Assignment

(20%) A production operation is making 150 units of a product by engaging five workers for 300 hours. However, 40 percent of the units appear to have various quality problems, and the company decides to sell them as seconds at a price of £50 each when a normal unit is sold for £150. To improve the situation, several initiatives are proposed, including a scheme where, for every improvement, 50 percent will be given to workers and the other 50 percent will be held by the company. This results in a significant drop in defects as now only 10 units are faulty out of an output of 130 units.

a) Compare the productivity after Bonus with the initial productivity. (10 %)
b) Determine the appropriate bonus per hour for the workers under the bonus scheme if the cost per piece is £70 both before and after the scheme. (10%)

PART A
Gross production = 150 units of a product by engaging five workers for 300 hours.
Defective units = 150*40% = 60 units // Sell them seconds at a price of £50
Normal units = 150-60 = 90 units // normal unit is sold for £150
Total revenue = 90*150+60*50 = 16,500 // For every improvement 50% will for workers and other 50% for company// After improvement only 10 products are faulted out of an output of 130 units.
Labor hours = 300 hours // labor working total hours of 5 workers
Initial productivity = Total revenue / Labour hours

= 16500/300
= pounds 55 per hour
Productivity after bonus = (120*150+10*50)/300
= pounds 61.67 per hour

Compare productivity after bonus with the initial productivity (10%):
Productivity after bonus=61.67 pounds per hour;
Initial productivity=55 pounds per hour(10%)
Initial productivity(10%)=5.5 pounds per hour
So, after comparing productivity after bonus is higher than initial productivity(10%) because Productivity after bonus is 61.67 pounds per hour and initial productivity(10%) is 5.5 pounds per hour.
Hence we can assume result improvement is necessary for generating higher productivity its also beneficial for workers and customers. It also reduces the rates of defective products. It will improve quality of the work.

PART B
Improvement = 120 – 90
= 30 units // Improvement is necessary for generating higher productivity
Total cost saved due to improvement = 30*70
= pounds 2,100 // cost per piece is £70.Because of Improvement will save a lot of cost. It’s beneficial both workers and company.
Total bonus payout = 2100*50%
= pounds 1,050 // For every improvement 50% will for workers and other 50% for company
Bonus per hour = 1050/300

= pounds 3.5 // Total bonus payout is 1050 and total work hours are 300.
Bonus per hour for the workers under the bonus scheme if the cost per piece is £70 both before and after the scheme. (10%).

If the cost per piece is 70 pounds so add 10% before the scheme and after the scheme of bonus.
Before the bonus per hour scheme
Total bonus payout=1050 pounds
After the bonus per hour scheme 3.5 pounds
So add both before and after 10%
Before bonus per hour scheme = 1050(10%) pounds
= 105 pounds
After the bonus per hour scheme= 3.5(10%) pounds
= 0.35 pounds per hour
So, if product price is 70 pounds worker will receive two times bonus before per hour bonus scheme and after bonus per hour scheme which is before 105 pounds and after 0.35 pounds beneficial in both conditions.

QUESTION NO. 2

PART A
The task time required = total time / demand
Total time = 40 hour or 40*60*60 = 144000 seconds.
Demand = 1000
Task time = 144
Using the shortest time first rule, we must assign the task that has the shortest duration among all available options. Once we run out of a workstation then we put them in the next workstation. This process is repeated as more tasks become available and until we finish assigning all the task to one or the other workstations. Each workstation can handle task time of 144 seconds.

The steps of the assignment is shown below:

Steps
Available task
Assigned task
Task duration
Assigned to WS
WS time remaining
Now available

