MGT2251 Hello Class,
Here are the solutions to both Form A and Form B
for Exam#2 (June 20, 2001). The principals covered
on these exams will be the material used to write the
comprehensive portion of the Final Exam.
RHB
MGT 2251 Summer Semester 2001
Exam #2 (Form P)
Professor Robert Burgess
Name ____________________________ SS#________________________
Please establish a code with any combination of four numbers and or letters.
This code will be used to post your exam grade on the Classweb. DO NOT
use your initials or your SS#. Write your code below and on the sheet
provided for your records.
Exam Code:
There are 17 questions worth a total of 100 points. The breakdown of
question type and points is provided below. Your score in this exam will
count for 20% of your semester grade.
Question Type
6 - True / False
6 - Multiple Choice
10 - Short Answer
Graphical IP Solution
Points
each
Question
2
3
5
20
TOTAL=
Total Points
for the
Section
12
18
50
20
100
Number Incorrect
Points Off
Partial Credit
Partial Credit
Total OFF=
Exam Score=
2. Please write your answer in the space provided after the questions. If
you need more space use the backside of the page, and refer to your
doing this on the front side of the page. There are 7 pages including
this top page. Make sure that you have all the pages.
3. Read the questions carefully and define your variables clearly.
4. Please be neat when the solution requires you to draw.
5. You must show all work to get full credit.
Page 1 of 1
MGT 2251 Summer Semester 2001
Exam #2 (Form P)
Professor Robert Burgess
Circle the correct answer (2 points each / 12 points total)
T or F
1. Integer linear programming is more difficult than standard linear
programming, because there are typically far more possible feasible solutions
to evaluate. (T page 162 of text )
T or F
2. Integer linear programming, like standard linear programming, yields a
great amount of sensitivity analysis information. (F page 168 of text )
T or F
3. Binary linear programming is a subset of integer linear programming, but
not a subset of standard linear programming. (F page 176 of text )
T or F
4. A rounded standard linear programming solution to a problem that has
integer requirements may not be optimal, but it may be feasible. (T page 167 of
text ).
T or F
5. One approach for solving an integer linear programming problem is simply
to enumerate all feasible points, and select the one yielding the "best" value
for the objective function; because the number of feasible integer points is
usually very small, even for small problems, that this approach is very
efficient for solving most models. (F page 167 of text ).
T or F
6. The feasible region for a two variable mixed integer linear programming
(MILP) model is a set of parallel lines. (T page 171 of text ).
-----------------------------------------------------------------------------------------------------------Write your answer on the line (3 points each / 18 points total)
______
7. An integer linear programming model is solved using linear programming
and the optimal solution to the linear programming model happens to be an
integer solution.
Statement #1. This solution is also optimal for the integer model.
Statement #2. Sensitivity analysis for the integer model can be done using
the results from the sensitivity report for the linear
programming model.
a. Statement #1 is true (pages 165 and 170 of text )
b. Statement #2 is true
c. Neither 1 nor 2 is true.
d. Both 1 and 2 are true.
e. Sub-optimal.
Page 2 of 2
MGT 2251 Summer Semester 2001
Exam #2 (Form P)
Professor Robert Burgess
______
8. Using standard sensitivity analyses such as that generated by Excel
Solver, LINDO, or WINQSB, you can determine:
a. How much the objective function will change per additional unit of a
resource. (page 60 of text)
b. How the optimal solution changes with changes to the right hand side
values.
c. If the optimal solution will change if the objective function coefficients
change simultaneously.
d. How the optimal solution is affected by changes to the left hand side
coefficients.
______
9. The theory of "complementary slackness" implies that:
a. If the reduced cost is not zero, then the value of the decision variable is
zero. (page 65 of text)
b. If a decision variable is zero, then its reduced cost must be positive.
c. If a decision variable is zero, then its reduced cost must be non-zero.
d. If a decision variable is zero, then its reduced cost must be zero.
______
10. A change to the right-hand-side value of what type of constraint in a
minimization problem necessarily changes the optimal solution?
a. A "less than or equal to" constraint.
b. A "greater than or equal to " constraint.
c. A binding constraint. (page 70 of text)
d. A redundant constraint.
