(a) How does the concept of wholesale price index work? Describe the major components of wholesale price index. Explain the…

WPI and agricultural indices The Wholesale Price Index is a fixed-base, base-weighted price index compiled in India by the Office of Economic Adviser, DPIIT, currently on 2011-12 base. It measures how wholesale prices of commodities move relative to the base period, using base-year weights so that the index is a Laspeyres-type aggregate: current value at base weights divided by base value, P01=Σp1q0/Σp0q0×100. Its three major groups are primary articles (~23 per cent), fuel and power (~13 per cent, including mineral oils such as crude petroleum), and manufactured products (~64 per cent). The index works by collecting prices for representative commodities, converting them into relatives, and aggregating with fixed weights; a rise in primary articles raises the index more if their weight is high, while fuel and power transmit cost changes to industry. Because weights are fixed, WPI isolates price movement from quantity shifts, though it may become less representative if the commodity mix changes. For example, a sharp rise in crude petroleum under fuel and power raises the index even if primary articles are stable.

For agriculture, area, production and yield indices are physical relatives. A base year is chosen; fixed-base indices compare each year with that year, while chain-base indices link successive years. For each crop i, area relative = A1i/A0i×100 and production relative = P1i/P0i×100. Crop relatives are aggregated, usually by a weighted arithmetic mean using base-year area or value weights; geometric aggregation is a theoretical alternative, but the yield index is normally derived as production index divided by area index×100, because yield is output per unit area. Thus the area index captures expansion of sown area, the production index captures output change, and the yield index isolates productivity change; if area rises but production rises less, the yield index falls. This keeps the three indices consistent.

G/M/1 and geometric arrival-point distribution In a G/M/1 queue, interarrival times are general with mean 1/λ, service times are exponential with rate μ, and there is one server. Stability requires λ/μ<1. Let aₖ be the probability that k services are completed during one interarrival time. If A*(s)=E[e^-sX] is the Laplace transform of the interarrival time, then a(z)=Σaₖ z^k=A*(μ(1-z)). Because a(z) is convex, a(0)>0, a(1)=1 and a'(1)=μ/λ>1, there is a unique σ∈(0,1) satisfying A*(μ(1-σ))=σ. The equation balances the probabilities of different numbers of services completed before the next arrival.

Let Yₙ be the number in the system just before the nth arrival. If Yₙ=i, after the arrival there are i+1 customers; during the next interarrival K services finish, so Yₙ₊₁=max(0,i+1-K). Hence pᵢ₀=Σₖ₌ᵢ₊₁∞aₖ and pᵢⱼ=aᵢ₊₁₋ⱼ for 1≤j≤i+1. The stationary equations are, for j≥1, πⱼ=Σᵢ≥j-1πᵢ aᵢ₊₁₋ⱼ=Σₖ≥0aₖπⱼ₋₁₊ₖ, and π₀=Σₖ≥1aₖΣₘ₌₀^k-1πₘ. Now put πⱼ=(1-σ)σ^j. For j≥1, the right side is (1-σ)σ^j-1Σaₖσ^k=(1-σ)σ^j-1a(σ)=(1-σ)σ^j=πⱼ. Summing the j≥1 equations and using Σπⱼ=1 gives the j=0 equation automatically. Since the chain is irreducible and positive recurrent, this normalized solution is the steady-state distribution; thus the arrival-point system is geometric. The geometric form follows because the memoryless service time makes the embedded chain one-step-up and the same root σ enters every balance equation. This distribution is the arrival-epoch analogue of the M/M/1 geometric queue-length distribution.

Life-table identity Here e(x) is the curtate expectation: the average number of complete years lived after age x, so e(x)=Σᵣ≥1l(x+r)/l(x), and e(x+1)=Σᵣ≥2l(x+r)/l(x+1). Equivalently, e(x)=(1-q)[1+e(x+1)], because a person aged x contributes one complete year only if surviving to x+1, and then contributes the complete years expected at x+1. Let S=Σᵣ≥2l(x+r). Then e(x)-e(x+1)=[l(x+1)+S]/l(x)-S/l(x+1) =l(x+1)/l(x)+S(1/l(x)-1/l(x+1)). Since l(x+1)=l(x)[1-q(x)], 1/l(x)-1/l(x+1)=-q(x)/l(x+1). Therefore e(x)-e(x+1)=1-q(x)-q(x)S/l(x+1)=1-q(x)-q(x)e(x+1) =1-q(x)[1+e(x+1)]. Rearranging gives q(x)=[1-(e(x)-e(x+1))]/[1+e(x+1)]. The denominator 1+e(x+1) appears because survival to x+1 adds one complete year to the future complete years. At the terminal age the sums vanish and the identity remains consistent, because e(x)=e(x+1)=0 and q(x)=1.