3/23/2017
Project Risk, Procurement and Integration Management Individual assignment
NAME : MAHIR M AHIR MALIK NSAIF ID : 110030766
TELOXY ENGINEERING (A)
Teloxy Engineering has received a one-time contract to design and build 10,000 units of a new product. During the proposal process, management felt that the new product could be designed and manufactured at a low cost. One of the ingredients necessary to build the product was a small smal l component that could be purchased purc hased for $60 in the marketplace, including in cluding quantity discounts. Accordingly, management budgeted $650,000 for the purchasing and handling of 10,000 components plus scrap. During the design stage, your engineering team informs you that the final design willrequire a somewhat higher-grade component that sells for $72 with quantity discounts. The new price is substantially higher than you had budgeted for. This will create a cost overrun. You meet with your manufacturing team to see if they can manufacture the component at a cheaper price than buying it from the outside. Your manufacturing team informs you that they can produce a maximum maxi mum of 10,000 units, just enough to fulfill ful fill your contract. con tract. The setup setu p cost co st will be $100,000 and the raw material cost is $40 per component. Since Teloxy has never manufactured this product before, manufacturing expects the following defects:
Percent defective 0 10 20 30 40 Probability of 10 20 30 25 15 Occurrence
All defective parts must be removed and repaired at a cost of $120 per part. 1. Using expected value, is it economically better to make or buy the component? 2. Strategically thinking, why might management opt for other than the most economical choice? After the calculation is done, we find that the cost per unit is 69.7-68.7 if we produce 1600018000, so the strategically it would be better to produce the parts our own, not only will it be cheaper (The cost per unit include repairs and removals etc.) but the company will also move to be a manufacturing company that can develop and expand and manufacture for other companies.
1
-
Unit 10,000 Defective
Probability
(Defective*Probability) (Defective*Probability)
0%
10%
0
10%
20%
2
20%
30%
6
30%
25%
7.5
40%
15%
6 21.5%
Setup cost = 100,000.
The raw material cost = $40 per component. Repaired a cost = $120 per part. Total raw material = setup cost * raw material per component = 100,000 * 40 = 400,000. Repairing cost = X units * sum (Defective*Probability) = 10,000 * 21.5% = 2150. Total Preparing cost = Preparing cost * Preparing a cost per part = 2150 * 120 = 258,000. Total cost = setup cost + raw material + total preparing cost = 100,000 + 400,000 + 258,000 = 758,000.
TELOXY ENGINEERING (B)
Your manufacturing team informs you that they have found a way to increase the size of the manufacturing run from 10,000 to 18,000 units in increments of 2000 units. However, the setup cost will be $150,000 rather than $100,000 for all production runs greater than 10,000 units and defects will cost the same $120 for removal and repair.
1. Calculate the economic feasibility of make or buy. 2. Should the probability of defects change if we produce 18,000 units as opposed to 10,000 units? 3. Would your answer to question 1 change if Teloxy management believes that followon contracts will be forthcoming? What would happen if the probability of defects changes to 15 percent, 25 percent, 40 percent, 15 percent, and 5 percent due to learning-curve efficiencies?