1
K3, J3
J3
3
1
141
K3,F7

2
K3, F7
F7
21
1
120
K3, C1

3
K3, C1
K3
60
1
60
K4, K9, J1, J2,C1

4
K4, K9,J1, J2,C1
J2
22
1
38
K4, K9,J1,C1, F3, F4

5
K4, K9, J1, C1, F3, F4
K4
24
1
14
K9, J1, C1, F3, F4

6
K9, J1, C1, F3, F4
K9
27
2
117
J1, C1, F3, F4, G4

7
J1, C1, F3, F4, G4
F3
32
2
85
J1, C1, F4, G4, E3

8
J1, C1, F4, G4, E3
J1
66
2
19
C1, F4, G4, E3, G5

9
C1, F4, G4, E3, G5
G5
29
3
115
C1, F4, G4, E3, E2

10
C1, F4, G4, E3, E2
E2
18
2
1
C1, F4, G4, E3

11
C1, F4, G4, E3
C1
78
3
37
F4, G4, E3, B5

12
F4, G4, E3, B5
G4
79
4
65
F4, E3, B5, F9

13
F4, E3, B5, F9
F4
92
5
52
E3, B5, F9, D6

14
E3, B5, F9, D6
D6
53
4
12
E3, B5, F9, D9

15
E3, B5, F9, D9
D9
37
3
0
E3, B5, F9,

16
E3, B5, F9
B5
108
6
36
E3, F9, A1, A2

17
E3, F9, A1, A2
A1
52
5
0
E3, F9, A2

18
E3, F9, A2
A2
72
7
72
E3, F9

19
E3, F9
E3
109
8
35
F9, D8

20
F9, D8
D8
78
9
66
F9

21
F9
F9
126
10
18
D7

22
D7
D7
72
7
0
B3

23
B3
B3
72
11
72
B7

24
B7
B7
18
10
0
A3

25
A3
A3
114
12
30
COMPLETE

The final workstation assignments, the idle time is shown below
WORKSTATION
TASK
IDLE TIME

1
J3, F7, K3, J3, K4
14

2
K9, F3, J1, E2
1

3
G5, C1, D9
0

4
G4, D6
12

5
F4, A1
0

6
B5
36

7
A2, A7
0

8
E3
35

9
D8
66

10
F9, B7
0

11
B3
72

12
A3
30

PART B
The total task time is 1462 seconds. This means that if we use a task time of 144 seconds then it is not possible to have a 100% balance. The best we can do is to use 11 workstations. This will leave an idle time of 144*11 – 1462 = 122 seconds. The efficiency can be at best 1462/(11*144) = 0.9229 or 92.29%

The balance can be improved by ensuring that each workstation production quota is met within the task time and doesn’t exceed it, Real time target should be set for each work station so that the exact work output at any given time can be known,There must be a steady and continuous flow of production, workstations that have the longest task time should be allocated more tasks, There should be a drastic reducing in the idle time. There should be no idle time in order to improve the balance.

QUESTION NUMBER 3

PART A
Annual demand, D = 250*24 = 6,000
Ordering cost, K = £30 per order
Unit carrying cost, h = £10 per annum
Unit cost, C = £100
Optimal order size = EOQ = (2.D.K/h)1/2 = SQRT(2*6000*30/10)
= 190 units (rounded off)
Total annual cost of ordering = (D/Q)*K = (6000/190)*30 = £947.37
Total annual carrying cost = (Q/2)*h = (190/2)*10 = £950
Total annual cost of purchase = D.C = 6000*100 = £600,000
So, total cost = 600,000 + 950 + 947.37 = £601,897.4

PART B
It has been found by some researchers (e.g. Lowe and Schwarz) that even when the EOQ model is too rudimentary and considers so many unrealistic assumptions, the model is highly insensitive to variability and errors in forecasting of demand. So, it can be useful even when the information in hand are not of a very high accuracy. EOQ model is that it gives explicit numbers specific to the business with respect to how much stock to hold, when to re-request it and the number of things to arrange. This smooths out the re-loading cycle and results in better client support as stock is accessible when required. The EOQ model accepts consistent interest of a business item and prompt accessibility of things to be re-supplied. It doesn’t represent occasional or monetary vacillations. It expects fixed expenses of stock units, requesting charges and holding charges. This stock model requires consistent checking of stock levels.

