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Final Exam Review Problems
Course: MGT 302, Spring 2012School: Miami University
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302 MGT Section E Final Exam Review Problems Chapter 10: 1, 3, 5 Chapter 11: 1, 2, 3, 12, 16, 17 Chapter 13: 1, 3, 5, 9, 10, 13, 24 Chapter 14: 4, 5, in-class exercise Chapter 10 1.) D = 10 gauges per hour L = 2 hours S = .20 C = 5 gauges K = DL(1+S)/C K = 10(2)(1+0.20) / 5 = 4.8 5 Kanban card sets 3.) D = 2,400 bottles/2 hours = 1200/60 minutes = 20 per minute L = 40 minutes S = .10 C = 120 bottles K =...
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302 MGT Section E
Final Exam
Review Problems
Chapter 10: 1, 3, 5
Chapter 11: 1, 2, 3, 12, 16, 17
Chapter 13: 1, 3, 5, 9, 10, 13, 24
Chapter 14: 4, 5, in-class exercise
Chapter 10
1.)
D = 10 gauges per hour
L = 2 hours
S = .20
C = 5 gauges
K = DL(1+S)/C
K = 10(2)(1+0.20) / 5 = 4.8 5 Kanban card sets
3.)
D = 2,400 bottles/2 hours = 1200/60 minutes = 20 per minute
L = 40 minutes
S = .10
C = 120 bottles
K = DL(1+S)/C
K = 2(40)(1+0.10) / 120 = 7.33 8 Kanban cards
5.)
D = 32 catalytic converters per hour
L = 1 hour
S =.125
C = 8 catalytic converters
K = DL(1+S)/C
K = 32(1)(1+0.125) / 8 = 4.5 5 Kanban cards
Chapter 11
1.
x
1
2
3
4
5
6
7
8
9
10
11
12
Total 78
Y
4200
4300
4000
4400
5000
4700
5300
4900
5400
5700
6300
6000
60200
=
=
b==
a= =
xy
4200
8600
12000
17600
25000
28200
37100
39200
48600
57000
69300
72000
418800
x2
y2
1
17640000
4
18490000
9
16000000
16 19360000
25 25000000
36 22090000
49 28090000
64 24010000
81 29160000
100 32490000
121 39690000
144 36000000
650 308020000
Y
(y-Y)2
3958.97 58093.360
4151.28 22117.028
4343.59 118053.912
4535.90 18468.113
4728.21 73872.452
4920.51 48625.904
5112.82 35036.160
5305.13 164128.863
5497.44 9493.754
5689.74
105.194
5882.05 174681.131
6074.36 5529.257
728205.128
6.5
5016.667
192.3077
3766.667
LINEST (Microsoft Excel function) will produce the same result: b = 192.3007. Now calculate
Y with 3766.67 + 192.3077 (Month) to get:
Month Forecast
13
14
15
16
17
18
19
20
21
22
23
24
b.
6266.67
6458.97
6651.28
6843.59
7035.90
7228.21
7420.51
7612.82
7805.13
7997.43
8189.74
8382.05
= 269.85
Therefore, 3 standard errors of the estimate would be 3(269.85) = 809.55 or 810
2.
a.
FJuly = .60(15) + .30(16) + .10(12) = 15.0
b.
FJuly = (15 + 16 + 12) / 3 = 14.3
c.
FJuly = FJune + (AJune FJune) = 13 + .2(15-13) = 13.4
d.
Total
x
1
2
3
4
5
6
21
=
=
b==
a= =
y
12
11
15
12
16
15
81
xy
12
22
45
48
80
90
297
x2
1
4
9
16
25
36
91
3.5
13.5
0.77
10.8
Y = a + bx = 10.8 + .77x
e.
FJuly, where July is the 7th month.
Y = a + bx = 10.8 + .77(7) = 16.2
3.
a.
Ft+1 = Ft + (At Ft), = .20
Month
Demand Forecast Absolute Deviation
January
100
80
20
February
94
84
10
March
106
86
20
April
80
90
10
May
68
88
20
June
94
84
10
Total
90
b. MAD = 90/6 = 15
12.
Month
April
May
June
July
August
September
Absolute
Forecast Actual Deviation RSFE deviation
250
200
-50
-50
50
325
250
-75
-125
75
400
325
-75
-200
75
350
300
-50
-250
50
375
325
-50
-300
50
450
400
-50
-350
50
Sum of
absolute
deviations
50
125
200
250
300
350
MAD
50.0
62.5
66.7
62.5
60.0
58.3
TS
-1
-2
-3
-4
-5
-6
For September, the MAD is 58.3 and the TS is 6. The model is performing poorly since
the tracking signal is 6 and moving in a downward direction.
16. a. FSeptember = (170 + 180 + 140)/3 = 163.3
b. FSeptember = .50(170) + .30(180) + .20(140) = 167.0
a. FJuly = FJune + (AJune FJune) = 130 + .3(140 - 130) = 133.00
FAugust = FJuly + (AJuly FJuly) = 133.00 + .3(180 133.00) = 147.10
FSeptember = FAugust + (AAugust FAugust) = 147.10 + .3(170 147.10) = 153.97
17. a. FOctober = (75 + 80 + 60 + 75)/4 = 72.5
b. FOctober = FSeptember + (ASeptember FSeptember) = 65 + .2(75 65) = 67.0
c.
