Discount Rate |
||||||||
10% |
20% |
30% |
40% |
50% |
||||
Year |
Return R |
Cost C |
Net Return= R-C |
PV |
PV |
PV |
PV |
PV |
2011 |
-£50,000.00 |
£0.00 |
-£50,000.00 |
-£50,000.00 |
-£50,000.00 |
-£50,000.00 |
-£50,000.00 |
-£50,000.00 |
2012 |
£24,000.00 |
£12,000.00 |
£12,000.00 |
£10,909.09 |
£10,000.00 |
£9,230.77 |
£8,571.43 |
£8,000.00 |
2013 |
£29,000.00 |
£12,000.00 |
£17,000.00 |
£14,049.59 |
£11,805.56 |
£10,059.17 |
£8,673.47 |
£7,555.56 |
2014 |
£32,000.00 |
£12,000.00 |
£20,000.00 |
£15,026.30 |
£11,574.07 |
£9,103.32 |
£7,288.63 |
£5,925.93 |
2015 |
£35,000.00 |
£12,000.00 |
£23,000.00 |
£15,709.31 |
£11,091.82 |
£8,052.94 |
£5,987.09 |
£4,543.21 |
2016 |
£40,000.00 |
£12,000.00 |
£28,000.00 |
£17,385.80 |
£11,252.57 |
£7,541.21 |
£5,206.16 |
£3,687.24 |
NPV |
£23,080.08 |
£5,724.02 |
-£6,012.58 |
-£14,273.22 |
-£20,288.07 |
Below is table to show the print out of Excel cell formula for the discounted cash flow analysis.
Discount Rate |
||||||||
0.1 |
0.2 |
0.3 |
0.4 |
0.5 |
||||
Year |
Return R |
Cost C |
Net Return= R-C |
PV |
||||
2011 |
-50000 |
0 |
=C6-D6 |
=E6/(1+$F$4)^A5 |
=E6/(1+$G$4)^A5 |
=E6/(1+$H$4)^A5 |
=E6/(1+$I$4)^A5 |
=E6/(1+$J$4)^A5 |
2012 |
24000 |
12000 |
=C7-D7 |
=E7/(1+$F$4)^A6 |
=E7/(1+$G$4)^A6 |
=E7/(1+$H$4)^A6 |
=E7/(1+$I$4)^A6 |
=E7/(1+$J$4)^A6 |
2013 |
29000 |
12000 |
=C8-D8 |
=E8/(1+$F$4)^A7 |
=E8/(1+$G$4)^A7 |
=E8/(1+$H$4)^A7 |
=E8/(1+$I$4)^A7 |
=E8/(1+$J$4)^A7 |
2014 |
32000 |
12000 |
=C9-D9 |
=E9/(1+$F$4)^A8 |
=E9/(1+$G$4)^A8 |
=E9/(1+$H$4)^A8 |
=E9/(1+$I$4)^A8 |
=E9/(1+$J$4)^A8 |
2015 |
35000 |
12000 |
=C10-D10 |
=E10/(1+$F$4)^A9 |
=E10/(1+$G$4)^A9 |
=E10/(1+$H$4)^A9 |
=E10/(1+$I$4)^A9 |
=E10/(1+$J$4)^A9 |
2016 |
40000 |
12000 |
=C11-D11 |
=E11/(1+$F$4)^A10 |
=E11/(1+$G$4)^A10 |
=E11/(1+$H$4)^A10 |
=E11/(1+$I$4)^A10 |
=E11/(1+$J$4)^A10 |
NPV |
=SUM(F6:F11) |
=SUM(G6:G11) |
=SUM(H6:H11) |
=SUM(I6:I11) |
=SUM(J6:J11) |
Decising whether or not this is a worthwhile investment, giving your reasons.
According to Francis (2004) and DCA2C (2017), investment is said to be viable if NPV > 0. Therefore, Shambles’ investment is worthwhile, as the NPV (23,080.08) is greater than zero. This shows that investment earns more than the discount rate (Murthy 2011).
The NPV at a discount rate of 10% is positive. This suggests that the internal rate of return is greater than the rate of return (De Reyck, Degraeve &Vandenborre, 2008).
As result, discounting future cash flows will give will produce better results, hence acceptable NPV.
