Q 4 – 1 a.Let T = number of TV advertisements

R = number of radio advertisements

N = number of newspaper advertisements

Max 100,000T + 18,000R + 40,000N

s.t.

2,000T + 300R + 600N ≦ 18,200 Budget

T ≦ 10 Max TV

R ≦ 20 Max Radio

N ≦ 10 Max News

-0.5T + 0.5R - 0.5N ≦ 0 Max 50% Radio

0.9T - 0.1R - 0.1N ≧ 0 Min 10% TV

T, R, N, 0≧

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 a. cont’d

Q 4 – 1 b. Computer Results

OPTIMAL SOLUTION

Objective Function Value = 1052000

Variable Value Reduced Costs

T 4.000 0.000

R 14.000 0.000

N 10.000 0.000

Constraint Slack/Surplus Dual Prices

1 0.000 51.304

2 6.000 0.000

3 6.000 0.000

4 0.000 11826.087

5 0.000 5217.391

6 1.200 0.000

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.

RIGHT HAND SIDE RANGE

Constraint Lower Limit Current Value Upper Limit

1 14750.000 18200.000 31999.996

2 4.000 10.000 No Upper Limit

3 14.000 20.000 No Upper Limit

4 0.000 10.000 12.339

5 -8.050 0.000 2.936

6 No Lower Limit 0.000 1.200

Q 4 – 10 a.

Let S = the proportion of funds invested in stocksB = the proportion of funds invested in bondsM = the proportion of funds invested in mutual fundsC = the proportion of funds invested in cash

Max 0.1S + 0.03B + 0.04M + 0.01C

s.t.

1S + 1B + 1M + 1C = 1

0.8S + 0.2B + 0.3M ≦ 0.4

1S ≦ 0.75

- 1B + 1M ≧ 0

1C ≧ 0.1

1C ≦ 0.3

S, B, M, C ≧ 0

Q 4 – 10 a. cont’dOPTIMAL SOLUTION

Objective Function Value = 0.054

Variable Value Reduced Costs

S 0.409 0.000

B 0.145 0.000

M 0.145 0.000

C 0.300 0.000

Constraint Slack/Surplus Dual Prices

1 0.000 0.005

2 0.000 0.118

3 0.341 0.000

4 0.000 -0.001

5 0.200 0.000

6 0.000 0.005

Q 4 – 10 a. cont’d

OBJECTIVE COEFFICIENT RANGES

Variable Lower Limit Current Value Upper Limit

S 0.090 0.100 No Upper Limit

B 0.028 0.030 0.036

M No Lower Limit 0.040 0.042

C 0.005 0.010 No Upper Limit

RIGHT HAND SIDE RANGES

Constraint Lower Limit Current Value Upper Limit

1 0.800 1.000 1.900

2 0.175 0.400 0.560

3 0.409 0.750 No Upper Limit

4 -0.267 0.000 0.320

5 No Lower Limit 0.100 0.300

6 0.100 0.300 0.500

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.

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.

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.

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.

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 – 1 (a) & (b)

0xx,3x,1x

9)3)(2()1)(3(3

123bBC

x

x

3

1

4

5

21

11bB

21

11B

*3

*1

*4

*2

1B

4

211

A – 1 (c)

10)5)(2()0)(3(5

023bBC

5

0

2

1

3

1

1

0

21

11

3

1)bb(BbB

1B

11

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 (a)

Min 10S + 8C + 25V + 15A + 6T

s.t. 50S + 100C + 120A + 80T ≧ 380

50S + 100C + 120A + 80T ≦ 420

-9S + 55C - 24A + 14T ≦ 0

20S + 50V + 2T ≧ 50

1S - 2A ≧ 0

3S + 8C + 1V + 2A + 25T = 15

All variables 0≧

3125.58Total

0T,6042.1A,0V,2708.0C,2083.3S *****

A – 2 (b)

Max 380u1 - 420u2 + 0u3 + 50u4 + 0u5 + 15u6

s.t. 50u1 - 50u2 + 9u3 + 20u4 + 1u5 + 3u6 ≤ 10

100u1 - 100u2 - 55u3 + 8u6 ≤ 8

50u4 + 1u6 ≤ 25

120u1 - 120u2 + 24u3 - 2u5 + 2u6 ≤ 15

80u1 - 80u2 - 14u3 + 2u4 + 25u6 ≤ 6

u1 ~ u5 ≧ 0, u6 : URS

A – 2 (c)

8125.1u0u

1875.1u0u

0u2250.0u

*6

*3

*5

*2

*4

*1

Since u1*, u5

*, u6* > 0, the 1st, 5th, and 6th

constraints are binding.

