13.6.1 Time Reversal Test

An index number method is said to satisfy time reversal test, if

$P_{01}\times P_{10}=1$

where P₀₁ and P₁₀ are the index numbers for two periods with base period and current period reverser.

In the words of A.M. Tuttle “If relative for a single series of any type (Price, quantity, etc.) are computed for the two periods ‘o’ and ‘n’, with period ‘o’ as a base, then recomputed for the two periods with period ‘n’ as a base, the two sets of relatives will always be proportional.”

Thus an ideal index number should work both ways, i.e., forward and backward. Fisher’s formula satisfies, the time reversal test:

According Fisher’s formula

$P_{01}=\sqrt{\frac{\mathrm{\Sigma }{p}_1{q}_0}{\mathrm{\Sigma }{p}_0{q}_0}·\frac{\mathrm{\Sigma }{p}_1{q}_1}{\mathrm{\Sigma }{p}_0{q}_1}}$

And where time is reversed, we get

$P_{10}=\sqrt{\frac{\mathrm{\Sigma }{p}_0{q}_1}{\mathrm{\Sigma }{p}_1{q}_0}.\frac{\mathrm{\Sigma }{p}_0{q}_1}{\mathrm{\Sigma }{p}_1{q}_1}}$

where P₀₁×P₁₀=1

where P₀₁=Price index for current year on the base year; P₁₀=Price index for base year on the basis of current year.

The following methods of constructing index numbers also satisfy this test:

Example 13.16: Given the sum of the products of prices and quantities for the current year 1 and base year 0 for five items as:

$\Sigma p_0{q}_0=782,\Sigma p_0{q}_1=1008,\Sigma p_1{q}_0=1084,\mathrm{and}\Sigma p_1{q}_1=1329$

On the basis of the given information show that the data satisfies time reversal test.

Solution:

$\begin{array}{ll}{P}_{01}\hfill & =\sqrt{\frac{\mathrm{\Sigma }{p}_1{q}_0}{\mathrm{\Sigma }{p}_0{q}_0}.\frac{\mathrm{\Sigma }{p}_1{q}_1}{\mathrm{\Sigma }{p}_0{q}_1}}\hfill \\ \hfill & =\sqrt{\frac{1084}{782}\times \frac{1329}{1008}}\times 100=\sqrt{135.19}\hfill \end{array}$

and

$\begin{array}{ll}{P}_{01}\hfill & =\sqrt{\frac{\mathrm{\Sigma }{p}_1{q}_0}{\mathrm{\Sigma }{p}_0{q}_0}·\frac{\mathrm{\Sigma }{p}_1{q}_1}{\mathrm{\Sigma }{p}_0{q}_1}}\times 100=73.97\hfill \\ P_{01}\times P_{10}\hfill & =\frac{135.19\times 73.97}{100\times 100}=1.00\hfill \end{array}$

∴ The given data satisfies time reversal test.

Example 13.17:

Now reverse the process taking 1982 as base year and 1981 as current year, and show that the two results are strictly consistent.

Solution:

We have n=6, Σ log R₁=11.8992, and Σ log R₀=12.1015

$\begin{array}{ll}\mathrm{Index}\mathrm{number}\mathrm{for}1982(1981\mathrm{as}\mathrm{base})\hfill & =\mathrm{Antilog}\frac{\mathrm{\Sigma }\log R_1}{n}\hfill \\ \hfill & =\mathrm{Antilog}\frac{11.8992}{6}\hfill \\ \hfill & =\text{Antilog}1.9832=96.20\hfill \end{array}$

$\begin{array}{ll}\mathrm{Index}\mathrm{number}\mathrm{for}1982(1981\mathrm{as}\mathrm{base})\hfill & =\mathrm{Antilog}\frac{\mathrm{\Sigma }\log R_0}{n}\hfill \\ \hfill & =\mathrm{Antilog}\frac{12.1015}{6}\hfill \\ \hfill & =\mathrm{Antilog}2.0169=104\hfill \\ \therefore P_{01}\times P_{10}\hfill & =\frac{96.20\times 104}{100\times 100}=1.00\hfill \end{array}$

The results are strictly consistent.

13.6.2 Factor Reversal Test

This rest was originated by Iring Fisher with the logic that a formula that correctly reflects price changes would also reflect quantity changes.

