1. Exam 1 Review/Instructions
370-Operations-Mgmt
2018/9/16
Exam 1 Review/Instructions
Content: Chapters 1, 2, 5, and 6S
Write your full name and ID on both the exam and the answer sheet.
Use a No.2 pencil to fill in the answer sheet.
Return the exam and the answer sheet when you finish the examination.
The duration of this examination is the same as one regular class duration.
All multiple-choice questions, 25 questions in total.
Closed books, notes, and computers. You may bring a 8.5 in*11in double-sided cheat sheet and a calculator.
Homework 1 is due in the discussion ahead of Exam 1 .

2. Chap 1: Introduction to Operations Management
370-Operations-Mgmt
2018/9/16
Chap 1: Introduction to Operations Management
What is Operations?
What is Operations Management?
Why study Operations Management?
Trends in Operations Management

3. Three Key Functions in an organization
370-Operations-Mgmt
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Three Key Functions in an organization
Operations
Finance
Marketing

4. Continuum of Types of Firms
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Continuum of Types of Firms
More like a manufacturing organization
More like
a service organization
Physical, durable product
Output that can be inventoried
Low customer contact
Long response time
Regional, national, or international markets
Large facilities
Capital intensive
Quality easily measured
Intangible, perishable product
Output that cannot be inventoried
High customer contact
Short response time
Local markets
Small facilities
Labor intensive
Quality not easily measured

5. Chap 2 Competitiveness, Strategy, and Productivity
Mission / Goals / Strategies / Tactics
Productivity
A measure of the effective use of resources, usually expressed as the ratio of output to input
Productivity =
Outputs
Inputs

6. Productivity Partial measures Multi-factor measures Total measure
output/(single input)
Multi-factor measures
output/(multiple inputs)
Total measure
output/(total inputs)

7. Measures of Productivity
Table 2.5
Partial Output Output Output Output measures Labor Machine Capital Energy
Multifactor Output Output
measures Labor + Machine Labor + Capital + Energy
Total Goods or Services Produced
measure All inputs used to produce them

8. Chap 5: Capacity Planning
370-Operations-Mgmt
2018/9/16
Chap 5: Capacity Planning
Explain capacity planning and its importance
Discuss ways of defining and measuring capacity.
Describe the determinants of effective capacity.
Steps for Capacity Planning
Identify capacity alternatives
Bottleneck operation analysis - handout
Conduct financial analysis
Cost-volume analysis - handout

9. Capacity Planning
Capacity is the upper limit or ceiling on the load that an operating unit can handle.
Capacity also includes
Equipment
Space
Employee skills
The basic questions in capacity planning are:
What kind of capacity is needed?
How much is needed?
When is it needed?

10. Efficiency and Utilization
Actual output
Efficiency =
Effective capacity
Utilization =
Design capacity
Both measures expressed as percentages

11. Efficiency/Utilization Example
Design capacity = 50 trucks/day
Effective capacity = 40 trucks/day
Actual output = 36 units/day
Actual output units/day
Efficiency = = = 90%
Effective capacity units/ day
Utilization = Actual output = units/day
= 72% Design capacity units/day

12. Production units have an optimal rate of output for minimal cost.
Figure 5.4
Production units have an optimal rate of output for minimal cost.
Minimum
cost
Average cost per unit
Rate of output
Minimum average cost per unit

13. Economies of Scale Economies of scale Diseconomies of scale
If the output rate is less than the optimal level, increasing output rate results in decreasing average unit costs
Diseconomies of scale
If the output rate is more than the optimal level, increasing the output rate results in increasing average unit costs

14. Cost-Volume Relationships
Figure 5.6c
Amount ($)
Q (volume in units)
= FC / (R-v)
BEP units
Profit
Total revenue
Total cost

15. Break-Even Problem with Step Fixed Costs
Textbook Page 195 Example 4
A manager has the option of purchasing one, two, or three machines.
Number of Total Annual Corresponding
Machines Fixed Costs Range of Output
$ 9, to 300
, to 600
, to 900
Variable cost is $10 per unit, and revenue is $40 per unit.
Determine the break-even point for each range.
QBEP1=FC1/(R-v)=9600/(40-10)=320
QBEP2=15,000/30=500, QBEP3=20,000/30 ≈667

16. Break-Even Problem with Step Fixed Costs
Figure 5.7b
b. If projected annual demand is between 580 and 660 units, how many machines should the manager purchase?
Answer: 2
machines TR $
BEP 3
BEP 2 TC TC
667 TC
500 Q
300
600
900

17. 370-Operations-Mgmt Linear Programming
2018/9/16
Chapter 6 Supplement
Linear Programming

18. Linear Programming
LP is a set of linear mathematical representations of constrained optimization problems.
It consists of 4 components:
Objective function
Decision variables
Constraints
Parameters

19. LP: Formulation Max (or Min) Objective subject to (s.t.): constraint 1…
Feasible solution: If a solution satisfies all the constraints, then it is feasible; Otherwise, i.e. it doesn’t not satisfy one or more of the constraints, then it is infeasible.

