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Lecture

# Solutions for Chapter 1,2 The professor gave us these solutions , very helpful - Winter 2010

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University of Waterloo

Management Sciences

MSCI 432

Binyamin Mantin

Fall

Description

Chapter-01
Ans-7 a) The focus of the current facility will be compromised.
b) The Operations V.P. may be taking a short-term view that could be
costly to the firm in the long run.
c) In order to maximize the current ROI, the CFO has chosen to limit
current investment. This could be a poor long-run strategy decision.
d) The reliability requirements for the second application are likely to be
much higher. The firm should consider producing these products in
different plants.
Ans-15 Time based competition refers to getting a product to market before
your competitors and getting to volume production before your competitors as
well. A company which has had several commercial successes by introducing
innovative products to the marketplace first is Gillette. Gillette has had
significant successes with its Atra and Sensor razors as well as the Good News
disposable razor line.
1
Ans- 22 a) 1. Start-up
2. Rapid growth
3. Maturation
4. Stabilization or decline
b) In most circumstances, it is inappropriate to operate very far off the
diagonal. For example, a firm that produces a varied product line
with a relatively low volume for each product would not invest the
capital required for the equipment to develop a continuous flow
production line. Similarly, a firm producing few products with high
volume would not choose a jumbled flow shop for manufacturing.
Ans- 25 a) I-I
IV-I)V
c) II-II (could be III-III depending on the product volume and diversity
of products produced)
II-I)II
I-Ie)
2
Ans-30a) ln (cum # units) ln (# hrs for next unit)
3.912 1.19
4.605 0.79
5.991 0
6.397 -.22
6.908 -.69
9.210 -1.61
b) 3.0
2.5
2.0
1.5
1.0
ln(hours)
0.5
0.0 1 2 3 4 5 6 7 8 9 10
‐0.5
‐0.1 ln(cumulative units)
‐15
3
c) The intercept is ln(a) => a = e2.7 = 14.88
ln(L) = -b ln(2) = (-.46)(.6931) = -.319
L = e-.319= .727
Hence, the first unit should require 14.88 hours and a 73% learning
curve accurately fits this data.
d) Exact values of the slope and intercept are
slope = -0.53834
intercept = 3.2319
based on the natural logarithms. Using these exact values, the first unit
should require e 3.2319 = 25.33 hrs.
Since ln(L) = -b ln(2) = (-.53834)(.693) = -.373
L = e -.37= .689
Hence, the first unit should require 25.33 hrs. and a 69% learning curve
should accurately fit this data based on the estimated least squares fit of
the data.
Ans-31 Using the exact least squares estimates from part (d), the learning
curve relationship is
Y(u) = 25.33u-.53834
Substituting u = 100,000 gives
Y(u) = (25.33)(100,000)-.53834 = .0515 hrs.
4
Ans-35 a) Let K = 30,000
c = 20
2
c1 = 85
Solve for x = K/(c-c ) = (30,000)/(65) = 461.54 ~ 462
1 2
c b) 1x = (85)(462) = $39,270.00.
c) x = (30,000)/(100-20) = (30,000)/(80) = 375
Ans- 36 Cum.
Year Sales Net Revenue ReNetnue
1 100 6,500 6,500
2 140 9,100 15,600
3 196 12,740 28,340
4 274 17,810 46,150
Hence, it will require slightly over 3 years to recoup the $30,000.00 initial
investment. Linear interpolation gives an exact answer of 3.11 years.
5
Ans-38 a) From Figure 1.14, a value of a = .58 gives a value of u very close to 1.0.
Solving x = u/r = 1/.12 = 8.33 years. The optimal size of each addition is
xD = (8.33)(3,000) = 25,000 tons.
b) f(y) = .0205y .58= (.0205)(25,000) .58- 7.29.
Hence, the cost of each new addition is $7.29 million.
c) At 12% interest the effective annu al discount rate is 1/(1+0.12) = .8929.
However, replacements are made once ever y 8.33 years giving a discount factor
8.33
per replacement of (.8929) = .3891. Hence, the present cost of the next four
replacements is
2 3 4
7.29[.39 + (.39) + (.39) + (.39) ] = 7.29[.6245] = $4.55 million
Ans-44 D = 8,000 tons/year, r = .10, a = .51
From Figure 1.14, we obtain u ≈ 1.25. It follows that
rx = 1.25, giving x = 1.25/.10 = 12.5 years.
6
Chapter-02
Ans-2Cycles have repeating patterns th

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