If P₀₁ and Q₀₁ are the price index number and quantity index number for the period t₁ corresponding to basic period t₀, then we must have

$P_{01}\times Q_{10}=V_{01}=\frac{\mathrm{\Sigma }{p}_1{q}_1}{\mathrm{\Sigma }{p}_0{q}_0}$

In other words an index number method is said to satisfy factor reversal test if the product of price index number and quantity index number, as calculated by the same method is equal to the value of the index number.

Fisher’s index number method is the only method which satisfies this test:

i.e.,

$\begin{array}{ll}{P}_{01}\times Q_{10}\hfill & =\sqrt{\frac{\mathrm{\Sigma }{p}_1{q}_1}{\mathrm{\Sigma }{p}_0{q}_0}·\frac{\mathrm{\Sigma }{p}_1{q}_1}{\mathrm{\Sigma }{p}_{0}q1}}\times \sqrt{\frac{\mathrm{\Sigma }{q}_1{p}_0}{\mathrm{\Sigma }{q}_0{p}_0}·\frac{\mathrm{\Sigma }{q}_1{p}_1}{\mathrm{\Sigma }{q}_0{p}_1}}\hfill \\ \hfill & =\sqrt{\frac{\mathrm{\Sigma }{p}_1{q}_0\times \mathrm{\Sigma }{p}_1{q}_1\times \mathrm{\Sigma }{q}_1{p}_0\times \mathrm{\Sigma }{q}_1{p}_1}{\mathrm{\Sigma }{p}_0{q}_0\times \mathrm{\Sigma }{p}_0{q}_1\times \mathrm{\Sigma }{q}_0{p}_0\times \mathrm{\Sigma }{q}_0{p}_1}}\hfill \\ \hfill & =\sqrt{\frac{\mathrm{\Sigma }{p}_1{q}_0\times \mathrm{\Sigma }{p}_1{q}_1\times \mathrm{\Sigma }{p}_0{q}_1\times \mathrm{\Sigma }{p}_1{q}_1}{\mathrm{\Sigma }{p}_0{q}_0\times \mathrm{\Sigma }{p}_0{q}_1\times \mathrm{\Sigma }{q}_0{p}_0\times \mathrm{\Sigma }{p}_1{q}_0}}\hfill \\ \hfill & =\sqrt{\frac{\mathrm{\Sigma }{p}_1{q}_1\times \mathrm{\Sigma }{p}_1{q}_1}{\mathrm{\Sigma }{p}_0{q}_0\times \mathrm{\Sigma }{p}_0{q}_0}}=\frac{\mathrm{\Sigma }{p}_1{q}_1}{\mathrm{\Sigma }{p}_0{q}_0}\hfill \\ \hfill & =\mathrm{True}\mathrm{value}\mathrm{ratio}\hfill \end{array}$

∴ Fisher’s method satisfies the factor reversal test.

Example 13.18: Compute Fischer’s ideal number from the following data and show that it satisfies factor reversal test:

Solution:

Calculation of Fisher’s index number

$\begin{array}{lll}\therefore \hfill & \sum p_0{q}_0=84,\hfill & \sum p_1{q}_1=141\hfill \\ \hfill & \sum p_1{q}_0=188,\hfill & \sum p_0{q}_1=67\hfill \end{array}$

$\begin{array}{ll}\text{Fisher}’s\mathrm{Ideal}\mathrm{index}\mathrm{number}=P_{01}\hfill & =\sqrt{\frac{\mathrm{\Sigma }{p}_1{q}_0}{\mathrm{\Sigma }{p}_0{q}_0}\times \frac{\mathrm{\Sigma }{p}_1{q}_1}{\mathrm{\Sigma }{p}_0{q}_1}}\times 100\hfill \\ \hfill & =\sqrt{\frac{188}{84}\times \frac{141}{67}}\times 100\hfill \end{array}$

∴ Fisher’s price index number for 1982 with base 1975=217.03

Fisher’s quantity index number for 1982 with base 1975

$\begin{array}{ll}{Q}_{01}\hfill & =\sqrt{\frac{\mathrm{\Sigma }{q}_1{p}_0+\mathrm{\Sigma }{q}_1{p}_1}{\mathrm{\Sigma }{q}_0{p}_0+\mathrm{\Sigma }{q}_0{p}_1}}\times 100\hfill \\ \hfill & =\sqrt{\frac{67}{84}\times \frac{141}{186}}\times 100=77.34\hfill \end{array}$

$\begin{array}{ll}{P}_{01}\times Q_{01}\hfill & =\frac{271.03}{100}\times \frac{77.34}{100}\hfill \\ \hfill & =2.1703\times 0.7734=1.67865\hfill \end{array}$ (13.2)

(13.2)

$V_{01}=\frac{\mathrm{\Sigma }{p}_1{q}_1}{\mathrm{\Sigma }{p}_0{q}_0}=\frac{141}{84}=1.6786$ (13.3)

(13.3)

From Eqs. (13.2) and (13.3) $P_{01}\times Q_{01}=\frac{\mathrm{\Sigma }{p}_1{q}_1}{\mathrm{\Sigma }{p}_0{q}_0}$

∴ Factor reversal test is satisfied.

