Opportunity Cost

The principle of opportunity cost ebIsinCaGñkdaMRsUvsalIenAelIépÞd

I 40 acre enaHGñkminGacdaMeBatenAelIépÞdITaMgenaHeT.

ebIsinCaGñkeFVIkarenAksidæan enaHGñkminGaceTAeFVIkarenATIRkug)aneTkñúgeBlEtmYy.

ebIsinCaGñkcMNayfvikareTAelIkarTijeRKOgynþ enaHGñkminGacvinieyaKfvikarTaMgenaHeTAelICIeT.

Examples

]TahrN¾fa eyIgcg;rktMél»kasénkareRbIR)as;dIsMrab;plitkmµRsUvR)aMg

eyIgGacTTYl)anR)ak;cMeNjsuT§ 80$ an acre

BIkareRbIR)as;épÞdITaMgenaHsMrab;plit kmµeBat 65$

BIplitkmµsENþkesog nig 45$

BIplitkmµRsUvsalI. tMél»kasénkareRbIR)as;épÞdIenaH

sMrab;plitkmµRsUvR)aMgKW 80$ an acre.

....Examples ]TahrN¾fa

Gñkcg;rktMél»kasénkMlaMgBlkmµsMrab;karsagsg;RTugRCUk

kMlaMgBlkmµnwg)anéføQñÜl 6 $ an

hour sMrab;karP¢Üras;erobcMdIsMrab;eFVI RsUvsalI nig 4.50$ an hour sMrab; karsuIQñÜllk;TMnij.

dUcenHtMél»kasénkMlaMgBlkmµsMrab;karsagsg;RTugRCUkenaHKW 6 $ an hour

mü:agvijeTotebIkarsagsg;RTugRCUkkñúgeBlEdlGñkKµankargareFVIenaHtMél»kas énkMlaMgBlkmµsMrab;karsagsg;RTugRCUkesµIsUnü.

....Examples As a third example, suppose you have a choice of

investing money in fertilizer or in machinery. You estimate that you can get 3000 $ in returns

from 2000 $ invested in fertilizer. The opportunity cost of spending 2000$ for

machinary would be the 3000$ you could get from using money for fertilizer.

The principle of oppertunity cost has many applications and is very

important in deciding where to invest capital for greatest returns Hypothetical Example of Marginal Returns From Units of 1000$

Invested in Various UsesAmount of invesment Return from 1000 $ invested in

Hogs Feeder cattleFertilizer

First 1000$ 1400$ 1200$ 1500$Second 1000$ 1325 1150 1300 Third 1000$ 1250 1100 1100Fourth 1000$ 1150 1050 1000Fifth 1000$ 1100 1000 950Sixth 1000$ 1000 950 900

Total return of 6000$ in one use 7225$ 6450$ 6750$

Returns from investing each unit of the 6000$ in the best alternative use = 7975 $

Each column shows the marginal returns from investing 1000$ in the use shown.

By investing funds in such a way as to equalize marginal returns we are able to maximize the total net income.

While doing so, we are also giving consideration to opportunity cost. For example, the opportunity cost of investing the fifth unit in hogs is the 1100$ return obtainable from investing it in the best alternative use, feeder cattle.

By following the principles of equimarginal returns and opportunity cost, we found how to invest 6000$

so that it will returns 7975$.We obtained this figure by adding the returns from the 3 units invested in hogs, the unit in feeder cattle, and the 2 units in fertilizer. If all of the funds were spent on hog production, the return would be 7225$, from cattle, it would be 6 450$, and from fertilizer, 6750 $.

If unlimited funds are available, we should invest them to the point where 1000$ will just return 1000$, or perhaps even where 100$ just returns 100$, if we can make such a determination. If the 1000 $ units do not include interest, an allowance should be made for that.

If we are borrowing money at 12 percent interest, we will need a return of 1 120 $ per unit. In that situation, we should invest another unit in hog production and another in feeder cattle.

