BJMP2033 PRODUCTION OPERATIONS MANAGEMENT
THE WORKSHOP 4.0 SEMESTER SEPTEMBER/A171
MATRIC NO: __________________ 14 DECEMBER 2017/8PM-10PM/DKG2/2
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BJMP2033/STML/A171
QUESTION 1 A. Explain the meaning of production. Production is something do to with a creation or manufacture of products and services. B. What is operations management? And list down THREE (3) (3) name of companies together with their products. Operations management (OM) is the set of activities that creates value in the form of products and services by transforming inputs into outputs. i – NESTLE; NESTLE; Milo, Nescafe, N escafe, Coffee Mate ii – MICROSOFT; MICROSOFT; Office Software Applications iii – TOYOTA; TOYOTA; Cars, Smart Car Technologies C. Based on the problem statement below, answer all the questions: Redha Insurance Sdn. Bhd. wants to evaluate its labor and multifactor productivity with a new computerized title-search system known as LOOK Up. The firm has a staff of four, each working eight hours per day (for a payroll cost of $640/day) and overhead expenses of $400 per day. Redha processes and close on eight titles each day. The LOOK Up system will allow the processing of 14 titles per day. Although the staff, their work hours, and pay are the same, the overhead expenses are now $800 per day. i – Labor Labor productivity with the manual system; 8 titles per day / 32 labor-hours = 0.25 titles per labor-hour ii – Labor Labor productivity with the LOOK Up system; 14 titles per day / 32 labor-hours = 0.44 titles per labor-hour iii – Multifactor Multifactor productivity with the manual system; 8 titles per day / $640 + $400 = 0.0077 titles per dollar iv - Multifactor productivity with the LOOK Up system; 14 titles per day / $640 + $800 = 0.0097 titles per dollar
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QUESTION 2 A. Complete the Gantt chart based on the information given in the table; ID Task Name Predecessors 1 Start 2 A 1 3 B 1 4 C 2 5 D 2 6 E 3,4 7 F 5 8 G 6 9 Finish 7,8
Duration (days) 0 2 3 4 2 3 4 3 0
Task Start A B C D E F G End
8
1
2
3
4
5
6
7
9
10
11
12
13
14
15
16
17
B. Based on the AON project network below, answers the following questions;
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activities and predecessors; i – Complete Complete the Jerai Paper Sdn. Bhd’s activities
Task Name A B C D E F G H I J K L M N O
Predecessors A A A A B C D E E, H B, D, G F, K I, J, K I L, M, N
Duration (days) 0 3 1 2 3 4 2 3 1 5 4 3 2 3 5
ii – Find Find the completion time (CT) for the above AON project network; CT= 17 days
iii – Show Show the critical path (CP) for the above AON project network; CP= A-D-H-J-M-O
iv – Explain Explain the meaning of Critical Path Analysis. Critical path analysis is the path through the network with the longest duration. The critical path may change from time- to- time as activities are completed ahead of or behind schedule. Critical path activities are identified as having zero float in the project.
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v – Based Based on the table below of Hydro Electric Manufacturing, answer all the questions; Task Name
Predecessors
Duration (weeks)
A B C D E F G H
A A, B C C D, E F, G
2 3 2 4 4 3 5 2
Time Estimate (weeks) a m b 1 2 3 1 4 1 1 2 3 1 4 7 1 4 7 1 4 1 3 3 15 1 2 3
i – Illustrate Illustrate the table via AON project network with node configuration.
C 2
2 2
4 4 E 4
A 2
0 0
2 2
S
F 3 D 4
B 3
5
0 1
2 4
4 4
8 8
4 7 12 15
H 2
6 8 G 5
3 4
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8 8
13 13
13 15 13 15
ii – Find Find the completion time and the critical path; CP= A-C-E-G-H
iii – Find Find the probability of the completing project with expected completion time of 18 weeks with the illustration. Task name A B C D E F G H
t 2 3 2 4 4 3 5 2
σ2
0.11 0 0.11 1.00 1.00 0 4 0.11
x= 18 weeks = 0.11+0.11+1.00+4+0.11= 5.33
σ2
=
σ
. = 2.31 √ .
µ= 15 weeks Z= x - µ /
σ
= 18 – 15 15 / 2.31 = 1.29 P (18) = 0.5+ 0.4015 = 90.15% the project can be completed within 18 weeks.
