Answer: 118 (Prices increased by 18%)
Example 3: A cooperative sells two types of beans. Base prices \(P_0 = [120, 150]\), current prices \(P_1 = [130, 160]\), base quantities \(Q_0 = [300, 400]\). Find Laspeyre’s index.
Given: \(P_0 = 120, 150\); \(P_1 = 130, 160\); \(Q_0 = 300, 400\)
$$ I_L = \frac{(130 \times 300) + (160 \times 400)}{(120 \times 300) + (150 \times 400)} \times 100 $$
$$ I_L = \frac{39000 + 64000}{36000 + 60000} \times 100 $$
$$ I_L = \frac{103000}{96000} \times 100 $$
$$ I_L = 1.0729 \times 100 = 107.29 $$
Answer: 107.29 (Prices increased by 7.29%)
Paasche’s index number uses current period quantities as weights, reflecting price changes considering the current consumption pattern. The formula is:
$$ I_P = \frac{\sum (P_1 \times Q_1)}{\sum (P_0 \times Q_1)} \times 100 $$
where \(Q_1\) = quantity in current period.
Example 1: A cooperative’s base prices \(P_0 = 50\), current prices \(P_1 = 60\), current quantities \(Q_1 = 1100\). Compute Paasche’s price index.
Given: \(P_0 = 50\), \(P_1 = 60\), \(Q_1 = 1100\)
$$ I_P = \frac{60 \times 1100}{50 \times 1100} \times 100 $$
$$ I_P = \frac{66000}{55000} \times 100 $$
$$ I_P = 1.2 \times 100 = 120 $$
Answer: 120 (Price increased by 20%)
Example 2: A cooperative sells three products: \(P_0 = [40, 60, 80]\), \(P_1 = [50, 70, 90]\), \(Q_1 = [210, 160, 90]\). Calculate Paasche’s price index.
Given: \(P_0 = 40, 60, 80\); \(P_1 = 50, 70, 90\); \(Q_1 = 210, 160, 90\)
$$ I_P = \frac{(50 \times 210) + (70 \times 160) + (90 \times 90)}{(40 \times 210) + (60 \times 160) + (80 \times 90)} \times 100 $$
$$ I_P = \frac{10500 + 11200 + 8100}{8400 + 9600 + 7200} \times 100 $$
$$ I_P = \frac{29800}{25200} \times 100 $$
$$ I_P = 1.1825 \times 100 = 118.25 $$
Answer: 118.25 (Prices increased by 18.25%)
Example 3: For two bean varieties, \(P_0 = [120, 150]\), \(P_1 = [130, 160]\), \(Q_1 = [310, 390]\). Find Paasche’s index.
Given: \(P_0 = 120, 150\); \(P_1 = 130, 160\); \(Q_1 = 310, 390\)
$$ I_P = \frac{(130 \times 310) + (160 \times 390)}{(120 \times 310) + (150 \times 390)} \times 100 $$
$$ I_P = \frac{40300 + 62400}{37200 + 58500} \times 100 $$
$$ I_P = \frac{102700}{95700} \times 100 $$
$$ I_P = 1.0733 \times 100 = 107.33 $$
Answer: 107.33 (Prices increased by 7.33%)
Fisher’s ideal index is the geometric mean of Laspeyre’s and Paasche’s indices, providing a balanced measure that mitigates bias. The formula is:
$$ I_F = \sqrt{I_L \times I_P} $$
where \(I_L\) = Laspeyre’s index, \(I_P\) = Paasche’s index.
Example 1: Given \(I_L = 120\) and \(I_P = 120\), compute Fisher’s ideal index.
Given: \(I_L = 120\), \(I_P = 120\)
$$ I_F = \sqrt{120 \times 120} $$
$$ I_F = \sqrt{14400} $$
$$ I_F = 120 $$
Answer: 120
Example 2: For a dairy cooperative, \(I_L = 118\), \(I_P = 118.25\). Calculate Fisher’s index.
Given: \(I_L = 118\), \(I_P = 118.25\)
$$ I_F = \sqrt{118 \times 118.25} $$
$$ I_F = \sqrt{13948.5} $$
$$ I_F = 118.11 $$
Answer: 118.11
Example 3: Given \(I_L = 107.29\), \(I_P = 107.33\) for bean prices, find Fisher’s index.
Given: \(I_L = 107.29\), \(I_P = 107.33\)
$$ I_F = \sqrt{107.29 \times 107.33} $$
$$ I_F = \sqrt{11511.68} $$
$$ I_F = 107.31 $$
Answer: 107.31
Marshall’s index number is the arithmetic mean of Laspeyre’s and Paasche’s indices, offering a simple average measure. The formula is:
$$ I_M = \frac{I_L + I_P}{2} $$
Example 1: Given \(I_L = 120\), \(I_P = 120\), compute Marshall’s index.
Given: \(I_L = 120\), \(I_P = 120\)
$$ I_M = \frac{120 + 120}{2} $$
$$ I_M = \frac{240}{2} = 120 $$
Answer: 120
Example 2: For a dairy cooperative, \(I_L = 118\), \(I_P = 118.25\). Calculate Marshall’s index.
Given: \(I_L = 118\), \(I_P = 118.25\)
$$ I_M = \frac{118 + 118.25}{2} $$
$$ I_M = \frac{236.25}{2} = 118.125 $$
Answer: 118.125
Example 3: Given \(I_L = 107.29\), \(I_P = 107.33\) for bean prices, find Marshall’s index.
Given: \(I_L = 107.29\), \(I_P = 107.33\)
$$ I_M = \frac{107.29 + 107.33}{2} $$
$$ I_M = \frac{214.62}{2} = 107.31 $$
Answer: 107.31
A cooperative’s base prices and quantities for three products are \(P_0 = [45, 70, 90]\) Ksh, \(Q_0 = [300, 200, 100]\) units. Current prices \(P_1 = [50, 75, 95]\) Ksh. Compute Laspeyre’s price index. (5 marks)
Using the data above, and current quantities \(Q_1 = [320, 180, 110]\), calculate Paasche’s price index. (5 marks)
Calculate Fisher’s ideal index using your answers from questions 1 and 2. (5 marks)
Calculate Marshall’s index using your answers from questions 1 and 2. (5 marks)
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Create a free accountThis chapter covered the essential formulae used for computing index numbers, providing the mathematical foundation necessary for their calculation. It detailed the computation methods of various index numbers, including Laspeyre's, Paasche's, Fisher's ideal, and Marshal's indexes, explaining how each formula accounts for price and quantity changes over time. Laspeyre's index uses base period quantities, while Paasche's index employs current period quantities, and Fisher's ideal index represents a geometric mean of the two. Marshal's index was also introduced as an alternative approach. The chapter concluded by exploring how index numbers assist in business decision making by measuring inflation, cost changes, and economic trends, enabling managers to make informed financial and operational choices. Understanding these concepts equips students with tools to analyze economic data effectively in practical business contexts.
A cooperative sold 100 bags of maize at Ksh 500 each in the base year. In the current year, the price per bag increased to Ksh 550. Calculate the Laspeyre’s price index number. (2 marks)
The quantity of sugar purchased by a cooperative in the base year was 200 kg at Ksh 80 per kg. In the current year, the cooperative bought 180 kg at Ksh 90 per kg. Calculate the Paasche’s price index number. (3 marks)
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