Academic Year: 2012/2013Instructors: Brenda Lynch and PJ Hunt Contact: email@example.com firstname.lastname@example.org
Income elasticityFormula % ∆ Qd %∆Y Example: Qd of coal is 9 ton/hour, then income rises from €475 to €525pw and Qd of coal rises to 11 ton/hour. Calculate Yd.
Change in Qd is 2, original quantity is 9 giving 2/9. Change in income is €50, original income is €475 giving €50/€475. The income elasticity of demand for coal is 2/9 = 2.11 50/475
Yd Greater than 1 (demand is elastic, normal good, Luxuries) Yd Positive and less than 1 (demand is inelastic, normal good, Necessities) Yd Negative (inferior good) Cross Price Elasticity Formula % ∆ Qda % ∆ Pb
And is the % change in demand for good A caused by the % change in price of good B.If good B is a substitute for good A, A’s demand will rise as B’s price rises and CED will be a positive figure.If good B is a compliment to good A, A’sdemand will fall as B’s price rises and CEDwill be negative.
The major determinant of CED is the closeness ofthe substitute or compliment. The closer it is thebigger will be the effect and hence the greaterthe CED – either positive or negative.
Example: Demand equation for a product is: Qd = 90 – 8P + 2Y + 2Ps/cwhere P = 10, Y = 20 and Ps/c = 9 Find Ed, Yd and CED. Is the demand for the product inelastic or elastic? Is it a normal or inferior good? Is the product a luxury or a necessity? Does the product have a close substitute or compliment?
First find a value for Qd. Substitute in the givenvalues for P, Y & Ps to the equation. Q = 90 – 8(10) + 2(20) + 2(9) Q = 68 Now calculate each elasticity using the following formulae
Ed = (ӘQ/ ӘP)(P/Q) = -8(10/68) = -1.18Inelastic or elastic? Yd = (ӘQ/ ӘY)(Y/Q) = 2(20/68) = 0.59 Necessity or luxury, normal or inferior? CED = (ӘQ/ ӘPs/c)(Ps/Q) = 2(9/68) = 0.26 Substitute or compliment, close or not?
Total Revenue (TR) Determine from the demand function Q = 100 - P that MR is zero when TR is maximised at unit elasticity. Rearrange equation to get P = 100 - Q TR = P*Q, TR = (100 – Q) * (Q). So TR = 100Q – Q2
MR = ∆TR / ∆Q = 100 – 2Q. TR is maximisedwhen MR = 0 Set MR = 100 – 2Q = 0. Q = 50 Recall: P = 100 – Q(50) P = 50 So TR is maximised when P = 50 and Q = 50
Ep = ∆Q/∆P * P/Q ∆Q/∆P = first derivative of P = 100 – Q = -1 EP = -1(50/50) = -1 so Ep = -1 where MR = 0 and TR is maximised.
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