emgt 501 hw #1 solutions chapter 2 - self test 18 chapter 2 - self test 20
DESCRIPTION
EMGT 501 HW #1 Solutions Chapter 2 - SELF TEST 18 Chapter 2 - SELF TEST 20 Chapter 3 - SELF TEST 28 Chapter 4 - SELF TEST 3 Chapter 5 - SELF TEST 6. Ch. 2 – 18 (a). Ch. 2 – 18 (b). (c) s 1 = 0, s 2 = 0, s 3 = 4/7. Ch. 2 – 20 (a). - PowerPoint PPT PresentationTRANSCRIPT
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EMGT 501
HW #1 SolutionsChapter 2 - SELF TEST 18
Chapter 2 - SELF TEST 20
Chapter 3 - SELF TEST 28
Chapter 4 - SELF TEST 3
Chapter 5 - SELF TEST 6
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Ch. 2 – 18(a)
Max 4x1 + 1x2 + 0s1 + 0s2 + 0s3
s.t.
10x1 + 2x2 + 1s1 = 30
3x1 + 2x2 + 1s2 = 12
2x1 + 2x2 + 1s3 = 10
x1, x2, s1, s2, s3 0
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Ch. 2 – 18(b)
x2
x10 2 4 6 8 10
2
4
6
8
10
12
14
Optimal Solution
x1 = 18/7, x2 = 15/7, Value = 87/7
(c)s1 = 0, s2 = 0, s3 = 4/7
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Ch. 2 – 20(a)
Let E = number of units of the EZ-Rider produced L = number of units of the Lady-Sport produced
Max 2400E + 1800L
s.t.
6E + 3L 2100 Engine time
L 280 Lady-Sport maximum
2E + 2.5L 1000 Assembly and testing
E, L 0
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Ch. 2 – 20(b)
0
L
Profit = $960,000
Optimal Solution
100
200
300
400
500
600
700
100 200 300 400 500E
Engine Manufacturing Time
Frames for Lady-Sport
Assembly and Testing
E = 250, L = 200
Number of Lady-Sport Produced
Num
ber
of E
Z-R
ider
Pro
duce
d
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Ch. 2 – 20(c)
The binding constraints are the manufacturing time and the assembly and testing time.
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Ch. 3 – 28(a)
Let A = number of shares of stock AB = number of shares of stock BC = number of shares of stock CD = number of shares of stock D
To get data on a per share basis multiply price by rate of return or risk measure value.
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Min 10A + 3.5B + 4C + 3.2D
s.t.
100A + 50B + 80C + 40D = 200,000
12A + 4B + 4.8C + 4D 18,000 (9% of 200,00)
100A 100,000
50B 100,000
80C 100,000
40D 100,000
A, B, C, D 0
Solution: A = 333.3, B = 0, C = 833.3, D = 2500Risk: 14,666.7Return: 18,000 (9%) from constraint 2
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Ch. 3 – 28(b) Max 12A + 4B + 4.8C +
4D
s.t.
100A + 50B + 80C +
40D
= 200,000
100A 100,000
50B 100,000
80C 100,000
40D
100,000
A, B, C, D 0
Solution: A = 1000, B = 0, C = 0, D = 2500Risk: 10A + 3.5B + 4C + 3.2D = 18,000Return: 22,000 (11%)
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Ch. 3 – 28(c)
The return in part (b) is $4,000 or 2% greater, but the risk index has increased by 3,333.
Obtaining a reasonable return with a lower risk is a preferred strategy in many financial firms. The more speculative, higher return investments are not always preferred because of their associated higher risk.
