As part of a new product feasibility, you are studying how much capacity should be built for making and packing a new product. Because a great deal of uncertainty exists about how much volume the new product will generate, you have decided to do a decision tree analysis to help sort out the best

Question

As part of a new product feasibility, you are studying how much capacity should be built for making and packing a new product. Because a great deal of uncertainty exists about how much volume the new product will generate, you have decided to do a decision tree analysis to help sort out the best course of action. At present, the estimates for market development (assume it takes two years to reach final market volume) are:
• Full development: 400K units per year. The probability of this happening is about 30%.
• Medium development: 280K units per year. The probability is about 60%.
• The product fails. The probability is about 10%.
Other data:
• The capital cost for the 400K-unit plant is $7.5M, and for the 280K unit plant, it is $6M. Assume it is spent in Year 0.
• Annual AT Profit for the 400K-unit plant is estimated to be $9/unit per year when making 400K units per year. When running at 280K units, the 400K-unit plant will lose $5/unit per year.
• If a 280K-unit plant is built and the market is 280K units per year, the profit will be $6/unit per year. If the market develops to 400K units per year, a contract manufacturer can be online at the end of Year 2 to produce the added units. That production will generate an AT profit of $5/unit per year.
• If the product fails, the plant will be mothballed and the equipment reused for other purposes. Assume this would happen at the end of Year 2. Assume losses of $2M per year for the first two years (until the plant is shutdown).
For the 400K-unit plant, the salvage value is $2.1M and the AT mothballing expenses are $0.5M. For the 280K-unit plant, the salvage value is $1.7M and AT mothballing costs are $0.4M.
• The Marketing and Sales Departments have estimated the product will have about a 10-year life, so you plan to use that as the economic life for your study.

• Use a hurdle (discount) rate of 15%.
• For simplicity, ignore the salvage value in Year 10, startup expenses, and working capital.

Accepted Answer

First, draw the decision tree (Figure 5.7).

Find the probability-weighted NPV for each of the possible decisions. For the decision to build the larger plant, this means finding the probability weighted NPVs for branches 1, 2, and 3. Similarly for the smaller plant, calculate branches 4, 5, and 6. For branch 1, the AT profit for Year 1 through Year 10 is:
AT profit/yr = $9/unit * 400K units/yr = $3.6M per year
Next, find the present value of the profit in Year 1 and multiply that by the probability for the branch (30%):
Probability-weighted P = ($3.6M)(P/F, 15%, 1)(0.3) = (3.6)(0.8696)(0.3) = $0.94M

The rest of the calculations are similar. Because a number of the cash flows are not uniform, the NPVs will be easier to calculate using a table similar to the one in Example 9. Table 5.5 is for the 400K-unit per year plant and Table 5.6 is for the 280K-unit plant.

TABLE 5.5
Example 10: Probabilty-Weighted NPV, 400 Unit/Year-Plant

				Branch 1				Branch 2				Branch 3
Year	P/F, 15%, n	Initial Capital	P, Capital	AT Profit	P	Probability	Probability Weighted P	AT Profit	P	Probability	Probability Weighted P	AT Profit	Salvage Value	Mothballing	P	Probability	Probability Weighted P	Total P, All Branches
0	1	–7.5	–7.5	—	—	0.3	—	—	—	0.6	—	—	—	—	—	0.1	—	–7.5
1	0.8696	—	—	3.6	3.13	0.3	0.94	–2.0	–1.74	0.6	–1.04	–2	—	—	–1.74	0.1	–0.17	–0.28
2	0.7561	—	—	3.6	2.72	0.3	0.82	–2.0	–1.51	0.6	–0.91	–2	2.1	0.5	–0.30	0.1	–0.03	–0.12
3	0.6575	—	—	3.6	2.37	0.3	0.71	–2.0	–1.32	0.6	–0.79	—	—	—	0.00	0.1	0.00	–0.08
4	0.5718	—	—	3.6	2.06	0.3	0.62	–2.0	–1.14	0.6	–0.69	—	—	—	0.00	0.1	0.00	–0.07
5	0.4972	—	—	3.6	1.79	0.3	0.54	–2.0	–0.99	0.6	–0.60	—	—	—	0.00	0.1	0.00	–0.06
6	0.4323	—	—	3.6	1.56	0.3	0.47	–2.0	–0.86	0.6	–0.52	—	—	—	0.00	0.1	0.00	–0.05
7	0.3759	—	—	3.6	1.35	0.3	0.41	–2.0	–0.75	0.6	–0.45	—	—	—	0.00	0.1	0.00	–0.05
8	0.3269	—	—	3.6	1.18	0.3	0.35	–2.0	–0.65	0.6	–0.39	—	—	—	0.00	0.1	0.00	–0.04
9	0.2843	—	—	3.6	1.02	0.3	0.31	–2.0	–0.57	0.6	–0.34	—	—	—	0.00	0.1	0.00	–0.03
10	0.2472	—	—	3.6	0.89	0.3	0.27	–2.0	–0.49	0.6	–0.30	—	—	—	0.00	0.1	0.00	–0.03
																	NPV=	–8.31

