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Seminar exercises The Product-mix Problem. Agnes Kotsis. Corporate system-matrix. 1.) Resource-product matrix Describes the connections between the company’s resources and products as linear and deterministic relations via coefficients of resource utilization and resource capacities.
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Seminar exercisesThe Product-mix Problem Agnes Kotsis
Corporate system-matrix 1.) Resource-product matrix Describes theconnections between the company’s resources and products as linear and deterministic relations via coefficients of resource utilization and resource capacities. 2.) Environmental matrix (or market-matrix): Describes the minimum that we must, and maximum that we can sell on the market from each product. It also desribes the conditions.
Resource-product matrix Product types Capacities Resources Resource utilization coefficients
Contribution margin • Unit Price - Variable Costs Per Unit = Contribution Margin Per Unit • Contribution Margin Per Unit x Units Sold = Product’s Contribution to Profit • Contributions to Profit From All Products – Firm’s Fixed Costs = Total Firm Profit
Partially convertible relations Non-convertible relations Resource-Product Relation types
Product-mix in a pottery – corporate system matrix e1: 1*T1+0,5*T2< 50 e2: 0,5*T1+1*T2< 50 e3: 0,1*T2< 10 p1, p2: 10 < T1< 100 p3, p4: 10 < T2< 100 ofF: 200 T1+200T2=MAX 200 200
Objective function • refers to choosing the best element from some set of available alternatives. X*T1 + Y*T2 = max weights(depends on what we want to maximize: price, contribution margin) variables(amount of produced goods)
Solution with linear programming 33 jugs and 33 plaits a per week Contribution margin: 13 200 HUF / week T1 e1 100 ofF e3 e1: 1*T1+0,5*T2< 50 e2: 0,5*T1+1*T2< 50 e3: 0,1*T2< 10 p1,p2: 10 < T1< 100 p3, p4: 10 < T2< 100 ofF: 200 T1+200T2=MAX 33,3 e2 100 T2 33,3
What is the product-mix, that maximizes the revenues and the contributionto profit!
Solution • T1: Resource constraint 2000/4 = 500 > market constraint 400 • T2-T3: Which one is the better product? Rev. max.: 270/2 < 200/1 thus T3 T3=(3000-200*2)/1=2600>1000 T2=200+1600/2=1000<1100 Contr. max.: 110/2 > 50/1 thus T2 T2=(3000-200*1)/2=1400>1100 T3=200+600/1=800<1000
T4: does it worth? Revenue max.: 1000/1 > 500 Contribution max.: 200 • T5-T6: linear programming e1: 2*T5 + 3*T6≤ 6000 e2: 2*T5 + 2*T6 ≤ 5000 p1, p2: 50 ≤ T5≤ 1500 p3, p4: 100 ≤ T6≤ 2000 cfÁ: 50*T5 + 150*T6 = max cfF: 30*T5 + 20*T6 = max
T5 e1 Contr. max: T5=1500, T6=1000 Rev. max: T5=50, T6=1966 3000 e2 2500 cfF 2000 2500 cfÁ T6