mb363 assignment 1 report
DESCRIPTION
MB363 Assignment 1 ReportTRANSCRIPT
MB363 Management Decision Tools
Assignment 1
NN NNNN NN NN NNAA AAAA AA AA AA
NN NNNN NN NN NNYY YYYY YY YY YY
AA AAAA AA AA AANN NNNN NN NN NN
GG GGGG GG GG GG TT TTTT TT TT TT
EE EEEE EE EE EECC CCCC CC CC CC
HH HHHH HH HH HHNN NNNN NN NN NN
OO OOOO OO OO OOLL LLLL LL LL LLOO OOOO OO OO OO
GG GGGG GG GG GGII IIII II II II CC CCCC CC CC CC
AA AAAA AA AA AALL LLLL LL LL LL UU UUUU UU UU UU
NN NNNN NN NN NNII IIII II II II VV VVVV VV VV VV
EE EEEE EE EE EERR RRRR RR RR RR
SS SSSS SS SS SSII IIII II II II TT TTTT TT TT TT
YY YYYY YY YY YY
Prepared By : Kor Xian Thong Ronnie
_(081723A15)
Sub-Group : 3
Supervisor : A/P Li Zhi-Feng, Michael
Date : 8th
October 2009
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Table of Contents Page
1)
Chapter 2: Linear Programming I
(a) Problem 2.16
2
(b) Problem 2.21
3
2) Chapter 3: Linear Programming II
(a) Problem 3.5
4
(b) Problem 3.16
7
(c) Problem 3.24
9
3) Chapter 5: Sensitivity Analysis for LP
(a) Problem 5.2
11
4) Chapter 8: Nonlinear Programming
(a) Problem 8.4
17
(b) Problem 8.10
19
5) Chapter 9: Decision Analysis
(a) Problem 9.7
22
(b) Problem 9.18
26
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Chapter 2: Linear Programming I Q1(a) Problem 2.16 Nutri-Jenny Weight Management Centre
(a) Linear Programming Model on Spreadsheet
(b) Notation:
B: Beef Tips
G: Gravy
P: Peas
C: Carrots
D: Dinner Roll
Choose the values of B, P, J, A, M and C so as to minimize
Total Cost, C=$(0.40B+0.35G+0.15P+0.18C+0.10D)
Subject to satisfying all the following constraints:
280 ≤ 54B + 20G + 15P + 8C + 40D ≤ 320
19B + 15G + 10D ≤ 96
15P + 350C ≤ 600
G + 3P + C ≤ 10
8B + P+ C + D ≤ 30
B ≥ 2
G ≥ 0.5B
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Q1(b) Problem 2.21 Learning Centre
(a) Linear Programming Model on Spreadsheet
(b) Notation:
B: Bread
P: Peanut Butter
J: Jelly
A: Apple
M: Milk
C: Cranberry Juice
Choose the values of B, P, J, A, M and C so as to minimize
Total Cost, C=$(0.06B+0.05P+0.08J+0.35A+0.20M + 0.40C)
Subject to satisfying all the following constraints:
300 ≤ 80B + 100P + 70J + 90A + 120M +110C ≤ 500
15B + 80P + 60M ≤ 0.30(80B + 100P + 70J + 90A + 120M +110C)
4J + 6A + 2M + 80C ≥ 60
4B + 3J + 10A + C ≥ 10
B ≥ 2
P ≥ 1
J ≥ 1
M + C ≥ 1
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Chapter 3: Linear Programming II Q2(a) Problem 3.5 Omega Manufacturing Company
(a) Resource Allocation Problem
The activities under consideration are:
Activity 1: Produce Product 1
Activity 2: Produce Product 2
Activity 3: Produce Product 3
The resources to be allocated to these activities are:
Resource 1: Available time (machine-hours per week) of Milling Machine
Resource 2: Available time (machine-hours per week) of Lathe
Resource 3: Available time (machine-hours per week) of Grinder
This problem has 3 resource constraints:
Constraint 1: Total Available time of Milling Machine = 500 Machine-Hours per Week
Constraint 2: Total Available time of Lathe = 350 Machine-Hours per Week
Constraint 3: Total Available time of Grinder = 150 Machine-Hours per Week
Side Constraint:
Constraint 4: Sales Potential for Product 3 = 20 units per week
(b) With the 3 products under consideration, these are the decisions to be made:
Decision 1: P1 = Number of units of Product 1 to produce
Decision 2: P2 = Number of units of Product 2 to produce
Decision 3: P3 = Number of units of Product 3 to produce
Overall Measure of Performance:
The objective is to maximize total profits earned from producing and selling certain
product-mix of Product 1, 2 and 3 per week.