______
11. Methods for solving for the optimal solution to an integer linear
programming model do not include:
a. enumeration of all feasible points.
b. the cutting plane method.
c. the branch-and-bound method.
d. rounding the standard linear programming solution. (page 167 of text)
______
12. The optimal value of the objective function to a standard linear
programming model with a maximization objective is 675. If integer
restrictions are now imposed on the decision variables, one can be sure that
that the optimal value of the objective function to the integer model will:
a. be less than 675.
b. equal 675.
c. not exceed 675. (page 167 of text)
d. not be less than 675.
Page 3 of 3
MGT 2251 Summer Semester 2001
Exam #2 (Form P)
Professor Robert Burgess
- ----------------------------------------------------------------------------------------------------------Short Answer (5 points each / 50 points total)
Healthy Diet Problem
My doctor has told me that my diet must change. The new basic food groups are protein, vitamins,
calories, and fiber. At present, the following four foods are available for consumption: granola bars, fat
free ice cream, water with vitamins added, and soyburgers. Each granola bar costs $1.00,. each scoop of fat
free ice cream costs $1.20, each bottle of water with vitamins costs $0.50, and each soyburgers costs
$1.25. Each day I must ingest at least 100 ounces of protein, 100% of the recommended vitamins, 1000
calories, and 8 ounces of fiber. The nutritional content per unit of each food is shown below. An LP model
can be used to satisfy my daily nutritional requirements at minimum cost. The solution is given below:
Protein
(ounces)
4
10
0
50
Granola Bar
Fat Free Ice Cream (1 scoop)
Water with Vitamins (1 bottle)
Soyburger (1/4 lb pre-cooked)
Vitamins
(%)
5
1
15
5
Calories
180
100
0
250
Fiber
(ounces)
2
4
0
5
Q SB:
Healthy Diet Problem LP
Variable -->
Minimize
Protein
Vitamins
Calories
Fiber
LowerBound
UpperBound
VariableType
GranolaBar
1.00
4
5
180
2
0
M
Continuous
Fat-free Ice Cream
1.20
10
1
100
4
0
M
Continuous
Water&Vitamins
.5
0
15
0
0
0
M
Continuous
Soyburger
1.25
50
5
250
5
0
M
Continuous
Direction
R. H. S.
>=
>=
>=
>=
100
100
1000
15
Q SB:
Combined Report for Healthy Diet Problem
Decision Variable
1
2
3
4
Total
Contribution
Reduced
Cost
Basis
Status
Allowable
Min. c(j)
Allowable
Max. c(j)
GranolaBar
Fat-free Ice
Cream
Water&Vitamins
Soyburger
0
0
1.0000
1.2000
0
0
0.0533
0.7333
at bound
at bound
0.9467
0.4667
M
M
5.3333
4.0000
0.5000
1.2500
2.6667
5.0000
0
0
basic
basic
0
0.1667
1.0714
1.3241
Function
(Min.) =
7.6667
Constraint
3
4
Unit Cost
or Profit
c(j)
Objective
1
2
Solution
Value
Left Hand
Side
Direction
Right Hand Side
Slack or
Surplus
Shadow
Price
Allowable
Min. RHS
Allowable
Max. RHS
Protein
Vitamins
Calories
Fiber
200.0000
100.0000
1,000.0000
20.0000
>=
>=
>=
>=
100.0000
100.0000
1,000.0000
15.0000
100.0000
0
0
5.0000
0
0.0333
0.0043
0
-M
20.0000
750.0000
-M
200.0000
M
5,000.0000
20.0000
NOTE:
Excel Solver INPUT, Answer Report, Sensitivity Report and Limits Report on the following page.
