Managerial Economics- Optimization Techniques

Managerial Economics (2)  √ Role and Scope of Managerial Economics  √ Mathematics Review  Basic Concepts and Tools for Economic Analysis

Optimal Decision: DMs Optimize 

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The optimal decision in managerial economics is one that brings the firm closest to this goal. Decision Makers Optimize 

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Jul-06

Practically in all managerial decisions the task of the manager is the same - each goal involves an optimization problem. The manager attempts either to maximize or minimize some objective function, frequently subject to some constraint(s). And, for all goals that involve an optimization problem, the same general economic principles apply! S. Karna/PG06/IILM/Mgrl Eco/ME Intro/3

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Basic Concepts:

Maximizing the Value of a Firm 

Value of a firm  

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Risk premium 

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Jul-06

Price for which it can be sold Equal to net present value of expected future profit Accounts for risk of not knowing future profits The larger the rise, the higher the risk premium, & the lower the firm’s value

S. Karna/PG06/IILM/Mgrl Eco/ME Intro/3

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Basic Concepts:

Maximizing the Value of a Firm 

Value of a firm =

T πt π1 π2 πT + + ... + =∑ 2 T t (1 + r ) (1 + r ) (1 + r ) t =1 (1 + r )

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Maximizing firm’s profit in each time period leads to maximizing value of firm 

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Jul-06

Only if Cost & revenue conditions are independent across time periods If say for example in mining industry more extraction in current time leads to higher cost of extraction in future then higher profit now will lead to lower profits in future. S. Karna/PG06/IILM/Mgrl Eco/ME Intro/3

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Economic Optimization 

Our first assumption 

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Economic agents (i.e., households, firms, managers, etc.) have an objective that they are trying to optimize.  

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Abstraction from Reality? A Simplification?

Individuals assumed to maximize utility. For-profit firms maximize profits and minimize costs. Not-for-profit firms may maximize output given the budget or minimize cost given the output Realistic?

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Economic Model of the Firm 

Theory of the firm: Goal is to maximize firm profits 

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Π = Total Revenue – Total Cost = TR - TC or Simply R - C 

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Use Π to represent profit

TR is determined by: sales and marketing strategy, pricing, economy, etc. TC is determined by: production methods, cost of capital, etc.

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Marginal Analysis 

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Marginal - change in the dependent variable caused by a one-unit change in an independent variable Marginal Revenue - change in total revenue associated with a one-unit change in output Marginal Cost - change in total cost associated with a one-unit change in output Marginal Profit - change in total profit associated with a one-unit change in output S. Karna/PG06/IILM/Mgrl Eco/ME Intro/3

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An Example with Profits Example: Given Demand Function: P = 60 – 2Q Derived Revenue Function: R = P*Q = 60Q-2Q2 Given Cost Function: C = 50Q - 12*Q2 + Q3 Derived Profit Function: Π = R-C = 10*Q + 10*Q2 – Q3 Dem and and Marginal Revenue Lines

70 60 50 40 30 20 10 0 -10

1

2

Price

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3

4

5

6

7

8

9

10

11 12 13

Marginal Revenue

14

15 16

17 18

Quantity Demanded

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Total, Marg. and Avg. Revenue, Cost & Profit at Different Outputs Outpu Total Marginal Average Total Marginal Average Total Marginal Average t

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Revenue

Revenue

Revenue

Cost

Cost

Cost

Profit

0

Profit

Profit

0

0

0

1

58

58

58

39

39

39

19

19

19

2

112

54

56

60

21

30

52

33

26

3

162

50

54

69

9

23

93

41

31

4

208

46

52

72

3

18

136

43

34

5

250

42

50

75

3

15

175

39

35

6

288

38

48

84

9

14

204

29

34

7

322

34

46

105

21

15

217

13

31

8

352

30

44

144

39

18

208

-9

26

9

378

26

42

207

63

23

171

-37

19

10

400

22

40

300

93

30

100

-71

10

11

418

18

38

429

129

39

-11

-111

-1

12

432

14

36

600

171

50

-168

-157

-14

13

442

10

34

819

219

63

-377

-209

-29

14

448

6

32

1092

273

78

-644

-267

-46

15

450

2

30

1425

333

95

-975

-331

-65

16

448

-2

28

1824

399

114

-1376

-401

-86

17

442

-6

26

2295

471

135

-1853

-477

-109

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Revenue Maximizing Output = 15; Maxm. Revenue= 450; Here MR=0 but Loss = 975 Maximization of Revenue 500

Total, Marginal and Average Revenue

400

300

200

100

0 0

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2

3

5

6

7

8

9

10 11 12 13 14 15 16 17

Output

-100

Total Revenue

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Marginal Revenue

Average Revenue

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Cost Per Unit or AC is Lowest when MC=AC at output =6 but Profit = 204 (Below Max) M inimization of Cost per unit 160