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Ch. 4 – 3 x1 = $ automobile loansx2 = $ furniture loansx3 = $ other secured loansx4 = $ signature loansx5 = $ "risk free" securities
Max 0.08x1 + 0.10x2 + 0.11x3 + 0.12x4 + 0.09x5
s.t.
x5 600,000 [1]
x4 0.10(x1 + x2 + x3 + x4)
or -0.10x1 - 0.10x2 - 0.10x3 + 0.90x4 0 [2]
x2 + x3 x1
or - x1 + x2 + x3 0 [3]
x3 + x4 x5
or + x3 + x4 - x5 0 [4]
x1 + x2 + x3 + x4 + x5 = 2,000,000 [5]
x1, x2, x3, x4, x5 0
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Automobile Loans (x1) = $630,000
Furniture Loans (x2) = $170,000
Other Secured Loans (x3) = $460,000
Signature Loans (x4) = $140,000
Risk Free Loans (x5) = $600,000
Solution
Annual Return $188,800 (9.44%)
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Ch. 5 – 6(a)
x 3
25
0
1
-1/2
-25
s 1
0
1
0
0
0
x 1
5
2
0
3
-5
x 2
1
2
20
0
-20
Basis
s 1
s 2
s 3
c B
0
0
0
Z -c j + z j
s 2
0
0
1
0
0
s 3
0
0
0
1
0
40
30
15
0
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Ch. 5 – 6(b)
Max 5x1 + 20x2 + 25x3 + 0s1 + 0s2 + 0s3
s.t.
2x1 + 1x2 + 1s1 = 40
2x2 + 1x3 + 1s2 = 30
3x1 -
1/2x3
+ 1s3 = 15
x1, x2, x3, s1, s2, s3, 0.
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Ch. 5 – 6(c) The original basis consists of s1, s2, and s3. It is the
origin since the nonbasic variables are x1, x2, and x3 and are all zero.
(d) 0
(e)x3 enters because it has the largest negative zj - cj and s2 will leave because row 2 has the only positive coefficient.
(f) 30; objective function value is 30 times 25 or 750.
(g) Optimal Solution: x1 = 10 s1 = 20x2 = 0 s2 = 0x3 = 30 s3 = 0z = 800.
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EMGT 501
HW #2Chapter 6 - SELF TEST 21
Chapter 6 - SELF TEST 22
Due Day: Sep 27
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s.t.
634Max 321 xxx
20211
3012
1515.01
321
32
321
xxx
xx
xxx
0 , , 321 xxx
Ch. 6 – 21Consider the following linear programming problem:
a. Write the dual problem.b. Solve the dual.c. Use the dual solution to identify the optimal solution to
the original primal problem.d. Verify that the optimal values for the primal and dual
problems are equal.
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Ch. 6 – 22A sales representative who sells two products is trying to determine the number of sales calls that should be made during the next month to promote each product. Based on past experience, representatives earn an average $10 commission for every call on product 1 and a $5 commission for every call on product 2. The company requires at least 20 calls per month for each product and not more than 100 calls per month on any one product. In addition, the sales representative spends about 3 hours on each call for product 1 and 1 hour on each call for product 2. If 175 selling hours are available next month, how many calls should be made for each of the two products to maximize the commission?
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a. Formulate a linear program for this problem.b. Formulate and solve the dual problem.c. Use the final simplex tableau for the dual problem to
determine the optimal number of calls for the products. What is the maximum commission?
d. Interpret the values of the dual variables.
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Duality Theory
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One of the most important discoveries in the early development of linear programming was the concept of duality.
Every linear programming problem is associated with another linear programming problem called the dual.
The relationships between the dual problem and the original problem (called the primal) prove to be extremely useful in a variety of ways.
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The dual problem uses exactly the same parameters as the primal problem, but in different location.
Primal and Dual Problems
Primal Problem Dual Problem
Max
s.t.
Min
s.t.
n
jjj xcZ
1
,
m
iii ybW
1
,
n
1jijij ,bxa
m
ijiij cya
1,
for for.,,2,1 mi .,,2,1 nj
for .,,2,1 mi for .,,2,1 nj ,0jx ,0iy
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In matrix notation
Primal Problem Dual Problem
Maximize
subject to
.0x .0y
Minimize
subject tobAx cyA
,cxZ ,ybW
Where and are row vectors but and are column vectors.
c myyyy ,,, 21 b x
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Example
Maxs.t.
Min
s.t.
Primal Problemin Algebraic Form
Dual Problem in Algebraic Form
,53 21 xxZ ,18124 321 yyyW
1823 21 xx
122 2 x41x
0x,0x 21
522 32 yy
33 3 y1y
0y,0y,0y 321
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Max
s.t.
Primal Problem in Matrix Form
Dual Problem in Matrix Form
Min
s.t.