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QUESTION 3 Chapter 2
A. Mustafa Al-Bakree Enterprise produces 355,800 units of cheese chapatti. The effective capacity for the production is half of million units. The production line operates 7 days per week with two shifts per day. The production line was designed to process the chapatti at rate of 18,000 in 30 minutes. Calculate: i - Design capacity. (7 days x 2 shifts x 12 hours) x (18,000 x 2) = 6,048,000 ii - What is the utilization rate for the company? Utilization = Actual Output/Design Capacity = 0.7116 iii – What What is the efficiency rate for the company? Efficiency = Actual Output/Effective Output/Effective Capacity = 0.0588
B. Miss Aleena Bakar is a quality supervisor at Omega Watches Sdn. Bhd. in Petaling Jaya, Selangor. She is responsible to investigate all samples arrived from the branches all around Malaysia. In week 7, she had found an average value of 8.4 fine scratches in every Omega TG-34 model produced by the Jitra factory. Then, by using 99% confidence, find the values of UCL and LCL for the c-control chart.
. ) √ . . )= 16.8 √ .
UCL= 8.4 + (3) ( = 17.09 LCL= 8.4 – (3) (3) (
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Chapter 2
C. Nilai Premium Glass Enterprise has a fixed cost of RM550,000 and produces sapphire glass for Citizen Watch Company in Japan. Usually the company have their inputs from the Ali & Co. for raw materials which approximately RM24.50 per unit and the company’s direct labor cost is RM55.24 per unit. If the selling price for the glass is RM121.88 per unit, calculate the questions; i - The break-even point in units. Fixed cost (cf) = 550,000 Variable cost (cv) = 24.50+55.24=79.74 24.50+55.24=79.74 BEP(units)= 550,000/ 121.88- 79.74= 13051.732 units
ii - The break-even point in (RM) including GST of 8%. Price= 131.63 Cv= 86.12 BEP(RM)= 550,000/ 1-(86.12/131.63) = RM 1590782.25
Chapter 4 D.Based D. Based on the basic information for a pastry shop that produces pizza dough as presented in the information below, answer all the questions; Business Weeks Cost of Goods Sold Total Asset Raw Material
= 48 = $11.3 million = $5.89 million = 0.56 million
MRO = $0.12 million Finished Products = $0.14 million
i - Calculate the inventory turnover. AAVI= RM+ FP+ MRO= 0.56+0.14+0.12= 0.82 million Inventory turnover= COGS/ AAVI = 11.3/0.82 = 13.78 ii - How many days of supply? Days of supply= AAVI/(COGS/WEEKS) = 0.82/ (11.3/48) = 3.48 WEEKS
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QUESTION 4 Chapter 5
A. Batik Irama Enterprise is located at Kota Kinabalu produces a superbrand of chocolate lava cake called the Premerro for Klang Valley. The firm receive an order from their customers twice a year with 3,000 units in first and second quarters, and 8% more in third quarter while 7% less in final quarter. The transport and handling costs are $6500 for a shipment. The storage fee is $2.36 per cake and paid twice a year. The firm intends to keep at 97% for a service satisfaction. The standard deviation is 1.8%. Cakes ordered be delivered within three weeks. The firm operates around a year. Calculate the following questions; (i) The optimal order order size. EOQ=
. = 7052.77 units
(ii) The (ii) The annual inventory cost. TC =
+ . . . = 16644.53
(iii)The number of orders per year. Number of orders per year = 9030/7052.77 = 1.28 per year (iv) The The order cycle time. Order cycle time = 365/ 1.28 = 285.16 times (v) The (v) The reorder point with safety stock stock
√
R= (d= 24.74) (L= 21) + (Z=1.89) (sd= 0.45)
=523.44 9
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Chapter 6
B. Complete the (i) transportation table below; Tempoh Pengeluaran
Tempoh Penggunaan Penggunaan Period of Use
1
Period of Production
Inv. Awalan Beg. Inventory
Lebih Masa
1
Overtime
Sub Kontrak Subcontract
Masa Biasa
2 0
Lebih Masa
60
450 80
Lebih Masa
90
92
94
60
65
Sub Kontrak
96
62
64
400
0 80
85
82
84
250
0
95
90
94
92
200
0 70
65
62
60
500 90
0 80
95
82
200 100
95
75
0 92
90
0
150
70
60
65
400 95
90
250 105
100
95
100
Permintaan
10
0 90
100
Subcontract
Demand
0 80
85
Overtime
4
86
0
Regular
Lebih Masa
150
0
250
Subcontract
Masa Biasa
84
0
Overtime
Sub Kontrak
66
150
Regular
3
64
82
0
Subcontract
Masa Biasa
62
750
700
850
Capacity
6
4
50
Overtime
Sub Kontrak
2
4
0
Regular
2
3
Kapasiti
Unused Capacity
300
Masa Biasa Regular
Kapasiti Tidak Guna
900
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700
300 500 300 250 400 250 200 500 200 150 400 250 200
3900
ii. Optimum production plan.
