Question

A farm is evaluating two options to expand its facilities to serve a new manufacturing plant. The new plant will require 2000 telephone lines this year and another 2000 lines in 10 years. The plant will operate for 30 years.

Option 1: Install now network with capacity to serve 4000 lines. This network will cost $27,000 and annual maintenance costs will be $1900.

Option 2: Provide a network with capacity to serve 2000 lines now and a second network to serve the other 2000 lines in 10 years.

Each network will cost $21000 and will have annual maintenance of $2000. Both options will last at least 30 years, and the cost of removing the cables is offset by their salvage value.

a) Draw a cash flow diagram for each Option.

b) Which option should be selected assuming 10% interest rate? Use the PW as decision criterion.

Answer 1

Solution:

There are two options for the farm to expand its facilities.

Number of telephone lines required in time 0 = 2000

Number of telephone lines required in time 10 = 2000

Number of years of operation (n)= 30

Rate of interest (i) = 10% = 0.1

Option 1:

Number of telephoone lines installed at time 0 = 4000

Cost of the network = $27,000

Annual mintenance cost of the network = $ 1900

Option 2:

Number of telephoone lines installed at time 0 = 2000

Number of telephoone lines installed at time 10 = 2000

Cost of each network = $21,000

Annual mintenance cost of the network = $ 2000

a) Cashflow diagrams:

b) Option 1:

Present worth = -27000 - 1900 { (P/F, i, n₁) x (P/F, i, n₃₀)}

= -27000 - 1900 { (P/F, 10%, 1) x (P/F, 10%, 30)}

= -27000 - 1900 ( 0.9091 x 0.0573) ...[Present factor values from compound interest tables]

= -27000 - (1900 x 0.05209143)

= -27000 - 98.973717

= - $27,098.97372

Option 2:

Present worth = -21000 -21000(P/F, i, n) - 2000 { (P/F, i, n₁) x (P/F, i, n₃₀)}

   = -21000 - 21000(P/F, 10%, 10) - 2000 { (P/F, 10%, 1) x (P/F, 10%, 30)}

= -21000 - (21000 x 0.3855) - 2000 ( 0.9091 x 0.0573) ..[Present factor values from compound interest tables]

    = -21000 - 8095.5 - (2000 x 0.05209143)

  = -21000 - 8095.5 - 104.18286

  = - $29,199.68286

Using Present worth as a decision criteria we observe that both the options have negative present worth. However, the option 1's negative present worth is less than that of option 2 ( - $27,098.97372 < - $29,199.68286) , which means that option 1 will also incur loss but relatively less than option 2. Hence, the farm should go ahead with Option1.