- 25 Marks
QMDM – APR 2024 – L2 – Q6 – Rural Bank Investment Proposals NPV
The management of a Rural Bank must decide between two investment proposals using Net Discounted Value (NPV) calculations at a 14% discount rate, explain the term, compute NPVs, and advise on selection.
Question
The Management of a Rural Bank must decide between two proposals, on the basis of the following information:
Proposal | Investment Now | Net Cash Inflow at the End of | 1991 | 1992 | 1993 |
---|---|---|---|---|---|
A | GHS 80,000 | GHS 95,400 | GHS 39,400 | GHS 12,000 | |
B | GHS 100,000 | GHS 35,000 | GHS 58,000 | GHS 80,000 |
Assume that on Projects of this type of the company can earn 14 percent per annum. (a) Explain briefly the term Net Discount Value in relation to the projects. (b) Calculate the Net Discounted Value of Proposal A. (c) Calculate the Net Discounted Value of Proposal B. (d) Using the values in (a) and (b), advise Management regarding the proposal that should be selected.
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- Tags: Decision Making, Discounted Cash Flow, Investment Appraisal, Net Present Value, NPV, proposals, rural bank
- Level: Level 2
QMDM – APR 2024 – L2 – Q4 – Exponential Smoothing in Time Series Forecasting for Mango Demand
Using monthly demand data for mangos over 15 months, explain exponential smoothing, compute and compare forecasts using alpha=0.1 and 0.4 with initial forecast of 500, describe plots of actual vs forecasts, and comment on suitability.
Kiki, the commercial mango seller has collected demand figures for mangos over the last 15 months in the table below:
(a) Explain briefly the term “Exponential Smoothing” in Time Series Analysis of the data above. [3 Marks]
(b) Use an initial forecast of 500 to compare Exponential Smoothing Forecasts with Smoothing Constant Values a = 0.1 and a=0.4.
(c) Plot the actual values of the time series and superimpose the forecast for the Smoothing Constant Values a= 0.1 and a=0.4 on the graph of the actual values.
(d) Comment on the suitability of the forecast from the Smoothing Constant Values a= 0.1 and a=0.4.
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