economic demand and supply test
Question 1
 Set up the transportation tableau and obtain an initial feasible solution using the cheapest cell method:
i)
To
From 
Muirness  Nettleham  Owlerton  Weekly Output 
Alphatown

11
170 
210 
220 
11 
Bravo City

4
180 
1
250 
9
240 
14 
Carlieville

220 
220 
12
210 
12 
Deltaborough

190 
8
160 
200 
8 
Weekly Requirement  15  9  21  45
45 
Calculate the shadow costs and the opportunity costs:
To
From 
Muirness
0 
Nettleham
70 
Owlerton
60 
Weekly Output 
Alphatown
170 
11
170 
30
210 
10
220 
11 
Bravo City
180 
4
180 
1
250 
9
240 
14 
Carlieville
50 
170
220 
100
220 
12
210 
12 
Deltaborough
90 
100
190 
8
160 
50
200 
8 
Weekly Requirement  15  9  21  45
45 
Trace a closed path for the cell having the largest negative opportunity cost and place alternate + and – signs at the corners.
To
From 
Muirness
0 
Nettleham
70 
Owlerton
60 
Weekly Output 
Alphatown
170 
11 –
170 
30 +
210 
10
220 
11 
Bravo City
180 
4 +
180 
1 –
250 
9
240 
14 
Carlieville
50 
170
220 
100
220 
12
210 
12 
Deltaborough
90 
100
190 
8
160 
50
200 
8 
Weekly Requirement  15  9  21  45
45 
Produce the new solution and test to see whether it is the optimum solution.
To
From 
Muirness
0 
Nettleham
40 
Owlerton
60 
Weekly Output 
Alphatown
170 
10 –
170 
1 +
210 
10
220 
11 
Bravo City
180 
5 +
180 
30 –
250 
9
240 
14 
Carlieville
150 
70
220 
30
220 
12
210 
12 
Deltaborough
120 
70
190 
8
160 
20
200 
8 
Weekly Requirement  15  9  21  45
45 
We still have a negative opportunity cost, so this is not an optimum solution.
Produce the new solution and test.
To
From 
Muirness
0 
Nettleham
40 
Owlerton
60 
Weekly Output 
Alphatown
170 
10 –
170 
1
210 
10 +
220 
11 
Bravo City
180 
5 +
180 
30
250 
9 –
240 
14 
Carlieville
150 
70
220 
30
220 
12
210 
12 
Deltaborough
120 
70
190 
8
160 
20
200 
8 
Weekly Requirement  15  9  21  45
45 
New Solution
To
From 
Muirness
0 
Nettleham
40 
Owlerton
50 
Weekly Output 
Alphatown
170 
1
170 
1
210 
9
220 
11 
Bravo City
180 
14
180 
30
250 
10
240 
14 
Carlieville
160 
60
220 
20
220 
12
210 
12 
Deltaborough
120 
70
190 
8
160 
30
200 
8 
Weekly Requirement  15  9  21  45
45 
All opportunity costs are positive, so we have found a unique, optimum solution.
(8 Marks)
 ii) The minimum delivery cost is £8,680 (1 x 170) + (14 x 180) + (1 x 210) + (8 x 160) + (9 x 220) + (12 x 210)
(2 Marks)
 No deliveries can be made from Alphatown to Nettleham. The initial solution is shown below:
i)
To
From 
Muirness  Nettleham  Owlerton  Weekly Output 
Alphatown