Fixed and Variable costs

Fixed costs Sometimes called overhead costs, are those costs

that do not change when production changes. In fact, they go on just the same whether one

produces anything or not. Interest and taxes are examples of costs that are

usually fixed. Since total fixed costs do not change, the fixed

cost per unit of production declines as more units are produced.

EX. Annual total fixed costs on a Rubber EstateCost item Amount

General charges-Salaries-Staff allowances

-Staff provident fund-bRmugTuk-Worker provident fund-Wage adjustment-Labour holidays-Food for function-Medical, Sanitation-All insurance-Office supplies- Commission paid-Visiting agent-Rents, tax-Watchmen-vehicle- Retirement gifts

Capital fixed costs

-Housing upkeep-karTMnukbMrug-Labour-lines upkeep-Minor buildings-Road, water supply

Total fixed costs

Farm capital investment inventory and schedule for calculating annual capital fixed costs

Capital item Item value $

Useful life

years

Depreciation

$

repairs

Operating

costs

Interest at 10 %

total

LandHouse, shedsTractorHand thresherCultivatorsLivestock gearBarn -CRgukRsUv

Fences-rbg Dam/pondWater pump Oxen

50 00015 00010 0002 0003 0006005 000

6 0008 0004 0003 000

-251065320

304085

-6001000330600200250

200200500600

-2004005010050200

400-200-

-------

--800200

50001500100020030060500

600800400300

-8001400380700250450

6002001500800

Total 106 600

4 480 1 600 1 000 10 660 7 080

Depreciation method Straight-line depreciation:

Assumes a constant flow of service with consequent wear and tear and /or obsolescence during the item’s useful life so that the value of the item will therefore decrease by a constant uniform annual amount over its life. This is the easiest method to apply. The formula for obtaining the annual straight-line depreciation amount, denoted by D, is

D=(PV-SV)/LPV: item’s present valueSV: expected salvage or residual value at the end of its useful lifeL: expected total years of lifeThe annual depreciation on a cultivator with the present cost of 1000$

and with has an expected total useful life of 10 years and can then be sold for 200$ as scrap would be

(1000-200)/10=80$Or it this has no salvage value,1000/10=100$

Sum of integers depreciation method

Implies a more rapid initial depreciation but slower later depreciation. The formula for sum of integers depreciation is

D=(PV – SV)(remaining year of life)/sum of integers of total life

n(n+1)/2D: is again depreciation for year t PV=o: is the expected replacement value of the

capital item at time zero, i.e. at the start of the depreciation analysis ( present value)

Ex. The cultivator with a life of 10 years and no

salvage value (sv=0), the sum of integers would be (1+2+3+…..+10) or 55 and the annual depreciation cost would be:

1 year: 1000*10/55=181.812 year: 1000*9/55=1643 year: 1000*8/55=1454 year: 1000*7/55=127…………………………..10 year: 1000*1/55=18

Variable or operating costs On the other hand, do change with production. The total annual operating cost will increase

with each increase in production.

Examples of variable cost items are fertilizer, fuel, seed, and hired labor.

Sometime it is difficult to dertermine whether an item is fixed or variable.

Machinary repairs tend to vary with production, indicating the characteristic of a variable cost. However, machines also rust with age, and that aspect illustraites a fixed cost characteristic.

That portion of an operator`s labor commited to farm work may be considered a fixed cost, but it tends to be a variable input if he is willing and able to get part time work off the farm.

Ex. The discussion that follows will be on the fixed and

variable costs of producing 100 acres of corn. Assume that the fixed costs amount to 12600$ per year,

as shown in a previous example, and that they include such things as interest and taxes on land, and labor and fixed costs of machinery for the minimum production shown.

The variable costs will include added labor and machinery costs, as well as fertilizer, for getting higher production.

These costs are shown in table 6.

