《运营管理》课后习题答案(2020年7月整理).pdf

5. Deposit/Withdrawal:
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Solutions_Problems_OM_11e_Stevenson
1) Deposit—place in an envelope (which you’ll find near or in the ATM) and insert it into the deposit slot
1.00 1.16 1.160 1.15 1.334
Total
4.332
8.
A = 24 + 10 + 14 = 48 minutes per 4 hours
A = 48 = .20 240
NT = 6(.95) = 5.70 min .
ST = 5.70x 1 = 7.125 min . 1 − .20
4. Output = OT = 420 min./day = 323.1 (rounds to 323)copiers / day CT 1.3 min . / cycle
b. 1. Total time = 4.6, CT = Total time = 4.6 = 2.3 minutes
N
2
2. Assign a, b, c, d, and e to station 1: 2.3 minutes [no idle time]
4. c
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a
b
d
h
a. l.
Solutions_Problems_OM_11e_Stevenson
2. Minimum Ct = 1.3 minutes
Task Following tasks
a
4
b
3
c
3
d
2
e
3
f
2
g
1
h
0
Work Station I
Eligible a
b,c,e, (tie)
c. Labor productivity increased by 31.25% ((21-16)/16).
Multifactor productivity increased by 4.5% ((.93-.89)/.89).
*Machine Productivity Before: 80 ÷ 40 = 2 carts/$1. After: 84 ÷ 50 = 1.68 carts/$1. Productivity increased by -16% ((1.68-2)/2)
Solutions_Problems_OM_11e_Stevenson
Chapter 02 - Competitiveness, Strategy, and Productivity
3.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
Week
Output
Worker Cost@ $12x40
Overhead Material Cost @1.5 Cost@$6
After: 84 4 = 21 carts per worker per hour.
b. Before: ($10 x 5 = $50) + $40 = $90; hence 80 ÷$90 = .89 carts/$1.
After: ($10 x 4 = $40) + $50 = $90; hence 84 ÷$90 = .93 carts/$1.
Effective capacity
Actual output = .8 (Effective capacity) Effective capacity = .5 (Design capacity) Actual output = (.5)(.8)(Effective capacity) Actual output = (.4)(Design capacity) Actual output = 8 jobs Utilization = .4
Assign f, g, and h to station 2: 2.3 minutes
3. Output = OT = 420 = 182.6 copiers / day CT 2.3
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Solutions_Problems_OM_11e_Stevenson
4.
Maximum Ct is 4.6. Output = 420 min./day = 91.30 copiers / day 4.6 min . / cycle
Total processing time by machine
Product 1
A 48,000
B 64,000
C 32,000
2
48,000 48,000 36,000
3
30,000 36,000 24,000
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Sol百度文库tions_Problems_OM_11e_Stevenson
4 Total
Work Station
Task
Task Time Time Remaining
Feasible tasks Remaining
I
F
5
9
A,D,G
A
3
6
B,G
G
6
–
–
II
D
7
7
B, E
B
2
5
C
C
4
1
–
III
E
4
10
H
H
9
1
–
IV
I
5
9
–
b. First rule: Largest positional weight. Assembly Line Balancing Table (CT = 14)
Taste
Ingredients √
Appearance
√
Texture/consistency
√
Handling √
Preparation
√ √ √
Chapter 04 - Strategic Capacity Planning for Products and Services
2.
Efficiency = Actualoutput = 80%
*refer to solved problem #2 Multifactor productivity dropped steadily from a high of 3.03 to about 2.84.
4. a. Before: 80 5 = 16 carts per worker per hour.
9.
a. Element PR OT NT
A
ST
1
1.10 1.19 1.309 1.15 1.505
2
1.15 .83 .955 1.15 1.098
3
1.05 .56 .588 1.15 .676
b.
x = .83 s = .034 z = 2.00 A = .01
n
=
zs ax
2
=
CT = Od perating time = 56 minutes per houer = 14 minutes per unit
h
i
Desired output 4 units per hour
5
6
Tasfk # of Following tasks gPositional Weight
A
4
23
Chapter 03 - Product and Service Design 6. Steps for Making Cash Withdrawal from an ATM
1. Insert Card: Magnetic Strip Should be Facing Down
2. Watch Screen for Instructions
Work Station
Task
Task Time Time Remaining
I
F
5
9
D
7
2
II
G
6
8
A
3
5
B
2
3
III
C
4
10
E
4
6
IV
H
9
5
I
5
–
Feasible tasks Remaining A,D,G –
A, E B,E –
E –
I
c. Efficiency = Total time = 45 = 80.36% CT x no. of stations 56
Total Cost
MFP (2) (6)
1
30,000
2,880 4,320
2,700
9,900
3.03
2
33,600
3,360 5,040
2,820
11,220
2.99
3
32,200
3,360 5,040
2,760
11,160
2.89
4
35,400
3,840 5,760
2,880
12,480
2.84
2) Withdrawal—lift the “Withdrawal Door,” being careful to remove all cash 6. Remove card and receipt (which serves as the transaction record)
8.
Technical Requirements Customer Requirements
60,000 186,000
60,000 208,000
30,000 122,000
NA
=
186,000 150,000
= 1.24
2
machine
208,000 NB = 150,000 = 1.38 2 machine
NC
=
122,000 150,000
= .81 1 machine
You would have to buy two “A” machines at a total cost of $80,000, or two “B” machines at a total cost of $60,000, or one “C” machine at $80,000. b. Total cost for each type of machine:
3. Select Transaction Options:
1) Deposit
2) Withdrawal
3) Transfer
4) Other
4. Enter Information:
1) PIN Number
2) Select a Transaction and Account
3) Enter Amount of Transaction
7.
1
5
4
3
8
7
6
2
Chapter 06 - Work Design and Measurement
3.
Element PR OT
NT
AFjob
ST
1
.90 .46 .414 1.15 .476
2
.85 1.505 1.280 1.15 1.472
3
1.10 .83 .913 1.15 1.050
4
Utilization = Actual output Design capacity
Design Capacity = Actualoutput = 8 = 20 jobs Effective capacity .4
10. a. Given: 10 hrs. or 600 min. of operating time per day. 250 days x 600 min. = 150,000 min. per year operating time.
A (2): 186,000 min 60 = 3,100 hrs. x $10 = $31,000 + $80,000 = $111,000
B (2) : 208,000 60 = 3,466.67 hrs. x $11 = $38,133 + $60,000 = $98,133
C(1): 122,000 60 = 2,033.33 hrs. x $12 = $24,400 + $80,000 = $104,400
II
d
III
f,g
IV
h
Assign A B C E D F G H
Time Remaining 1.1 0.7 0.4 0.3 0.0 0.5 0.2 0.1
Idle Time
0.3 0.0 0.2 0.1 0.6
3. Idle percent = (idle time) = .6 = 11.54 percent N x CT 4(1.3)
B
3
20
C
2
18
D
3
25
E
2
18
F
4
29
G
3
24
H
1
14
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Solutions_Problems_OM_11e_Stevenson
I
0
5
a. First rule: most followers. Second rule: largest positional weight. Assembly Line Balancing Table (CT = 14)
Buy 2 Bs—these have the lowest total cost. Chapter 05 - Process Selection and Facility Layout
3.
3
2
4
a
b
c
Desired output = 4
Operati7ng time = 56 minutes
4
9
5