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(c) Decisions: P1: Number of Product 1 produced and sold P2: Number of Product 2 produced and sold
P3: Number of Product 3 produced and sold
Resource Constraints:
Total number of machine-hours per week used for Milling machine ≤ 500
Total number of machine-hours per week used for Lathe ≤ 350
Total number of machine-hours per week used for Grinder ≤ 500
Side Constraint:
Total units of Product 3 produced per week ≤ Sales potential for Product 3 (20 units/week)
Overall Measure of Performance:
Maximize Total Profit = Sum of Profits earned from selling certain product mix of Product
1,2 and 3.
Total Profit = 50P1 + 20P2 + 25P3
(d)
Excel Equation for Output Cell (Milling Machine): SUMPRODUCT (Machine Hours for milling machine used by each product, Units Produced) Excel Equation for Output Cell (Lathe): SUMPRODUCT (Machine Hours for Lathe used by each product, Units Produced) Excel Equation for Output Cell (Grinder): SUMPRODUCT (Machine Hours for Grinder used by each product, Units Produced)
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(e) Maximize Profit = 50P1 + 20P2 + 25P3
Subject to
Milling Machine: 9P1 + 3P2 + 5P3 ≤ 500
Lathe: 5P1 + 4P2 ≤ 350
Grinder: 3P1 + 2P3 ≤ 150
And P3 ≤ 20
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Q2(b) Problem 3.16 Fagersta Steelworks
(a) Notation:
M1S1: Amount of iron ore shipped per month from Mine M1 to Storage Facility S1
M1S2: Amount of iron ore shipped per month from Mine M1 to Storage Facility S2
M2S1: Amount of iron ore shipped per month from Mine M2 to Storage Facility S1
M2S2: Amount of iron ore shipped per month from Mine M2 to Storage Facility S2
S1P: Amount of iron ore shipped per month from Storage Facility S1 to Steel Plant P
S2P: Amount of iron ore shipped per month from Storage Facility S2 to Steel Plant P
Resources:
Resource 1: Amount of iron ore produced by Mine M1 = 40 tons
Resource 2: Amount of iron ore produced by Mine M2 = 60 tons
Fixed Requirement Constraints:
Requirement 1: Total amount of iron ore shipped out of Mine M1 = Total Produced by
Mine M1
Requirement 2: Total amount of iron ore shipped out of Mine M2 = Total Produced by
Mine M2
Requirement 3: Steel Plant P must receive 100 tons of iron ore.
Resource Constraints:
Constraint 1: Amount of iron ore shipped from Mine M1 to Storage Facility S1 ≤ 30
tonnes
Constraint 2: Amount of iron ore shipped from Mine M1 to Storage Facility S2 ≤ 30
tonnes
Constraint 3: Amount of iron ore shipped from Mine M2 to Storage Facility S1 ≤ 50
tonnes
Constraint 4: Amount of iron ore shipped from Mine M2 to Storage Facility S2 ≤ 50
tonnes
Constraint 5: Amount of iron ore shipped from Storage Facility S1 to Steel Plant P ≤ 70
tonnes
Constraint 6: Amount of iron ore shipped from Storage Facility S2 to Steel Plant P ≤ 70
tonnes
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(b)
(c) Minimize Total Shipping Cost = 2000M1S1 + 1700M1S2 + 1600M2S1 + 1100M2S2 +
400S1P + 800S2P
Subject to the following constraints:
1. Fixed Requirement Constraints:
M1S1 + M1S2 = 40
M2S1 + M2S2 = 60
S1P + S2P = 100
2. Resource Constraints:
M1S1 ≤ 30, M1S2 ≤ 30 (Mine M1)
M2S1 ≤ 50, M2S2 ≤ 50 (Mine M2)
S1P ≤ 70 (Storage Facility S1)
S2P ≤ 70 (Storage Facility S2)
3. Nonnegativity Constraints:
M1S1 ≥ 0, M1S2 ≥ 0, M2S1 ≥ 0, M2S2 ≥ 0, S1P ≥ 0, S2P ≥ 0
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Q2(c) Problem 3.24 Day Care for Preschoolers
(a) Resource Constraints:
Requirement 1: Each food choice needs to provide no more than 600 calories.
Requirement 2: Each food choice should not have more than 30% of the calories that
comes from fats.
Benefits Constraints:
Requirement 3: Each foold choice needs to provide at least 400 calories.
Requirement 4: Each food choice needs to have at least 60mg of Vitamin C.
Requirement 5: Each food choice needs to provide at least 12g of protein.