Page 4 of 4
MGT 2251 Summer Semester 2001
Exam #2 (Form P)
Professor Robert Burgess
Excel Solver INPUT
Healthy Diet Problem
Input Data
X1
X2
X3
X4
GranolaBar Fat-free Ice Cream Water$Vitamins Soyburger Total Used Constraint Total Available
4
10
0
50
64.00
>=
100
5
1
15
5
26.00
>=
100
180
100
0
250
530.00
>=
1000
2
4
0
5
11.00
>=
15
Total Cost
Cost per Ad Type $
1.00 $
1.20 $
0.50 $
1.25 $
3.95
Protein
Vitmins
Calorie
Fiber
Food Purchased
1
1
1
1 <--Initial "Guessed Solution"
Excel Solver Answer Report
Target Cell (Min)
Cell
Name
$F$11 Cost per Ad Type Total Cost
Original Value
$
3.95
Adjustable Cells
Cell
Name
$B$14 Food Purchased GranolaBar
$C$14 Food Purchased Fat-free Ice Cream
$D$14 Food Purchased Water$Vitamins
$E$14 Food Purchased Soyburger
Original Value Final Value
1
0
1
0
1 5.333333333
1
4
Constraints
Cell
Name
$F$6 Protein Total Used
$F$7 Vitmins Total Used
$F$8 Calorie Total Used
$F$9 Fiber Total Used
Cell Value
200.00
100.00
1,000.00
20.00
Final Value
$
7.67
Formula
$F$6>=$H$6
$F$7>=$H$7
$F$8>=$H$8
$F$9>=$H$9
Status
Slack
Not Binding 100.00
Binding
0.00
Binding
0.00
Not Binding
5.00
Excel Solver Sensitivity Report
Adjustable Cells
Cell
$B$14
$C$14
$D$14
$E$14
Name
Food Purchased GranolaBar
Food Purchased Fat-free Ice Cream
Food Purchased Water$Vitamins
Food Purchased Soyburger
Final
Value
0
0
5.333333333
4
Reduced
Objective
Allowable
Allowable
Cost
Coefficient
Increase
Decrease
0.053333333
1
1E+30 0.053333333
0.733333333
1.2
1E+30 0.733333333
0
0.5 0.571428571
0.5
0
1.25 0.074074074 1.083333333
Constraints
Cell
$F$6
$F$7
$F$8
$F$9
Name
Protein Total Used
Vitmins Total Used
Calorie Total Used
Fiber Total Used
Final
Value
200.00
100.00
1,000.00
20.00
Shadow
Constraint
Price
R.H. Side
0.00
100
0.03
100
0.00
1000
0.00
15
Allowable
Increase
100
1E+30
4000
5
Allowable
Decrease
1E+30
80
250
1E+30
Excel Solver Limits Report
Target
Cell
Name
$F$11 Cost per Ad Type Total Cost
Cell
$B$14
$C$14
$D$14
$E$14
Adjustable
Name
Food Purchased GranolaBar
Food Purchased Fat-free Ice Cream
Food Purchased Water$Vitamins
Food Purchased Soyburger
$
Value
7.67
Value
Lower
Limit
0
0
5.333333333
4
0
0
5.333333333
4
Page 5 of 5
Target
Result
7.67
7.67
7.67
7.67
Upper Target
Limit Result
#N/A #N/A
#N/A #N/A
#N/A #N/A
#N/A #N/A
MGT 2251 Summer Semester 2001
Exam #2 (Form P)
Professor Robert Burgess
13. What is the cost of my new healthy diet?
$7.6667 per day
14. What do I eat daily and in what amounts?
5.3333 bottles of water & vitamins and 4.000 soyburgers
15. Granola bars have gone on sale at Big Lots for $0.95 each. How does this change my diet?
Depends on the value of the reduced cost for variable granola bars.
Since the reduced cost is currently 0.0533 cents, granola bars would have to
cost (1.00-0.0533)=94.27 cents to be included in our minimum cost diet.
Since 94.27 is smaller than 95 cents, I would not change my diet.
16.
What is the maximum price I would pay per soyburger to continue buying my
current diet?
(Soyburgers = X4) Minimum=16.67 cents Current =$1.25 Maximum=$1.3241
Depends on the range for the coefficients for variable Soyburgers. Answer is $1.3241!
17. I want to decrease my diet cost even further. What diet restrictions currently prevent me
from doing so?