,5,32
1
x
xZ
18
12
4
,
2
2
0
3
0
1
2
1
x
x
.0
0
2
1
x
x .0,0,0,, 321 yyy
5,3
2
2
0
3
0
1
,, 321
yyy
18
12
4
,, 321 yyyW
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Primal-dual table for linear programmingPrimal Problem
Coefficient of: RightSide
Rig
ht
Sid
eDu
al P
rob
lem
Co
effi
cien
to
f:
my
y
y
2
1
21
11
a
a
22
12
a
a
n
n
a
a
2
1
1x 2x nx
1c 2c ncVI VI VI
Coefficients forObjective Function
(Maximize)
1b
mna2ma1ma
2b
mb
Coe
ffic
ient
s fo
r O
bjec
tive
Fun
ctio
n(M
inim
ize)
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One Problem Other Problem
Constraint Variable
Objective function Right sides
i i
Relationships between Primal and Dual Problems
Minimization Maximization
Variables
Variables
Constraints
Constraints
0
0
0
0
Unrestricted
Unrestricted
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The feasible solutions for a dual problem are
those that satisfy the condition of optimality for
its primal problem.
A maximum value of Z in a primal problem
equals the minimum value of W in the dual
problem.
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Rationale: Primal to Dual Reformulation
Max cxs.t. Ax b x 0
L(X,Y) = cx - y(Ax - b) = yb + (c - yA) x
Min yb
s.t. yA c
y 0
Lagrangian Function )],([ YXL
X
YXL
)],([
= c-yA
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The following relation is always maintained
yAx yb (from Primal: Ax b)
yAx cx (from Dual : yA c)
From (1) and (2), we have
cx yAx yb
At optimality
cx* = y*Ax* = y*b
is always maintained.
(1)
(2)
(3)
(4)
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“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)
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Any pair of primal and dual problems can be
converted to each other.
The dual of a dual problem always is the primal
problem.
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Min W = yb,
s.t. yA c
y 0.
Dual ProblemMax (-W) = -yb,
s.t. -yA -c
y 0.
Converted to Standard Form
Min (-Z) = -cx,
s.t. -Ax -b
x 0.
Its Dual Problem
Max Z = cx,
s.t. Ax b
x 0.
Converted toStandard Form
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Mins.t.
64.06.0
65.05.0
7.21.03.0
21
21
21
xx
xx
xx
0,0 21 xx
21 5.04.0 xx
Mins.t.
][y 64.06.0
][y 65.05.0
][y 65.05.0
][y 7.21.03.0
321
-221
221
121
xx
xx
xx
xx
0,0 21 xx
21 5.04.0 xx
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Maxs.t.
.0,0,0,0
5.04.0)(5.01.0
4.06.0)(5.03.0
6)(67.2
3221
3221
3221
3221
yyyy
yyyy
yyyy
yyyy
Maxs.t.
.0, URS:,0
5.04.05.01.0
4.06.05.03.0
667.2
321
321
321
321
yyy
yyy
yyy
yyy
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Question 1: Consider the following problem.
,342 Maximize 321 xxxZ
8023
4022
60243
321
321
321
xxx
xxx
xxxsubject to
and
.0,0,0 321 xxx
(b) Work through the simplex method step by step in tabular form.
(c) Use a software package based on the simplex method to solve
the problem.
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Question 2:
For each of the following linear programming models, give your recommendation on which is the more efficient way (probably) to obtain an optimal solution: by applying the simplex method directly to this primal problem or by applying the simplex method directly to the dual problem instead. Explain.
(a) Maximize
subject to
,7410 321 xxxZ
202
90
4025
2532
2523
321
321
321
321
321
xxx
xxx
xxx
xxx
xxx
and
.0,0,0 321 xxx
(b) Maximize
subject to
,4352 54321 xxxxxZ
157564
6323
54321
54321
xxxxx
xxxxx
and
.5,4,3,2,1,0 jforx j
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Question 3:
Consider the following problem.
Maximize
subject to
,2 321 xxxZ
1
122
321
321
xxx
xxx
and
.0,0,0 321 xxx
(a) Construct the dual problem.
(b) Use duality theory to show that the optimal solution
for the primal problem has .0Z