Requirement 6: Each food choice needs to have at least twice as much peanut butter as
jelly.
Requirement 7: Each food choice should have at least 1 cup of liquid (milk and/or juice).
Fixed Requirement Constraints:
Requirement 7: Each food choice should have exactly 2 slices of bread.
(b)
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(c) Notation:
B: Bread
P: Peanut Butter
S: Strawberry Jelly
G: Graham Cracker
M: Milk
J: Juice
Choose B, P, S, G, M, J to minimize
Total Cost C = 0.05B + 0.04P + 0.07S + 0.08G + 0.15M + 0.35J
Subject to the following constraints:
1. Resource Constraints:
70B + 100P + 50S + 60G + 150M + 100J ≤ 600
10B + 75P + 20G + 70M ≤ 0.30(70B + 100P + 50S + 60G + 150M + 100J)
2. Benefits Constraints:
70B + 100P + 50S + 60G + 150M + 100J ≥ 400
3S + 2M + 120J ≤ 60
3B + 4P + G + 8M + J ≤ 12
P ≥ 2S
M + J ≥ 1
3. Fixed Requirement Constraints:
B = 2
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Chapter 5: Sensitivity Analysis for LP Q3 Problem 5.2
(a)
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(b) With the unit profit of activity 2 at a constant $5 per unit,
When the unit profit changes for activity 1 changes from $2 to $1, the optimal solution
changes from (6,2) to (0,4).
When the unit profit changes for activity 1 changes from $2 to $3, the optimal solution changes from (6,2) to (10,0).
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(c) With the unit profit of activity 1 at a constant $2 per unit,
When the unit profit changes for activity 2 changes from $5 to $2.50, the optimal solution
changes from (6,2) to (10,0).
When the unit profit changes for activity 2 changes from $5 to $7.50, the optimal solution
changes from (6,2) to (0,4).
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(d)
Unit Profit Optimal Units Used Total
for Activity 1 Activity 1 Activity 2 Profit
6 2 $22.00
$1.00 0 4 $20.00
$1.20 0 4 $20.00
$1.40 0 4 $20.00
$1.60 0 4 $20.00
$1.80 6 2 $20.80
$2.00 6 2 $22.00
$2.20 6 2 $23.20
$2.40 6 2 $24.40
$2.60 10 0 $26.00
$2.80 10 0 $28.00
$3.00 10 0 $30.00
Unit Profit Optimal Units Used Total
for Activity 2 Activity 1 Activity 2 Profit
6 2 $22.00
$2.50 10 0 $20.00
$3.00 10 0 $20.00
$3.50 10 0 $20.00
$4.00 6 2 $20.00
$4.50 6 2 $21.00
$5.00 6 2 $22.00
$5.50 6 2 $23.00
$6.00 0 4 $24.00
$6.50 0 4 $26.00
$7.00 0 4 $28.00
$7.50 0 4 $30.00
For Activity 1, the allowable range for the unit profit of Activity 1 is from $1.80 to $2.40.
For Activity 2, the allowable range for the unit profit of Activity 2 is from $4.00 to $5.50.
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(e)
(f)
From the Graphical Linear Programming and Sensitivity Analysis module, when the unit
profit of Activity 2 is kept constant at $5.00, the allowable range for unit profit of Activity 1
is from $1.67 to $2.50. When the unit profit of Activity 1 is kept constant at $2.00, the
allowable range for unit profit of Activity 1 is from $4.00 to $6.00.