11
170 
210 
220 
11 
Bravo City

4
180 
1
250 
9
240 
14 
Carlieville

220 
220 
12
210 
12 
Deltaborough

190 
8
160 
200 
8 
Weekly Requirement  15  9  21  45
45 
To
From 
Muirness
0 
Nettleham
70 
Owlerton
60 
Weekly Output 
Alphatown
170 
11 –
170 
210 
10 +
220 
11 
Bravo City
180 
4 +
180 
1
250 
9 –
240 
14 
Carlieville
150 
70
220 
0
220 
12
210 
12 
Deltaborough
90 
100
190 
8
160 
50
200 
8 
Weekly Requirement  15  9  21  45
45 
New solution:
To
From 
Muirness
0 
Nettleham
70 
Owlerton
60 
Weekly Output 
Alphatown
170 
2 –
170 
210 
9 +
220 
11 
Bravo City
180 
13 + 180  1 –
250 
10
240 
14 
Carlieville
150 
60
220 
10 +
220 
12 –
210 
12 
Deltaborough
90 
100
190 
8
160 
60
200 
8 
Weekly Requirement  15  9  21  45
45 
To
From 
Muirness
0 
Nettleham
70 
Owlerton
60 
Weekly Output 
Alphatown
170 
1
170 
210 
10
220 
11 
Bravo City
180 
14 180 
250 
10
240 
14 
Carlieville
150 
60
220 
1
220 
11
210 
12 
Deltaborough
90 
100
190 
8
160 
60
200 
8 
Weekly Requirement  15  9  21  45
45 
To
From 
Muirness
0 
Nettleham
60 
Owlerton
50 
Weekly Output 
Alphatown
170 
1
170 
210 
10
220 
11 
Bravo City
180 
14 180  10
250 
10
240 
14 
Carlieville
160 
60
220 
1
220 
11
210 
12 
Deltaborough
100 
90
190 
8
160 
50
200 
8 
Weekly Requirement  15  9  21  45
45 
(8 Marks)
This gives a minimum delivery cost of:
£8,700 (1 x 170) + (14 x 180) + (1 x 220) + (8 x 160) + (10 x 220) + (11 x 210)
(2 Marks)
 c) If demand is less than supply we invent a fictitious destination (a ‘dummy’ destination) to take the surplus supply. Since nothing will actually be transported to the dummy destination the transportation costs to the dummy destination are usually equal to zero. However, if there is a penalty cost for overproduction (e.g. storage cost) it can be incorporated as a transportation cost to the dummy destination.
If demand exceeds supply we invent a fictitious source (a ‘dummy’ source) to make up the deficiency. The transportation costs from this source are usually equal to zero, since no actual deliveries will be made from the dummy source. However, if there is a penalty cost for falling short of the demand at a particular destination, this can be incorporated as a transportation cost from the dummy source to this destination.
(5 Marks)
Total 25 Marks
Question 2
a)  
i.  EOQ =  √2cod  √2 x 7500 x 50  = 750 units  
ch  0.20 x 6.67  
d = Annual demand = 7500  
co = set up cost = £50  
Ch = Holding cost = 0.20  
P = Cost Price = £6.67  
Order frequency =  annual demand  =  7500  = 10  
EOQ  750  
11 orders per year or an order every 5 weeks.  
(6 Marks)  
ii.  ROL = expected demand during lead time + (Z x standard deviation of demand)  
Standard Deviation during lead time = √3×50² = 86.6  
Demand during lead time = 150 x 3 = 450.  
450 + (1.64 x 86.6) = 592 units.  
Safety stock = (Z x standard deviation) = 1.64 x 86.6 = 142 units.  
Cost of safety stock = 142 units x ip = 142 x (0.2 x 6.67) = 142 x 1.33 = £188.86  
(6 Marks)  
b)  
i.  POQ =  
√2CD  
Ip (1 d/r)  
d = 30,000  
co = 200  
p = 17.50  
Ch = 0.15  
R = 35,000  
ch = 2.625 1 – d/r = 0.1429  
√2x200x30,000  = 5,657 Production Order Quantity (POQ)  
2.625 x 0.1429  
Stock Cycle Length = Q/d = 5,656/30,000 = 0.189 years.  
0.189 x 50 = 9.45 = 9.5 weeks in practice  
30,000/50 x 9.5 = 5,700 units every 9.5 weeks  
Total Annual Cost  
Chq/2 (1d/r) + cod/Q + pd  
2.625 x 5700/2 x 0.1429 + 200 x 30,000/5700 + 17.50 x 30,000  
1,069.07 + 1,052.01 + 525,000 = £527,121  
(7 Marks)  
ii.  √2cd/ip  
d = 30,000  
co = 130.00  
ch = 0.15  
p = 18.50  
√2 x 30,000 x 130 / 0.15 x 18.50 = 1,676.55 units 1,677  
Total Annual Cost  
chq/2 + Cod/Q +pd  
0.15 x 18.50 x 1,677/2 + 30,000 x 130/1,677 + 30,000 x 18.50  
2,326.84 + 2,325.58 + 555,000 = £ 559,652.42  
Buying in this component is not a worthwhile investment. Total cost of buying is £32,530.78 a year more expensive than producing in house.  
(6 marks)  
Total Q3 25 Marks  
Question 3
 Interpret range chart first
Variability of the process is in control – the range chart has no points outside of the warning limits and there are no patterns (runs, etc) to show the process range is out of control.
The means chart has sample 13 outside of the upper action limit.
However, samples 4 to 9 have means which are consistently increasing. Therefore, the process is out of control at sample 9, with an indication of an upward shift in the mean. The process should have been investigated at sample 9.
(5 Marks)
 Lower warning and action limits are often omitted as points below lower limits indicate an improvement in the process variability.
They may be included so that an improvement is evident, so that reasons for the improvement in the process can be investigated.
(2 Marks)
C_{p} = 1.23 C_{p} >1, therefore the process is potentially capable.
C_{pk} = 1.11 C_{pk} >1, therefore the process is capable in practice.
C_{p} = and does not take centring of the process in to account.
C_{pk} = min {,} and takes centring of the process in to account.
(5 Marks)
 Use samples 1 – 10 to set up the chart.
Sample Number  1  2  3  4  5  6  7  8  9  10 
Number Defective  2  0  1  3  1  1  3  0  3  2 
Sample size (n) = 50
The total % of defects over the period, p = = 0.032 = 3.2%.
The centre line is drawn at 3.2%
The warning limits are drawn at p ± 1.96 = 3.2 ±
= 3.2 ± 4.88 = (1.68, 8.08)
The action limits are drawn at p ± 3.09 = 3.09 ±
= 3.2 ± 6.69 = (4.49, 10.89)
The lower warning and action limits are negative, therefore omit.
Use the chart to monitor samples 1120
Sample Number  11  12  13  14  15  16  17  18  19  20 
Number Defective  1  2  1  2  3  4  3  2  2  1 
Percent Defective  2  4  2  4  6  8  6  4  4  2 
Students should draw chart and plot samples 11 – 20.
(10 Marks)
 No points breach the upper action or warning limits. However, from sample 14 to 19, there is a run of 6 points above the centre line. This means that the process can be considered out of control at sample 19, so the process should be stopped and the assignable causes of variation investigated. An upward shift in the % defective is indicated.
(3 Marks)
Total 25 Marks
Question 4
a)
Month  Price  Forecast  Error  Error^{2}  T 
May 14  322  –  –  –  – 
Jun 14  360  322  38.0  1444.0  0.12 
Jul 14  350  329.6  20.4  416.2  0.06 
Aug 14  312  333.7  21.7  470.0  0.06 
Sep 14  340  329.3  10.7  113.6  0.03 
Oct 14  331  331.5  0.5  0.2  0.00 
Nov 14  348  331.4  16.6  276.2  0.05 
Dec 14  349  334.7  14.3  204.4  0.04 
Jan15  398  337.6  60.4  3652.6  0.18 
Feb 15  374  349.7  24.3  592.9  0.07 
Mar 15  404  354.5  49.5  2448.2  0.14 
Apr 15  368  364.4  3.6  12.8  0.01 
May 15  365.1 
(7 Marks)
b)
RMSE = = 29.6
95% confidence interval for May 2015 = 365.1 ± 1.96 x 29.6 = 307.1 to 423.1
(4 Marks)
c)
T > 0.15 in Jan 2015. Therefore, the model is unsatisfactory here. T values then fall below 0.15.
(4 Marks)
d)
C5: = B4
C6: = B5*$C$2 + C5*(1$c$2)
D5: = B5 – C5
F5: =ABS(D5/C5)
E18: =SQRT(Sum(E5:E15)/11) or SQRT(AVERAGE(E5:E15))
(7 Marks)
 Restart the series in Feb 2015 by inputting the formula =B12 in cell C13.
(3 Marks)
Total: 25 Marks
Question 5
a)  
Activity  Mean  Standard Deviation  
A  5  0.67  
B  4  0.67  
C  5  1  
D  4  0.67  
E  3  0.67  
F  7  1  
G  5  0.67  
H  6  0.67  
I  8  1  
J  7  0.67  
Mean =  O + 4M + P  Standard Deviation =  P – O  
6  6  
5 marks  