The Langley County School District is trying to

determine the relative efficiency ofits three high schools. In particular,it wants to evaluate Roosevelt High.

The district is evaluating performances on SAT scores, thenumber of seniors finishing highschool, and the number of studentswho enter college as a function of thenumber of teachers teaching seniorclasses, the prorated budget for senior instruction,

and the number of students in the senior class.

Data Envelopment Analysis

• Input

Roosevelt Lincoln Washington

Senior Faculty 37 25 23

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

Senior Enrollments 850 700 600

• Output

Roosevelt Lincoln Washington

Average SAT Score 800 830 900

High School Graduates 450 500 400

College Admissions 140 250 370

• 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 Analysis

• Objective Function

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

MIN E

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)

Data Envelopment Analysis

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

OBJECTIVE FUNCTION VALUE = 0.765

VARIABLE VALUE E 0.765 W1 0.000 W2 0.500

W3 0.500

Data Envelopment Analysis

• 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 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

Table 1.1 : 1 input – 1 output Case

Company A B C D E F G HEmployees 4 3 3 2 8 6 5 5Output 3 2 3 1 5 3 4 2Output/Employee 0.75 0.667 1 0.5 0.625 0.5 0.8 0.4

Case : 1 input – 1 output

0

Out

put

Employees

D

B

A

G

H

F

E

C

Efficiency Frontier

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

0

Out

put

Employees

C

Efficiency Frontier

Figure 1.2 : Regression Line and Efficiency Frontier

Regression Line

D

B

A

G

H

F

E

Table 1.2 : Efficiency

Company A B C D E F G HEfficiency 0.75 0.667 1 0.5 0.625 0.5 0.8 0.4

1 of employeeper Sales

another of employeeper Sales0

C

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

0

Out

put

Employee

D

C

Efficiency Frontier

Figure 1.3 : Improvement of Company D

D2

D1

Table 1.3 : 2 inputs – 1 output Case

Company A B C D E F G H IEmployees 4 4 2 6 7 7 3 8 5Offices 3 2 4 2 3 4 4 1 3Sales 1 1 1 1 1 1 1 1 1

Case : 2 inputs – 1 output

0

Off

ices

/Sal

es

Employees/Sales

DB

A

G

H

F

E

C

Efficiency Frontier

Figure 1.4 : 2 inputs – 1 output Case

I

Production Possibility Set

0

Off

ices

/Sal

es

Employees/Sales

B

C

Figure 1.5 : Improvement of Company A

AA1

A2

C and B :A of set referenceR

OAOAA of efficiency The 2

Table 1.4 : 1 input – 2 outputs Case

Company A B C D E F GOffices 1 1 1 1 1 1 1Customers 1 2 4 4 5 6 7Sales 6 7 6 5 2 4 2

Case : 1 input – 2 outputs

0

Sal

es/O

ffice

Customers/Office

B

F

C

Figure 1.6 : 1 input – 2 outputs Case

G

A

A1

D

E1

E

Efficiency Frontier

Production Possibility Set

1

1

OEOEE of efficiency The

OAOAA of efficiency The

Table 1.5 : Example of Multiple inputs–Multiple outputs Case

Company A B C D E F G H I J K LEmployees 10 26 40 35 30 33 37 50 31 12 20 45Offices 8 10 15 28 21 10 12 22 15 10 12 26Customers 23 37 80 76 23 38 78 68 48 16 64 72Sales 21 32 68 60 20 41 65 77 33 36 23 35