= 405/6 = 67.5
= 21/6 = 3.5
b = = 3.86
a = = 67.5 3.86(3.5) = 54.00
Y = a + bx = 54.0 + 3.86x
d. FOctober = 54.00 + 3.86(7) = 81.01
Chapter Cu 13
1. = $10 - $4 = $6
Co = $4 - $1.50 = $2.50
, NORMSINV(.7059)=0.541446
Should purchase 250 + .541446 (34) = 268.4 or 268 boxes of lettuce.
3.
= 22.36 22
5. Service level P = .98, = 150, T = 4 weeks, L = 3 weeks, = 30 per week, and I =500 pounds.
= 79.4
From Standard normal distribution, z = 2.05
150(4+3) + 2.05(79.4) 500 = 712.77 713pounds
9.
a. = 100 units
10.
= 3120 units
= 180 units
From Standard normal distribution, z = 2.05
= 300(4) + (2.05)180 = 1200.0 + 369 = 1569
If safety stock is reduced by 50 percent, then ss = 185 units.
, = 1.03, so the service probability is 84.8%
13.
= 2,000 units
= 80 units
From Standard normal distribution, z = 2.05
= 250(4) + (2.05)80 = 1000 + 164 = 1164
If safety stock is reduced by 100 units, then ss = 64 units.
, = .80
From Standard normal distribution, z = .80, service probability is 79%
24.
= 31.62 32 refrigerators
10 refrigerators
From Standard normal distribution, z = 1.88
= (500/365)(7) + (1.88)10 = 9.59 + 18.8 = 28.39 28 refrigerators
Order 32 refrigerators when the on-hand inventory level reaches 28 refrigerators.
Chapter 14
4.
Level
0
Z
A(2)
B(4)
1
D(4)
Period
Item
Z
LT=
2
Q=
L4L
1
Gross requirements
Scheduled receipts
On hand from prior period
Net requirements
Planned order receipts
Planned order releases
Item
A
LT=
1
Q=
L4L
Gross requirements
Scheduled receipts
On hand from prior period
Net requirements
Planned order receipts
Planned order releases
Item
B
LT=
1
Q=
L4L
Gross requirements
Scheduled receipts
On hand from prior period
Net requirements
Planned order receipts
Planned order releases
Item
C
LT=
1
Q=
L4L
Gross requirements
Scheduled receipts
On hand from prior period
Net requirements
Planned order receipts
Planned order releases
Item
D
Gross requirements
Scheduled receipts
2
2
E(2)
C(3)
3
3
4
5
6
LT=
1
Q=
L4L
On hand from prior period
Net requirements
Planned order receipts
Planned order releases
Item
E
LT=
3
Q=
L4L
Gross requirements
Scheduled receipts
On hand from prior period
Net requirements
Planned order receipts
Planned order releases
800
5.
Level
A
C
B(3)
E(2)
Period
1
1
2
B
D
E(2)
F
Item
A
LT=
2
Q=
L4L
0
D
E(2)
F
2
3
D(2)
4
F
3
4
5
6
7
Gross requirements
Scheduled receipts
On hand from prior period
0
10
1
0
10
Net requirements
Planned order receipts
Planned order releases
Item
B
LT=
1
Q=
L4L
Gross requirements
50
Scheduled receipts
On hand from prior period
Net requirements
Planned order receipts
Planned order releases
Item
C
LT=
1
Q=
50
5
0
0
50
50
60
Gross requirements
Scheduled receipts
On hand from prior period
10
10
1
0
10
Net requirements
Planned order receipts
Planned order releases
Item
D
LT=
2
Q=
50
Gross requirements
5
0
Scheduled receipts
On hand from prior period
0
0
5
0
5
0
Net requirements
Planned order receipts
Planned order releases
Item
E
LT=
1
Q=
50
60
100
100
Gross requirements
Scheduled receipts
60
1
0
0
50
220
200
On hand from prior period
50
100
1
0
0
Net requirements
220
400
Planned order receipts
Planned order releases
Item
F
LT=
1
Q=
L4L
4
0
0
4
0
0
Gross requirements
Scheduled receipts
On hand from prior period
50
150
200
Net requirements
Planned order receipts
Planned order releases
0
200
2
0
0
2
0
0
2
0
0
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EEL 6814NEURAL NETWORKS FOR SIGNAL PROCESSING (3)Department of Electrical and Computer Engineering, University of FloridaGraduatePrereq: Knowledge of adaptive signal processing.Nonlinear signal processing and neural networks. Gradient descentlearnin
University of Florida - EEL - 6814
The MRMI Algorithm The batch mode adaptation algorithm for the rotation matrix, which is parameterized in terms of Givens rotations, can be summarized as follows. 1. Whiten the observations cfw_z1 , . . . , z N using W to produce the samples cfw_x1 , . .