Operation and Production Mgt Assignment.pdf

2k13/BBA Seetal Daas (seetal.daas@gmail.com) BBA-2k13
1 Operation and Production Management
FINDING PRODUCTIVITY
Problem#1:
a- 3 employees produce 6 hundred insurance policies in a week. They work 8 hours per
day, 5 days per week.
b- A team of workers makes 4 hundred units of product which is valued by it standard cost
of $10 each, the accounting department reports that for this job the actual cost are $400
per labor, $1000 for material cost and $300 for overheads.
Solution
a- Data: Employees=3
Working hours=8
Days=5
No.of Policies=600
P=
𝑂𝑢𝑡𝑝𝑢𝑡
𝐼𝑛𝑝𝑢𝑡
=
600
3×8×5
=
600
120
= 5 𝑃𝑜𝑙𝑖𝑐𝑖𝑒𝑠 𝑝𝑒𝑟 ℎ𝑜𝑢𝑟.
b- Data:No. of Units=400
Standard Cost=$10
Labor cost=$400
Material Cost=$1000
Overhead=$300
P=
=
𝑈𝑛𝑖𝑡𝑠 ×𝑐𝑜𝑠𝑡 𝑜𝑓 𝑢𝑛𝑖𝑡𝑠
𝐿.𝐶+𝑀.𝐶+𝑂𝐻𝐶
=
400×10
400+1000+300
=
4000
1700
= 2.35
Problem#2:
BBA-III students are going to commence a new project in Badin City. The appointed members
are working in 2 consecutive shifts Mr. Adeel and Seetal work in morning shift 3 hours daily
and 4 day per week and producing 700 Research Proposal, while Mr.Ilamdin and Vikio work in
evening shift to produce 800 insurance policies. They work 7 hours per day and 4 days per
week. The estimated worth of their overhead proposals is 700 by the people of Badin. Calculate
the Productivity.
Data
Shift-1: hours=3.days=4=12
Output=700
Shift-2: hours=7.days=4=28
Output=800
P=
=
700+800
12+28
=
1500
40
= 37.5

Problem#3:
Students tuition at a university is $100 per semester credit hour. The state supplement school
revenue by matching student tuition dollar for dollar. Average cost size for a typical 3 credit
course is 50 students. Labor cost $4000 per class, material cost is $20 per student per class,
overhead cost is $25000 per class.
A-Calculate productivity b-if instructor works an average of 14 hours per week for 16 weeks
for each 3 credit class of 50 students, what is labor productivity ratio?
Solution:
Data
Student fees=$100
State supplement=$100
No.of students=50
Credit hours=3
Labor cost=$4000
Material cost=20/student
Overhead cost=$2500
A--P=
𝑂𝑢𝑝𝑢𝑡
=
50×3×100+100
4000+20(50)+25000
=
150×200
5000+25000
=
30000
30000
= 1
B--Productivity Ratio=
=
30000
14×16
=
30000
224
=133.92 per hours.
Problem# 4:
The output of a process is valued at 100 units. The cost of labor is $50 per hour including
benefits. The accounting department provided the following information about the process for
past four week.
Week 1 Week 2 Week 3 Week 4
Units produced 1124 1310 1092 981
Labor Cost $ 12735 14842 10602 9526
Material Cost $ 21041 25523 20442 18364
Overhead Cost $ 8992 10480 8732 7848
Solution:
Productivity Week 1=
𝑜𝑢𝑡𝑝𝑢𝑡
𝑖𝑛𝑝𝑢𝑡
=
𝑈𝑛𝑖𝑡𝑠 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑×100
𝐿.𝐶+𝑀.𝐶+𝑂.𝐻
=
1124×100
12735+21041+8992
=
112400
42768
= 2.628
=
=
1310×100
14842+25523+10480
=
131000
50845
= 2.576