Case : Multiple inputs – Multiple outputs

1.1,

21

22221

11211

21

22221

11211

snss

n

n

mnmm

n

n

yyy

yyy

yyy

Y

xxx

xxx

xxx

X

nsnm

sjj2j1

rj

mjj2j1

ij

j

y,,y,y

DMU th j theofoutput th r theofamount The :y

x,,x,x

DMU th j theofinput th i theofamount The :x

n), 2, 1,j ( UnitMakingDecision th j The : DMU

0u,,u,u,0v,,v,v

2.1n,,2,1j1xvxv

yuyusubject to

xvxvxv

yuyuyuEMaximize

s21m21

mjmj11

sjsj11

mkmk22k11

sksk22k11

Ratio model

s,,rru

m,,iiv

r

i

21output th the toassigned weight The :

21 input th the toassigned weight The :

0u,,u,u,0v,,v,v

xvxvyuyu

1xvxvsubject to

yuyuEMaximize

s21m21

mjmj11sjsj11

mkmk11

sksk11

n,,1j,xvyu:jRm

1iij

*i

s

1rrj

*rk

*** θ,u,v :Solution OptimalAn

R : A Reference Set

0u and 0v

1xv

0yuxvsubject to

yuMaximize

ri

m

1iiki

s

1rrjr

m

1iiji

s

1rrkr

Primal Problem

edunrestrict:E and 0w

ywy

0xwxsubject to

EMinimize

j

rkn

1jjrj

ikn

1jjij

Dual Problem

rkn

1jjrj

yr

n

1jjijik

xi

ywys

and wxExs

Slack

*yrrkrkrkrk

*xiik

*ikikik

syyyy

sxxxx

n,,1j,0jR *jk

Reference Set:

Table 1.6 : 2 inputs – 1 output Case

1x2xy

DMU A B C D E FInput 4 4 4 3 2 6

2 3 1 2 4 1Output 1 1 1 1 1 1

Example Problem

D,CR,833.0u,167.0v,167.0v

0u,0v,0v

1v2v4

F0uvv6

E0uv4v2

D0uv2v3

C0uvv4

B0uv3v4

A0uv2v4subject to

uMaximize

A**

2*1

21

21

21

21

21

21

21

21

Primal Problem

833.0E,0ww,667.0w,333.0w,0ww

URS:E,F,,B,Aj0w

1wwwwww

0E2ww4w2ww3w2

0E4w6w2w3w4w4w4subject to

EMinimize

**F

*E

*D

*C

*B

*A

j

FEDCBA

FEDCBA

FEDCBA

Dual Problem

0

D

F

E

Figure 1.7 : Efficiency of DMU A

A

A1

C

yx2

yx1

Efficiency Frontier

URS:E and 0w

ywy

0Exwxsubject to

EMinimize

j

rkn

1jjrj

ikn

1jjij

Original Ratio Model

URS:E and 0w

1w

ywy

0Exwxsubject to

EMinimize

j

n

1jj

rkn

1jjrj

ikn

1jjij

Model Under Variable RTS

edunrestrict: and 0u,0v

1xv

0yuxvsubject to

yuMaximize

ri

m

1iiki

s

1rrjr

m

1iiji

s

1rrkr

Dual Problem

0

Out

put

Input

b

a

c

Efficiency Frontier of Ratio model

Figure 2.1 : Efficiency Frontier and Production Possibility Set

d

(A)

(C)

(B)

Efficiency Frontier of VRTS model

MIN E

s.t. Weighted outputs > Unit k’s output (for each measured output)

Weighted inputs < E [Unit k’s input] (for each measured input)

Sum of weights = 1

E, weights > 0

Data Envelopment Analysis

Final Project:International Competitiveness in the

Semiconductor Industry: An Application of DEA

Tim DekkerMEMGT

New Mexico Institute of Mining and Technology

DEA Efficiency Results From Data Rev 2No. DMU Score 2002 Score 2003 Score 2004

1 Intel 0.761493304 1 12 Texas Instr 0.602603992 0.80699724 0.8904771293 STMicroelectronics 0.749494732 0.92533591 0.9852604474 Motorola 0.217143536 0.233115775 AMD 0.670097193 0.91555231 0.9548385926 Micron Technology 0.61023256 0.82163709 0.9291214867 QUALCOMM 0.486643612 0.86431345 0.7519564218 National Semiconductor 1 1 19 LSI Logic 0.569611642 0.70882038 1

