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Q 4 – 1 a. Let T = number of TV advertisements R = number of radio advertisements N = number of newspaper advertisements. Q 4 – 1 a. cont’d. Optimal Solution: T = 4, R = 14, N = 10 Allocation: TV 2,000(4) = $8000 Radio 300(14) = $4,200

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q 4 1 a

Q 4 – 1 a.

Let T = number of TV advertisements

R = number of radio advertisements

N = number of newspaper advertisements

q 4 1 a cont d

Q 4 – 1 a. cont’d

Optimal Solution: T = 4, R = 14, N = 10

Allocation: TV 2,000(4) = $8000

Radio 300(14) = $4,200

News 600(10) = $6,000

Objective Function Value

(Expected number of audience):

100,000(4) + 18,000(14) + 40,000(10)

=1,052,000

q 4 1 b cont d

Q 4 – 1 b. cont’d

The dual price for the budget constraint is 51.30. Thus, a $100 increase in budget should provide an increase in audience coverage of approximately 5,130. The RHS range for the budget constraint will show this interpretation is correct.

q 4 10 a

Q 4 – 10 a.

Let S = the proportion of funds invested in stocks

B = the proportion of funds invested in bonds

M = the proportion of funds invested in mutual funds

C = the proportion of funds invested in cash

q 4 10 a cont d2

Q 4 – 10 a. cont’d

From computer results, the optimal allocation among the four investment alternatives is

Stocks 40.0%

Bonds 14.5%

Mutual Funds 14.5%

Cash 30.0%

The annual return associated with the optimal portfolio is 5.4%

The total risk = 0.409(0.8) + 0.145(0.2)

+ 0.145(0.3) + 0.300(0.0) = 0.4

q 4 10 b

Q 4 – 10 b.

Changing the RHS value for constraint 2 to 0.18 and resolving using computer, we obtain the following optimal solution:

Stocks 0.0%

Bonds 36.0%

Mutual Funds 36.0%

Cash 28.0%

The annual return associated with the optimal portfolio is 2.52%

The total risk = 0.0(0.8) + 0.36(0.2)

+ 0.36(0.3) + 0.28(0.0) = 0.18

q 4 10 c

Q 4 – 10 c.

Changing the RHS value for constraint 2 to 0.7 and resolving using computer, we obtain the following optimal solution:

Stocks 75.0%

Bonds 0.0%

Mutual Funds 15.0%

Cash 10.0%

The annual return associated with the optimal portfolio is 8.2%

The total risk = 0.75(0.8) + 0.0(0.2)

+ 0.15(0.3) + 0.10(0.0) = 0.65

q 4 10 d

Q 4 – 10 d.

Note that a maximum risk of 0.7 was specified for this aggressive investor, but that the risk index for the portfolio is only 0.67. Thus, this investor is willing to take more risk than the solution shown above provides. There are only two ways the investor can become even more aggressive: increase the proportion invested in stocks to more than 75% or reduce the cash requirement of at least 10% so that additional cash could be put into stocks. For the data given here, the investor should ask the investment advisor to relax either or both of these constraints.

q 4 10 e

Q 4 – 10 e.

Defining the decision variables as proportions means the investment advisor can use the linear programming model for any investor, regardless of the amount of the investment. All the investor advisor needs to do is to establish the maximum total risk for the investor and resolve the problem using the new value for maximum total risk.

a 2 a

A – 2 (a)

Let S = Tablespoons of Strawberry

C = Tablespoons of Cream

V = Tablespoons of Vitamin

A = Tablespoons of Artificial sweetener

T = Tablespoons of Thickening agent

a 2 c

A – 2 (c)

Since u1*, u5*, u6* > 0, the 1st, 5th, and 6th constraints are binding.

data envelopment analysis
Data Envelopment Analysis

The Langley County School District is trying to

determine the relative efficiency of

its three high schools. In particular,

it wants to evaluate Roosevelt High.