The NPV of Shamble returns and cost at the discounted rates are £ 5,724.02, -£ 6,012.58, -£ 14,273.22 and -£ 20,288.07, respectively. These results are as shown in the tables above. Computations were done on the Microsoft Excel application.
Below is a graphical representation of NPV against discount rate
On the graph above, NPV of zero is located where the interpolation line intersects with the . The discount rate at this point is 28%. This is the internal rate of return (IRR) of the investment, which is attained when (Benninga, Benninga &Czaczkes 2000). IRR will help Shambles Retailer to measure its profitability. The IRR of Shambles Retailer is greater than 10 % ( rate of return giving a positive NPV), this suggest is likely to make higher a profit if the rate of return is kept below 28% but above 10%.
Exercise 2: Linear Programming Optimization Model
Let be number of Pegasus Toys and be number of Phoenix Toys
First, the number of hours needs to sew or stuff per day is computed, which will be given by,
To conduct optimization, the graph of the above constraints is plotted. Here, and of the inequalities are determined. They are determined by substitution,
where is replaced with 0 when is computed, and is replaced with 0 when is computed (Berenson e ta l 2012). Below is a list of intercepts for the two inequalities.
. Below is a graphical representation of the four constraint inequalities.
R is the feasible region. The points that bound the feasible region are . These are points below the two lines and but above . To obtain the points that optimize profit, the trial and error method will be applied, where points , furthest away from the origin are considered (Francis 2004). The values of are substituted in the profit objective function, , to compute the profits. The point that leads to the highest profit value is considered as the profit optimal
0 |
5 |
6 |
10 |
7 |
|
12 |
7 |
6 |
0 |
4.4 |
|
Profit value |
120 |
110 |
108 |
80 |
100 |
Comparing the five profit values, profit is optimized when are . Therefore to maximize profit, Shambles will have to maintain zero production of Pegasus and produce 12 units of Phoenix toy.
Pegasus |
Phoenix |
Total |
Max Capacity |
||
Decision Variable |
0 |
12 |
|||
Profit |
£ 8 |
£ 10 |
£ 120 |
||
Constraint |
|||||
Sewing |
60 |
40 |
480 |
600 |
|
Stuffing |
40 |
50 |
600 |
600 |
Print out of Cell formula
Pegasus |
Phoenix |
Total |
Max Capacity |
||
Decision Variable |
0 |
12 |
|||
Profit |
8 |
10 |
=D37*D36+E37*E36 |
||
Constraint |
|||||
Sewing |
60 |
40 |
=D40*D36+E40*E36 |
600 |
|
Stuffing |
40 |
50 |
=D41*D36+E41*E36 |
600 |
A print-out of the Solver dialogue box with optimization conditions and constraints.
The following is a diagram Excel Solver dialogue.
Optimal profit value is 120,
This is attained when the company fails to produce Pegasus toys but maximize profit by producing 12 units of Phoenix.
Exercise Three: GPSS Simulation Model
The sales counter next to the soft toy display in Shambles receives a customer every 3-5 minutes.
Most of these customers (90%) are buying toys and are dealt with by the cashier in 2-4 minutes. The remaining 20% of customers come to open accounts that require an account manager. These
customers wait for the account manager, who spends 20-30 minutes serving them.
Solution
GENERATE 4,0
QUEUE system
QUEUE Arrival
SEIZE Cashier
DEPART Arrival
ADVANCE 4,2
RELEASE Cashier
DEPART Arrival
TRANSFER 0.9,Cashier,Manager
Cashier TERMINATE 1
Manager TERMINATE 1
There are 3 entries and eleven block type
Exercise Four: Association Rule Mining
Mining all the association rules for the data in the table below, given that support, 66% (0.66)
Tid |
Items bought |
10 |
|
20 |
|
30 |
To work out this problem data is analyzed first, to determine the frequencies of occurrence of sets of items bought (Han, Pei & Kamber 2011). The table below shows the sets of items bought with their respective frequencies.
Set of items bought |
Frequency |
2 |
|
2 |
|
3 |
|
1 |
|
2 |
|
2 |
|
1 |
These frequencies are used to determine the degree of support and confident of items set
( Agrawal& Srikant1994). Second, the possible association rules are determined. Finally, with the degree of support and confidence are subjected to the association rule conditions. “Given the set of transactions T, the goal of association rule is to find all the rules that satisfy, support ≥ min sup threshold and a confidence ≥ min conf threshold”( Tan, Steinbach& Kumar 2005). For case in question, the conditions, , will have to be met. Below are possible Association rules, and computation of their Support and Confidence.From the above possibilities, the association rules that meet the requirements are:Their computed degree of support and confidence are greater than the minimum threshold of support (0.66) and confidence (0.34).