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(g)
Units Used Unit Profit for Activity 2
(6,2) $2.50 $3.00 $3.50 $4.00 $4.50 $5.00 $5.50 $6.00 $6.50 $7.00 $7.50
$1.00 (6,2) (0,4) (0,4) (0,4) (0,4) (0,4) (0,4) (0,4) (0,4) (0,4) (0,4)
$1.20 (6,2) (6,2) (6,2) (0,4) (0,4) (0,4) (0,4) (0,4) (0,4) (0,4) (0,4)
$1.40 (10,0) (6,2) (6,2) (6,2) (0,4) (0,4) (0,4) (0,4) (0,4) (0,4) (0,4)
Unit Profit $1.60 (10,0) (10,0) (6,2) (6,2) (6,2) (0,4) (0,4) (0,4) (0,4) (0,4) (0,4)
for Activity 1 $1.80 (10,0) (10,0) (10,0) (6,2) (6,2) (6,2) (0,4) (0,4) (0,4) (0,4) (0,4)
$2.00 (10,0) (10,0) (10,0) (6,2) (6,2) (6,2) (6,2) (0,4) (0,4) (0,4) (0,4)
$2.20 (10,0) (10,0) (10,0) (10,0) (6,2) (6,2) (6,2) (6,2) (6,2) (0,4) (0,4)
$2.40 (10,0) (10,0) (10,0) (10,0) (10,0) (6,2) (6,2) (6,2) (6,2) (6,2) (0,4)
$2.60 (10,0) (10,0) (10,0) (10,0) (10,0) (10,0) (6,2) (6,2) (6,2) (6,2) (6,2)
$2.80 (10,0) (10,0) (10,0) (10,0) (10,0) (10,0) (10,0) (6,2) (6,2) (6,2) (6,2)
$3.00 (10,0) (10,0) (10,0) (10,0) (10,0) (10,0) (10,0) (6,2) (6,2) (6,2) (6,2)
(h) From the plot formed in the Graphical Linear Programming and Sensitivity Analysis
module, it can be seen that for any objective function line to cross the optimum point, the
gradient of the function has to lie between and not inclusive of 0.333, to and not inclusive
of 0.50. From the data obtained from using two-dimensional Solver Table, the highlighted
data are the unit profit of activity 1 and activity 2 that corresponds to the optimal solution
of (6,2), and this values of respective unit profit satisfies the condition of
0.333≤Gradient≤0.50. Outside of this range, the optimal solution changes. Hence the
data obtained from the two-dimensional Solver Table is proved to be correct.
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Chapter 8: Nonlinear Programming Q4(a) Problem 8.4
(a)
Production Rate (R) Profit per Day (P) P=$100R-$5R^2 % Error
0 $0.00 $0.00 0
1 $95.00 $95.00 0
2 $184.00 $180.00 2.1739
3 $255.00 $255.00 0
4 $320.00 $320.00 0
(b)
Production Rate (R) Profit per Day (P) P=$104R-$6R^2 % Error
0 $0.00 $0.00 0
1 $95.00 $98.00 3.15789
2 $184.00 $184.00 0
3 $255.00 $258.00 1.17647
4 $320.00 $320.00 0
(c) From the calculations, it can be seen that the approximation P=$100R - $5R2 fits closely
to the actual data (Profit per Day), except at R=2 with a % difference of 2.1739%. Hence it fits the graph for 80% of the data and thus provides a close approximation of P.
As for the approximation P=$104R - $6R2, it fits the data at 3 close points and does not fit at 2 points, namely at R=1 with a % difference of 3.158% and R=3 with a % difference of 1.176%. Overall % difference is slightly larger at 4.334%. Also, It only fits the graph for 60% of the data. Thus, it does not provide a close approximation of P.
Hence it can be concluded that the approximation P=$100R - $5R2 provides better fit to all the data.
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(d) y = -5.5714x2 + 102.29x – 0.3429
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Q4(b) Problem 8.10 Dorwyn Company
(a)
(b)
Production Rate
Door Gross Profit
($’000) Marketing Costs
($’000) Profit ($’000)
Incremental Profit ($’000)
0 0 0 0 -
1 4 1 3 3
2 8 8 0 -3
3 12 27 -15 -15
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(c)
Production Rate
Window Gross Profit Marketing Costs Profit Incremental Profit
0 0 0 0
1 6 2 4 4
2 12 8 4 0
3 18 18 0 -4
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(d)
Dorwyn management should produce the product mix of producing 1 door and 1 window.
(e) The solution based on separable programming approximation from part d is the same as
the solution obtained in part a for exact nonlinear programming.
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Chapter 9: Decision Analysis Q5(a) Problem 9.7
(a)
Maximax Criterion
State of Nature
Alternate S1 S2 S3
Maximum in
Row
A1 220 170 110
220 ← Maximax
A2 200 180 150
200
From Maximax criterion, Alternative A1 should be chosen because because it has the maximum of the maximum payoffs from each alternative.
(b)
Maximin Criterion
State of Nature
Alternate S1 S2 S3
Minimum in
Row
A1 220 170 110
110
A2 200 180 150
150 ←Maximin
From Maximin criterion, Alternative A2 should be chosen because it has the maximum of the minimum payoffs from each alternative.
(c)
Maximum Likelihood Criterion
State of Nature
Alternate S1 S2 S3
A1 220 170 110 ←2. Maximum
A2 200 180 150
Prior Probabilities 0.6 0.3 0.1
↑
1.
Maximum
From Maximum Likelihood criterion, Alternative A1 should be chosen because State S1 has the highest prior probabilities, and A1 has the maximum payoff for that state.