Vitamins and Calories are tight or binding.
18.
How much could I save if I reduced the vitamin requirement keeping my current
diet?
Decreasing the requirement for Vitamins by one unit will reduce the cost of my diet by 3.33 cents.
I can decrease this constraint down to 20% (from 100%) and keep the same basis (water and soyburgers).
My cost would be:
New Optimal Objective Value = Old Optimal Objective Value + (Shadow Price)*( )
New Optimal Objective Value = $7.6667 + ($0.0333)*( 20 - 100) = $ 5.0027
SAVINGS of $2.644
19. How much could I save if I reduced the calorie requirement keeping my current diet?
Decreasing the requirement for Calories by one unit will reduce the cost of my diet by 0.0043 cents.
I can decrease this constraint down to 750 calories (from 1000) ounces and keep the same basis.
My cost would be:
New Optimal Objective Value = Old Optimal Objective Value + (Shadow Price)*( )
New Optimal Objective Value = $7.6667 + ($0.0043)*(750 - 1000) = $6.5917
SAVINGS of $1.075
20. Healthy diets are expensive. I can only afford $5.00 per day. How much would soyburgers
need to cost before I can afford the current healthy diet?
Current Optimal Objective Value = (5.3333*$0.50) + (4.000*$1.25) = ($2.667) + ($5.00) = $7.6667
Current Optimal Objective Value = $2.6667 + 4.000*(new cost of soyburgers)=$5.00
Solving for new cost of soyburgers = ($5.00 - $2.6667) / 4 = ($2.3333) / 4 = $0.583325
Since the allowable minimum cost of a soyburger (maintaining same diet basis) is $0.1667, I can afford it!
21.
I am trying to grow my muscle mass and the current minimum of 100 ounces of
protein is not enough. I want to increase my protein intake to 150 ounces per day.
How does this affect my current diet and its cost?
The current solution is giving you a diet with 200 ounces of protien calories.
So your diet and your cost stay the same.
22. If daily fiber intake is increased to 22 ounces, what is the new optimal z-value?
Depends on the value of ; if (20-15) = 5, then basis remains optimal and can use shadow price.
In that range, shadow price is zero, so there would be no cost change and optimal z-value would be $7.6667.
But, =(25-15)= +7 is not in range so cannot use the shadow price. Must resolve model to get exact answer.
Page 6 of 6
MGT 2251 Summer Semester 2001
Exam #2 (Form P)
Professor Robert Burgess
Graphical Solution (20 points each / 20 points total)
17. Using graphical techniques, find the optimal solution to the following integer
program. Also find the optimal solution to the LP relaxation of the integer program
below. Be sure and label the feasible region, all extreme points, all the constraints
and the optimal isoprofit line. Also, designate which constraints are binding.
Max z =
C1:
C2:
C3:
C4:
C5:
C6:
C7:
Max
3X1 +
6X1 +
20X1 +
2X1 +
X1
z=
3X1 +
X1
2X2 = 9 @ point (3,0), integer
2X2 18
8X2 25
6X2 14
0
X2 0
integer
X2 integer
Binding Constraint
Non-Binding Constraint
Non-Binding Constraint
Non-Binding Constraint (X2 axis)
Binding Constraint (X1 axis)
Binding Constraint
Binding Constraint
2X2 = 10.5 @ point (2.5,1.5), LP relaxation
Five Feasible Points
"Z"
X1 X2 Obj FV
1
1
1
5000
2
1
2
7000
3
2
0
6000
4
2
1
8000
5
3
0
9000
Page 7 of 7
MGT 2251 Summer Semester 2001
Exam #2 (Form P)
Professor Robert Burgess
THE LP RELAXATION SOLUTION @ POINT (2.5, 1.5)
ALL INTEGER SOLUTION
THE FIVE FEASIBLE POINTS ARE CIRCLED BELOW:
Feasible Points
4.5
4
3.5
LP Relaxation Solution:
Z=10.5 @ point (2.5,1.5)
X2
3
2.5
2
C3
1.5
1
C2
0.5
C1
0
0
1
2
3
X1
Page of 8 8
4
max Z = 9
5
MGT 2251 Summer Semester 2001
Exam #2 (Form Q)
Professor Robert Burgess
Name ____________________________ SS#________________________
Please establish a code with any combination of four numbers and or letters.