=
=
1092×100
10602+20442+8732
=
98100
39776
= 2.745
=
=
981×100
9526+18364+7848
=
98100
35738
= 2.744
Problem#5:
A- Data
Season Ticket Price: 192$
Overhead Cost: 25000$
Students:75
Labor Cost:6500$
Credit Course/hours: 3
Students fees: 200$
State support: 100$
Material Cost: 25$/student
Solution;
Productivity:
=
75×3×100+200+192
6500+25(75)+25000
=
110700
33375
= 3.316
B- Labor Productivity (20 hours/week for 16 weeks for each three-credit class of 75)
Labor Productivity=
=
110700
20×16
=
110700
320
= 345.94/hours
Problem 6:
Data
Hours: 40/week
Standard Cost:120 each
Units Produced: 2500
Employees:70
Hours:72/week
Units Produced: 4000
Standard Cost:144 each
Solution;
a- Productivity:
2500×120
40×70
=
3,00,000
2800
= 107.14 Ans.
b- Labor Productivity Ratio:
4000×144
72×70
=
576000
5040
= 114.28/ℎ𝑜𝑢𝑟𝑠 Ans.

BREAKEVEN
Problem#1:
A hospital considering a new product to be offered at $200 per patient; the fixed cost per year
would be $1 lac and total variable cost is $100 per patient. What is the Breakeven quantity, what
will be total contribution to the profits?
Data
Fixed cost=100,000
Price=$200
Variable cost=$100
No.of Patient=1500
A-Breakeven quantity?
B-Total Contribution to the Profit(TCP)?
Solution:
A--B.E=
𝐹𝑖𝑥𝑒𝑑 𝑐𝑜𝑠𝑡
𝑃𝑟𝑖𝑐𝑒−𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝐶𝑜𝑠𝑡
=
100,000
200−100
=
100,000
100
= 1000 Ans.
B--TCP=Revenue−Expenses=𝑃𝑄 − (𝐹 + 𝑉𝑄)=200×1500−(100000 + 100 × 1500)
TCP=300,000−100,000 −150,000=300,000−250,000=50,000 Ans.
Problem#2:
Data Solution:
A-Variable Cost=6 A--B.E=
𝐹𝑖𝑥𝑒𝑑 𝑐𝑜𝑠𝑡
𝑝𝑟𝑖𝑐𝑒−𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒 𝑐𝑜𝑠𝑡
=
60000
18−6
=
60000
12
= 5000 Ans.
Fixed Cost=60000
Price=18
B.E=?
B-Sales=10000 B—TCP=PQ−(F+VQ)=14(10000)−{60000+6× 10000}
Price=14 TCP=140,000−120,000=20,000 Ans.
TCP=?
C-Sales=15000 C1—TCP=PQ−(F+VQ)=14(15000)−{60000 + 6 × 15000}
1-Price=14 TCP=210,000−(60000+90000)=210,000−150,000=60,000 Ans.
2-Price=12.50 C2—TCP=PQ−(F+VQ)=12.50(15000)−{60000+6×15000}
TCP=? TCP=187500−(60000+90000)=187500−150,000=37500 Ans.

Problem#3:
Data
Q=17500
Selling Cost=22
Variable Cost=18
Fixed Cost=80,000
A—B.E?
B—Which alternative yield the more profit.1—V=85% reduce or 2—S=30% increase.
Solution:
A—B.E=
𝐹
𝑃−𝑉𝐶
=
80,000
22−18
=
80,000
4
=20,000 Ans.
B—By calculating the profits alternative 1.
TPM=PQ−(F+VQ)=22(17500)−(80,000+2.7×17500)=38,5000−(80,000+47250)
TPM=38,5000−80,000−47,250=38,5000−127,250=2,57,750 Ans.
By calculating the profit of alternative 2.
TPM= PQ−(F+VQ)=22(17,500) –(80,000+18×22,750)
TPM=500500−80,000−40,9500=500500−489500=11,000
Answer:A1˃A2
Management will decide to reduce cost by 85% because it yield more.
Problem#4:
Data
Fixed Cost=$10600
Variable Cost=$1.60 per year
Quantity Sold=800
Price?
Solution:
Driving formula for getting Price formula.
Q=
𝐹
𝑃−𝑉
=˃Q(P-V)=F =˃ QP-QV=F =˃ P=
𝑄𝑉−𝐹
𝑄
=˃ We get this equation P=
𝑉−𝐹
𝑄
P=
1.60−10600
800
=13.25 Answer.