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(d)
Bayes' Decision Rule
Payoff Table State of Nature
Expected
Alternate S1 S2 S3
Payoff
A1 220 170 110
194
A2 200 180 150
189
PriorProbabilities 0.6 0.3 0.1
From Bayes’ Decision Rules, Alternative A1 should be chosen as it has larger expected payoff than A2.
(e)
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(f)
From TreePlan, Alternative A1 should be chosen.
(g)
Probability of S3 remains fixed:
Probability of S1 Action Expected Payoff
A1 194
0.3 A2 183
0.35 A2 184
0.4 A2 185
0.45 A1 186.5
0.5 A1 189
0.55 A1 191.5
0.6 A1 194
0.65 A1 196.5
0.7 A1 199
From Sensitivity Analysis using Solver Table, it can be seen that the action changes from
Alternative A2 to Alternative A1 at values of probabilities of S1 above 0.4. Hence the best
alternative changes to A1 as probabilities of S1 increases above 0.40. By redoing the
analysis with an increment of 0.01, it is found that the best alternative changes when the
probability of S1 increases above 0.43.
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(h)
Probability of S2 remains fixed:
Probability of S1 Action Expected Payoff
A1 194
0.3 A2 174
0.35 A2 176.5
0.4 A2 179
0.45 A2 181.5
0.5 A2 184
0.55 A1 188.5
0.6 A1 194
0.65 A1 199.5
0.7 A1 205
From Sensitivity Analysis using Solver Table, it can be seen that the action changes from
Alternative A2 to Alternative A1 at values of probabilities of S1 above 0.5. Hence the best
alternative changes to A1 as probabilities of S1 increases above 0.50. By redoing the
analysis with an increment of 0.01, it is found that the best alternative changes when the
probability of S1 increases above 0.51.
(i)
Probability of S1 remains fixed:
Probability of S2 Action Expected Payoff
A1 194
0 A2 180
0.05 A2 181.5
0.1 A2 183
0.15 A1 185
0.2 A1 188
0.25 A1 191
0.3 A1 194
0.35 A1 197
0.4 A1 200
From Sensitivity Analysis using Solver Table, it can be seen that the action changes from
Alternative A2 to Alternative A1 at values of probabilities of S2 above 0.1. Hence the best
alternative changes to A1 as probabilities of S2 increases above 0.10. By redoing the
analysis with an increment of 0.01, it is found that the best alternative changes when the
probability of S1 increases above 0.13.
(j) Based on the 3 sensitivity analysis done above, the desired alternative changes from
Alternative 1 to Alternative 2 when the prior probabilities of the given state of nature
deviates more than 0.10 from the original value. Since the true probabilities of the states
of nature should within 10% of given prior probabilities, I would choose Alternative 1.
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Q5(b) Problem 9.18 Telemore Company
(a) Decision Alternatives:
• Develop and market the new product
• Not to develop and market the new product
States of Nature:
• Successful launch of new product
• Unsuccessful launch of new product
(b)
From Bayes’ Decision Rule, it is found that launching the new product has a higher
expected payoff of $400 000, hence the decision chosen is to launch the new product.
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(c)
EP (without marketing survey) = 0.667 x $400,000 + 0.333 x $0
= $266,667
EP (with marketing survey) = $1,000,000
EVPI = EP (with marketing survey) - EP (without marketing survey
= $1,000,000 - $266,667
= $7,333,333
Since Cost ($100,000) < EVPI ($7,333,333), it is worthwhile to conduct marketing survey.
(d)
P(S|FSS) = P(S)•P(FSS|S) / [P(S)•P(FSS|S) + P(US)•P(FSS|US)] = 0.8(2/3) / [0.8(2/3) + 0.2(1/3)]
= 16/19 (0.889)
P(US|FSS) = P(US)•P(FSS|US) / [P(S)•P(FSS|S) + P(US)•P(FSS|US)] = 0.2(1/3) / [0.8(2/3) + 0.2(1/3)]
= 3/19 (0.111)
P(S|USS) = P(S)•P(USS|S) / [P(S)•P(USS|S) + P(US)•P(USS|US)] = 0.3(2/3) / [0.3(2/3) + 0.7(1/3)]
= 4/11 (0.462)
P(US|USS) = P(US)•P(USS|US) / [P(S)•P(USS|S) + P(US)•P(USS|US)] = 0.7(1/3) / [0.3(2/3) + 0.7(1/3)]
= 7/11 (0.538)
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(e) Optimal Policy:
Conduct the marketing survey.
If the result is favourable, launch the new product.
If the result is unfavourable, do not launch the product.
The expected payoff (which includes the cost of marketing survey) is $520 000.
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(f) Spider Graph
Tornado Diagram