Samuel Coy
BUS 355B
Homework Assignment 2
As a student of the Dr. Robert B. Pamplin Jr. School of Business Administration I have read and strive
to uphold the University’s Code of Academic Integrity and promote ethical behavior. In doing so, I
pledge on my honor that I have not given, received, or used any unauthorized materials or assistance
on this examination or assignment. I further pledge that I have not engaged in cheating, forgery, or
plagiarism and I have cited all appropriate sources.
Jacob Neary, Kenny King, Taylor Jones, Samuel Coy
_________________________________________________
Student Signature
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BUS 355B
Homework Assignment 2
Problem 1:
Let:
x be the first home mortgage
y be the second home mortgage
z be the commercial loan
w be the automobile loan
v be the home improvement loan
r be the risk-free securities
1.1.
max
s.t.
.06 x  .08 y  .11z  .09w  .1v  .04r
x
y
z
w v
r  90
A
r  9
v  8
B
C
 .6 x  y  z  w  v 
x y
D
x

2y
E
v
 .4 x
F

G
wv
z
 .5 x  y 
z
x, y, z , w, v, r
H
 0
1.2.
From the information given in the problem, constraints A, B, and C are known constraints.
Constraint D is currently an unknown variable as a result of being non linearity, variables on both
sides, and divisibility. To become a usable constraint, it must be manipulated to…
D
.4 x  .4 y  .6 z  .6w  .6v  0
Constraint E is also an unknown variable due to divisibility, variables on both sides, and linearity. To
be usable, the constraint must be manipulated to…
E
x  2y  0
Constraint F is unknown due to variables on both sides, divisibility, linearity. To be usable, the
constraint must read…
F
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v  .4 x  0
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Homework Assignment 2
Constraint G is also unknown because variables are on both sides. To be usable, the constraint should
read...
G
w  v  z 0
Constraint H is an unknown variable because of variables on both sides, divisibility, linearity. The
constraint should read…
H
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z  .5 x  .5 y  0
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BUS 355B
Homework Assignment 2
Problem 2:
Let:
Telephone
x1
x2
x3
x4
x5
x6
Mailings
x7
x8
x9
x10
x11
x12
Single, Under 55, No kids under 18
Married, Under 55, No kids under 18
Single, 55, with kids under 18
Married, 55, with kids under 18
Single, Over 55, No kids under 18
Married, Over 55, No kids under 18
Single, Under 55, No kids under 18
Married, Under 55, No kids under 18
Single with kids under 18
Married with kids under 18
Single, Over 55, No kids under 18
Married, Over 55, No kids under 18
2.1.
10 x1  11x 2  12 x3  15x 4  7 x5  9 x6  x7 x8  1.5x9  1.5x10  x11  x12
min
s.t. x3  x4  .6(2000)
A
x9  x10  .548,000
B
x1  x3  x5  x7  x9  x11
 .3
x1  x2  x3  x4  x5  x6  x7  x8  x9  x10  x11  x12
C
x2  x4  x6  x8  x10  x12
 . 25
x1  x2  x3  x4  x5  x6  x7  x8  x9  x10  x11  x12
D
x1  x2  x3  x4  x5  x6  x7  x8  x9  x10  x11  x12  0
2.2. We have determined that constraints A and B are linear and reliable constraints as a result of
given information, thus they are composed of known values.
Constraint C is unusable in its original form because it does not follow the divisibility and linearity
requirements. To be a usable constraint, it must be manipulated to read…
C
.7 x1  .3x2  .7 x3  .3x4  .7 x5  .3x6  .7 x7  .3x8  .7 x9  .3x10  .7 x11  .3x12  0
Constraint D is also unable because it is composed of an unknown value and it does not follow the
divisibility or linearity requirement. To be usable, the constraint must read…
D
 .25x1  .75x2  .25x3  .75x4  .25x5  .75x6  .25x7  .75x8  .25x9  .75x10  .25x11  .75x12  0
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BUS 355B
Homework Assignment 2
Problem 3:
max 10 x1 
x2
3x1