10 Cypress 0.669665641 0.73797854 111 Fairchild 1 1 112 Mosaid Technologies 0.740146702 1 113 TSMC 1 1 114 Toshiba 0.210183267 0.27463386 0.34229277115 NEC Electronics 0.194462855 0.29446832 0.29093448316 Matsushita 0.068516963 0.086749598 0.08492488617 Sharp 0.228450337 0.31905201 0.30241332418 Hynix Semiconductor 0.519758507 0.79389776 0.97508488419 SANYO 0.185975212 0.20765726 0.12396147720 Fujitsu 0.12261189 0.11582924 0.10815617821 Rohm 0.867790264 0.61887155 0.61486083822 Via Technologies 1 1 123 Samsung 0.410945897 0.55344242 0.767531753

Weights used in the Analysis (2003) No. DMU Score V(1) V(2) V(3) V(4) U(0) U(1) U(2)

1 Intel 1 0.42644 0 0 0.188805 -3.29E-03 0 0.1778832 Texas Instr0.806997 0 0 0.299084 0.997193 -2.13E-02 0.534118 03 STMicroelectronics0.925336 0 0 0.354227 1.181049 -0.025256 0.632595 04 Motorola 0.233116 0 6.16E-02 0 0.506816 -1.12E-02 0.252286 05 AMD 0.915552 0.688273 0 0 2.294629 -2.75E-02 1.146893 06 Micron Technology0.821637 0 0 0.684147 2.281055 -4.88E-02 1.221782 07 QUALCOMM 0.864313 8.529142 0 0 0 -2.27E-03 0 1.6443558 National Semiconductor1 1.810097 0 0 6.177444 0 0 2.6165959 LSI Logic 0.70882 13.3663 0 0 0 -3.56E-03 0 2.576923

10 Cypress 0.737979 4.822213 0 0 8.512714 -8.91E-02 0 4.78469111 Fairchild 1 1.339373 15.63514 0 0 0 3.423204 012 Mosaid Technologies1 513.1354 0 0 0 0.248727 0 95.7476913 TSMC 1 4.078316 0.201364 0 0 0 0.816043 014 Toshiba 0.274634 0 8.17E-02 0 0.328684 -0.011731 0.186009 015 NEC Electronics0.294468 0 0.106336 0 0.405664 -1.45E-02 0 0.23375716 Matsushita 8.67E-02 0 5.60E-02 0 0.213542 -7.66E-03 0 0.1230517 Sharp 0.319052 7.36E-02 0.181132 0 0.799577 -1.84E-02 0 0.46057918 Hynix Semiconductor0.793898 0 0.614209 0 2.343158 -8.40E-02 0 1.35020519 SANYO 0.207657 0 0.175443 0 0.6693 -0.023993 0 0.38567320 Fujitsu 0.115829 0 0.102369 0 0.41183 -0.014699 0.233063 021 Rohm 0.618872 0 0.026985 1.138374 2.148457 -9.83E-03 0 1.19446522 Via Technologies1 43.2892 0 0 0 0 7.871678 023 Samsung 0.553442 0 1.64E-02 0.118873 0.477883 -9.50E-03 0.256124 0

Final Ranking for Firms in 2004

Company Region Score 2004 Final Ranking

2004 Intel Americas 1.8 5

Texas Instr Americas .86 10 STMicroelectronics Europe .83 12

AMD Americas .94 8 Micron Technology Americas .84 11

QUALCOMM Americas .71 14 National Semiconductor Americas 1.98 2

LSI Logic Americas .97 7 Cypress Americas .79 13 Fairchild Americas 2 1

Mosaid Technologies Americas 1.5 6 TSMC Asia/Pacific 1.83 4

Toshiba Japan .27 18 NEC Electronics Japan .23 19

Matsushita Japan .0731 22 Sharp Japan .28 17

Hynix Semiconductor Asia/Pacific .9 9 SANYO Japan .12 20 Fujitsu Japan .097 21 Rohm Japan .6 15

Via Technologies Asia/Pacific 1.96 3 Samsung Asia/Pacific .53 16

Home Work

• Problem 5-2

• Problem 5-3

• Problem 5-4

• Due Day: Sep 23

“Complementary slackness Conditions” are

obtained from (4)

( c - y*A ) x* = 0

y*( b - Ax* ) = 0

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

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

(5)

(6)

Fundamental Insight

Z

Z

RHS

Row0

Row1~N

1BcB

1B

bBcB1

bB 1

X

BX

SX

1

0 AB 1

cABcB 1

18-1

BV

x1 x2 x3 s1 s2 s3 RHS

0 2 0 1 2 0 220

s3 0 4 0 -2 7 1 80

x3 0 2 1 -1 3 0 30

x1 1 0 0 1 -2 0 20

CB = [0, 4, 5]

021

031

172

B 1

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

18-1 a.