QUESTION NUMBER 4

PART A
Annual demand, D = 200*12 = 2400
Unit carrying cost, h = 3.333 (per annum? why not given?) Ordering cost, K = 10
,First, compute the EOQ as follows:
EOQ = (2.D.K/h)1/2 = SQRT(2*2400*10/3.333) = 120 pounds
Now check the total relevant cost for each vendor for each feasible price brackets stating above the EOQ as follows:
Vendor 1 Annual Cost

Ordering
Carrying
Purchase
Total

Q
P
(D/Q)*K
(Q/2)*h
D*P

120
33.55
$200
$200
$80,520
$80,920

150
32.35
$160
$250
$77,640
$78,050

300
31.15
$80
$500
$74,760
$75,340

500
30.75
$48
$833
$73,800
$74,681

Vendor 2 Annual Cost

Ordering
Carrying
Purchase
Total

Q
P
(D/Q)*K
(Q/2)*h
D*P

120
34
$200
$200
$81,600
$82,000

150
32.8
$160
$250
$78,720
$79,130

300
31.6
$80
$500
$75,840
$76,420

500
30.5
$48
$833
$73.200
$74,081(min)

Vendor 3 Annual Cost

Ordering
Carrying
Purchase
Total

Q
P
(D/Q)*K
(Q/2)*h
D*P

120
33.75
$200
$200
$81,000
$81,400

200
32.5
$120
$333
$78,000
$78,453

400
31.1
$60
$667
$74,640
$75,367

Vendor 4 Annual Cost

Ordering
Carrying
Purchase
Total

Q
P
(D/Q)*K
(Q/2)*h
D*P

120
34.25
$200
$200
$82,000
$82,600

200
33
$120
$333
$79,200
$79,653

400
31
$60
$667
$74,400
$75,127

So,
Vendor-2 should be opted for an order qty, Q = 500
PART B
Economic order quantity is the ideal order quantity a should purchase to minimize the the inventory costs such as holding costs, shortage costs and order costs. Quality of the product is good which is why EOQ formula will be applied because it assumes the demand of the product. Total cost is a key factor in the management of the inventories in an organization.Few other factors that should be considered beside total cost are:
Safety with which the good is being produced and being shipped : The goods should be manufactured and produced following the right safety measures in an hazard free and Environment. When the goods are being shipped, it Should be handled with care, the quality of the products should also be preserved.
The shipping time or the distance of shipping : This is the distance of the vendors from the plant. The shipping time and distance should be taking into consideration. The product must be able to last and maintain it’s quality under the time frame at which it is being shipped.
The quality of the product : The product must be of very good quality and standard, it must not be of a lesser standard or quality.

QUESTION NUMBER 5

PART A

Table:

Samples
Observation
Mean
Range

1
2
3
4
5

1
10
9
10
9
12
10
3

2
9
9
11
11
10
10
2

3
13
9
10
10
9
10.2
4

4
10
10
11
10
10
10.2
1

5
12
10
9
11
10
10.4
3

6
10
10
8
12
9
9.8
4

7
10
11
10
8
9
9.6
3

8
13
10
8
10
8
9.8
5

9
8
8
12
12
9
9.8
4

10
10
12
9
8
12
10.2
4

10
3.3

Mean of sample means, x double-bar = 10.
Mean of ranges, R-bar = 3.3
Z=3, = 1.36

X-bar chart:

Upper Control Limit, UCLR = x double-bar + z. x-bar = 10 + (3 x 0.61) = 11.83
Lower Control Limit, LCL = x double-bat – z. x-bar 10- (3 x 0.61/1 = 8.17
Central Line, CL x = x double-bar = 10

Number of observations in subgroup, n = 5; Factor for- X- bar chart. Ay = 0.38;

Factor for R -chart (Lower control limits, D = 0; Factor for R-chart (upper control limit) D₄ = 2.11

R-chart:

Upper Control Limit, UCL R = (3D/ A₂) . (/n)=(3 x 2.11) /( 0.581 x 0.61) = 6.66

Lower Control Limit, LCL R = (3D₃/A₂).(/n) = (3 x 0) / 0.58) x 0.61 = 0
Central Line, CL R = R-bar = 3.3
PART B
This means ranges of the= observations of all the 10 samples lie with in the Upper Control Limit and Lower Control Limit of X-bar chart and R-chart respectively.
If the plotted point falls within the control limits, then the process is assumed to be in control, otherwise it is out of control. None of these processes here are above our upper control limit and or below are lower control limit.So there is no cause of variation to look for in this situation, as there is no evidence of any processes going out of control at this particular point in time
Hence, the process is in control since the plotted points fall within the control limit.

The X-bar chart and R-chart are given below:

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