2 x2
 12
2 x1

2 x2
 16 B
x1

x2

2
x1 , x2

0
s.t.
A
C
1. Show your graph of the feasible region.
45.0
40.0
iso-profit: 10X1 + 1X2 =
33.2
35.0
A: 3X1 + 2X2 = 12
30.0
B: 2X1 + 2X2 = 16
25.0
C: 1X1 + -1X2 = 2
X2
20.0
15.0
10.0
5.0
0.0
-2
-5.0
0
2
4
6
8
10
X1
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BUS 355B
Homework Assignment 2
2. Use the iso-profit line method to identify the optimal solution (be sure to show any work with
simultaneous equations).
3 x1  2 x 2  12
3x1  2 x 2  12
3x1  2 x 2  12
x1  x 2  2
 2( x1  x 2  2)
  2 x1  2 x 2   4
x1  3.2
5 x1  16
x1  x2  2
3.2  x2  2
The optimal solution is point, (3.2, 1.2)
 x2   1.2
x2  1.2
3. Find the optimal value.
Maximize: 10x1 + x2 = 10(3.2) + 1.2 = 32+ 1.2= 33.2
The optimal value is 33.2
4. Calculate slack/surplus for every constraint.
Constraint A :
3x1  2 x2  12
3 3.2   2 1.2   12
9.6  2.4  12
12  12
Surplus : 12 12  0
Filename: Document1
Constraint B :
Constraint C :
2x1  2 x2  16
2 3.2  2 1.2  16
x1  x2  2
3.2  1.2  2
6.4  2.4  16
22
8.8  16
Surplus :16  8.8  7.2
Surplus : 2  2  0
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BUS 355B
Homework Assignment 2
Problem 4:
min 3 x1

4 x2
s.t.

x2
4 x1 
4 x2
 320 B

5 x2
 300 C
x1 , x2

6 x1
3 x1
 120
A
0
1. Show your graph of the feasible region.
140.0
120.0
iso-profit: 3X1 + 4X2 = 270
A: 6X1 + 1X2 = 120
100.0
B: 4X1 + 4X2 = 320
80.0
C: 3X1 + 5X2 = 300
X2
60.0
40.0
20.0
0.0
-20
0
20
40
60
80
100
-20.0
X1
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BUS 355B
Homework Assignment 2
2. Use the iso-profit line method to find the optimal solution (be sure to show any work with
simultaneous equations) and calculate the corresponding optimal value.
3 x1  4 x 2
4 x1  4 x 2  3 2 0
3x1  5 x 2  3 0 0
4 x1  4 x 2
5(4 x1  4 x 2  3 2) 0

20 x1  20 x 2  1600
4( 3x1  5 x 2  3 0) 0

12 x1  20 x 2  1200
3 x1  5 x 2
20 x1  20 x 2  1600

12 x1  20 x 2  1200
4(50)  4 x 2  320
200  4 x 2  320
4 x 2  120
x 2  30
8x1  400
x1  50
3x1  4 x 2
350  430 
150  120  270
Optimal Solution: (50, 30); 270
3. Calculate slack/surplus for every constraint.
Constraint A :
6x1  x2  120
Constraint B :
Constraint C :
4x1  4 x2  320
3x1  5 x2  300
6 50   30  120
4 50   4 30   320
3 50   5 30   300
300  30  120
200 120  320
150 150  300
330  120
320  320
300  300
Slack : 210
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Slack :0
Slack : 0
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BUS 355B
Homework Assignment 2
4.
Calculate the shadow price and range of feasibility for each constraint.
Range of Feasibility
Constraint A :
(0, 80) : 60   80  LEPA
0  80  LEPA
80  LEPA
Allowable Decrease B  330  80  250
50, 30 :650  30 UEPA
300  30 UEPA
Allowable Increase B  330
Rof A  (257.776, 400)
Constraint B :
(11.111, 53..333) : 411.111  4 53.333  LEPB
44.444  213.332  LEPB
257.776  LEPB
Allowable Decrease B  320  257.776  62.224
100, 0 : 4100  40 UEPB
400 UEPB
Allowable Increase B  400  320  80
Rof B  (257.776, 400)
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BUS 355B
Homework Assignment 2
Constraint C :
(80, 0) : 380  5 0  LEPC
240  LEPC
Allowable Decrease C  300  240  60
8, 72 :38  572  UEPC
384  UEPC
Allowable Increase C  384  300  84
Rof C  (240, 384)
Shadow Price
Constra int A :
3x1  5 x2  300