Problem#5:
Process 1-Data
Fixed Cost=$30,000
Variable Cost=$600
Process 2-Data
Fixed Cost=$120,000
Variable Cost=$900
Difference in Total Cost?
Solution:
Process 1-
TC=F+VQ=30,000+600(800)
TC1=30,000+480,000=780,000
Process 2-
TC=F+VQ=120,000+900(800) =120,000+720,000
TC2=840,000
Difference in Total Cost
TC=TC2−TC1=840000−780000=60000 Ans.
Problem#6: Solution
Data—Case 1: TPM=PQ−(F+VQ)
Variable Cost=$5 TPM=10(30000)−(140000+5×30000)
Fixed Cost=$14,0000 TPM=300,000−140000−150000=300,000−290,000
Price=$10 TPM=10,000 Ans.
Q=30000
TPM=?
Data—Case 2: TPM=PQ−(F+VQ)
Variable Cost=5+1=$6 TPM=11(50,000) – (60,000+6×50,000)
Fixed Cost=$60,000 TPM=550,000−60,000−300,000
Price=$10+1=$11 TPM=550,000−360,000
Q=50,000 TPM=190,000 Ans.
TPM=?

PREFERENCE MATRIX
Problem#1:
The table below shows the performance criteria, weight and score (1 from worst and 10 from
best) for a new product; thermal storage conditioner. If the management want to introduce just
one new product at highest total score of any other product ideas is 800, should the firms pursue
making air conditioner?
Performance Criteria A-Weight B-Score
Material Potential 30 8
Unit Profit Margin 20 10
Operation Compatibility 20 6
Competitive advantages 15 10
Investment Requirement 10 2
Profit Risk 5 4
Solution:
Weight Average
240+200+120+150+20+20=750
The company would go and launch because it has more weight score than AC. Therefore, B˃A.
Possible Future Demand
Rows are called=Alternatives and Columns are called=Events
Alternatives Low High
Small Facility 200 270
Large Facility 160 800
Do nothing 0 0
A-Maximin—An alternative worst payoff/lower number in row of payoff matrix, he is pessimist
so takes negative views.
Alternative Payoff
Small Facility 200
Large Facility 160
B-Maximax—He sees optimist so sees things in positive view.
Alternative Payoff
Small Facility 270
Large Facility 800
C-Laplace—it is calculated through 1/n.(n=no of rows)
Small Facility (0.5)200=100 + (0.5)270=135 = 235
Large Facility (0.5)160=80 + (0.5)800=400 = 480
D-Minimax Regret— it is calculated by given payoff minus best payoff.
Alternative Low High Max
Small Facility 200−200=0 800−270=530 530
Large Facility 200−160=40 800−800=0 40

HOMEWORK
PRODUCTIVITY RATIO
Problem #1:
Coach Bajourn Toulouse led the Big Red Herrings to several disappointing football seasons.
Only better recruiting will return the Big Red Herring to winning from. Because of the current
state of the programmers Boehning University fan are unlikely to support increases in the 192
season ticket price. Improved recruitment will increase overhead costs to 30,000 per class
section from the current 25,000 per class section. The university’s budget plan is to cover
recruitment costs by increasing the average class size to 75 students. Labor costs will increase to
6,500 per three credit course, tuition will be 200 per semester credit, which is matched by state
support of 100 per semester credit.
a-find out the Productivity Ratio?
Solution:
Data
Ticket Price=120
Increase O.H=30,000
Current O.H=25,000
Students=75
Labor Cost=6500 per three credit course
Material Cost=25 per student
Tuition=200 per semester credit
State Support=100 per semester credit
Credit hours=3
P==
=
75×3(200+100)
6500+(25×75)+30,000
=
225×300
6500+1875+30,000
=
67500
38375
= 1.75
b-If Instructor work an average of 20 hours per week for 16 weeks for each three-credit class of
75 students, what is the Labor Productivity?
Solution
L.P==
=
67500
20×16
= 210.93

