economic demand and supply test

Question 1

  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)

 

 

 

 

 

 

 

 

  1. 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)

 

  1. 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)

 

 

 

  1. 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 over-production (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 (1-d/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

 

  1. 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)

 

  1. 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)

 

 

Cp = 1.23   Cp >1, therefore the process is potentially capable.

Cpk = 1.11   Cpk >1, therefore the process is capable in practice.

Cp  =     and does not take centring of the process in to account.

Cpk = min {,} and takes centring of the process in to account.

(5 Marks)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. 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 11-20

 

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)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

  1. 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 Error2 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)

 

  1. 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

 

 

 

 

 

 

 

 

 

 

Network Diagram

 

 

 

 

 

 

 

 

 

 

 

Estimated total project time = 42 weeks

Critical Path = A, C, D, F, H, I, J

 

 

 

  1. 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