The district is evaluating

performances on SAT scores, the

number of seniors finishing high

school, and the number of students

who enter college as a function of the

number of teachers teaching senior

classes, the prorated budget for senior instruction,

and the number of students in the senior class.

slide20
Input

RooseveltLincolnWashington

Senior Faculty 37 25 23

Budget ($100,000's) 6.4 5.0 4.7

Senior Enrollments 850 700 600

slide21
Output

RooseveltLincoln Washington

Average SAT Score 800 830 900

High School Graduates 450 500 400

College Admissions 140 250 370

data envelopment analysis1
Data Envelopment Analysis
  • Decision Variables

E = Fraction of Roosevelt's input resources required by the composite high school

w1 = Weight applied to Roosevelt's input/output resources by the composite high school

w2 = Weight applied to Lincoln’s input/output resources by the composite high school

w3 = Weight applied to Washington's input/output

resources by the composite high school

data envelopment analysis2
Data Envelopment Analysis
  • Objective Function

Minimize the fraction of Roosevelt High School's input resources required by the composite high school:

MIN E

data envelopment analysis3
Data Envelopment Analysis
  • Constraints

Sum of the Weights is 1:

(1) w1 + w2 + w3 = 1

Output Constraints:

Since w1 = 1 is possible, each output of the composite school must be at least as great as that of Roosevelt:

(2) 800w1 + 830w2 + 900w3> 800 (SAT Scores)

(3) 450w1 + 500w2 + 400w3> 450 (Graduates)

(4) 140w1 + 250w2 + 370w3> 140 (College Admissions)

slide25
Input Constraints:

The input resources available to the composite school is a fractional multiple, E, of the resources available to Roosevelt. Since the composite high school cannot use more input than that available to it, the input constraints are:

(5) 37w1 + 25w2 + 23w3< 37E (Faculty)

(6) 6.4w1 + 5.0w2 + 4.7w3< 6.4E (Budget)

(7) 850w1 + 700w2 + 600w3< 850E (Seniors)

Non-negativity of variables:

E, w1, w2, w3> 0

data envelopment analysis4
Data Envelopment Analysis

OBJECTIVE FUNCTION VALUE = 0.765

VARIABLEVALUE

E 0.765

W1 0.000

W2 0.500

W3 0.500

slide27
Conclusion

The output shows that the composite school is made up of equal weights of Lincoln and Washington. Roosevelt is 76.5% efficient compared to this composite school when measured by college admissions (because of the 0 slack on this constraint (#4)). It is less than 76.5% efficient when using measures of SAT scores and high school graduates (there is positive slack in constraints 2 and 3.)

data envelopment analysis5
Data Envelopment Analysis
  • (1) Relative Comparison
  • (2) Multiple Inputs and Outputs
  • (3) Efficiency Measurement (0%-100%)
  • (4) Avoid the Specification Error between Inputs and Outputs
  • (5) Production/Cost Analysis
slide29
Case : 1 input – 1 output

Table 1.1 : 1 input – 1 output Case

slide30
Efficiency Frontier

E

G

Output

F

C

A

H

B

D

0

Employees

Figure 1.1:Comparison of efficiencies in 1 input–1 output case

slide31
Efficiency Frontier

E

G

Output

F

C

A

Regression Line

H

B

D

0

Employees

Figure 1.2 : Regression Line and Efficiency Frontier

slide32
Table 1.2 : Efficiency

1 = C > G > A> B > E > D = F > H = 0.4

slide33
Efficiency Frontier

Output

C

D2

D1

D

0

Employee

Figure 1.3 : Improvement of Company D

slide34
Case : 2 inputs – 1 output

Table 1.3 : 2 inputs – 1 output Case

slide35
Production Possibility Set

G

F

C

A

I

Offices/Sales

E

D

B

Efficiency Frontier

H

0

Employees/Sales

Figure 1.4 : 2 inputs – 1 output Case

slide36
C

A

Offices/Sales

A1

A2

B

0

Employees/Sales

Figure 1.5 : Improvement of Company A

slide37
Case : 1 input – 2 outputs

Table 1.4 : 1 input – 2 outputs Case

slide38
A1

B

C

A

Efficiency Frontier

D

F

Sales/Office

Production Possibility Set

E1

G

E

0

Customers/Office

Figure 1.6 : 1 input – 2 outputs Case

slide39
Case : Multiple inputs – Multiple outputs

Table 1.5 : Example of Multiple inputs–Multiple outputs Case

slide47
Example Problem

Table 1.6 : 2 inputs – 1 output Case

slide50
Efficiency Frontier

E

A

D

A1

F

C

0

Figure 1.7 : Efficiency of DMU A

slide54
Efficiency Frontier of Ratio model

(B)

d

c

Efficiency Frontier of VRTS model

Output

b

(C)

a

(A)

0

Input

Figure 2.1 : Efficiency Frontier and Production Possibility Set

data envelopment analysis6
Data Envelopment Analysis

MIN E

s.t. Weighted outputs > Unit k’s output

(for each measured output)

Weighted inputs

(for each measured input)

Sum of weights = 1

E, weights > 0

final project international competitiveness in the semiconductor industry an application of dea