Total, Marginal and Average Cost

140 120 100 80 60 40 20 0 0

1

Total Cost

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2

3

4 Output5

Marginal Cost

6

7

8

Average Cost

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Maximum Total Profit 217 is at Output = 7 Total Profit =Total Rev enue - Total Cost 700

Total Revenue, Cost and Profit

600 500 400 300 200 100 0 -100 0

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2

3

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5

6

7

8

9

10

11

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Output

-200 -300

Total Cost

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Total Revenue

Total Profit

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Maximum Profit =217 at output level of 7 units at this Marg. Profit = MR - MC = 0 250

Total, Marginal and Average Profit

200

150

100

50

0 0

1

3

4

5

6

7

8

Output

-50

Total Profit

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2

Marginal Profit

Average Profit

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Profit Maximizing Output = 7 where MR = MC i.e. MC curve cuts MR from below M a rg in a l P ro fi t = M R - M C

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80 60

Marginal Revenue, Cost and Profit

40 20

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0 0

1

-20

2

3

4

5

O u tp u t

6

7

8

9

-40

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-60

M a rg in a l R e ve n u e M a rg in a l C o s t M a rg in a l P ro fit

Jul-06

If MR>MC: increase output, increase profit If MR<MC: increase output, decrease profit MR=MC: profit maximization assured

S. Karna/PG06/IILM/Mgrl Eco/ME Intro/3

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Optimization Using Calculus 

If y = f(x), the maxima or minima of y exists for that value of x (say at x=x*) where 

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The second and sufficient condition for maxima or minima is 

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First derivative, dy/dx = 0

For maxima, 2nd derivative should be negative i.e. d2y/dx2 < 0 at x= x* For minima, 2nd derivative should be positive i.e. d2y/dx2 > 0 at x= x*

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Example: Maximize R= 60Q-2Q2 

Here dR/dQ = 60 – 4Q (dR/dQ is basically MR)   

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Jul-06

For maxima or minima dR/dQ=0 or 60-4Q=0 i.e. Q*=60/4 = 15 2nd derivative, d2R/dQ2 = -4 <0 therefore maximum of R exists at Q=15 The maximum revenue, Rmax= 60xQ*-2Q*2 =60x15 -2x152 = 900-450 = 450. (What’s the profit at this output?)

S. Karna/PG06/IILM/Mgrl Eco/ME Intro/3

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Marginal Revenue, Cost and Profit 

Marginal Revenue, Cost & Profit:

dR R MR = and AR = = P dQ Q dC C MC = and AC = dQ Q Profit, Π = R − C dΠ d (R - C) dR dC ∴ Marginal Profit = = = − dQ dQ dQ dQ Marginal Profit = MR - MC Jul-06

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Optimizing: Maximizing Revenue 

For maximum revenue, dR (i) = 0 i.e. MR = 0 dQ i.e. Slope of revenue curve = 0 d 2R d  dR    (ii) <0 ⇒ <0 2   dQ dQ  dQ  d(MR) Or, <0 dQ i.e. Slope of MR curve should be negative

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Optimizing: Minimizing Cost 

For minimizing cost per unit (i.e. C/Q)

d ( AC ) (i) = 0 i.e. Slope of avg. cost curve = 0 dQ (Check out that this means AC = MC ) 2

d ( AC ) (ii) >0 2 dQ (Check out : this is possible when Slope of MC > Slope of AC. 1st and 2nd Condition together means that AC is minimum at level of output at which MC curve cuts AC curve from below.) Jul-06

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Optimizing: Maximizing Profit or Minimizing Loss dΠ dR dC (i) Marginal Profit = = − =0 dQ dQ dQ ⇒ MR - MC = 0 or MR = MC, and dΠ d(MR) d ( MC ) (ii) <0 ⇒ − <0 2 dQ dQ dQ i.e. Slope of MR curve < Slope of MC curve 2

Graphically this means that profit maximizing output is given by the point where MC curve intersects the MR curve from below Jul-06

S. Karna/PG06/IILM/Mgrl Eco/ME Intro/3

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Applying Derivatives for Cost Optimization: Last Numerical Example 

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Cost Function given was: C = 50Q - 12*Q2 + Q3 Firms normally will optimize (i.e. minimize) cost per unit. So objective function here is: AC = 50 - 12*Q + Q2 

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Jul-06

For minimum, d(AC)/dQ = -12 + 2Q =0 Or, Q = 12/2=6 Checking out 2nd derivative d(AC)/dQ = 2 which is +ve so we have minimum at Q* = 6. Calculating minimum AC, ACMin = 50-12*6+62 = 14

S. Karna/PG06/IILM/Mgrl Eco/ME Intro/3

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Maximizing Profit: Applying Derivative on Last Numerical Example 

