Supply Chain Challenges And Forecasting Techniques
Vendor management or Vendor list selection
Three possible challenges that can be identified from the given supply chain diagram are Vendor management or Vendor list selection, Optimization of Transportation costs and Forecasting of Inventory Stocks.
Selection of the appropriate or best vendors can have far reaching consequences on the supply chain. It can influence the cost of operations, risk and quality of the products. The task of choosing the right vendor from the available options in the market is therefore crucial. This requires keen market insight. The objective is mainly to select those who would offer transparency in business, will be easy to communicate with when it comes to planning strategy and most importantly offer quality products (Bozarth and Handfield 2019) .
Yet another issue is to identify the best options whereby transportation costs could be reduced in terms of time and capital. The challenge in this case is to identify the best routes in times of cost as well as time efficiency. Transportation of goods forms a key aspect of supply chain management which in turn may affect product quality and delivery as well as the over all costs (Christopher 2016).
Inventory management deals with the challenge to plan and procure goods for business. An efficient supply chain takes into account the demand of goods in the market and plans the supply to meet the demand. It is crucial to be able to tell how much the demand for a product will be on the market and hence how much stock one must keep in the inventory to ensure that the inventory is neither surplus nor falls short of the supply (Christopher, 2016).
Consumer Market Surveys are tools to gain market insights by companies from customers themselves. The survey involves asking people about their purchases and preferences. The surveys can be conducted via telephones or through personal interviews or even by questionnaires that the consumer is requested to fill out and submit either via mail or online. The responses are then analyzed using statistical tools and methods to gather market insight and validate conjectures regarding the market.
The Naïve forecasting method is the simplest forecasting technique of all. The method operates under the assumption that what happens today is expected to happen tomorrow as well. This means it assumes that the observations of the current time point as the forecast for the subsequent time period. It disregards any causality or influence of observations from time points preceding the current time point. So if observed Y at time point t is Yt then as per this method the forecast for time point t+1 is Yt (Montgomery, Jennings and Kulahci 2015).
Optimization of Transportation costs
The simple linear regression technique identifies two variables, one dependent and the other independent based on whom a model establishing the linear relationship between the two is identified. Simple linear regression is denoted by a straight line where the Y axis denotes the dependent and X denotes the independent variable. The regression line is the line of best fit which is line for which sum of squared errors from each data point (defined by a pair of observed (x, y)) is least. Let Y= a+ bX, be the regression line. Then this equation gives a value of Y for each value of X. This is how the forecasting works (Chatterjee and Hadi 2015).
Period |
Naïve Forecast |
Error |
Absolute Error |
Percent Error |
Squared Error |
|
January |
10 |
N/A |
N/A |
N/A |
N/A |
N/A |
February |
12 |
10 |
2 |
2 |
16.66667 |
4 |
March |
16 |
12 |
4 |
4 |
25 |
16 |
April |
13 |
16 |
-3 |
3 |
23.07692 |
9 |
May |
17 |
13 |
4 |
4 |
23.52941 |
16 |
June |
19 |
17 |
2 |
2 |
10.52632 |
4 |
July |
15 |
19 |
-4 |
4 |
26.66667 |
16 |
August |
20 |
15 |
5 |
5 |
25 |
25 |
September |
22 |
20 |
2 |
2 |
9.090909 |
4 |
October |
19 |
22 |
-3 |
3 |
15.78947 |
9 |
November |
21 |
19 |
2 |
2 |
9.52381 |
4 |
December |
19 |
21 |
-2 |
2 |
10.52632 |
4 |
0.818182 |
3 |
17.76332 |
10.09091 |
|||
BIAS |
MAD |
MAPE |
MSE |
|||
|
Standard error (Square Root of MSE) =3.176619 |
Data on sales of cosmetics for the year 2017 for a cosmetic company was used to forecast units sold per month using the technique showed in the given diagram. The computation was done by finding the sum of product of observed value with the weights 0.2, 0.3 and 0.5 respectively for the preceding three years of the current forecast.
The following table shows the data and the corresponding forecasts.
Month |
Actual Units sold (in 000) |
Three Period Moving Average |
weights |
|
January |
155 |
0.2 |
||
February |
168 |
0.3 |
||
March |
159 |
0.5 |
||
April |
162 |
161 |
||
May |
178 |
162 |
||
June |
177 |
169 |
||
July |
189 |
174 |
||
August |
198 |
183 |
||
September |
205 |
191 |
||
October |
215 |
200 |
||
November |
220 |
209 |
||
December |
210 |
216 |
||
January |
214 |
214 |
(forecasted) |
|
February |
214 |
214 |
(forecasted) |
|
March |
213 |
213 |
(forecasted) |
|
April |
214 |
214 |
(forecasted) |
|
May |
214 |
214 |
(forecasted) |
|
June |
214 |
214 |
(forecasted) |
The forecast for the next January is 214 which is an increase from that observed in January of current year. Computing further forecasts using forecasted values, values are seen to become stable.