20. Example: LP Formulation
370-Operations-Mgmt
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Example: LP Formulation
X1: number of tables produced
X2: number of chairs produced
Max 7*X1+5*X2
S.t. 4X1+3X2<=240 (Carpentry time)
2X1+ X2<= (Painting time)
X1, X2>=0
Carpentry time=210 hours
Painting time= 90 hours
X1=30, X2=30
Profit=$360
(Feasible)
Carpentry time= 240 hours
Painting time=120 hours
X1=60, X2=0
Profit=$420
(Infeasible)

21. Graphical Linear Programming
Set up objective function and constraints in mathematical format
Plot the constraints
Identify the feasible solution space
Plot the objective function
Determine the optimum solution

22. Find Optimal Solution Approach 1: Isoprofit Line Method
Let profit equal some arbitrary but small dollar amount.
Graph the profit line.
Move the profit line farther from the 0 origin to reach the highest possible profit.
Approach 2- Corner Point Method
Optimal Solution to any LP problem will lie at a corner point, or extreme point, of the feasible region.
Calculate the values of the variables at each corner
Find the maximum profit or optimal solution at one of the corner points.

23. Feasible Region Approach 1: Isoprofit Line Method Objective Function
4X1+3X2=240
Objective Function
Optimal Solution
2X1+ X2=100
Feasible Region

24. Approach 2: Corner Point Method
370-Operations-Mgmt
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Approach 2: Corner Point Method X2
Corner Point A:
X1=0,
X2=80
100
Profit = 7X1+5X2 =400
Corner Point B:
X1=0,
X2=0 A 80
Profit=7*0+5*0=0 60
Corner Point C:
X1=50,
X2=0
Profit=7*50+5*0=350 D 40
Corner Point D:
X1=30,
X2=40
Feasible
Region
Profit=7*30+5*40=410 20
Feasible
Region
D is optimal solution B 20 40 C 60 80
100 X1

26. What is Sensitivity Analysis
Sensitivity analysis is a means of assessing the impact of potential changes to the parameters (the numerical values) of an LP model.
Change in the objective function coefficient
Change in the right-hand side (RHS) values of constraints
Change in constraint coefficients (not required)

27. Determine the range of optimality
[Coefficient-allowable decrease, Coefficient+allowable increase]
e.g. Max 7X1+5X2
Range of optimality for the objective function coefficient of X1: [7-0.33, 7+3] = [6.67,10],
for X2: [5-1.5, ] = [3.5, 5.25].

28. Determine the range of optimality
Sensitivity Analysis - objective function coefficient
Determine the range of optimality
[Coefficient-allowable decrease, Coefficient+allowable increase]
e.g. Max 7X1+5X2, opt. solution: (X1,X2)=(30,40)
Range of optimality for the objective function coefficient of X1:
[7-0.33, 7+3] = [6.67,10],
-If “Max 8X1+5X2”, then 6.67 ≤ 8 ≤ 10, so opt. (X1,X2)=(30,40), not changed
- If “Max 11X1+5X2”, then 11 > 10, so (X1, X2)=(50,0), changed

29. Sensitivity Analysis Example - RHS of constraints
Shadow price: Amount by which the value of the objective function would change with a one-unit change in the RHS value of a constraint.
Range of feasibility: Range of values for the RHS of a constraint over which the shadow price remains the same.
-The range is determined by
[RHS-allowable decrease, RHS+allowable increase].
For Carpentry time: [240-40,240+60]=[200, 300]
Original cons. is “4X1+3X2<=240”, opt. profit =410.
Within the range of feasibility : [200, 300], If “4X1+3X2<=200 ”, opt. profit = *( ) = 350;
If “4X1+3X2<=250 ”, opt. profit = *( ) = 425;
If “4X1+3X2<=300 ”, opt. profit = *( ) = 500;
Out of the range of feasibility:
If “4X1+3X2<=160 ”, opt. X1=40, X2=0, profit = 280 ;
Using the shadow price of 1.5, the opt. profit = *( ) = 290 which is wrong!