6 x1  x2  121
311.296  5 x2  300
33.888  5 x2  300
5 x2  266.111
x2  53.222
Filename: Document1
3x1  5 x2  300
  5 (6 x1  x2  121)
3 x1  5 x 2  300
  30 x1  5 x2   605
x1  11.296
 27 x1   305
3x1  4 x2 
311.296  453.222 
Shadow Pr ice :
246.78  270  23.22  0
33.888  212.888  246.78
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BUS 355B
Homework Assignment 2
Constra int B :
3x1  5 x 2  300

4 x1  4 x 2  321
 4 ( 3x1  5 x 2  300)

5 (4 x1  4 x 2  321)
350.625  5 x 2  300
 20 x1  20 x 2  1605
x1  50.625
8 x1  405
3x1  4 x 2 
151.875  5 x 2  300
350.625  429.625 
5 x 2  148.125
151.875  118.5  270.375
x 2  29.6256
Shadow Pr ice :
270.375  270  .375
Constra int C :

 12 x1  20 x 2   1200
4 x1  4 x 2  320
 5( 4 x1  4 x2  320)
3x1  5 x 2  301
 4 (3x1  5 x2  301)
449.5  4 x2  320
198  4 x2  320
4 x2  122
x2  30.5
3x1  4 x2 
349.5  430.5 
 20 x1  20 x2   1600
 12 x1  20 x2  1204
x1  49.5
 8 x1   396
Shadow Pr ice :
270.5  270  .5
148.5  122  270.5
5. Calculate the range of optimality for each decision variable.
Range of Optimality for x1
3 x1  4 x 2
4 x1  4 x 2
3 x1  5 x 2
x
4 a
a1  1  / 1
x2 4 4
a2 
x1 3 a 2
 /
x2 5 4
a1  4
a2 
3
20
4 3 12
* 
 2.4
1 5
5
Range of Optimality for x1 = (2.4, 4)
Range of Optimality for x2
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BUS 355B
Homework Assignment 2
3 x1  4 x 2
4 x1  4 x 2
3 x1  5 x 2
a1 
x 2 4 a1
 /
x1 4 3
a2 
a1 
12
3
4
a2 
x2 5 a2
 /
x1 3 3
15
5
3
Range of Optimality for x2 = (3, 5)
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Homework Assignment 2
Problem 5:
Let:
x
y
z
w
v
r
Under 30 and border Mexico
Between 31 and 50 and border Mexico
Over 51 and border Mexico
Under 30 and do not border Mexico
Between 31 and 50 and do not border Mexico
Over 51 and do not border Mexico
min 7.5 x  6.8 y  5.5 z  6.9w  7.25v  6.1r
s.t. x  y  z  w  x  r  2300
A
x  w  900
B
y  v  600
C
v  200
D
.85 x  .85 y  .85 z  .15w  .15v  .15r  0
 .2 x  .2 y  .8 z  .2w  .2v  .2r  0
E
F
x, y, z , w, v, r  0
Minimize
Household Surveyed
Households whose heads are 30 and under
between 31 and 50
BETween 31 and 50 and line in nbs
15% live in BM
20% 51+ and BM
Non negativity
Filename: Document1
x
0
7.5
1
1
y
400
6.8
1
z
800
5.5
1
w
900
6.9
1
1
1
0.85
-0.2
1
0.85
-0.2
1
0.85
0.8
1
-0.15
-0.2
1
v
200
7.25
1
r
0
6.1
1
1
1
-0.15
-0.2
1
-0.15
-0.2
1
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14780
2300
900
600
200
855
340
2300
>= 2300
>= 900
>= 600
>= 200
>=
0
>=
0
>=
0
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Samuel Coy
BUS 355B
Homework Assignment 2
Microsoft Excel 12.0 Sensitivity Report
Worksheet: [BUS355B_Coy_Samuel_HW1_Analysis_1.xlsm]Sheet1
Report Created: 4/21/2011 2:23:32 PM
Adjustable Cells
Cell
$B$2
$C$2
$D$2
$E$2
$F$2
$G$2
Name
x
y
z
w
v
r
Final
Reduced
Objective Allowable
Allowable Range of Optimality
Value
Cost
Coefficient Increase
Decrease
LEP
UEP
0 0.599999999
7.5
1E+30 0.599999999
6.9
1E+30
400
0
6.8
0.45
1.3
5.5
7.25
800
0
5.5
0.6
5.5 2.31E-14
6.1
900
0
6.9 0.599999999
1.4
5.5
7.5
200
0
7.25
1E+30
0.45
6.8
1E+30
0
0.6
6.1
1E+30
0.6
5.5
1E+30
Constraints
Cell
$H$4
$H$5
$H$6
$H$7
$H$8
$H$9
$H$10
Name
Household Surveyed
Households whose heads are 30 and under
between 31 and 50
BETween 31 and 50 and line in nbs
15% live in BM
20% 51+ and BM
Non negativity
Final
Value
2300
900
600
200
855
340
2300
Shadow
Constraint Allowable
Allowable Range of Feasibility
Price
R.H. Side
Increase
Decrease
LEP
UEP
5.5
2300
1E+30
425
1875
1E+30
1.4
900
340
900 -3.2E-10
1240
1.3
600
340
400
200
940
0.45
200
400
200
-1E-10
600
0
0
855
1E+30
-1E+30
855
0
0
340
1E+30
-1E+30
340
0
0
2300
1E+30
-1E+30
2300
1. MRA would like to lower its costs. Which of the constraints, if relaxed, would allow the
biggest cost reductions?
a. The constraint that relates to the how many households surveyed would have the largest