Table 6. Examples of fixed, variable, and marginal costs in producing corn

Total Total Costs per bushel .

production variabble fixed ä variable total marginal ö(bushel) costs

2 000 3 500 $ 6,30 $ 1,75 $ 8,05 $ ......4 000 5 600 3,15 1,40 4,55 1,056 000 7 200 2,10 1,20 3,30 0,808 000 8 700 1,58 1,09 2,65 0,7510 000 11 200 1,26 1,12 2,38 1,2512 000 14 300 1,05 1,19 2,24 1,5514 000 19 900 0,90 1,42 2,32 2,80

ä - Based on total fixed costs of 12600$ö - Change in variable costs divided by change in production

Where is the most prifitable point in this operation when the price of producing corn equal 2.5 $ per bushel? And 3.0$ per bushel?

Graphic

0

1

2

3

4

5

6

7

8

9

2000 4000 6000 8000 10000 12000 14000

Total cost

Marginal cost

Resource substitution

When we talked about the principles of diminishing returns, equimarginal returns, and opportunity cost, we were concerned with decisions about how much of a single resource to use and where to invest that resource.

When we talked about the concept of fixed and variable costs, we considered all fixed resources as a group and all variable resource as a group.

Now we want to see what happens when two or more resources can be used in varying amounts to produce a fixed amount of product and also when one resource will substitute entirely for another.

We can illustraite the substitution of two resource in varying amounts by data on feeding grain and hay to dairy cows.

These data, published by the University of Kentusky, are shown in table 7.

Table7: Combinations of Grain and Forage for outputs of

10 000 pounds of Milk by good cows a/ Feed

combinationPounds of hay

Pounds of grain

Pounds of grain replaced by 500 poundsof hay

Pounds of grain replaced by one pounds of hay (G/H)

 

ABCDEFGHIJKL

5 0005 5006 0006 5007 0007 5008 0008 5009 0009 500 10000

10 500

6 1255 4504 9254 4504 0503 7003 4103 1602 9252 7002 5402 400

- -675525470400350290250235225160140

- -1,35 1,05 0,94 0,80 0,70 0,58 0,500,470,450,320,28

back to 27 3337 31

Note that, as more hay is added, it replaced less and less grain. This is known as a decreasing rate of substitution. One pound of hay first replaces 1,35 pounds of grain, then 1,05, and so on until the figure finally decreases to 0,28.

The measure, pounds of grain replaced by one pound of hay, is also called the marginal rate of sustitution of hay for grain and is sometimes expressed as MRS hay/grain.

Still another way of expressing it would be ∆G/∆H (change in grain/change in hay).

We suggest that you set up the substitution ratio so that you always have:

change in replaced resource / change in added resource Then your results should clearly illustratete the

decreasing rate of substitution mentioned earlier.

In this type of problem, we are trying to find the lowest cost of producing a gaven amount of product.

In our example, the product is 10000 pounds of milk, and the substituting resources are grain and hay.

We are trying to find the least cost combination of grain and hay that will produce 10000 pounds of milk.

We can, of course, find the costs for each feed combination and then select the one that costs less.

At a price of 5,0 cents per pounds for the grain mixture and 2 cents per pound for hay, the costs for various feed combinations would be as follows:

Feed Pounds of grain replaced Total cost of grain combination by one pounds of hay and hay

A -- 406,25 $B 1,35 382,50C 1,05 366,25D 0,94 352,50E 0,80 342,50F 0,70 335,00G 0,58 330,50H 0,50 328,00I 0,47 326,25J 0,45 325,00 K 0,32 327,00L 0,28 330,00

37 3338

We find that Feed Combination J has the lowest cost.

However, such calculations take considerable time, and we still have not illustrated a principle for solving problems of this type.

Let us do so by trying to find a point where the substitution ratio is equal to the price ratio.

The formular for this condition is: G = PH , or in general terms :

H PG Resource B = Price of A

Resource A Price of B

The ratio PH is equal to 2 or 0,40. PG 5

Let us fit this ratio into the column of figuers representing G .

H

Notice that, in going from I to J, we replace 0, 45 pound of grain with one pound of hay.

This point is near the price ratio of hay to grain but is still above that figuer in the column .

The next , lowest substitution rate, 0,32, is too low to match the price ratio of 0,40.