Production Plan Period
Demand
Regular Production
Overtime
Subcontract
Ending Inventory
1
750
500
300
0
300+(500+300)-750=350 300+(500+300)-75 0=350
2
700
400
250
0
350+(400+250)-700=300 350+(400+250)-70 0=300
3
850
500
200
0
300+(500+200)-850=150 300+(500+200)-85 0=150
4
900
400
250
100
150+(400+250+100)-900=0 150+(400+250 +100)-900=0
TOTAL
3200
1800
1000
100
800
iii. Calculate the plan production cost. 1800 x $60 + 1000 x $80 + 100 x $90 + 800 x $2= $198,600
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Chapter 6
C. Prepare TWO (2) strategies (2) strategies in aggregate planning and compute its total cost: Strategy 1 - Use chase demand with supplement by using inventory, overtime and sub-contracting as needed.
JULY 24,000
AUG 32,000
SEPT 56,000
OCT 72,000
Current work force Average monthly output per worker Maximum overtime Inventory holding cost Regular wage rate Overtime wage rate Subcontracting Subcontracting wage rate Hiring cost Firing cost Beginning inventory Mth
JULY
Demand
24,000
Regular Production
OT
NOV 64,000
DEC 52,000
8 workers 2,000 cases per month 1,000 cases per month $0.30 per case per month $20.00 per hour $30.00 per hour $40.00 per hour $1,200 per worker $3,600 per worker 4,000
S/k
Ending Inventory
4000+ 20,000
Workers
H
10
2
F
AUG
32,000
32,000
16
6
SEPT
56,000
56,000
28
12
OCT
72,000
72,000
36
8
NOV
64,000
64,000
32
4
DEC
52,000
52,000
26
6
Total
300,000
296,000
-
Total Cost: 296,000($20) +28($1200) +10($3600) = $5,989,600 12
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Strategy 2 - Use a level production strategy with supporting of overtime (maximum of 6,000 cases), subcontracting, and inventory as needed. JULY 24,000
AUG 32,000
SEPT 56,000
OCT 72,000
Current work force Average monthly output per worker Inventory holding cost Regular wage rate Overtime wage rate Subcontracting Subcontracting wage rate Hiring cost Firing cost Beginning inventory Mth
Demand
JULY
24,000
Regular Production 4000+ 46,000
OT
NOV 64,000
DEC 52,000
8 workers 2,000 cases per month $0.30 per case per month $20.00 per hour $30.00 per hour $40.00 per hour $1,200 per worker $3,600 per worker 4,000
S/k
Ending Inventory
Workers
H
26,000
23
15
25
2
-
17
AUG
32,000
50,000
44,000
SEPT
56,000
50,000
38,000
OCT
72,000
50,000
16,000
NOV
64,000
50,000
2000
DEC
52,000
50,000
0
Total
300,000
296,000
126,000
Total Cost: 296,000($20) +126,000($0.30) +17($1,200) = $5,978,200 Which strategy is economical? Strategy 2 13
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F
Chapter 8
QUESTION 5 A volatile rate of demand makes the Sweet CC Enterprise make use 15% of safety stock in the factory. A carrier of fresh pandan coconut may contain X units to be processed in several stations. The chopping station can process 220 coconuts within 30-minute. Then the basket of chopped coconuts is transferred to a carving station in 20-minute. After that, the carving coconuts were dispatched to pressing station in 15-minute before it be transported by small lorry to cold room warehouse which takes about 12 minutes. The department issues six Kanban for the process. Solve for the X?