Example 13.19: Show that Fisher’s ideal index satisfies both the time reversal test as well as the Factor reversal test using the data given below:

Solution:

We have

$\begin{array}{cc}\sum p_1{q}_0=1900,& \sum p_0{q}_0=1360\\ \sum p_1{q}_1=1880,& \sum p_0{q}_1=1344\end{array}$

Time Reversal Test:

$\begin{array}{ll}{P}_{01}\hfill & =\sqrt{\frac{\mathrm{\Sigma }{p}_1{q}_0}{\mathrm{\Sigma }{p}_0{q}_0}\times \frac{\mathrm{\Sigma }{p}_1{q}_1}{\mathrm{\Sigma }{p}_0{q}_1}}\hfill \\ \hfill & =\sqrt{\frac{1900}{1360}\times \frac{1880}{1344}}\hfill \end{array}$

$\begin{array}{ll}{P}_{10}\hfill & =\sqrt{\frac{\mathrm{\Sigma }{p}_0{q}_1}{\mathrm{\Sigma }{p}_1{q}_1}\times \frac{\mathrm{\Sigma }{p}_0{q}_0}{\mathrm{\Sigma }{p}_1{q}_0}}\hfill \\ \hfill & =\sqrt{\frac{1344}{1880}\times \frac{1360}{1900}}\hfill \end{array}$

$\therefore P_{01}\times P_{10}=\sqrt{\frac{1900}{1360}\times \frac{1880}{1344}\times \frac{1344}{1880}\times \frac{1360}{1900}}$

$\therefore P_{01}\times P_{10}=1$

Hence Time reversal test is satisfied.

Factor Reversal Test:

$\begin{array}{ll}{Q}_{01}\hfill & =\sqrt{\frac{\mathrm{\Sigma }{q}_1{p}_0}{\mathrm{\Sigma }{q}_0{p}_0}\times \frac{\mathrm{\Sigma }{q}_1{p}_1}{\mathrm{\Sigma }{q}_0{p}_1}}\hfill \\ \hfill & =\sqrt{\frac{1344}{1360}\times \frac{1880}{1900}}\hfill \end{array}$

$\begin{array}{ll}\therefore P_{01}\times Q_{01}\hfill & =\sqrt{\frac{1900}{1360}\times \frac{1880}{1344}\times \frac{1344}{1360}\times \frac{1800}{1900}}\hfill \\ \hfill & =\sqrt{{\left({\displaystyle \frac{1880}{1360}}\right)}^2}\hfill \\ \hfill & =\frac{1880}{1360}\hfill \end{array}$

and

$\frac{\mathrm{\Sigma }{p}_1{q}_1}{\mathrm{\Sigma }{p}_0{q}_0}=\frac{1880}{1360}$

$\therefore P_{01}\times Q_{01}=\frac{\mathrm{\Sigma }{p}_1{q}_1}{\mathrm{\Sigma }{p}_0{q}_0}$

∴ Factor reversal test is satisfied.

∴ Fisher’s Ideal Index satisfies, Time reversal test as well as Factor reversal test.

Unit Test:

An index number method is said to satisfy unit test if it is not changed by a change in the measuring units of some items under consideration. All index number method, except simple aggregative method satisfies this test.

13.6.3 Circular Test

This test is an extension of the time reversal test and is applicable when the indexes for more than 2 years are given. Suppose an index number is computed for the period 1, on the base period 0, another index number is computer for period 2 on the base period 1, and another index number is computer for the period 3 on the base period 2, and so on, their product should be equal to 1.

Symbolically:

$P_{01}·P_{12}·P_{23}\dots P_{(n–1)n}·P_n=1$

The following methods satisfy circular test:

Fisher’s Ideal formula does not satisfy circular test.

13.8.1 Utility of Consumer Price Index Number

13.8.2 Formulas for Constructing Consumer Price Index

There are two methods for constructing consumer price index numbers.