2
Defective
Probability
(Defective * Probability)
0%
15%
0
10%
25%
2.5
20%
40%
8
30%
15%
4.5
40%
5%
2 17%
Units
Setup cost
Total raw material
12,000
150,000
(12,000*40) =480,000
14,000
150,000
(14,000*40)= 560,000
16,000
150,000
(16,000*40) =640,000
18,000
150,000
(18,000*40) =720,000
To Calculate Number of DEFECTED UNITS Units
Rate of Defect
Defected Units
12000
17%
2040
14000
17%
2380
16000
17%
2720
18000
17%
3060
To Calculate COST OF REPAIR FOR DEFECTED UNITS Units
Defected Units
Total preparing cost= (Defected Units * $120 for removal and repair)
12,000
2040
244,800
14,000
2380
285,600
16,000
2720
326,400
18,000
3060
367,200
3
(Setup cost + Material + Repair)
Cost Per Unit
874800
874800/12000=72.9
995600
995600/14000=71.1
1116400
1116400/16000=69.7
1237200
1237200/18000=68.7
Typical values for the exponent K are:
Learning curve % 100
K 0.000
95
0.074
90
0.152
85
0.235
80
0.322
75
0.415
70
0.515
3. A company working on a 75 percent learning curve has decided that the production standard should be 85 hours of production for the 100th unit. How much time should be required for the first unit? If the first unit requires more hours than you anticipated, does this mean that the learning curve is wrong? T1=Tx(x)k K= 0.415, Tx= 75, X=100th unit T1 = 75(100)0.415 T1 = 574.67 4
4. A company has just received a contract for 700 units of a certain product. The pricing department has predicted that the first unit should require 2,250 hours. The pricing department believes that a 75 percent learning curve is justified. If the actual learning curve is 77 percent, how much money has the company lost? Assume that a fully burdened hour is $65. What percentage error in total hours results from a 2 percent increase in learning curve percentage? %75
%80
T1 = 2250 hrs
T1 = 2250 hrs
X = 700
X = 700
Tx =?
Tx =?
Tx = 2250(700)−. = 148.41
Tx = 2250(700)−. = 272.93
Tc = 148.42\(1-0.415) = 253.69
Tc = 272.93(1-0.322) = 402.55
Tc to all units = 700*253.69 = 177583
Tc for all units = 700*402.55 = 281785
Percentage difference (281785 – 177583)/177583 177583)/177583 = 0.58 = 58% 5. A company has decided to bid on a follow-on contract for 500 units of a product. The company has already produced 2,000 units on a 75 percent learning curve. The 2000th unit requires 80 hours of production time. If a fully burdened hour is $80 and the company wishes to generate a 12 percent profit, how much should be bid? 2000
units
2500 units
Tx = 80 hrs
T1 = 1875 hrs
K = 0.415
K = 0.415
T1 = ??
X = 2500 Tx??
1 = 80 ∗ (2000). = 1875 hrs
1 = 1875 ∗ (2500)−. = 72.12 hrs
Tc one unit = 80\(1-0.415) = 136.75
Tc to one unit = 72.12\(1-0.415) = 124.61
Total = 135.75 * 2000 = 237,400
Total = 124.61 * 2500 = 311,600
Difference = 311,600 – 237,700 237,700 = 38200 Difference cost = 38200 * 80 = 3056000 Total profit = 3056000 * 1.12 = 3427720
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6. An arable farmer is thinking of sowing scientifically modified crops next year. She believes that if she did so her profits would be $75,000, compared to $50,000 if she sowed unmodified crops. A neighbouring farmer has made it clear that if his crops are contaminated he will demand compensation. The arable farmer guesses the probability of contamination to be 35%. In the event that the neighbouring n eighbouring farmer claims clai ms compensation there is a 25% chance ch ance that the arable farmer would have to pay $25,000 in compensation and a 75% chance she would have to pay $50,000 in compensation. a). Construct a decision tree and use it to advise the arable farmer b). An expert puts the probability that there will be contamination of the crops of the neighbouring farmer at 60%. Should the arable farmer change her strategy in the light of this information? Modified CROPS:
We will calculate the modified crops first on 25% and 75% = (0.75*50,000) + (0.25*25,000) = 43,750$ Then we will calculate the second on 65% and 35% = (0.35*43,750) + (0.65*75,000) = 640625$ So we will take the modified crops
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The second tree We will calculate the modified crops first on 25% and 75% = (0.75*50,000) + (0.25*25,000) = 43,750$ Then we will calculate the second on 60% and 40% = (0.40*43,750) + (0.60*75,000) = 62,000$ So we will take the modified crops
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