Model answers to all the questions will appear on April 4.
The questions in the first section are worth 4 points each; the questions in the second section are worth 15 points each. Total time for the exam is three hours.
Assume the inflation rate is zero unless the question says it isn't.
The new company has $1.5M; 0.5M at 10% interest, 0.5M from investors who expect 15% growth, and 0.5M from you, where you also expect 15% growth. So the WCC is (1*0.15 + 0.5*0.1)/1.5.
Your MARR should be at least as great as the WCC.
The acid-test ratio is the ratio of a company's current assets, including cash and accounts receivable, but not including inventory, to the company's current indebtedness. The desirable value for this ratio varies from industry to industry; it gives an indication of the company's abilitiy to pay its debts if credit becomes short (as it might, for example, following a fall in stock prices.)
`
(d)
If the proposals being considered are the ONLY ones available, you need to consider which makes the best use of your TOTAl investment funds. It's better to get a medium rate of return on a large sum than an excellent rate of return on a small sum, if the latter choice leaves a lot of funds uninvested. To take this into account, you should do an incremental analysis.
`
PERT is concerned with estimating the time to completion of a project, CPM with making the best cost/time trade-off.`
A simplex is a closed, convex polygon in n dimensions. The simplex method is an algorithm that visits the vertices of such a simplex in turn, searching for the optimal solution. `
The shadow price of a resource is the premium it would be worth paying for additional supplies of that resource. Its value is conveniently given by the coefficient of the associated slack variable in the objective function of the final tableau of the simplex method.
The CLT allows us to conclude that the overall completion time of a project, being the sum of the randomly distributed completion times of the activities on the critical path, will itself be a random variable with a normal distribution, its mean the sum of the activity means, its variance the sum of the activity variances. `
The critical path is the sequence of activities required for completion of a project such that delay to any one of the activities will delay completion of the project.
It will be exactly twice as long.
The present cost of the purchase is 800,000*CCTF. The value of the CCTF factor depends on the depreciation rate, according to the formula provided in the cribsheet. So you need to calculate 800,000(CCTF(20%)-CCTF(30%)).
Your pre-tax MARR is 25%. Which strategy will you choose?
To solve this, you need to compare the PW of the first strategy with the expected value of the PW of the second strategy, taking a four-year study period. We can speed up the calculation of the PW of the first strategy by lumping together the decline in sales and the discounting of future income: PW(1) = 200,000*1.2*(P/A,0.45,4), where the factor of 1.2 is needed to get the decline in sales and the discounting of future income in step.
To calculate the expected PW of the second option, construct a probability tree, and find the present worth of each of the leaves of the tree. Then apply the formula
E(PW)=Sum(probability of outcome i * PW of outcome i)
The bank charges 10% interest per year. The bonds are sold now for $1000, with a promise to redeem them for $1,500 four years in the future. If you use internal funds for this project, you will have to divert them from other potential investments within the company. Your pre-tax MARR is 15%. How should you raise the money?
The cash flow diagram is in the notes. The second half of the question is just asking for the cheapest source of capital. Your internal funds cost 15%, the bank loan costs 10%, and the issue of bonds costs i%, where
1000(1+i)4=1,500
Solving for i, it turns out to be just over 10%. So you should borow from the bank.
A new machine of advanced design can perform the same function as the current machine for an operating cost of just $12,000 per year. It will cost $24,000 to purchase, and will have a salvage value of $6,000 after 5 years of use.
The company has a 52% effective tax rate, and its after-tax MARR is 20%. Both machines are in Class 8: declining balance depreciation at 20%. Should the machine be replaced?
This calls for an after-tax analysis, with a study period of 5 years. We compare the PW of each option.
Option 1: keep old machine:
PW = -19,000(1-t)(P/A,i,5)+1,000.CCTF.(P/F,i,5)
Option 2: Buy new machine
PW = -12,000(1-t)(P/A,i,5) - (24,000-8,000).CCTF + 6,000.CCTF.(P/F,i,5)
First cost | $17,000,000 |
Painting every 6 years | $1,000,000 |
Deck resurfacing every 10 years | $3,000,000 |
Structural overhaul after 15 years | $4,000,000 |
Annual maintenance | $300,000 |
The tunnel is expected to cost $24,000,000 and will require repaving every 10 years at a cost of $2,000,000. If both designs are expected to last 30 years with minimal salvage value, determine the maximum equivalent annual amount for maintenance that could be permitted for the tunnel while holding the total equivalent annual cost equal to that of the bridge. The interest rate is 8%.
