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Question 16 of 43

Q.(a) Suppose that the quantity demanded Qd=29−2p−5dpdt+d2pdt2Q_d = 29 - 2p - 5\dfrac{dp}{dt} + \dfrac{d^2p}{dt^2} and quantity supplied Qs=5+4pQ_s = 5 + 4p, where ′p′'p' is the price. Find the equilibrium price for market clearance.

(OR)
(b) The average number of phone calls per minute into the switch board of a company between 10.0010.00 a.m. and 2.302.30 p.m. is 2.52.5. Find the probability that during one particular minute there will be,
(i) no phone at all,
(ii) exactly 33 calls,
(iii) at least 55 calls. (e−2.5=0.08208)(e^{-2.5} = 0.08208)
Tamil Nadu DgeTamil Nadu HSC (DGE) Commerce Board 2020Subjective· 5mImportance★★★★★
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(a) Setting Qd=QsQ_d = Q_s yields p′′−5p′−6p=−24p'' - 5p' - 6p = -24; the market-clearing (steady-state) price is p=4p = 4. (b) Poisson with λ=2.5\lambda = 2.5 gives P(0)=0.0821P(0) = 0.0821, P(3)=0.2138P(3) = 0.2138, P(≥5)=0.1089P(\ge 5) = 0.1089.

Part (a) — Equilibrium price

Market clearance requires Qd=QsQ_d = Q_s:

29−2p−5dpdt+d2pdt2=5+4p.29 - 2p - 5\frac{dp}{dt} + \frac{d^2p}{dt^2} = 5 + 4p.

Rearranging:

d2pdt2−5dpdt−6p=−24.\frac{d^2p}{dt^2} - 5\frac{dp}{dt} - 6p = -24.

The equilibrium (market-clearing) price is the steady-state value where pp is constant, so dpdt=d2pdt2=0\dfrac{dp}{dt} = \dfrac{d^2p}{dt^2} = 0:

−6p=−24 ⇒ p=4.-6p = -24 \ \Rightarrow\ p = 4.

(The full solution p(t)=Ae6t+Be−t+4p(t) = A e^{6t} + B e^{-t} + 4 has particular/equilibrium part p=4p = 4.)

Part (b) — Poisson distribution

Mean λ=2.5\lambda = 2.5, e−2.5=0.08208e^{-2.5} = 0.08208; P(X=x)=e−λλxx!P(X = x) = \dfrac{e^{-\lambda}\lambda^x}{x!}.

(i) No calls: P(0)=e−2.5=0.08208.P(0) = e^{-2.5} = 0.08208.

(ii) Exactly 33 calls:

P(3)=e−2.5(2.5)33!=0.08208×15.6256=0.2138.P(3) = \frac{e^{-2.5}(2.5)^3}{3!} = \frac{0.08208 \times 15.625}{6} = 0.2138.

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