TABLE 5.6
Example 10: Probabilty-Weighted NPV, 280 Unit/Year-Plant

				Branch 4				Branch 5				Branch 6
Year	P/F, 15%, n	Initial Capital	P, Capital	AT Profit	P	Probability	Probability Weighted P	AT Profit	P	Probability	Probability Weighted P	AT Profit	Salvage Value	Mothballing	P	Probability	Probability Weighted P	Total P, All Branches
0	1.0000	–6	–6	—	—	0.3	—	—	—	0.6	—	—	—	—	—	0.1	—	–6.00
1	0.8333	—	—	1.68	1.40	0.3	0.42	1.68	1.40	0.6	0.84	–2	—	—	–1.67	0.1	–0.17	1.09
2	0.6944	—	—	1.68	1.17	0.3	0.35	1.68	1.17	0.6	0.70	–2	–1.7	–0.4	–2.85	0.1	–0.28	0.77
3	0.5787	—	—	2.28	1.32	0.3	0.40	1.68	0.97	0.6	0.58	—	—	—	0.00	0.1	0.00	0.98
4	0.4823	—	—	2.28	1.10	0.3	0.33	1.68	0.81	0.6	0.49	—	—	—	0.00	0.1	0.00	0.82
5	0.4019	—	—	2.28	0.92	0.3	0.27	1.68	0.68	0.6	0.41	—	—	—	0.00	0.1	0.00	0.68
6	0.3349	—	—	2.28	0.76	0.3	0.23	1.68	0.56	0.6	0.34	—	—	—	0.00	0.1	0.00	0.57
7	0.2791	—	—	2.28	0.64	0.3	0.19	1.68	0.47	0.6	0.28	—	—	—	0.00	0.1	0.00	0.47
8	0.2326	—	—	2.28	0.53	0.3	0.16	1.68	0.39	0.6	0.23	—	—	—	0.00	0.1	0.00	0.39
9	0.1938	—	—	2.28	0.44	0.3	0.13	1.68	0.33	0.6	0.20	—	—	—	0.00	0.1	0.00	0.33
10	0.1615	—	—	2.28	0.37	0.3	0.11	1.68	0.27	0.6	0.16	—	—	—	0.00	0.1	0.00	0.27
																	NPV=	0.37

The calculations show that building the smaller plant is best economically because its NPV is $0.37M, as compared to –$8.31M for the larger plant. From this data, one can also see that the probability weighted ROI for the small plant is just above 15% (the discount rate) because the NPV is slightly positive.
Also note that if one only considered the large plant at full market development (an optimistic assumption), the NPV would be $10.6M and the ROI around 47%.
Had company decision-makers not seen the decision tree and were only given the optimistic estimate, they would have mistakenly authorized the large plant. This points out the value of decision trees whenever a great deal of uncertainty exists in the project assumptions.

Question 5.10: As part of a new product feasibility, you are studying how m...

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