Supply Chain Management Assignment: Cases on Optimization and Analysis

This document presents a comprehensive solution to a Supply Chain Management assignment, addressing various analytical techniques. The assignment covers four cases. Case 1 focuses on linear programming, including model building, computer solutions, and sensitivity analysis for advertising decisions and make-or-buy scenarios. Case 2 explores decision models using decision trees and the Analytical Hierarchy Process (AHP) for supplier selection. Case 3 delves into probability distributions in logistics and applies them to a grocery retailer's delivery problem. Finally, Case 4 examines hypothesis testing and regression analysis, including a fleet analysis and a fashion retailer's sales forecast, with interpretations and model validation.

RUNNING HEADER: SUPPLY CHAIN MANAGEMENT 1
Supply Chain Management
Student’s name:
Student’s ID:
Institution:

Supply Chain Management 2
Case 1: Linear Programming Model Building and Computer Solution [25 marks]
Q1.1 Cranfield Grill
a. Identify the decision variables.
The decision variable for this function is exposure rating per Ad.
b. Build the objective function and the constraints.
Maximize Z = 90x1 + 25x2 + 10x3
s.t
x1 + x2 + x3 < 279,000
x1 > 140,000
x2 < 99,000
x3 > 30,000
c. Transfer your model to Excel and solve. Please write in your report the total exposure
and the total number of potential new customers reached for each ad medium.
Ad costs:
X1 = 140,000
X2 = 99,000
X3 = 30,000
Number of Ads:
X1 = 140,000/10,000 = 14
X2 = 99,000/3,000 = 33
X3 = 30,000/1000 = 30
Number of customers:

Supply Chain Management 3
X1 = 14; 4000 * 10 + 4 * 1,500 = 46,000
X2 = 33; 2000*15 + 18 * 1200 = 51,600
X3 = 30; 1000 * 20 + 10 * 800 = 28,000
Total = 125,600 customers
d. How would the total exposure change if Cranfield Grill had an addition £10,000?
The total exposure change of the ds if Cranield Grilll had an additiona £10,000 would not
change.
e. Please discuss how sensitive Cranfield Grill’s solution is to exposure ratings.
The Cranfield Grill’s solution is not sensitive is too sensitive to exposure ratings since a 1%
increase in exposure rating results in a 1% increase in exposure ratings.
f. How would the solution change if Cranfield Grill’s objective was to maximise the
total number of new customers reached?
The solution of the variable would remain the same. However, the objective function
would change where it increase.
g. Which objective function (max exposure versus max new customers reached) makes
more sense? Why?
The max esposure objective functions makes more sense since it gives a validation of the
esense of an Ad in its potentiall to reach the maximum number of customers.
Q1.2 Cranfield Materials

Supply Chain Management 4
a. Formulate and solve a linear programming model for this make-or-buy decision
problem. How many of each component should be manufactured and how many
should be purchased?
The number of support and straps to be manufactured are 14,824 and 2,162 units each.
However, frames were to be purchased and not mafactured and would require 5,000.
b. What is the total cost of manufacturing and purchasing plan?
Manufacturing cost = 14,824 * 11.5 + 2,162 * 6.5 = 187,454
Purchasing cost = 5,000 * 38 = 190,000
Total cost = 377,454
c. For which departments should Cranfield Materials seek additional processing time?
What would be the effect of one additional hour on the total cost?
The shapping department needs to be given an additional processing time. The additional
hour will result in a minimization of total cost since it is running below full capacity.
d. Cranfield Materials identified another supplier for frames with a cost of £45 / frame.
Should they pursue this opportunity? Why or why not?
Cranfield should not pursue this opportunity since it will increase the cost of production of
frames.
e. What are some of the limitations of this approach?
The limitaons of linear progrmming is that increasing production by a single process will
require the quantity of all inouts to be increased in a fixed proportion. Moreover, it is possible
that the objective functna nd the constrainrs may not be directly specified by linear in
equality equations.

Supply Chain Management 5
Case 2: Decision Models
Q2.1 Decision Trees
a. Use a decision tree analysis to aid in this undertaking. Hint: Expecting a sequential
decision analysis with probabilities estimated using the Bayes’ Rule.
0.1
Fraudulent 5 -200000
2500 -200000
Container 2
0 0.9
Non fraudulent 6 25000
25000
No container 3 0
10500 0
1
4 0.95
Fraudulent 9 -210000
-2E+05 -200000
Positive 7
0 0.05
Non fraudulent 10 15000
20500 25000
Test container 4
-10000 8 0.02
Fraudulent 11 -210000
20500 -200000
Negative 8
0 0.98
Non fraudulent 12 15000
25000
b. Calculate the Expected Value of Perfect Information
The expected value of perfect information is 10,500
c. Calculate the Expected Value of Sample Information
The ecpected value of sample information is:
10,500 – 2,500 = 8,000

Supply Chain Management 6
d. Under what circumstances would this approach be preferable?
The decision tree anlaysis approach is preferable when there are options to a problem which
need to be challenged.
Q2.2 Analytical Hierarchy Process
a. Identify the weights of the main criteria.
CTQ'S Economic Social Environmental
Economic 1 4 3
Social 0.25 1 2
Environmental 0.33 0.5 1
Column total 1.58 5.5 6
CTQ'S Normalize D Score Economic
Normalize D Score
Social
Normalize D Score
Environmental
Weights/
average
Economic 0.63 0.73 0.50 0.62
Social 0.16 0.18 0.33 0.22
Environmental 0.21 0.09 0.17 0.16
Column total 1 1 1 1.00
The weights of economic, social and environmental criteria are 0.62, 0.22 and 0.16.
b. Check the consistency of the decision maker in their assessment of the main criteria.
CTQ'S Consistency measure
Economic 1.99
Social 0.69
Environmental 0.47
c. Complete the AHP analysis and produce a rank order of the eight suppliers.