This code will be used to post your exam grade on the Classweb. DO NOT
use your initials or your SS#. Write your code below and on the sheet
provided for your records.
Exam Code:
There are 17 questions worth a total of 100 points. The breakdown of
question type and points is provided below. Your score in this exam will
count for 20% of your semester grade.
Question Type
6 - True / False
6 - Multiple Choice
10 - Short Answer
Graphical IP Solution
Points
each
Question
2
3
5
20
TOTAL=
Total Points
for the
Section
12
18
50
20
100
Number Incorrect
Points Off
Partial Credit
Partial Credit
Total OFF=
Exam Score=
2. Please write your answer in the space provided after the questions. If
you need more space use the backside of the page, and refer to your
doing this on the front side of the page. There are 7 pages including
this top page. Make sure that you have all the pages.
3. Read the questions carefully and define your variables clearly.
4. Please be neat when the solution requires you to draw.
5. You must show all work to get full credit.
Page 1 of 1
MGT 2251 Summer Semester 2001
Exam #2 (Form Q)
Professor Robert Burgess
Circle the correct answer (2 points each / 12 points total)
T or F
1. Integer linear programming is less difficult than standard linear
programming, because there are typically far fewer possible feasible solutions
to evaluate. (F page 162 of text)
T or F
2. Integer linear programming, unlike standard linear programming, does not
yield a great amount of sensitivity analysis information. (T page 168 of text)
T or F
3. Binary linear programming is a subset of integer linear programming,
which in turn is a subset of standard linear programming. (T page 176 of text)
T or F
4. A rounded standard linear programming solution to a problem that has
integer requirements may not be optimal, but at least will be feasible. (F page
167 of text).
T or F
5. One approach for solving an integer linear programming problem is simply
to enumerate all feasible points, and select the one yielding the "best" value
for the objective function; however the number of feasible integer points is
usually so large, even for small problems, that this approach is inefficient for
solving most models. (T page 167 of text).
T or F
6. The feasible region for a two variable mixed integer linear programming
(MILP) model is a set of disjoint points. (F page 171 of text).
-----------------------------------------------------------------------------------------------------------Write your answer on the line (3 points each / 18 points total)
______
7. Using standard sensitivity analyses such as that generated by Excel
Solver, LINDO, or WINQSB, you can determine:
a. How much the objective function will change per additional unit of a
resource. (page 60 of text)
b. How the optimal solution changes with changes to the right hand side
values.
c. If the optimal solution will change if the objective function coefficients
change simultaneously.
d. How the optimal solution is affected by changes to the left hand side
coefficients.
______
8. The theory of "complementary slackness" implies that:
a. If the reduced cost is not zero, then the value of the decision variable is
zero. (page 65 of text)
b. If a decision variable is zero, then its reduced cost must be positive.
c. If a decision variable is zero, then its reduced cost must be non-zero.
d. If a decision variable is zero, then its reduced cost must be zero.
Page 2 of 2
MGT 2251 Summer Semester 2001
Exam #2 (Form Q)
Professor Robert Burgess
______
9. A change to the right-hand-side value of what type of constraint in a
minimization problem necessarily changes the optimal solution?
a. A "less than or equal to" constraint.
b. A "greater than or equal to " constraint.
c. A binding constraint. (page 70 of text)
d. A redundant constraint.
______
10. Methods for solving for the optimal solution to an integer linear
programming model do not include:
a. enumeration of all feasible points.
b. the cutting plane method.
c. the branch-and-bound method.
d. rounding the standard linear programming solution. (page 167 of text)
______
11. The optimal value of the objective function to a standard linear
programming model with a maximization objective is 675. If integer
restrictions are now imposed on the decision variables, one can be sure that
that the optimal value of the objective function to the integer model will:
a. be less than 675.
b. equal 675.
c. not exceed 675. (page 167 of text)
d. not be less than 675.