Final Project:International Competitiveness in the Semiconductor Industry: An Application of DEA

Tim Dekker

MEMGT

New Mexico Institute of Mining and Technology

home work
Home Work
  • Problem 5-2
  • Problem 5-3
  • Problem 5-4
  • Due Day: Sep 23
slide61
“Complementary slackness Conditions” are obtained from (4)

( c - y*A ) x* = 0

y*( b - Ax* ) = 0

xj* > 0 y*aj = cj , y*aj> cjxj* = 0

yi* > 0 aix* = bi , ai x*

(5)

(6)

slide62
Fundamental Insight

Z

RHS

Z

1

Row0

0

Row1~N

slide63

18-1

Max 5x1+6x2+4x3+0s1+0s2+0s3

CB = [0, 4, 5]

18 1 a

18-1 a.

Replace “5” by “c1”. Since x1 is basic,

then cBB-1A2 – c2≥ 0 becomes

18 1 a cont

18-1 a. cont.

cBB-1As1 – cs1≥ 0 becomes

18 1 a cont1

18-1 a. cont.

cBB-1As2 – cs2≥ 0 becomes

18 1 b

18-1 b.

Replace “6” by “c2”. Since x2 is non-basic,

then cBB-1A2 – c2≥ 0 becomes

18 1 c

18-1 c.

Replace “4” by “cs1”. Since s1 is non-basic,

then cBB-1As1 – cs1≥ 0 becomes

18 3 a b c
= 1

= 2

= 0

18-3 a, b, c

The dual prices for the first, second, third constraints

18 15 a b

18-15 a, b

The optimal simplex tableau

The optimal solution: x1 = 4, x2 = 0.5

The value of the objective function: 75

18 15 c

18-15 c

The optimal simplex tableau

If a dual variable is positive, then its corresponding constraint is binding. So, the first and second constrains binding (due to CSC).

18 15 d

18-15 d

The redundant (non-binding) is Constraint 3.

x1+x2+2x3+s3=6 is replaced by x1=4, x2=0.5, x3=0 as follows:

4+0.5+2(0)+s3=6.

Hence, s3=1.5.

18 15 e
Dual prices:

= 7.5

= 15

= 0

18-15 e

The optimal simplex tableau

18 15 e1

18-15 e

Increasing the RHS of constraint 2 would have the

greatest positive effect on the objective function.

18 15 f

18-15 f

Replace “15” by “c1”. Since x1 is basic,

then cBB-1A3 – c3≥ 0 becomes

18 15 f cont

18-15 f. cont.

cBB-1As1 – cs1≥ 0 becomes

18 15 f cont1

18-15 f. cont.

cBB-1As2 – cs2≥ 0 becomes

18 15 f cont2

18-15 f cont.

Replace “30” by “c2”. Since x2 is basic,

then cBB-1A3 – c3≥ 0 becomes

18 15 f cont3

18-15 f cont.

cBB-1As1 – cs1≥ 0 becomes

18 15 f cont4

18-15 f cont.

cBB-1As2 – cs2≥ 0 becomes

18 15 f cont5

18-15 f cont.

Replace “20” by “c3”. Since x2 is non-basic,

then cBB-1A3 – c3≥ 0 becomes

18 15 f cont6

18-15 f cont.

The optimal solution will not change as long as the objective function coefficients stay in these intervals.

18 15 g1

18-15 g

For b1

Therefore -4 ≤ ∆b1 ≤ 2

Range: (4-4 ≤ b1 ≤ 4+2) So, 0 ≤ b1 ≤ 6

18 15 g cont

18-15 g cont.

For b2

Therefore -1 ≤ ∆b2 ≤ 3

Range: (3-1≤ b2 ≤ 3+3) So, 2 ≤ b2 ≤ 6

18 15 g cont1

18-15 g cont.

For b3

Therefore -1.5 ≤ ∆b3

Range: (6-1.5≤ b3) So, 4.5 ≤ b3

18 17 b

18-17 b

The optimal simplex tableau

Optimal solution: y1 = 0.3, y2 = 0, y3 = 1.8

x1 = 0, x2 = 25, x3 = 125, x4 = 0.

18 17 c

18-17 c

Dual variables y1 = 0.3, y2 = 0, y3 = 1.8indicate the increase level of the objective function by one unit increase in each RHS.

If we increase Machine A by one hour, the profit will be increased by 0.3 ($/hour).

Similarly, a profit increase of Machine C will be 1.8 ($/hour).

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