Aggregate expenditure method: In this method base period quantities are used as weights. The formula is given by:

$\mathrm{Consumer}\mathrm{price}\mathrm{index}\mathrm{number}=\frac{\mathrm{\Sigma }{p}_1{q}_0}{\mathrm{\Sigma }{p}_0{q}_0}\times 100$

where “0” and “1” suffixes stand for base period and current period, respectively.
Example 13.20: Compute the cost of living index from the following data using aggregate expenditure method:

Item	Consumer quantity in the given year	Price in base year	Price in given year
Rice	2½ Quintal×12	12	25
Pulses	3 kg×12	0.4	0.6
Oil	2 kg×12	1.5	22
Clothing	6 m×12	0.75	1
Housing	–	20 P.M.	30 P.M.
Miscellaneous	–	10 P.M.	15 P.M.

Solution:

Item	q1	p0	p1	p1q1	p0q1
Rice	30	12	25	750	360
Pulses	36	0.4	0.6	21.6	14.4
Oil	24	1.5	2.2	52.6	36
Clothing	72	0.75	1.0	72	54
Housing	12	20	30	360	240
Miscellaneous	12	10	15	180	120
Total				1436.2	824.4

$\therefore \mathrm{\Sigma }{p}_1{q}_1=1436.4,\mathrm{\Sigma }{p}_0{q}_1=824.4$

$\begin{array}{ll}\mathrm{Cost}\mathrm{of}\mathrm{living}\mathrm{Index}\mathrm{number}\hfill & =\frac{\mathrm{\Sigma }{p}_1{q}_0}{\mathrm{\Sigma }{p}_0{q}_0}\times 100\hfill \\ \hfill & =\frac{1436.2}{174.24}\times 100=174.24\hfill \end{array}$

Family budget method: In this method the expenditure on different commodities in the base period are used as weights.
According to the method

$\mathrm{Consumer}\mathrm{price}\mathrm{index}\mathrm{number}=\frac{\mathrm{\Sigma }Pw}{\mathrm{\Sigma }w}$

where P=price relative=$P_{1/p_0}\times 100$; P₀=price of commodity in the base period; P₁=price of commodity in the current period; and w=p₀q₀.
Note:

$\mathrm{We}\mathrm{can}\mathrm{write}\mathrm{consumer}\mathrm{price}\mathrm{index}\mathrm{number}=\frac{\mathrm{\Sigma }{p}_1{q}_0}{\mathrm{\Sigma }{p}_0{q}_0}\times 100$

or

$\mathrm{C}\mathrm{.P}.\mathrm{index}\mathrm{number}=\frac{\mathrm{\Sigma }IV}{\mathrm{\Sigma }V}$

where $I=\mathrm{Relatives},V=\mathrm{Weights}$.
Example 13.21: Construct cost of living index for 1982 based on 1975 from the following data:

Group	Group Index No. For 1982 (based on 1975)	Weight
Food	122	32
Housing	140	10
Clothing	112	10
Fuel and light	116	6
Miscellaneous	106	42

Solution:

Group	Index No.	Weights, I	Weighted relative, IV
Food	122	32	3904
Housing	140	10	1400
Clothing	112	10	1120
Fuel and Light	116	6	696
Miscellaneous	106	42	4452
Total		100	11,572

$\mathrm{Index}\mathrm{number}=\frac{\mathrm{\Sigma }IV}{\mathrm{\Sigma }V}=\frac{11572}{100}=115.72$

13.9.1 Advantages of using Chain Base Method

In this method introduction of new items and deletion of old ones can be done without hazards and hassles. Because of this the chain base index (C.B.I.) numbers are used in consumer and wholesale price indices. By using chain base method, comparison is possible between any two successive periods. Index numbers calculated by the chain base method are free from the effect of seasonal variations. The chain base method brings homogeneity error almost to zero.

13.9.2 Limitation

The chain base indices are not suitable for long-range comparisons.