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

then cBB-1A2 – c2 ≥ 0 becomes

)1( c00

06

2

2

4

]0,c212,c4[

06

2

2

4

021

031

172

]c,4,0[

1

11

1

18-1 a. cont.

cBB-1As1 – cs1 ≥ 0 becomes

)2( 4c0c4

00

0

0

1

]0,c212,c4[

00

0

0

1

021

031

172

]c,4,0[

11

11

1

18-1 a. cont.

cBB-1As2 – cs2 ≥ 0 becomes

6c4)3(),2(),1(

)3( c60c212

00

0

1

0

]0,c212,c4[

00

0

1

0

021

031

172

]c,4,0[

1

11

11

1

18-1 b.

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

then cBB-1A2 – c2 ≥ 0 becomes

8c0c8

0c

2

2

4

]0,2,1[

0c

2

2

4

021

031

172

]5,4,0[

22

2

2

18-1 c.

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

then cBB-1As1 – cs1 ≥ 0 becomes

1c0c1

0c

0

0

1

]0,2,1[

0c

0

0

1

021

031

172

]5,4,0[

1s1s

1s

1s

18-3 a, b, c

The dual prices for the first, second, third

constraints

BV

x1 x2 x3 s1 s2 s3 RHS

0 2 0 1 2 0 220

s3 0 4 0 -2 7 1 80

x3 0 2 1 -1 3 0 30

x1 1 0 0 1 -2 0 20

*1y = 1

*2y = 2

*3y = 0

18-3 d

Let

bbb , then

25

25

70

5

5

10

20

30

80

0

0

5

021

031

172

20

30

80

bBbB)bb(B 111BX

BV

x1 x2 x3 s1 s2 s3RHS

0 2 0 1 2 0 220

s3 0 4 0 -2 7 1 80

x3 0 2 1 -1 3 0 30

x1 1 0 0 1 -2 0 20

18-3 d cont.

225

5220

0

0

5

)0,2,1(220

bBcbBc 1B

1B

bBc 1B

18-3 e

210

)10(220

0

0

10

)0,2,1(220

bBcbBc

10

40

100

10

10

20

20

30

80

0

0

10

021

031

172

20

30

80

bBbB)bb(B

1B

1B

111

BX

bBc 1B

18-15 a, b

BV

x1 x2 x3 s1 s2 s3 RHS

0 0 2.5 7.5 15 0 75

x1 1 0 1 1 0 0 4

x2 0 1 0.25 -0.25 0.5 0 0.5

s3 0 0 0.75 -0.75 -0.5 1 1.5

The optimal simplex tableau

The optimal solution: x1 = 4, x2 = 0.5

The value of the objective function: 75

18-15 c

BV

x1 x2 x3 s1 s2 s3 RHS

0 0 2.5 7.5 15 0 75

x1 1 0 1 1 0 0 4

x2 0 1 0.25 -0.25 0.5 0 0.5

s3 0 0 0.75 -0.75 -0.5 1 1.5

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

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

BV

x1 x2 x3 s1 s2 s3 RHS

0 0 2.5 7.5 15 0 75

x1 1 0 1 1 0 0 4

x2 0 1 0.25 -0.25 0.5 0 0.5

s3 0 0 0.75 -0.75 -0.5 1 1.5

Dual prices: *1y = 7.5

*2y = 15

*3y = 0

The optimal simplex tableau

18-15 e

0

1

0

0

0,15,5.7bBc

15

0

1

0

0,15,5.7bBc

5.7

0

0

1

0,15,5.7bBc

31

B

21

B

11

B

Increasing the RHS of constraint 2 would have the greatest positive effect on the objective function.

