MANAGERIAL ECONOMICS

DEFINITION• PROVIDES LINK BETWEEN

ECONOMIC THEORY AND DECISION SCIENCES IN ANALYSIS OF MANAGERIAL DECISION MAKING.

• THE DIAGRAMME BELOW SHOWS HOW MANAGERIAL ECONMICS DIFFER FROM MICROECONOMICS:

• MICROECONOMICS IS LAGERLY DESCRIPTIVE, THAT IS, IT ATTEMPTS TO DESCRIBE HOW THE ECONOMY WORKS WITHOUT INDICATING HOW IT SHOULD OPERATE.

A LINK BETWEEN ECONOMIC THEORY AND THE DECISION SCIENCE IN ANALYSIS OF MANAGERIAL ECONOMICS

MANAGERIAL ECONOMICS, WHICH

APPLIES AND EXTENDS ECONOMICS AND THE

DECISION

SCIENCES TO

SOLVE MANAGEMENT PROBLEMS.

PROBLEM

FACED BY DECISION

MAKERS IN MANAGEMENT

DECISION SCIENCES

SOLUTIONS TO DECISION PROBLEMS FACED BY

MANAGERS.

ECONOMIC

THEORY

RELATIONSHIPS

• MANAGERIAL ECONOMICS IS LARGELY PRESCRIPTIVE, THAT IS, IT ATTEMPTS TO ESTABLISH RULES AND TECHNIQUES TO FULLFILL SPECIFIC GOALS.

• MANAGERIAL ECONOMICS ON THE OTHER HAND USES OPTIMIZATION TECHNIQUES, SUCH AS DIFFERENTIAL CALCULUS AND MATHEMATICAL PROGRAMMING, TO DETERMINE OPTIMAL COURSES OF ACTION FOR DECISION MAKERS.

THE BASIC PROCESS OF DECISION MAKING

ESTABLISH OR IDENTIFY THE OBJECTIVES In making any decision, you as the decision maker

should determine what the organisation or individuals objective. Unless you know what it is you trying to achieve, there is no sensible way to make the decision.

DEFINE THE PROBLEM One of the difficult part of decision making is to

determine exactly what the problem is. To solve short-comings in business, you should identify the problem brought about the short-comings otherwise, you as a manager, you will have little chance of solving the problem.

THE BASIC PROCESS OF DECISION MAKING

IDENTIFY POSSIBLE SOLUTIONSOnce the problem is defined, try to construct and

identify possible solutions, which includes, effective production and marketing of its products based on existing designs, as well as redesigning its entire products line.

SELECT BEST POSSIBLE SOLUTIONHaving identified the set of alternative possible

solutions, evaluate each one and determine which is best, give the objective of the organisation.

THE BASIC PROCESS OF DECISION MAKING

IMPLEMENT THE DECISION

This is of crucial importance stage, because the particular solution which has been chosen might positively or negatively affect the business processes if the implementation process was not followed properly or effectively.

THEORY OF THE FIRM

• The firms main aim is to make profit, at its maximum level. This is determined by looking at the expected profit of the firm, which can be expressed as an equation:

• The value of the firm equals:

• Present value of expected future profit

• =𝜋1

1+𝑖+

𝜋2

(1+𝑖)2+ …… . . +

𝜋𝑛

(1+𝑖)𝑛

THEORY OF THE FIRM

• =σ𝑡=𝑖𝑛 𝜋𝑡

(1+𝑖)𝑡

• Where 𝜋𝑡 is the expected profit in year t. ί is the interest rate, and t goes from 1(next year) to n ( the last year in the planning horizon)

• Please note that, this equation will be used in all upcoming classes to determine profits and losses/cost to company.

OPTIMIZATION TECHNIQUES

• Theses refers to the basic elements of differential calculus, including the rules of differentiation and use of derivative to maximize a function such as profit or minimize a function such as cost.

• Differential tells us what changes will occur in one variable called, the dependent variable when a small marginal change is made in another variable called, the independent variable.

OPTIMIZATION TECHNIQUES

• While we want to maximize the profit of our firm or minimize the cost of production, such maximization or minimization is often subject to constraints such as, producing certain amount to adhere to a contract or utilizing a certain amount of labor in a union agreement.

