91Ó°ÊÓ

The Oklahoma Pipeline Company projects the following pattern of inflows from an investment. The inflows are spread over time to reflect delayed benefits. Each year is independent of the others.

Year 1

Cash

Inflow Probability

55 ............... 0.40

70 ............... 0.20

85 ............... 0.40

Year 5

Cash

Inflow Probability

40 .............. 0.30

70 .............. 0.40

100 .............. 0.30

Year 10

Cash

Inflow Probability

20 .............. 0.40

70 .............. 0.20

120 .............. 0.40

The expected value for all three years is \(70.

a. Compute the standard deviation for each of the three years.

b. Diagram the expected values and standard deviations for each of the three years in a manner similar to Figure 13-6.

c. Assuming 6 percent and 12 percent discount rates, complete the following table for present value factors:

Year

PVIF

6%

PVIF

12% Difference

1 .............. 0.943 0.893 0.050

5 ..............

10 ..............

d. Is the increasing risk over time, as diagrammed in part b, consistent with the larger differences in PVIFs over time as computed in part c?

e. Assume the initial investment is \)135. What is the net present value of the investment at a 12 percent discount rate? Should the investment be accepted?

Step by step solution

01

Computation of expected cash flow-

Probability (a)	Cash flow (b)	Expected Cash flow (a*b)
Year 1
0.40	$55	$22
0.20	$70	$14
0.40	$85	$34
		$70
Year 5
0.30	$40	$12
0.40	$70	$28
0.30	$100	$30
		$70
Year 10
0.40	$20	$8
0.20	$70	$14
0.40	$120	$48
		$70

02

Computationj of total probability-

Probability (a)	Cash flow(in thousands) (b)	Probability*(sales-Expected sales)2	Total
Year 1
0.4	$55	0.40*(55-70)2	$90
0.2	$70	0.20*(70-70)2	$0
0.4	$85	0.40*(85-70)2	$90
			$180
Year 5
0.3	$40	0.30*(40-70)2	$270
0.4	$70	0.40*(70-70)2	$0
0.3	$100	0.30*(100-70)2	$270
			$540
Year 10
0.4	$20	0.40*(20-70)2	$1000
0.2	$70	0.20*(70-70)2	$0
0.4	$120	0.40*(120-70)2	$1000
			$2000

03

Step 3:a. Computation of standard deviation-

$\begin{array}{rcl}\mathbf{Standarddeviationyear}\mathbf{1}\mathbf{cashflow}& \mathbf{=}& \sqrt{\mathbf{Probability}\mathbf{×}{\left(\mathrm{sales}-\mathrm{Expectedsales}\right)}^{\mathbf{2}}}\\ & \mathbf{=}& \sqrt{\mathbf{180}}\\ & \mathbf{=}& \mathbf{13}\mathbf{.}\mathbf{4164}\\ & & \end{array}$

$\begin{array}{rcl}\mathbf{Standarddeviationyear}\mathbf{5}\mathbf{cashflow}& \mathbf{=}& \sqrt{\mathbf{Probability}\mathbf{×}{\left(\mathrm{sales}-\mathrm{Expectedsales}\right)}^{\mathbf{2}}}\\ & \mathbf{=}& \sqrt{\mathbf{540}}\\ & \mathbf{=}& \mathbf{23}\mathbf{.}\mathbf{2380}\\ & & \end{array}$

$\begin{array}{rcl}\mathbf{Standarddeviationyear}\mathbf{10}\mathbf{cashflow}& \mathbf{=}& \sqrt{\mathbf{Probability}\mathbf{×}{\left(\mathrm{sales}-\mathrm{Expectedsales}\right)}^{\mathbf{2}}}\\ & \mathbf{=}& \sqrt{\mathbf{2000}}\\ & \mathbf{=}& \mathbf{44}\mathbf{.}\mathbf{7214}\\ & & \end{array}$

04

Step 4:(b) Decision tree

Alternative	Expected Sales	Expected value	Standard deviation
Year 1	Low	$70	13.4164
Year 2	Moderate	$70	23.2380
Year 3	High	$70	44.7214

05

Step 5:(c) Table for present value factors 

Year	PVIF 6%	PVIF 12%	Difference
1	0.943	0.893	0.050
5	0.747	0.567	0.180
10	0.558	0.322	0.236

06

Step 6:(d)Analysis

The difference between present value factor of Year 10 is higher in comparison to Year 1 and Year 5. The Standard deviation of year 10 is also higher in comparison to year 1 and year 5. This states that the risk is higher in Year 10 investment. The risk remains with the high difference between present vale factor of 6% and present value factor of 12%.

07

Step 7:Computation of Present value

Year	Cash flow	PVIF 12%	Present value
Year 1	$70	0.893	62.51
Year 5	$70	0.567	39.69
Year 10	$70	0.322	22.54
Present value of cash inflows	$124.74

08

(e) Computation of Net present value-

$\begin{array}{rcl}\mathbf{Netpresentvalue}& \mathbf{=}& \mathbf{Presentvalueofinflow}\mathbf{-}\mathbf{Investment}\\ & \mathbf{=}& \mathbf{\$}\mathbf{124}\mathbf{.}\mathbf{74}\mathbf{-}\mathbf{\$}\mathbf{135}\\ & \mathbf{=}& \mathbf{-}\mathbf{\$}\mathbf{10}\mathbf{.}\mathbf{26}\\ & & \end{array}$

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