The substitution ratio of 0,45 best matches the price ratio of 0,40.

The MRS represented by the change from I to J is nearest to the price ratio but is still above it.

Therefore, it pays to make the change from I to J, and J is the least cost combination.

The following data shows where the price ratio fits in

Feed combination Grain

Hay

GH 0,50I 0,47 PH = 2,0 = 0,40J 0,45 PG 5,0K 0,32

When we find the point indicated by the formula, we know that we have found the lowest possible cost.

If we move upward in the table, the cost will be higher; if we move downward, it will also be higher.

If the substitution ratio is exactly equal to the price ratio, we have a '' toss-up'' situation. we are trading dollars, because the value of the hay added is just equal to the value of the grain saved.

If the price ratio had been 0,45, I and J would have been equal in cost, lower than any other points in the table.

But since 0,40 is below the substitution ratio of 0,45, J costs less than I.

If the price ratio had been 0,33., J would still be the least cost combination because the MRS of hay for grain must be equal to or just above the price ratio.

31

As long as the MRS stays above the price ratio following a change., it indicates that the value of the replaced resource was greater than the value of the added resource, and the cost of the resource combination has decreasesed.

Here are additional examples. If PH = 0,25 PG

Such as at prices of 1 cent per pound for hay and 4 cents per pound for grain. Feed combination L would be most economical.

If PH = 1,0 PG

such as at prices of 2,5 cents per pound for hay and 2,5 cents per pound for grain, Feed Combination C would be most economical.

However, if PH = 1,1 PG

such as at price of 3,3 cents per pound for hay and 3 cents per pound for grain, Feed Combination B would be most economical. It would not pay to make the change to C indicated by an MRS below the price ratio. ( Refer back to Table.7).

31

We will use a different example to show how to solve a resource substitution proplem. Let us say that we can produce $4000 worth of a product with any of the following combinations of labor and capital:

Capital Labor(units) (units)

1 8,02 4,53 2,84 2,05 1,46 1,27 1,1

Also, let us say that the price of capital is $200 per units and the price of the labor is $400.

What is the least combination of labor and capital in this situation?

Sometimes you may think of examples where the substitution rate does not appear to vary, no matter how much of one or the other resource is used. For example, if one unit of B always replaces two units of A, this is know as a constant rate of substitution.

Ex. A specific example of a constant rate of substitution

is the substitution of a larger machine for a smaller one. A 6-plow tractor may substitute for a 4-plow tractor at the rate of 1,33; that is, the 6-plow tractor does 1,33 times as much work as the 4- plow tractor. It makes no difference whether the larger tractor replaces the smaller one for one month's work, two month's work, or all of it; the larger one continues to substitute for the smaller one at the same rate.

Combinations of elemental phosphate and nitrogen that will produce 3500kg/ha corn

Combination N(kg) P (kg)

123456789

1011

556065707580859095

100105

625040272014108520

sMNYr³ etI Combination mYyNaEdlmantMélefakCageK kñúgkrNIEdl Nitrogen éfø 5 dollar

kñúg 1 Kg and

Phosphate 3 dollar kñúg 1 Kg?

]TahrN_eTAelIkMriténkareRbIR)as;RbePTCIepSgKñarbs;ksikrenAe

xtþéRBEvgPhilipine (kg/ha) Urea (kg/ha) TinñplRs

Uv (kg/ha)

2525

25100

560600

50505050

2550100150

584684674780

757575757575

255075100125150

50162076560025602400

100 255075100150

4138013627711120

125 2550125167250

66715831414300450

150 255075100150

14401100102910801597

175 5075125175

160011401200600

200 255075100200

4009207729601680

225 100 1527

250 5075250

9001800480

275 25 320

300 50100200

21827202400

450 200 1333

kñúgkrNI CI Philipine éfø 1200r

Urea 1000r nig RsUvéfø 600r kñúg 1 KILÚ etIksikrKYreRbI combination mYyNaEdlmanRbsiT§PaB esdækic©x<s;CageK?