effect on the cost if it was relaxed. Currently the shadow price (the amount the
objective will change when one more unit is added) of the households surveyed is 5.5.
This number means that for every extra unit added to the households surveyed the cost
would increase by 5.5 dollars. If they would really like to cut costs they would
lower/relax this constraint.
2. If MRA wants to keep its costs below $15,000, how many fewer households would it have to
survey? Is this a feasible reduction?
a. According to solver, MRA is able to keep the cost under $15,000 while keeping within
their stated constraints. The minimum that MRA will have to spend is $14,780.
3. MRA’s client is considering eliminating the requirement that a minimum of 200 households
with head of household age 31 to 50 in non-border states be surveyed. Is this a feasible
change? If so, discuss the impact of eliminating this constraint.
a. This is a feasible change. When one works out this problem, one finds that the range of
feasibility is from -1.02E^-10 (which is almost zero) to the upper end point of 600. This
means that it would be feasible to get rid of this constraint. If this constraint was gotten
rid of, then that would also mean that the other numbers would change. For instance
this constraint is one of the more costly constraints. It costs $7.25 in order to survey one
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Homework Assignment 2
person between ages 31 and 50 who lives in a non-bordering state. If there was no
longer a constraint on having to survey these people (and if one was only concerned
with cost and not the range of their study) then one would most likely choose to survey
the people that cost less. Some of these options could be those people who are above
the age of 51 and are living in a state bordering Mexico.
4. MRA is concerned that their cost of surveying households in border states whose heads are
between 31 and 50 years of age may be understated. By how much can the cost increase
without changing the cost minimizing survey strategy?
a. If one does a calculation to find the range of optimality (by using solver and by opening
up a sensitivity report) then one can see this allowable increase. The Range of
Optimality tells one how much the co-efficient (in this case the price of the survey for
each person) can change without changing the graph. It shows how much it can change
without having to change the survey strategy. When one does this, one finds that the
range of optimality for the households in border states who are between 31 and 50 years
of age has a lower endpoint of $5.5 per household and an upper endpoint of $7.25 per
household. This means that the cost of this group can increase 45 cents or to $7.25
without changing the cost minimizing survey strategy.
5. The cost of surveying households in non-border states whose heads are 51 or older may be
overstated (because this demographic is more likely to be successfully surveyed on the first
attempt). At what cost would MRA want to survey more of these households?
a. Currently the optimal amount of households in this category surveyed (according to
solver and our calculation) was zero. In order for this number to change, the price
would have to drop below the range of optimality. Once this happened then the optimal
solution would change. Based on this and the lower endpoint of the range of optimality
for this section ($5.5 or a 60 cent decrease from the current status) the price would have
to drop below $5.50 in order for the optimal solution to change. If it dropped below this
number then it is very likely that it would then become more optimal to survey some
people in this demographic.
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Homework Assignment 2
Problem 6:
max
s.t.