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Profit Function we have here Π = R – C = (60Q-2Q2) – (50Q - 12*Q2 + Q3) Π = 10*Q + 10*Q2 – Q3 For Maximization, taking first derivative of above 

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Jul-06

dΠ /dQ = 10 + 20Q – 3Q2 = 0 ⇒ Q = 7.1 ≈ 7 d2Π /dQ2 = 20 – 6Q at Q=7.1, d2Π /dQ2 = 20- 6 x 7 = -22 < 0 Therefore we have maximum at Q= 7, and maximum profit, Π Max= 10x7 – 10x72 – 73= 217

S. Karna/PG06/IILM/Mgrl Eco/ME Intro/3

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Generalization: Using Marginal Analysis to Find Optimal Activity Level A* 

If marginal benefit > marginal cost 

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If marginal cost > marginal benefit 

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Activity should be decreased to reach highest net benefit

Optimal level of activity 

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Jul-06

Activity should be increased to reach highest net benefit

When no further increases in net benefit are possible Occurs when MB = MC

S. Karna/PG06/IILM/Mgrl Eco/ME Intro/3

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Constraints on Optimization 

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Resource Constraints – Limited Availability Output Quantity and Quality Constraints Legal Constraints Environment Constraints

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Constrained Optimization: Equimarginal Principle 

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Jul-06

The ratio MB/P represents the additional benefit per additional Re spent on the activity Ratios of marginal benefits to prices of various activities are used to allocate a fixed amount of fund among activities

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Constrained Optimization: Equi-marginal Principle 

To maximize or minimize an objective function subject to a constraint 

Ratios of the marginal benefit to price must be equal for all activities

MBA MBB MBZ = = ... = PA PB PZ 

Jul-06

Constraint must be met

S. Karna/PG06/IILM/Mgrl Eco/ME Intro/3

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Constrained Maximization: An Example 

Suppose that a firm’s production function is by Q = KL2 and the costs are given by C = wL + rK, where K = capital, L = labor and w and r their per-unit costs. a) Suppose that w = Rs.40, r = Rs.10 and that the desires to produce 2000 units of output. How much capital and labor should be used, if the firm wants to produce at minimum cost? b) Suppose now that instead of having the objective of producing 2000 units the firm decides to produce with a total cost budget of Rs.1800. How much capital and labor should be used to maximize output?

Jul-06

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Solution: Constrained Maximization Example a) Objective Function 

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Subject to Q = KL2 = 2000 Since Q = KL2 

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Minimize C = wL + rK

MPL = dQ/dL = 2KL and MPK = dQ/dK = L2

Cost will be minimum 

when MPL/w = MPK/r ⇒ MPL/MPK=w/r …(1)

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MPL/MPK= 2KL/ L2 =2K/L & w/r = 40/10 =4

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Putting into eq. (1), we get 2K/L = 4

i.e. K = 2L Jul-06

S. Karna/PG06/IILM/Mgrl Eco/ME Intro/3

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Solution: Constrained Maximization Example (Contd.)  

Therefore Q = KL2 = 2L L2 = 2L3 Since Q = 2000, therefore 2000 = 2L3

Or, L3 = 1000 

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Jul-06

=> L =10

Now, K = 2L = 2 x 10 = 20

Optimal combination is 20 units of K and 10 units of L. (The least cost C = wL + rK = 40x10 + 10x20 = 600 i.e. Rs. 600)

S. Karna/PG06/IILM/Mgrl Eco/ME Intro/3

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Solution: Constrained Maximization Example (Contd.) b) Objective Function: Maximize output Q = KL2 Subject to C = wL + rK = 1800  As in (a), the condition for maximizing output MPL/w = MPK/r yields K = 2L 

Putting K=2L, w= 40 and r =10 in cost constraint,    

Jul-06

wL + r 2L = 1800 40L + 10 x 2L =1800 L = 1800/60 = 30 And therefore K = 2L = 2 x 30 = 60 S. Karna/PG06/IILM/Mgrl Eco/ME Intro/3

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Solution: Constrained Maximization Example (Contd.) 

Jul-06

Optimal combination is K=60 units and L= 30 units. (Maximum output, Qmax = KL2 = 60 x 302 = 54000 units)

S. Karna/PG06/IILM/Mgrl Eco/ME Intro/3

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Problem Set #1:

(DUE Wed, July 15th )

1. Find the derivatives of: a. b. c. d. e.

Y Y Y Y Y

= = = = =

√X 100 1/X2 2X + 3X2 LogX/3X2

2. Profit (Π )= -25 + 75Q - 5Q2 a. Compute the optimal profit maximizing output rate b. Demonstrate that you have found a maximum (not a minimum) c. Calculate the total profit at the profit maximizing output Jul-06

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