Example 13.22: From the following index numbers prepare new one by (1) taking the year 1982 as base and (2) using chain base method:

Solution:

1. Index numbers with 1982 as base (base 1982=100)

Year	Index number	Base changed to 1982 (1982=100)	Index number (1982=100)
1979	100	$(100/250)\times 100$	40
1980	110	$(110/250)\times 100$	44
1981	175	$(175/250)\times 100$	70
1982	250	100	100
1983	300	$(300/250)\times 100$	120
1984	400	$(400/250)\times 100$	160

2. Index numbers using chain base method

Year	Index number	Conversion	Chain base Index number
1979	100	100	100.00
1980	110	$(110/100)\times 100$	110.00
1981	175	$(175/110)\times 100$	159.09
1982	250	$(250/170)\times 100$	142.86
1983	300	$(300/250)\times 100$	120.00
1984	400	$(400/300)\times 100$	133.33

13.10.1 Base Conversion

(a) Conversion of F.B.I. to C.B.I.

To convert F.B.I. numbers to C.B.I. numbers, the following procedure is adopted:

1. The first years’ index number is taken as 100.

2. For subsequent years, the index number is obtained by the formula current year’s C.B.I. number.

$=\frac{\text{Current}\text{year}\text{'}\text{s}\text{F}\mathrm{.B}\mathrm{.I}.}{\mathrm{Previous\; year\; F}\mathrm{.B}\mathrm{.I}.}\times 100$

Example 13.23: From the F.B.I. numbers given below compute C.B.I. numbers.

Solution:

Year	F.B.I.	Conversion	C.B.I.
1975	376	–	100.0
1976	392	$(392/376)\times 100$	104.3
1977	408	$(408/392)\times 100$	104.1
1978	380	$(380/408)\times 100$	93.1
1979	392	$(392/380)\times 100$	103.2
1980	400	$(400/392)\times 100$	102.0

Example 13.24: Compute C.B.I. numbers for the following series of F.B.I. numbers:

Solution: It is not possible to compute the C.B.I. number for the year 1981 (since the C.B.I. for 1981 is the index number for 1981 with 1980 as the base year).

Year	F.B.I. (1975-100)	Conversion	C.B.I.
1981	210	–	–
1982	220	$(220/210)\times 100$	104.762
1983	250	$(250/220)\times 100$	113.636
1984	400	$(400/250)\times 100$	160
1985	300	$(300/400)\times 100$	75
1986	400	$(400/300)\times 100$	133.333

Conversion of C.B.I. to F.B.I.

The following steps are involved:

1. The first years’ index number is taken what the C.B.I. is, but if it is from the base it is taken equal to 100.

2. In the subsequent years, the index is obtained by the formula:

$\mathrm{Current}\text{year}\text{'}\mathrm{s}\mathrm{F}\mathrm{.B}\mathrm{.I}.=\frac{(\mathrm{Current}\text{year}\text{'}\mathrm{s}\mathrm{C}\mathrm{.B}\mathrm{.I}.)\times (\mathrm{Preceding}\text{year}\text{'}\mathrm{s}\mathrm{F}\mathrm{.B}\mathrm{.I}.)}{100}$

Example 13.25: From the C.B.I. numbers given below, construct F.B.I. numbers with 1970 as base:

Solution:

Year	C.B.I.	Conversion	F.B.I.
1978	80	–	80
1979	140	$(80/100)\times 140$	112
1980	130	$(80/100)\times (140/100)\times 100$	145.6
1981	100	$(80/100)\times (140/100)\times (130/100)\times 100$	145.6
1982	110	$(80/100)\times (140/100)\times (130/100)\times (300/400)\times 100$

13.10.2 Base Shifting

We sometimes shift the base from one period to another period. When the given indices are referred to a long-back period and one wants the indices based on recent base period, we shift the base.

We also shift the base to make two given series comparable. The method is to divide the indices of the other years by the index of the year selected as base and multiplying the quotient by 100.

Symbolically,

$\mathrm{New}\mathrm{base}\mathrm{index}\mathrm{number}=\frac{\mathrm{Old}\mathrm{index}\mathrm{number}\mathrm{of}\mathrm{current}\mathrm{year}}{\mathrm{Index}\mathrm{number}\mathrm{of}\mathrm{new}\mathrm{base}\mathrm{year}}$

Example 13.26: Following are the wholesale price index numbers from 1975 to 1980 to base 1970:

Find the wholesale price indices to the base 1977?

Solution: Index number of 1977 is 185.4.

∴ The wholesale index numbers to the base 1977

Year	By formula	Index numbers
1975	$(175.8/185.4)\times 100$	94.82
1976	$(172.4/185.4)\times 100$	92.99
1977	$(185.4/185.4)\times 100$	100
1978	$(185.0/185.4)\times 100$	99.79
1979	$(206.2/185.4)\times 100$	111.22
1980	$(246.8/185.4)\times 100$	133.12