18-15 f

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

then cBB-1A3 – c3 ≥ 0 becomes

)1( 5.12c05.12c

020

2

1

1

]0,15,5.7c[

020

2

1

1

15.075.0

05.025.0

001

]0,30,c[

11

1

1

18-15 f. cont.

cBB-1As1 – cs1 ≥ 0 becomes

)2( 5.7c05.7c

00

0

0

1

]0,15,5.7c[

00

0

0

1

15.075.0

05.025.0

001

]0,30,c[

11

1

1

18-15 f. cont.

cBB-1As2 – cs2 ≥ 0 becomes

1

1

1

1

c5.12)3(),2(),1(

)3( c00

00

0

1

0

]0,15,5.7c[

00

0

1

0

15.075.0

05.025.0

001

]0,30,c[

18-15 f cont.

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

then cBB-1A3 – c3 ≥ 0 becomes

)4( 20c05c25.0

020

2

1

1

]0,c5.0,c25.015[

020

2

1

1

15.075.0

05.025.0

001

]0,c,15[

22

22

2

18-15 f cont.

cBB-1As1 – cs1 ≥ 0 becomes

)5( 60c0c25.015

00

0

0

1

]0,c5.0,c25.015[

00

0

0

1

15.075.0

05.025.0

001

]0,c,15[

22

22

2

18-15 f cont.

cBB-1As2 – cs2 ≥ 0 becomes

60c20)6(),5(),4(

)6( c00c25.0

00

0

1

0

]0,c5.0,c25.015[

00

0

1

0

15.075.0

05.025.0

001

]0,c,15[

2

22

22

2

18-15 f cont.

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

then cBB-1A3 – c3 ≥ 0 becomes

33

3

3

c5.220c5.22

0c

2

1

1

]0,15,5.7[

0c

2

1

1

15.075.0

05.025.0

001

]0,30,15[

18-15 f cont.

The optimal solution will not change as long as

the objective function coefficients stay in these

intervals.

1c5.12 60c20 2

5.22c3

18-15 g

0

0

b

121

43

021

41

001

5.1

5.0

4

0

0

b

bB

1

11

18-15 g

Therefore -4 ≤ ∆b1 ≤ 2

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

4 + ∆b1(1) ≥ 0 → ∆b1 ≥ -4

0.5 + ∆b1(-0.25) ≥ 0 → ∆b1 ≤ 2

1.5 + ∆b1(-0.75) ≥ 0 → ∆b1 ≤ 2

For b1

18-15 g

0

b

0

121

43

021

41

001

5.1

5.0

4

0

b

0

bB

2

21

18-15 g cont.

Therefore -1 ≤ ∆b2 ≤ 3

Range: (3-1≤ b2 ≤ 3+3) So, 2 ≤ b2 ≤ 6

4 + ∆b2(0) ≥ 0 → no restriction

0.5 + ∆b2(0.5) ≥ 0 → ∆b2 ≥ -1

1.5 + ∆b2(-0.5) ≥ 0 → ∆b2 ≤ 3

For b2

18-15 g

3

3

1

b

0

0

121

43

021

41

001

5.1

5.0

4

b

0

0

bB

18-15 g cont.

Therefore -1.5 ≤ ∆b3

Range: (6-1.5≤ b3) So, 4.5 ≤ b3

4 + ∆b3(0) ≥ 0 → no restriction

0.5 + ∆b3(0) ≥ 0 → no restriction

1.5 + ∆b3(1) ≥ 0 → ∆b3 ≥ -1.5

For b3

18-17 a

Min 550y1 + 700y2 + 200y3

s.t.

1.5y1 + 4y2 + 2y3 ≥ 4

2y1 + 1y2 + 3y3 ≥ 6

4y1 + 2y2 + 1y3 ≥ 3

3y1 + 1y2 + 23 ≥ 1

y1, y2 , y3 ≥ 0

18-17 b

BVy1 y2 y3 s1 s2 s3 s4 RHS

0 425 0 0 25 125 0 525

y3 0 0 1 0 -0.4 0.2 0 1.8

s1 0 -3.25 0 1 -0.65 -0.05 0 0.05

s4 0 0.5 0 0 -0.5 -0.5 1 3.5

y1 1 0.5 0 0 0.1 -0.3 0 0.3

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

Dual variables y1 = 0.3, y2 = 0, y3 = 1.8 indicate 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).