FUNCTIONAL RELATIONSHIP

• To understand this chapter, you must know how economic relationships are expressed. Frequently, the relationship between two or more economic variables can be represented by a table, graph or in a form of an equation. For example: Q = f(p), where Q are numbers of units sold and P represent the price per unit.

FUNCTIONAL RELATIONSHIP

• The equation above should be read as, number of units sold is a function of price, which means that, the number of units sold depends on price.

• The number of units in this case, is the dependent variable and price is the independent variable.

FUNCTIONAL RELATIONSHIP

• GIVEN THAT, Q = 200 – 5p, calculate the number of units sold, given that, the product was sold at a price of $10.

• This is a very simple calculation as p is replaced by $10 as the price and your answer will be Q=200-5(10) = 150 units.

RELATIONSHIP BETWEEN OUTPUT AND PROFIT

• STUDENTS WILL BE EXPECTED TO KNOW HOW TO CALCULATE MARGINAL REVENUE AND AVERAGE PROFIT.

OPTIMISATION TECHNIQUES

• DERIVATIVES• Using ∆(called Delta) to denote change in the

independent variable can be expressed as ∆X, and a change in the dependent variable can be expressed as ∆Y. thus, the marginal value of Y can be estimated by𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑌

𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑋=

Δ𝑦

Δ𝑥

• For Example, if a two – unit increase in X results in a one unit increase in Y, ∆X = 2 and ∆Y = 1, this means that, the marginal value of Y is about ½ that is, the depended variable Y increases by about ½ if the independent variable X increases by 1.

OPTIMISATION TECHNIQUES

• Unless the relationship between Y and X can be represented as a straight line graphically, the value of Δ𝑦

Δ𝑥𝑖𝑠 𝑛𝑜𝑡 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡.

• Derivatives of Constant• Managers wants to know how much to optimize the

performance of their organizations. If Y is some measure of the organisational performance and X is a variable under a particular managers control, he or she would like to know the value of X that maximizes Y.

• If the dependent variable Y is a constant, its derivative with respect to X is always zero. That is if Y=a ( where a is a constant)

𝑑𝑦

𝑑𝑥= 0.

OPTIMIZATION TECHNIQUES

• Derivatives of power functions

• A power function can be expressed as 𝑌 = 𝑎𝑥𝑏, where a and b are constant, if the relationship between X and Y is of this kind, the derivative of Y with respect to X equals b times. Students are expected to read further on this and practice.

OPTIMIZATION TECHNIQUES

• EXAMPLE: a new company did a study to estimate the effects of advertising expenditures on the sales of their product and found that the relationships between advertising expenditure and sales in two district were: 𝑆1 = 10 + 5𝐴1 and 𝑆2 = 12 + 4𝐴2 − 0.5𝐴2

2 .• Where 𝑆1 is the sales of the product (in millions of dollars

per year) in the first district and 𝑆2 is its sales of the product in the second district.

• Determine the amount of additional sales that an extra dollar of advertising would generate in each district.

• To answer this question, the derivatives of sales with respect to advertising must be calculated for each district.

OPTIMIZATION TECHNIQUES

• THEREFORE: 𝑑𝑆1

𝑑𝐴1= 5 − 3𝐴1

•𝑑𝑆2

𝑑𝐴2= 4 − 𝐴2

this means, in each district, the effective on sales of an extra dollar of advertising expenses, depend on the amount spend on advertising.

OPTIMIZATION TECHNIQUES

• Suppose that $0.5 million was being spend on advertising in the first district and $1 million was being spend on advertising in the second district.

•𝑑𝑆1

𝑑𝐴1= 5 − 3 0.5 = 3.5

•𝑑𝑆2

𝑑𝐴2= 4 − 1 = 3, this means an extra dollar of

advertising generated an extra $3.50 of sales in the first district and extra $3.00 sales in the second district.

OPTIMIZATION TECHNIQUES

• RECOMMENDATIONS: its recommended that, if the company wants to boost the total sales of its product, more should be spend on advertising in the first district and less in the second district.

OPTIMIZATION TECHNIQUES

• MAXIMIZATION AND MINIMIZATION PROBLEMS.

• Having determined how to find the derivative of Y with respect to X, we now take up the way to determine the value of x that maximises or minimizes Y.