d = 220 x 2 = 440 units L = 20 + 15 + 12 minutes = 47 / 60 = 0.78 S = 0.15 (440 x 0.78) = 51.48 C = X units Kanban = 6
N=d L + S/C 6= (440 x 0.78) + 51.48/X X= (440 x 0.78) + 51.48/6 = 351.78
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QUESTION 6 A. Based on the product structure three, complete the MRP tables and calculate the each cost when ordering cost is $1500 and the carrying cost is $0.22 per unit. i. A – MULTIPLE MULTIPLE 25 Part Name: T
MRP Record Lead Time = 1 week Safety Stock = 30 Lot Size = MULT25
Week 11
Week 12
Week 13
Week 14
Week 15
140
Gross Requirements Scheduled Receipts
20
50
Available Inventory 50
70
120
120 30
30
60
40
50
90
Planned Order Receipts
50
100
50
Week 17
Week 18
50 40
Net Requirements
Planned Order Releases
Week 16
80
30
30
Week 17
Week 18
50
30
30
50
100
(70+120+30+60+40+80+30+30) x $0.22= $3101.20 COST FOR QQ: 2 x $1500 + (70+120+30+60+40+80+30+30)
ii. B – MINIMUM MINIMUM 50 Part Name: E
MRP Record Lead Time = 2 weeks Safety Stock = 30 Lot Size = MIN 50
Week 11
Week 12
Gross Requirements
Week 13
Week 14
Week 15
140
Scheduled Receipts
20
50
Available Inventory 50
70
120
120 30
30
Week 16
60
40 30
70
Net Requirements
50
90
50
30
Planned Order Receipts
50
90
50
50
90
50
Planned Order Releases
50
50
(70+120+30+60+30+70+30+50) x $0.22 = $6101.20 COST FOR QQ: 4 x $1500 + (70+120+30+60+30+70+30+50)
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iii. C – PERIODIC PERIODIC ORDER QUANTITY 2 Part Name: L
MRP Record Lead Time = 2 weeks Safety Stock = 30 Lot Size = POQ 2
Week 11
Week 12
Gross Requirements Scheduled Receipts
20
50
Available Inventory 50
70
120
Net Requirements Planned Order Receipts Planned Order Releases
130
Week 13
Week 14
Week 15
Week 16
Week 17
Week 18
140
80
180
50
150
30
60
30
40 110
30
80
70
50
180
110
130
230
140
230
140
(70+120+110+30+80+70+60+30) x $0.22= $4625.40 COST FOR QQ: 3 x $1500 + (70+120+110+30+80+70+60+30)
iv. D – FIXED FIXED ORDER QUANTITY Part Name: U
MRP Record Lead Time = 2 weeks Safety Stock = 30 Lot Size = FOQ120
Week 11
Week 12
Gross Requirements Scheduled Receipts
20
50
Available Inventory 50
70
120
Net Requirements Planned Order Receipts Planned Order Releases
120
120
Week 13
Week 14
Week 15
Week 16
Week 17
Week 18
140
80
180
50
150
30
130
100
40 100
110
50
50
40
100
140
120
120
120
240
120
40
240
(70+120+100+110+50+40+130+100) x $0.22= $6158.40 COST FOR QQ: 4 x $1500 + (70+120+100+110+50+40+130+100)
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B. Based on the statement below, construct a product structure tree. Product X is built from two components of BET and COM. The BET is built from two components of DELT and two components of COLT. The COM is built from ELT and three components of FEM. The COLT is built from three sub-components sub-components of ERA and five sub-components of FES.
X
BET
2 DELT
3 ERA
COM
2 COLT
ELT
5 FES
THE END
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3 FEM
Appendix 1
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Appendix 2 N = d L + S C
D/Q opt
Cost of goods sold Avg Aggregate Value of Inventory
Y = (I)(%g1)(%g2)(%g3)… (I)(%g1)(%g2)(%g3)…
S = LS-ES R = dL Avg agg value of inventory (COGS)/(365 days)
Good quality x 100 (I)(unit processing cost) + (reworked units)(rework cost)
Y= (I)(%G) + (I)(1-%G) (%R)
Product Cost = (K d)(I) + (K r r)(R) ) (R) Y
Output/Labour
Business Days D/Qopt
No. production production runs = D/Q
Output/Material
CoD Qopt
+
CcQopt + PD 2
v = __cf_____ p - cv
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