15T
2.4Q  3.0 R 
4.0T

960
minutes Sorting
2.5Q  1.8R

2.4T

607
minutes Crushing


24T
 3, 600
14Q
12Q

11R
18R
Q, R, T
net profit

minutes Packing
0
Let Q, R, and T been number of pounds of each type of bottle.
Outcomes
Profit
Sorting
Crushing
Packing
Non-negativity
Q
190
14
2.4
2.5
12
1
Bottle Types
R
0
11
3
1.8
18
1
T
55
15
4
2.4
24
1
3485
676
607
3600
245
<=
<=
<=
>=
960
607
3600
0
Microsoft Excel 12.0 Sensitivity Report
Worksheet: [BUS355B_Coy_Samuel_HW1_Analysis_1 (3).xlsm]Problem 6
Report Created: 4/20/2011 6:08:59 PM
Adjustable Cells
Final Reduced Objective Allowable
Allowable Range of Optimality
Cell
Name
Value
Cost Coefficient Increase
Decrease
LEP
UEP
$B$3 Outcomes Q
190
0
14
1.625
6.5
7.5
15.625
$C$3 Outcomes R
0
-0.25
11
0.25
1E+30
-1E+30
11.25
$D$3 Outcomes T
55
0
15
13 0.333333333 14.66667
28
Constraints
Cell
Name
$E$5 Sorting
$E$6 Crushing
$E$7 Packing
Filename: Document1
Final Shadow Constraint Allowable
Allowable Range of Feasibility
Value Price
R.H. Side
Increase
Decrease
LEP
UEP
676
0
960
1E+30
284
676
1E+30
607
5
607
143
247
360
750
3600
0.125
3600 2089.811321
686.4
2913.6 5689.811
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Samuel Coy
BUS 355B
Homework Assignment 2
1. Interpret the business meaning of the objective function.
a. The business meaning for the objective function is that it tells one what the net profit is.
The simple goal of any business is to maximize revenue while lowering costs. The net
profit tells you the answer to how well you are doing this. This objective function tells
you the net profit and from this net profit a business can tell how well it is doing. This
equation tells us that whenever we add one extra unit of T, R, and Q, T has the greatest
impact on the final total. Q has the second most and R has the least.
2. Interpret the business meaning of the first constraint.
a. This constraint sheds light on the sorting process of the company. This equation tells
how much time is spent on sorting the different products. In addition it also tells us
what products take longer to sort. For instance the T unit takes the longest to sort
followed by the R unit and the Q unit.
3. Which decision variables are in the solution (i.e. which are non-zero)? What are their optimal
values (don’t forget to specify units of measure)?
a. In order to find this we ran the solver program. When one runs this program they find
that the Q and T units are both non-zero while the R unit is a zero. This means that the
optimal solution for the best combination of these units, under the current constraints is
190 units of Q, 55 units of T and zero units of R. This tells you one that R is not really
useful for the company. The company (again under these current constraints) would do
better without any units of R. Q seems to be the most widely used.
4. Find the range of optimality for the decision variables that are in the optimal solution. Find the
range of insignificance (i.e. the range of optimality where the decision variable remains “out
of” the optimal solution) for any other variables.
a. The range of optimality for the outcome Q has a lower endpoint of 7.5 and an upper
endpoint of 15.625. currently the coefficient is at 14. This means that for the Q
outcome it can drop by 6.5 units and still be in the optimal range. It can also increase
by 1.625 units and still not change the optimal solution. The T outcome currently has a
coefficient of 15. It also has a range of optimality with a lower endpoint of 14.66667
and an upper endpoint of 28. This means that one could increase this coefficient by 13
and the amount of units would still be the optimal solution. This also means that if the