Problem #2:
Compact Disc players are produced on an automated assembly line process. The standard cost of
Compact disc player is 150 per unit, (labor cost=30, material cost=70 and overhead cost=50).
The sales price is 300 per unit.
a-To achieve a 10% multifactor productivity improvement by reducing materials cost only, by
what percentage must those costs be reduced?
Solution
Data: Standard cost=150
Labor Cost=30
Material Cost=70
Overhead Cost=50
Sales Price=300
P==
=
300
30+70+50
= 2
➢ For 10% productivity improvement
➢ Multiplying=2×0.2=2.2, then dividing 300/2.2=136 and 300/2=150
➢ The cost of input must be deceased by: 150 ̶ 136=14
➢ 14 reductions in material cost= 14/70=0.2=20%
➢ To improve 10% productivity 20% material cost will be reduced.
b-To achieve a 10% multifactor productivity improvement by reducing labor costs only, by what
percentage must those costs be reduced?
Solution
M.P=
=
300
30+70+50
= 2
➢ 14 reductions in labor cost= 14/30=0.46=46%
➢ To improve 10% productivity, 46% labor cost will be reduced.
c-To achieve a 10% multifactor productivity improvement by reducing overhead cost only, by
what percentage must those costs be reduced?
Solution
M.P=
=
300
30+70+50
= 2

➢ 14 reductions in overhead cost= 14/50=0.28=28%
➢ To improve productivity 10%, 28% overhead cost will be reduced.
Problem #3:
The output of a process is valued at 100 per unit. The cost of labor is 50 per hour including
benefits. The accounting department provided the following information about the process for
the past four weeks.
W1 W2 W3 W4
Unit Produced 1,124 1310 1,042 981
Labor Cost 12,735 14,842 10,603 9,526
Material Cost 21,041 24,523 20,442 18,364
Overhead 8,992 10,480 8,736 7,848
a-Use the multifactor productivity ratio use to see whether recent process improvements bad any
effect and, if so, when the effect was noticeable.
Solution
Data: Process of valued: 100 per unit
Labor Cost=850 per hour
P=
=W1=
1124×100
47768
=
112400
47768
= 2.35
P=
=W2=
1310×100
50845
=
131000
50845
= 2.57
P=
=W3=
1092×100
10603+20442+8736
=
109200
39781
= 2.75
P=
=W4=
981×100
19210
=
98100
19210
= 5.11
b-Has labor productivity changed? Use the labor productivity ratio to support your answer.
Solution (data same as above)
P=
=W1=
112400
850
= 132.235
P=
=W2=
13,1000
850
= 154.117
P=
=W3=
109200
850
= 128.470
P=
=W4=
98100
850
= 115.41
Problem #4:
The Big Black Bird Company (BBBC), has a large order for special plastic-lined military
uniform to be used in an urgent military operation. Working the normal two shifts of 40 hours,

the BBBC production process usually produces 2500 uniforms per week at a standard cost of
120 each. 70 employees work the first shift and 30 in second. The contract price is 200 per
uniform because of the urgent need, BBBC is authorized to use around-the-clock production, 6
days per week. When each of the two shifts works 72 hours per week, production increases to
4000 uniforms per week but at a cost of 144 each.
Did the productivity ratio increase, decrease or remain the same? If it changed, by what
percentage did it change?
Solution
Data: Working Hours=40
Produce=2500 per week
S.C=120 each
Employees=70+30=100
Contract price= 200 per each
Day per week=6
Produce=4000 per week
S.C=144
First Process
P=
=
2500×120
40(100)
=
300,000
4000
= 75
Second Process
P=
=
4000×144
7200
=
676000
7200
= 80
75(100)/80=6.25
Productivity Ratio Increases=6.25%
b-Did the labor productivity ratio increase, decrease or remain the same? If it changed, by what
percentage did it change?
Solution
First Process
Labor Productivity=
2500
4000
= 0.625=6.25
Second Process
Labor Productivity=
4000
7200
= 0.555=5.55
5.55(100)/6.25=11.2
Labor Productivity decrease=11.2%