______
12. An integer linear programming model is solved using linear programming
and the optimal solution to the linear programming model happens to be an
integer solution.
Statement #1. This solution is also optimal for the integer model.
Statement #2. Sensitivity analysis for the integer model can be done using
the results from the sensitivity report for the linear
programming model.
a. Statement #1 is true (pages 165 and 170 of text)
b. Statement #2 is true
c. Neither 1 nor 2 is true.
d. Both 1 and 2 are true.
e. Sub-optimal.
Page 3 of 3
MGT 2251 Summer Semester 2001
Exam #2 (Form Q)
Professor Robert Burgess
- ----------------------------------------------------------------------------------------------------------Short Answer (5 points each / 50 points total)
Healthy Diet Problem
My doctor has told me that my diet must change. The new basic food groups are protein, vitamins,
calories, and fiber. At present, the following four foods are available for consumption: granola bars, fat
free ice cream, water with vitamins added, and soyburgers. Each granola bar costs $1.00,. each scoop of fat
free ice cream costs $1.20, each bottle of water with vitamins costs $0.50, and each soyburgers costs
$1.25. Each day I must ingest at least 100 ounces of protein, 100% of the recommended vitamins, 1000
calories, and 8 ounces of fiber. The nutritional content per unit of each food is shown below. An LP model
can be used to satisfy my daily nutritional requirements at minimum cost. The solution is given below:
Protein
(ounces)
4
10
0
50
Granola Bar
Fat Free Ice Cream (1 scoop)
Water with Vitamins (1 bottle)
Soyburger (1/4 lb pre-cooked)
Vitamins
(%)
5
1
15
5
Calories
180
100
0
250
Fiber
(ounces)
2
4
0
5
Q SB:
Healthy Diet Problem LP
Variable -->
Minimize
Protein
Vitamins
Calories
Fiber
LowerBound
UpperBound
VariableType
GranolaBar
1.00
4
5
180
2
0
M
Continuous
Fat-free Ice Cream
1.20
10
1
100
4
0
M
Continuous
Water&Vitamins
.5
0
15
0
0
0
M
Continuous
Soyburger
1.25
50
5
250
5
0
M
Continuous
Direction
R. H. S.
>=
>=
>=
>=
100
100
1000
15
Q SB:
Combined Report for Healthy Diet Problem
Decision Variable
1
2
3
4
Total
Contribution
Reduced
Cost
Basis
Status
Allowable
Min. c(j)
Allowable
Max. c(j)
GranolaBar
Fat-free Ice
Cream
Water&Vitamins
Soyburger
0
0
1.0000
1.2000
0
0
0.0533
0.7333
at bound
at bound
0.9467
0.4667
M
M
5.3333
4.0000
0.5000
1.2500
2.6667
5.0000
0
0
basic
basic
0
0.1667
1.0714
1.3241
Function
(Min.) =
7.6667
Constraint
3
4
Unit Cost
or Profit
c(j)
Objective
1
2
Solution
Value
Left Hand
Side
Direction
Right Hand Side
Slack or
Surplus
Shadow
Price
Allowable
Min. RHS
Allowable
Max. RHS
Protein
Vitamins
Calories
Fiber
200.0000
100.0000
1,000.0000
20.0000
>=
>=
>=
>=
100.0000
100.0000
1,000.0000
15.0000
100.0000
0
0
5.0000
0
0.0333
0.0043
0
-M
20.0000
750.0000
-M
200.0000
M
5,000.0000
20.0000
NOTE:
Excel Solver INPUT, Answer Report, Sensitivity Report and Limits Report on the following page.