Exercise Five: Artificial Neural Networks
There are four training input/target pairs for a two-class problem:The initial weight vector given is
A two-input perceptron with hard limit activation function is used to solve this classification problem. Write down the step-by-step solution to train the classifier by the basic perceptron learning.
Solution
Steps involved are:
Inputs |
Desired Output |
|
2 |
0 |
1 |
1 |
2 |
1 |
-1 |
3.5 |
0 |
0 |
-1 |
0 |
According to (Yegnanarayana, 2009) output ( ) is given by To determine the value of , decision boundary is considered, the output is equated to 0. Therefore, the output will be given by Since the decision boundary passes through the point (0.5, 0) as illustrated in the chart above, then the computation of will be determined by The hardline transfer function forces neural output a 1 if the input reaches a threshold, otherwise, its output is 0. This allows the neuron to make a decision (Mitchell 1997).
Designing a DSS to help decision-makers run the 2012 London Olympics.
Linear programming optimization model will be the model of DSS to be utilized. According to Dantzig ( 2016 ) and Zoutendijk (1960), is among the common decision making support model that is utilized by most decision-makers, in the production industry to make a production plan. Therefore, since Olympic competition will require equipment, mostly medals for the winners, this model will be critical as it will help Olympic in deciding the number of medals that can be produced at minimal cost and time, to meet the demand of the whole event. To design DSS, the following projections well made.
Projections
The following table shows the whole information concerning the production of medal for the Olympics.
Time in minutes |
|||||
Medal |
Cost of Production ($ ) |
Pressing &Cleaning |
Re-Heating |
Engraving and & Packaging |
Time for producing One medal |
Gold |
600 |
180 |
60 |
120 |
360 |
Silver |
500 |
150 |
60 |
90 |
300 |
Bronze |
400 |
120 |
30 |
90 |
240 |
To find credible solutions for the above established Olympic project, Solver in the Ms excel is used to determine the cost-optimal point.
References
Agrawal, R. and Srikant, R., 1994, September. Fast algorithms for mining association rules. In Proc. 20th int. conf. very large data bases, VLDB (Vol. 1215, pp. 487-499).
Berenson, M., Levine, D., Szabat, K.A. and Krehbiel, T.C., 2012. Basic business statistics: Concepts and applications. Pearson higher education AU.
Benninga, S., Benninga, S.Z. and Czaczkes, B., 2000. Financial modeling. MIT press.
DCA2C, D.D., 2017. BUSINESS MATHEMATICS AND STATISTICS.
De Reyck, B., Degraeve, Z. and Vandenborre, R., 2008. Project options valuation with net present value and decision tree analysis. European Journal of Operational Research, 184(1), pp.341-355.
Dantzig, G., 2016. Linear programming and extensions. Princeton university press.
Dantzig, G.B. and Thapa, M.N., 2006. Linear programming 1: Introduction. Springer Science & Business Media
Francis, A., 2004. Business mathematics and statistics. Cengage Learning EMEA.
Han, J., Pei, J. and Kamber, M., 2011. Data mining: concepts and techniques. Elsevier.
Murthy, G., 2011. Project Appraisal Techniques.
Mitchell, T.M., 1997. Artificial neural networks. Machine learning, 45, pp.81-127.
Rosen, K.H., 2017. Handbook of discrete and combinatorial mathematics. Chapman and Hall/CRC
Russell, S.J. and Norvig, P., 2016. Artificial intelligence: a modern approach. Malaysia; Pearson Education Limited,
Remer, D.S. and Nieto, A.P., 1995. A compendium and comparison of 25 project evaluation techniques. Part 1: Net present value and rate of return methods. International journal of production economics, 42(1), pp.79-96.
Tan, P.N., Steinbach, M. and Kumar, V., 2005. Introduction to data mining. 1st.
Yegnanarayana, B., 2009. Artificial neural networks. PHI Learning Pvt. Ltd.
Zoutendijk, G., 1960. Methods of feasible directions: a study in linear and non-linear programming. Elsevier.
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