Estimated total project time = 42 weeks
Critical Path = A, C, D, F, H, I, J
 c) ACDFHIJ = 42 weeks
A  B  C  D  E  F  G  H  I  J  
St.Dev  0.67  0.67  1  0.67  0.67  1  0.67  0.67  1  0.67 
Var  0.45  0.45  1  0.45  0.45  1  0.45  0.45  1  0.45 
√ 0.45 + 1 + 0.45 + 1 + 0.45 + 1 + 0.45 = 2.19
95% confidence interval = µ ± Zα
42 ± (1.96 x 2.19)
42 + 4.2924 = 46.2924
42 – 4.2924 = 37.7076
95% confidence that the project will not take less than 37.707 weeks and not more than 46.29 weeks.
d)  A B E G I J 
µ = 32 weeks  
α = 1.80 (calculated using variances)  
43 – 32 /1.80 = 6.11  
Q = 1  
A C D E G I J  
µ = 37 weeks  
α = 2.06 (calculated using variances)  
43 – 37/2.06 = 2.91  
Q = 0.4982 + 0.5 = 0.9982  
A B F H I J  
µ = 37 weeks  
α = 1.95 (calculated using variances)  
43 – 37/1.95 = 3.08  
Q = 0.4990 + 0.5 = 0.9990  
A C D F H I J  
µ = 42 weeks  
α = 2.19 (calculated using variances)  
43 – 42 / 2.19 = 0.46  
Q = 0.1772 + 0.5 = 0.6772  
1 x 0.9982 x 0.9990 x 0.6772 = 0.6753  
1 – 0.6753 = 0.3247 32.47%  
The probability that the project will take more than 43 weeks is 32.47%. 
Total 25 marks