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).