• Given Y = -50+100X-5𝑥2

»𝑑𝑦

𝑑𝑥= 10 − 10𝑥

» therefore., if this derivative equals zero,

» 100-10x = 0

• X=10

OPTIMIZATION TECHNIQUES

• GIVEN y = -1+9x-6𝑥2 + 𝑥3

• Find the values of output that maximizes or minimizes profit.

•𝑑𝑦

𝑑𝑥= 9 − 12𝑥 + 3𝑥2 = 0

•𝑑2 𝑦

𝑑𝑥2= 12 + 6𝑥

• If x=1 𝑑2𝑦

𝑑𝑥2= −12 + 6 1 = −6

OPTIMIZATION TECHNIQUES

• Since the second derivative is negative, profit is maximized.

• If x=3, 𝑑2𝑦

𝑑𝑥2= −12 + 6 3 = 6

• Since the second derivative is positive, profit is minimum, when output equals 3 million units.

CONSTRAINED OPTIMIZATION

• Managers of firms and other organisations generally face constraints that limit the options available to them. A production manager may want to minimize his or her firms costs but may not be permitted to produce less than is required.

• Suppose that, a company produces two products and that its total cost equals

CONSTRAINED OPTIMIZATION

• TC =4𝑄12 + 5𝑄2

2 + 𝑄1 + 𝑄2

• Where 𝑄1, equals its output per hour of the first product and 𝑄2 equals its output per hour of the second product. Because of the commitments to customers, the number produced of both products combined cannot be less than 30 per hour.

• TC=TC =4𝑄12 + 5𝑄2

2 + 𝑄1 + 𝑄2

CONSTRAINED OPTIMIZATION

• 𝑄1 + 𝑄2 = 30

• SOLVE FOR 𝑄1, 𝑄1 = 30 − 𝑄2• THEREFORE; TC = 4(30 − 𝑄2)

2 + 5𝑄22 − (30 −

𝑄2)𝑄2

• = 4(900 − 60𝑄22 ) + 5𝑄2

2 − 30𝑄2 − 𝑄22

• TC = 3600 − 270𝑄2 + 10𝑄22

• Find the value of 𝑄2 that minimizes TC, by obtaining the derivative of TC with respect to 𝑄2and we must set it equals to zero.

CONSTRAINED OPTIMIZATION

•𝑑𝑇𝐶

𝑑𝑄2= −270 + 20𝑄2 = 0

• 20𝑄2 = 270

• 𝑄2 = 13.5

• To ensure that, this is a minimum, we obtain the second derivative

•𝑑2𝑇𝐶

𝑑𝑄2=

20, 𝑠𝑖𝑛𝑐𝑒 𝑡ℎ𝑖𝑠 𝑖𝑠 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒,𝑤𝑒 ℎ𝑎𝑣𝑒 𝑓𝑖𝑛𝑑 𝑎 𝑚𝑖𝑛𝑖𝑚𝑢𝑚.

CONSTRAINED OPTIMIZATION

• Find the value of 𝑄1 that minimizes total cost

• 𝑄1 + 𝑄2 = 30

• 𝑄1 = 30 − 𝑄2• = 30 − 13.5

• =16.5

• This means, in order for the company to minimize the total cost, it should produce 16.5 units of the first product and 13.5 units of the second product per hour.

• What is the actual Total Cost?

PRICE ELASTICITY OF DEMAND

• Price Elasticity of demand is defined as the percentage change in quantity demanded resulting from a 1 percent change in price.

• Example: given the hypothetical demand function for school uniforms

• 𝑄 = −700𝑝 + 200𝐼 − 500𝑆 + 0.01𝐴

• Assuming that per capital disposable Income(I) is $13 000, the average price of software(S) is $400, and advertising expenditure(A) is $50 million.

PRICE ELASTICITY OF DEMAND

• Calculate price elasticity of demand

• 𝑄 = 2900000 − 700𝑝

• Where p=$3000

• 𝑄 = 2900000 − 700 3000=800 000

– Evaluate the partial derivative given 2 900 000 –700p.

–𝜕𝑄

𝜕𝑝= −700

PRICE ELASTICITY OF DEMAND

• To obtain the price elasticity of demand, we

multiply 𝑑𝑄

𝑑𝑝by P/Q

• −700 Τ3000800 000

• =-2.62 as the price elasticity of demand.

• What does it tell us?