Page 4 of 4
MGT 2251 Summer Semester 2001
Exam #2 (Form Q)
Professor Robert Burgess
Excel Solver INPUT
Healthy Diet Problem
Input Data
X1
X2
X3
X4
GranolaBar Fat-free Ice Cream Water$Vitamins Soyburger Total Used Constraint Total Available
4
10
0
50
64.00
>=
100
5
1
15
5
26.00
>=
100
180
100
0
250
530.00
>=
1000
2
4
0
5
11.00
>=
15
Total Cost
Cost per Ad Type $
1.00 $
1.20 $
0.50 $
1.25 $
3.95
Protein
Vitmins
Calorie
Fiber
Food Purchased
1
1
1
1 <--Initial "Guessed Solution"
Excel Solver Answer Report
Target Cell (Min)
Cell
Name
$F$11 Cost per Ad Type Total Cost
Original Value
$
3.95
Adjustable Cells
Cell
Name
$B$14 Food Purchased GranolaBar
$C$14 Food Purchased Fat-free Ice Cream
$D$14 Food Purchased Water$Vitamins
$E$14 Food Purchased Soyburger
Original Value Final Value
1
0
1
0
1 5.333333333
1
4
Constraints
Cell
Name
$F$6 Protein Total Used
$F$7 Vitmins Total Used
$F$8 Calorie Total Used
$F$9 Fiber Total Used
Cell Value
200.00
100.00
1,000.00
20.00
Final Value
$
7.67
Formula
$F$6>=$H$6
$F$7>=$H$7
$F$8>=$H$8
$F$9>=$H$9
Status
Slack
Not Binding 100.00
Binding
0.00
Binding
0.00
Not Binding
5.00
Excel Solver Sensitivity Report
Adjustable Cells
Cell
$B$14
$C$14
$D$14
$E$14
Name
Food Purchased GranolaBar
Food Purchased Fat-free Ice Cream
Food Purchased Water$Vitamins
Food Purchased Soyburger
Final
Value
0
0
5.333333333
4
Reduced
Objective
Allowable
Allowable
Cost
Coefficient
Increase
Decrease
0.053333333
1
1E+30 0.053333333
0.733333333
1.2
1E+30 0.733333333
0
0.5 0.571428571
0.5
0
1.25 0.074074074 1.083333333
Constraints
Cell
$F$6
$F$7
$F$8
$F$9
Name
Protein Total Used
Vitmins Total Used
Calorie Total Used
Fiber Total Used
Final
Value
200.00
100.00
1,000.00
20.00
Shadow
Constraint
Price
R.H. Side
0.00
100
0.03
100
0.00
1000
0.00
15
Allowable
Increase
100
1E+30
4000
5
Allowable
Decrease
1E+30
80
250
1E+30
Excel Solver Limits Report
Target
Cell
Name
$F$11 Cost per Ad Type Total Cost
Cell
$B$14
$C$14
$D$14
$E$14
Adjustable
Name
Food Purchased GranolaBar
Food Purchased Fat-free Ice Cream
Food Purchased Water$Vitamins
Food Purchased Soyburger
$
Value
7.67
Value
Lower
Limit
0
0
5.333333333
4
0
0
5.333333333
4
Page 5 of 5
Target
Result
7.67
7.67
7.67
7.67
Upper Target
Limit Result
#N/A #N/A
#N/A #N/A
#N/A #N/A
#N/A #N/A
MGT 2251 Summer Semester 2001
Exam #2 (Form Q)
Professor Robert Burgess
13. What is the cost of my new healthy diet?
$7.6667 per day
14. What do I eat daily and in what amounts?
5.3333 bottles of water & vitamins and 4.000 soyburgers
15. Granola bars have gone on sale at Big Lots for $0.95 each. How does this change my diet?
Depends on the value of the reduced cost for variable granola bars.
Since the reduced cost is currently 0.0533 cents, granola bars would have to
cost (1.00-0.0533)=94.27 cents to be included in our minimum cost diet.
Since 94.27 is smaller than 95 cents, I would not change my diet.
16.
What is the maximum price I would pay per soyburger to continue buying my
current diet?
(Soyburgers = X4) Minimum=16.67 cents Current =$1.25 Maximum=$1.3241
Depends on the range for the coefficients for variable Soyburgers. Answer is $1.3241!
17. I want to decrease my diet cost even further. What diet restrictions currently prevent me
from doing so?
Vitamins and Calories are tight or binding.
18.
How much could I save if I reduced the vitamin requirement keeping my current
diet?
Decreasing the requirement for Vitamins by one unit will reduce the cost of my diet by 3.33 cents.