coefficient were to be dropped to below 14.66667 then the optimal solution would
change. The range of insignificance is any value that lies outside of the range of
optimality. If the number of the coefficients above was outside the previously stated
range of optimality then the optimal solution would change. In addition if R had a
coefficient greater then 11:25 then it would also change the optimal solution.
5. Identify the marginal value of a minute of each processing phase.
a. The marginal value of a minute of each processing phase can be calculated by finding
the shadow price of each processing unit. When one finds this shadow price, they can
see the marginal benefit of having an additional minute. For sorting the shadow price is
zero. This means that there is no benefit for adding an additional minute. For the
crushing process there is a shadow price of 5. This is the largest shadow price of the
three processing units and shows that for every extra minute that is added one can gain
5 units. Lastly the Packing process has a shadow price of 0.125 which means that for
every additional unit added, one only gains 0.125 addition units.
6. Over what ranges of the processing capacity are these marginal values valid?
Filename: Document1
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Samuel Coy
BUS 355B
Homework Assignment 2
a. In order to find what ranges of the processing capacity are the marginal values valid,
one must find the range of feasibility. This range tells the range that is feasible. In this
range the sorting department can increase their sorting virtually to infinity. With
crushing, this is not the case however. With crushing they can decrease their final value
to as low as 360 and can increase it to 750 without changing the range of feasibility.
With Packing it can decrease it to 2913.6 and can raise it to 5689.811 without changing
the range of feasibility.
7. By how much would revenue decrease if sorting capacity was reduced to 900 minutes? How
much would revenue increase if sorting capacity was increased to 1,269 minutes?
a. According to the data in excel, the revenue will not decrease if sorting was reduced to
900 minutes. This is because there is an allowable decrease of 284. If the sorting
capacity was increased to 1,269 minutes, the revenue still would not increase. This is a
result of an infinite allowable increase.
8. How would increasing unit revenue by $2 per pound of T processed affect total revenue?
a. It would be allowable because it would still fall under the range or optimality for the T
outcome. Because of this the revenue would increase by 110. There are $55 pounds of
T and an additional $2 would therefore increase total revenue by $110.
9. Would a $1 increase per pound of R processed have any impact on the optimal solution?
Explain.
a. A $1 increase would have an impact on the optimal solution. Currently the range of
optimality for R only allows an increase of $0.25. Therefore an increase of $1 would
make R go beyond the range of optimality. This would therefore change the optimal
solution.
10. Assuming the cost of resources is already built into the net unit profit figures, which resource
would you choose to increase to maximize net profit? By how much could net profit be
increased as a result? Explain.
a. I would choose to increase T to maximize net profit. T has the highest potential to
increase by the most without changing the optimal solution. I can only increase Q by
1.625 and I can only increase R by 0.25 but I can increase T by 13. 13 is a lot to be able
to increase it by without changing the optimal solution. This would have an effect on
raising an additional $715 of profit.
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