# 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