I can decrease this constraint down to 20% (from 100%) and keep the same basis (water and soyburgers).
My cost would be:
New Optimal Objective Value = Old Optimal Objective Value + (Shadow Price)*( )
New Optimal Objective Value = $7.6667 + ($0.0333)*( 20 - 100) = $ 5.0027
SAVINGS of $2.644
19. How much could I save if I reduced the calorie requirement keeping my current diet?
Decreasing the requirement for Calories by one unit will reduce the cost of my diet by 0.0043 cents.
I can decrease this constraint down to 750 calories (from 1000) ounces and keep the same basis.
My cost would be:
New Optimal Objective Value = Old Optimal Objective Value + (Shadow Price)*( )
N ew Optimal Objective Value = $7.6667 + ($0.0043)*(750 - 1000) = $6.5917
SAVINGS of $1.075
20. Healthy diets are expensive. I can only afford $5.00 per day. How much would soyburgers
need to cost before I can afford the current healthy diet?
Current Optimal Objective Value = (5.3333*$0.50) + (4.000*$1.25) = ($2.667) + ($5.00) = $7.6667
Current Optimal Objective Value = $2.6667 + 4.000*(new cost of soyburgers)=$5.00
Solving for new cost of soyburgers = ($5.00 - $2.6667) / 4 = ($2.3333) / 4 = $0.583325
Since the allowable minimum cost of a soyburger (maintaining same diet basis) is $0.1667, I can afford it!
21.
I am trying to grow my muscle mass and the current minimum of 100 ounces of
protein is not enough. I want to increase my protein intake to 150 ounces per day.
How does this affect my current diet and its cost?
The current solution is giving you a diet with 200 ounces of protien calories.
So your diet and your cost stay the same.
22. If daily fiber intake is increased to 22 ounces, what is the new optimal z-value?
Depends on the value of ; if (20-15) = 5, then basis remains optimal and can use shadow price.
In that range, shadow price is zero, so there would be no cost change and optimal z-value would be $7.6667.
But, =(25-15)= +7 is not in range so cannot use the shadow price. Must resolve model to get exact answer.
Page 6 of 6
MGT 2251 Summer Semester 2001
Exam #2 (Form Q)
Professor Robert Burgess
Graphical Solution (20 points each / 20 points total)
17. Using graphical techniques, find the optimal solution to the following integer
program. Also find the optimal solution to the LP relaxation of the integer program
below. Be sure and label the feasible region, all extreme points, all the constraints
and the optimal isoprofit line. Also, designate which constraints are binding.
Max z =
C1:
C2:
C3:
C4:
C5:
C6:
Max
3X1 +
6X1 +
20X1 +
2X1 +
X1
z=
3X1 +
X1
2X2 = 9.333 @ point (2,1.6667), mixed-integer
2X2 18
8X2 25
6X2 14
0
X2 0
integer
Non-Binding Constraint
Non-Binding Constraint
Binding Constraint
Non-Binding Constraint (X2 axis)
Non-Binding Constraint (X1 axis)
Binding Constraint
2X2 = 10.5 @ point (2.5,1.5), LP relaxation
If X2 is continuous, then there are three candidate points
(e.g.- the highest point on each parallel line)
"Z"
X1 X2 Obj FV
0 2.3 4666.7 not feasible at intercept X 2
1
1
2
7000
2
2 1.7 9333.3
3
3
0
9000
Page 7 of 7
MGT 2251 Summer Semester 2001
Exam #2 (Form Q)
Professor Robert Burgess
THE LP RELAXATION SOLUTION @ POINT (2.5, 1.5)
MIXED INTEGER SOLUTION
FEASIBLE LINES ARE IN RED, THE THREE POINTS TO CHECK ARE CIRCLED
BELOW:
Feasible "Lines"
4.5
4
3.5
LP Relaxation Solution:
Z=10.5 @ point (2.5,1.5)
X2
3
2.5
2
C3
1.5
1
C2
0.5
C1
0
0
1
2
3
X1
Page 8 of 8
4
max Z = 9.333
5