C-Did weekly profit increase, decrease or remain the same?
Solution
First Process
Profit=
200×2500
4000
= 125
Problem #05:
Natallia Attire makes fashionable garments. During a particular week employees worked 360
hours to produce a batch of 132 garments of which 52 were seconds (meaning that they were
flawed). Seconds are sold for 90 each at Attire’s factory outlet store. The remaining 80 garments
are sold to retail distribution, at 200 each. What is the labor productivity ratio of this
manufacturing process?
Solution
Data: Produce=132
Working hours=360
First sell=52 per 90 each
Second sell=80 per 200 each
Labor Productivity=
(52×90)+(80×200)
360
=
20680
360
= 57.44
BREAKEVEN ANALYSIS
Problem #1:
Mary William, owner of Willians products, is evaluating. Whether to introduce a new product
line. After thinking through the production process and the cost of raw materials and new
equipment, Willians estimates the variable costs of each unit produced and sold at 6 and the
fixed cost per year at 60,000.
a-if the selling price is set at 18 each, how many units must be produced and sold for Willians to
breakeven? Use both graphic and algebraic approaches to get your answers.
Solution
Data: Variable Cost=6per unit
Fixed Cost=60,000 per year
Selling Price=18 each
Breakeven=
𝐹
𝑃 ̶ 𝑉
=
60,000
18−6
= 5000
b-Willians forecasts sales of 10,000 units for the first year if the selling price is set at 14 each.
What would be the total contribution to profits from this new product during the fiscal year?

Data
Variable Cost=6
Fixed Cost=60,000
Selling Price=14 each
Sales=10,000 units
TCP=PQ ̶ (F+VQ)=14×10,000 ̶ {60,000+6(10,000)}
TCP=140,000 ̶ 60,000 ̶ 60,000=20,000
c-if the selling price is at 1250, Willians forecasts that first year sales would, increase to 15,000
units which pricing strategy 14 or 12.50 would result in the greater total contribution to profit?
Solution
Data
Variable Cost=6
Fixed Cost=60,000
Selling Price=14 or 12.50
Units=15,000
a-TCP=PQ ̶ (F+VQ)
TCP=14(15000) ̶ {60,000+6(15,000)} =210,000 ̶ 60,000 ̶ 90,000=210,000 ̶ 150,000=60,000
b-TCP=PQ ̶ (F+VQ)
TCP=12.50(15000) ̶ { 60,000+6(15,000)}=187,500 ̶ 150,000=37,500
Problem #02:
Solution (a)
Data:Q=17,500
Selling Cost=22 per unit
Variable Cost=18 per unit
Fixed Cost=80,000
B.E=
𝐹
𝑃 ̶ 𝑉
=
80,000
22 ̶ 18
= 20,000
Solution (b)—85% reducing
A1=TCP=22(17500) ̶ {80,000+2.7(17500)}=38,5000 ̶ 80,000 ̶ 47,250=25,7750
A2=TCP=22(17500) ̶ {80,000+18(22750)}=500,500 ̶ 49500=11,000
A1˃A2 : Management should decide to reduce cost by 85% because it yield more profit.

Problem #03:
Data: Fixed Cost=10,600
Variable Cost=6.70
Q=800
Price=?
Solution
Q=
𝐹
𝑃 ̶ 𝑉
= ˃ Q(P ̶ V)=F
QP-QV=F , P=QV+F/Q
P=V ̶ F/Q=6.70 ̶ 10,600/800=-6.55
Problem #4:
Data
Q=800 units
First Process
Fixed Cost=30,000
Variable Cost=600
Second Process
Fixed Cost=120,000
Variable Cost=900
Solution
TC1=F+VQ=30,000+600(800) =30,000+48,0000=780,000
Problem #05:
a-1:Data:Fixed Cost=140,000
Selling Price=10 per unit
Q=30,000 units
Solution
TPM=PQ ̶ (F+VQ)
TPM= 10(30,000) ̶ {140,000+5(30,000)}=300,000 ̶ 140,000 ̶ 50,000
TPM=300,000 ̶ 290,000=10,000

a-2-Data: Fixed Cost=60,000
Variable Cost=5+1=6 per unit
Selling Price=10+1=11 per unit
Q=50,000 units
Solution
TPM=PQ ̶ (F+VQ)
TPM=11(50,000) ̶ {60,000+6(5,000)}=55,0000 ̶ 60,000 ̶ 300,000
TPM=55,0000 ̶ 360,000=190,000
Process Two
TC2= 120,000+900(800)=120,000+720,000=840,000
ΔTC=TC2 ̶ TC1= 840,000 ̶ 780,000=60,000
b-Data: Fixed Cost=200,000
Selling Price=11 per unit
Q=45,000
Solution
TPM= PQ ̶ (F+VQ)
TPM= 11(45,000) ̶ {200,000+6(45,000)= 495,000 ̶ 470,000= 25,000
New equipment should implement because it yield more margin profit.

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