Since the unknown is an equivalent annual cost, we should calculate the equivalent annual cost for each option, using a 30-year study period. We assume that no periodic maintainance is done in the final year. For the bridge, we have:
EUAC(bridge) = (A/P,0.08,30) * (17 + 1(P/A,i6,4) + 3(P/A,i10,2) +4(P/F,0.08,15)) + 0.3
EUAC(tunnel)=(A/P,0.08,30) * ( 24 +2(P/F,0.08,10)+2(P/F,0.08,20)) + x
This can then be solved for x. Note that i6 and i10 are the effective interest rates for 6 and 10-year periods respectively.
We take a five-year study period and consider all cashflows to occur at year's end. For the second position, we note that the inflation rate exactly matches the raises, so the present worth is just
PW(aerospace)=52,000(P/A,0.15,5)
For the faculty position, the initial payment is 45,000, and the present worth of payments in subsequent years changes by a factor of 1.08/(1.05)(1.15) every year. This can be simplified to a factor of 1/(1+x). So the present worth of the position is
PW(faculty)=45,000(P/A,x,5)
Activity | Initial State | Final State | Duration | Description |
---|---|---|---|---|
1 | 1 | 2 | 2 | Overall Design |
2 | 2 | 3 | 2 | Design Casings |
3 | 2 | 4 | 4 | Write Chip Specifications |
4 | 3 | 5 | 6 | Purchase Moulder |
5 | 4 | 6 | 2 | Order Chips |
6 | 4 | 7 | 3 | Write Battery Specifications |
7 | 5 | 8 | 9 | Produce Moulds |
8 | 6 | 9 | 11 | Receive Chips |
9 | 7 | 10 | 2 | Order Batteries |
10 | 8 | 9 | 13 | Receive Moulds |
11 | 9 | 11 | 0 | Dummy |
12 | 10 | 12 | 8 | Receive Batteries |
13 | 11 | 13 | 7 | Install Chip in Casing |
14 | 12 | 13 | 0 | Dummy |
15 | 13 | 14 | 10 | Install Battery in Casing |
16 | 14 | 15 | 12 | Test |
17 | 15 | 16 | 2 | Package and Ship |
The first step is to plot a network showing the states and the linking activities. Then mark the activity times on the network. Then take a pass through the nextwork, noting that the earliest we can reach state 1 is at time 0, the earliest we can reach state 2 is at time 2, and so on, till we find the earliest that we can reach the final state, state 17. Now take a backward pass through the network, noting the latest time we can leave each state if the final state is not to be delayed. The difference between the earliest and latest times for each state is the slack for that state, and states with no slack lie on the critical path.
Colony | Useable Land (Hectares) | Water Allocation (Hectare metres/hectare) |
---|---|---|
Lowell | 400 | 600 |
Bradbury | 600 | 800 |
Clarke | 300 | 375 |
Crops adapted for Martian conditions include sour potatoes, bladder-root, and transgenic hemp. There is an upper limit on the total area that can be devoted to each crop by the three colonies together, imposed by the Martian Board of Agriculture. This limit, and the different profitability and water requirement of the crops, are given by the following table:
Crop | Maximum Quota (Hectares) | Water Consumption (Hectare metres/hectare) | Net Return |
---|---|---|---|
Sour Potatoes | 600 | 3 | 400 |
Bladder-root | 500 | 2 | 300 |
Transgenic Hemp | 325 | 1 | 100 |
The three colonies are developing a joint plan to maximize the profitability of their crops. By selecting suitable decision variables, set up a linear programming problem whose solution will give the most profitable crop mix for the three colonies together. You should put the problem into suitable form for input to a standard simplex solution package, but do not attempt to solve it.
The hardest part is choosing the right decision variables. There should be nine, representing the acreage of each crop grown at each colony:
xl,s, xl,b,xl,h, xb,s,xb,b, xb,h,xc,s, xc,b,xc,h
The objective function is then
Z = 400*(xl,s+xb,s+xc,s) + 300*(xl,b+xb,b+xc,b) + 100*(xl,h+xb,h+xc,h)
There are a number of constraints:
Maximum quotas:
xl,s+xb,s+xc,s <= 600
xl,b+xb,b+xc,b <=500
xl,h+xb,h+xc,h <= 325
Water consumption
3*xl,s+2*xl,b+xl,h <= 600
3*xb,s+2*xb,b+xb,h <= 800
3*xc,s+2*xc,b+xc,h <= 375
Useable Land
xl,s+xl,b+xl,h <= 400
xb,s+xb,b+xb,h <= 600
xc,s+xc,b+xc,h <= 300
These can be converted to standard form by adding slack variables.