The same data as the previous part was used to forecast using Moving average method in this part. The moving averages were computed by taking weighted averaging using given weights over the past three years of the time point for which forecast is being made. The data therefore does not have forecasts for the first three time points.
Month |
Actual Units sold (in 000) |
Weights |
Three Month Weighted Moving average |
|||
January |
155 |
0.76 |
||||
February |
168 |
0.18 |
Calculation of Weighted Moving Average (Period=3) |
|||
March |
159 |
0.06 |
Sum/ Sum of weights(=1) |
|||
April |
162 |
157.58 |
117.8 |
30.24 |
9.54 |
|
May |
178 |
166.02 |
127.68 |
28.62 |
9.72 |
|
June |
177 |
160.68 |
120.84 |
29.16 |
10.68 |
|
July |
189 |
165.78 |
123.12 |
32.04 |
10.62 |
|
August |
198 |
178.48 |
135.28 |
31.86 |
11.34 |
|
September |
205 |
180.42 |
134.52 |
34.02 |
11.88 |
|
October |
215 |
191.58 |
143.64 |
35.64 |
12.3 |
|
November |
220 |
200.28 |
150.48 |
36.9 |
12.6 |
|
December |
210 |
207.7 |
155.8 |
38.7 |
13.2 |
|
Next Period |
|
215.6 |
163.4 |
39.6 |
12.6 |
|
Sum of Weights |
1 |
|
The forecast for the January of next year is 215.6. So there has been positive change or increase in units sold when compared to observation of January of current year.
The completed regression analysis for the given diagrams and tables is given in the following table. The estimate of slope and intercept was computed using the formula as provided in the example:
Actual Value (Y) |
Period number X |
XY |
X^2 |
||||
74 |
1995 |
147630 |
3980025 |
Slope= |
b= |
10.53571429 |
|
79 |
1996 |
157684 |
3984016 |
Intercept= |
a= |
-20951.5 |
|
80 |
1997 |
159760 |
3988009 |
||||
90 |
1998 |
179820 |
3992004 |
||||
105 |
1999 |
209895 |
3996001 |
||||
142 |
2000 |
284000 |
4000000 |
||||
122 |
2001 |
244122 |
4004001 |
||||
Total |
692 |
13986 |
1382911 |
27944056 |
|||
X |
Predicted Y |
|
|||||
2002 |
141 |
The trend line was thus found to be given by the equation:
y= 10.54x- 20951.5
(b) The following analysis was based on collected data, on distance travelled by a vehicular unit and the fuel consumed by the same. The fuel consumed is the dependent variable and the distance travelled is the independent variable. The following table shows the computations.
Distance Travelled (X) |
Fuel Consumed (Y) |
XY |
X^2 |
Y^2 |
|
|
75000 |
12.5 |
937500 |
5625000000 |
156.25 |
75300 |
9.5 |
717142.9 |
5670090000 |
90.7029478 |
|
75600 |
12.0 |
907200 |
5715360000 |
144 |
|
75900 |
11.1 |
843333.3 |
5760810000 |
123.45679 |
|
76200 |
9.4 |
714375 |
5806440000 |
87.890625 |
|
76550 |
11.3 |
864274.2 |
5859902500 |
127.471384 |
|
76820 |
9.0 |
691380 |
5901312400 |
81 |
|
77110 |
10.0 |
771100 |
5945952100 |
100 |
|
77460 |
13.0 |
1004111 |
6000051600 |
168.038409 |
|
77690 |
8.2 |
638167.9 |
6035736100 |
67.4744898 |
|
SUM |
763630 |
105.9774919 |
8088584 |
5.8321E+10 |
1146.28465 |
r |
-0.315 |
||||
Slope= |
b= |
-0.00055 |
|||
Intercept= |
a= |
52.6730 |
The regression equation was found as y=52.6730 – 0.00055x. So increase in distance travelled would correspond to a decrease in fuel consumed by 0.00055 units.
Reference
Bozarth, C.C. and Handfield, R.B., 2019.operations and supply chain management.
Chatterjee, S. and Hadi, A.S., 2015. Regression analysis by example. John Wiley & Sons.
Christopher, M., 2016. Logistics & supply chain management. Pearson UK.
Montgomery, D.C., Jennings, C.L. and Kulahci, M., 2015. Introduction to time series analysis and forecasting. John Wiley & Sons.