Internet Economics Basics of Demand Side Shi-Chung Chang Yi-Nung Yang

Outline Law of demand Factors Affecting demand Elasticity of Demand Complements and Substitutes Cross Elasticity of Demand Utility Maximization Constraints to U(.) maximization Market Demand

Law of Demand Definition: A simple demand function Quantity demanded is inversely related to price, cetris paribus. i.e., Qd ↑(↓) <=> P↓(↑) A simple demand function Qd = D(P, W), W: other factors affecting demand  Qd / P<0, implied by the law of demand. e. g., W =>所得、職業、性別、其他商品價格...

Factors Affecting demand Qc: 上網連線時間(min), Pc: 連線價格 ($/min) Own price: Pc↑ => Qc↓ 所得: Y↑ => Qc↑ 職業: 電腦族上網需求較大 姓別: 男生上網時數較長 Price of PC modem Price of Cable modem

Elasticity of Demand (1) 衡量價格變動對需求量變動的影響 需求彈性= (需求量變動的百分比/價格變動百分比) ||>1 => elastic 富有彈性, ||<1 => inelastic缺乏彈性, ||=1 => 單位彈性 unit elastic,

Elasticity of Demand (2) 價格彈性之簡單應用 已知市內電話之需求彈性為 -1.2, 若中華電信宣佈電話費率將調降10%, 則預期之通話量將? 增加12%

Complements and Substitutes e.g., Modem vs 連線服務, PC vs Modem 特性: Qm↑ <=> Qc↑ Pm↓=> Qc ? Qc↑ Substitutes 替代財 e.g., Cable connection vs dialup connection 特性: QA↑ <=> Qc↓ PA↓=> Qc ? Qc↓

Cross Elasticity of Demand (1) 交叉彈性 衡量 j 產品價格變動造成 i 產品需求量之影響 數學定義

Cross Elasticity of Demand (2) 互補財交叉彈性 CM = (PM/QC) ( QC /  PM) < 0 e.g., 撥接需求對數據機之交叉彈性若為 -1.2, 表示當數據機價格下降50%時,撥接需求會? 增加60% 替代財交叉彈性 CA = (PA/QC) ( QC /  PA) > 0 e.g., 撥接需求對Cable連線價格之交叉彈性若為2, 表示當Cable連線價格下降50%時,撥接需求會? 減少100%

Summary Rule 1: 必需商品之需求彈性小 Rule 2: 替代品多, 需求彈性愈大 若撥接費用貴的話…, 多在學校上網 少在家裡上網 Rule 2: 替代品多, 需求彈性愈大 若Hinet太貴的話…(P is higher) 我就用Seednet (Decrease in Qhinet is larger)

Utility Maximization Utility maximization is economic approach to human behavor Key assumptions more is better marginal utility is diminishing e.g., 很餓時, 吃第一支炸雞很不錯, 但吃到第5支時.. Utility function: U(Q) MU=U’(Q)>0 MU/  Q = U”(Q)<0

Constraints to U(.) maximization “資源有限而欲望無窮” e.g., 所得限制 Budget constraint PC : 連線價格 ($/min), QC: 你所消費之連線時間 m : 其他的花費 (看電影, 吃東西…) M: 收入總額 PC QC + m <= M Assume an interior solution exists: PC QC + m = M

Constrained U(.) maximization Max. U(QC , m) s.t. PC QC + m = M Solution 1: Substitute m = M - PC QC, into U(QC, m): Max U(QC, M - PCQC) 求最佳連線時數 Q*C:

Largrange-Multiplier Method Max. U(QC , m) s.t. PC QC + m = M Solution 2: Let L = U(QC, m) +(M - PCQC - m) 求最佳連線時數 Q*C:

Interpretations of Consumer equil. 花在撥接連線 “最後” 1元所得到的邊際效用 = 花在其他費用上“最後” 1元所得到的邊際效用 解出最佳連線時數 Qc 就可以找出 “需求函數” Note: 由效用極大之假設可以導出 “需求函數 What if PC=0, QC=?

Log form of U(.): An example Max. U(QC , m)=a log(Qc) + log(m) s.t. PC QC + m = M Let L = U(QC, m) +(M - PCQC - m)

Market Demand 市場需求 = 所有個別消費者需求之總和 Qd = Qd1+ Qd2+ Qd3+ Qd4+... P P P Qd QC Individual 1 + P QC Individual 2 = P QC Market demand Qd Qd1 Qd2