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Unit 6: Capital Budgeting
Buying new machinery Notes
Purchase cost 49000
Less: realization from old machine 5000
Annual equivalent cost for 10 years 44000/6145 7160
Running & Maint. cost per machine 14000/net of tax 7000
Tax benefit of depreciation pa. 49000/10/× 50 (2450)
Total annual cost 11710
Difference in annual cost in buying 796
Since annual cost of buying is less than that of repairing, one should go for buying option.
Second solution:
1. To repair existing machine
Cost of repair immediately net of tax. ( 19,000 × 50%) 9,500
Running & Maint. Cost of 5 years ( 20,000 × 3.791) 37,910
Total net present value of after tax cash outflows for 5 years. 47,410
Hence net equivalent cash outflows p.a. 47,410/3.791 12,506
2. To buy new machine
Purchase cost of new machine 49,000
Less: Sale proceeds of old machine 5,000 44,000
Tax benefit on depreciation p.a. ( 49000/10×50%) (2,450)
Running & Maint. Cost p.a. (50% of 14000) 7,000
Net cash outflow for 10 years ( 4550×6.145) 4,550 27,960
Net cash outflows for 10 years. 71,960
Hence net equivalent cash outflows pa. 71,960/6.145 11,710
Since, net equivalent cash outflow p.a. for buying a new machine 11,710 is less than net
equivalent outflows of 12,506 for repairing of an existing machine. Therefore, it is advisable
that the company should go for buying a new machine.
!
Caution Project with Unequal Lives: Where one is considering more than one project
(mutually exclusive projects) with different project lives, one should consider the equivalent
annual value method. Under this method, work out the following:
1. The total net present value of after tax cash flows of each project during the project
life.
2. Divide the NAV of cash flows by the annual factor corresponding to the life of the
project at the given cost of capital, the result and figure in the equivalent annual net
present value. (EANPV).
The decision criteria, in the case of revenue expanding proposal, is the maximization of
EANPV and minimization of equivalent annual cost of in the case of cost reduction
proposal. This is illustrated in example above.
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