Supply Chain Management 7
Scores by rank
Supplier 7 0.15
Supplier 3 0.14
Supplier 6 0.13
Supplier 8 0.13
Supplier 2 0.13
Supplier 4 0.12
Supplier 1 0.11
Supplier 5 0.09
Thus, supplir 7 should be considered.
d. What other criteria do you think they should have considered?
Under the social criteria, aother criteria that should have been considered is social
responsibility.
Case 3: Probability Distributions and Sampling
Q3.1 Probability Distributions in Logistics and Supply Chain Management
In logistics, a regression model is used in predicting demand based on historical trends.
Another example of an application of probability theory in logistics is decision trees whch is
applied to determine possible consequences with regards to chance of event outcomes, utlity
and resource costs. The third example of an application of probability theory is the analytic
hierarchy process which is used in organizaing and analysing comlex decisions in logistics.
Q3.2 Grocery Retailer’s Delivery Problem
a. Plot the histogram of distance from store per order

Supply Chain Management 8
2565 2685 2805 2925 3045 3165 3285 3405 3525 3645 More
b. State the mean and the standard deviation of the sample along with five number
summary
Mean 3027.2
Standard
deviation 224.5
Minimum 2446
First quartile 2876
Median 3028.5
Third quartile 3019
Maximum 3600
c. What should be the range of the autonomous robot, if the retailer plans for it to deliver
4 orders without having to recharge 99% of the time?
Coonfidence interval of number of orders at 99%:

Supply Chain Management 9
The range of orders to be delivered with 99% confidence is from 2969.3 to 3085.1.
Thus, the range of autonomous robots is: 2969.3/4 – 3085.1/4
= 743 to 772 autonomous robots
Case 4: Hypothesis Testing and Regression
Q4.1 A Logistics Company’s Fleet
a. State the null and the alternative hypotheses.
H0: There is no difference in average miles per gallon between the two groups.
H1: There is a difference in average miles per gallon between the two groups.
b. Conduct the appropriate test at alpha = 0.05 significance level to establish whether the
training worked.
SUMMARY
Groups Count Sum Average Variance
MPG of drivers who did the training. 77 391.41 5.08 2.43
MPG of drivers who did not do the training. 83 382.14 4.60 2.83
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 9.170489 1 9.17 3.48 0.06 3.90
Within Groups 416.2303 158 2.63
Total 425.4008 159
It can be seen that there is no statistical difference of the ANOVA model (p > 0.05). Thus, we
fail to reject the nul hypotheisis and conclude that there is no difference in the average miles
per gallon between the two groups. Thus, the did not work.
c. Would an extremely risk averse organisation/manager make the same decision?

Supply Chain Management 10
Thus, a risk averse organisation/manager will not opt to take the risk since the miximization
of miles travelled per gollon of fuel is minmal.
d. Discuss the caveats in this approach.
The limitations of the ANOVA method/approach is that all the populaon means from each
dara groups are considered roughly equal while all the variance from each data grop must be
equal. As a result, in real-world applications, these conditions/assumptions are rarely met.
Q4.2 A Fashion Retailer’s Sales Forecast
Develop a multiple regression model to explain the variability in net sales. Interpret the
regression results.
Table 1: Multiple Regression Output
Regression Statistics
Multiple R 0.92
R Square 0.85
Adjusted R Square 0.85
Standard Error 326.88
Observations 567
ANOVA
df SS MS F Significance F
Regression 4 331440304.4 82860076.09 775.48 0.00
Residual 562 60049911.52 106850.3764
Total 566 391490215.9
Coefficients Standard Error t Stat P-value Lower 95% Upper 95%
Intercept -243.44 32.52 -7.49 0.00 -307.32 -179.56
Number of Customers 1.43 0.19 7.72 0.00 1.07 1.80
Store Area 0.33 0.04 8.79 0.00 0.26 0.40
Sales Staff 78.61 2.92 26.96 0.00 72.88 84.33
Store Type -471.02 34.72 -13.57 0.00 -539.22 -402.83

Supply Chain Management 11
From table 1, it can be seen that 85% of the variability is accounted for by factors in the
model. On the other hand, 15% of the variability is accounted for by factors which are not in
the model.
The regression model was proved to be statistically significant (p < 0.05). Conversely, the
coeffcients of the model were found to be statistically signifnicat (p < 0.05). Holding all other
factors constant, a unit increase in either the number of customers, the store area of the sales
staff increases the net sales by either 1.43, 0.33 or 78.61 units respectively. It can also be
noted that the nets sales reduces by 471.02 units when the store is located in the high street
compared to stores located in shoping malls.
a. The company is expecting to open a new, 200 square-meters store in a shopping mall
to be staffed with 6 sales people. Assuming they will expect 500 customers to visit the
store per day, what net sales can they expect?
y = -243.44 + 1.43*number of customers + 0.33*store area + 78.61*sales staff – 471.02*store
type
y = -243.44 + 1.43*500 + 0.33*200 + 78.61*6 – 471.02*0
y = 1009.22
Thus, the expected net sales will be 1,009.22.
b. Check whether the assumptions of multiple linear regression are met.
Assumption 1: The indepednt variable (net sales) is continouos measure.
Assumption 2: there are more than one indepedent variables.

Supply Chain Management 12
c. Is multiple regression an appropriate approach to address the fashion retailer’s
problem? Explain why or why not?
The multiple regression si an appropriate approach in addressing the fashion retailer’s
problem since it helps understanding whether net sales can be predicted based on number of
customers, store area, number of sales staff and store type.