QUANTITATIVE TECHNIQUES QUESTION ONE (a) The main purpose of input-output analysis is to analyze the interdependence of various segments of an economy. According to this technique, an economy is dividend into two broad sectors, the sources, or output or producers‟ sector and the destinations or input or users‟ sectors. The producers‟ sector includes: (i) The section that manufacturers goods (ii) The section that provides services. These goods and services are intended for: (i) Industrial usage and are called Industrial goods and services. (ii) Final consumption and these are termed consumer goods and services. The output for each industry from each industry is either for industrial use or for final consumption. Apart from industrial outputs, the other output constitutes payment for services rendered. These services are obtained from: - Government - Imports - Depreciation of capital goods - Household services e.g. wages and salaries. The model aims at indicating how products produced from different sectors of the economy are consumed by the industrial sector and the final consumption sector. The model therefore indicates how the independent sectors of economy interact. The interaction of these sectors is a necessary information for planning purposes so as adequate stock of goods can be produced by the economy. A device to assist in such planning in the technical coefficients. (b) (i) The technical coefficients are calculated for each industry to help in the planning or projections. A technical coefficient is the ratio for which the denominator is the total input for an industry and the numerator is one of the inter-industrial elements. In this case, the technical coefficients in input-output matrix are constructed as below:
Transcript
QUANTITATIVE TECHNIQUES
QUESTION ONE
(a) The main purpose of input-output analysis is to analyze the interdependence of various segments of an economy. According to this technique, an economy is dividend into two broad sectors, the sources, or output or producers‟ sector and the destinations or input or users‟ sectors. The producers‟ sector includes: (i) The section that manufacturers goods (ii) The section that provides services. These goods and services are intended for: (i) Industrial usage and are called Industrial goods and services. (ii) Final consumption and these are termed consumer goods and services.
The output for each industry from each industry is either for industrial use or for final consumption. Apart from industrial outputs, the other output constitutes payment for services rendered. These services are obtained from: - Government - Imports - Depreciation of capital goods - Household services e.g. wages and salaries. The model aims at indicating how products produced from different sectors of the economy are consumed by the industrial sector and the final consumption sector. The model therefore indicates how the independent sectors of economy interact. The interaction of these sectors is a necessary information for planning purposes so as adequate stock of goods can be produced by the economy. A device to assist in such planning in the technical coefficients.
(b) (i) The technical coefficients are calculated for each industry to help in the
planning or projections. A technical coefficient is the ratio for which the denominator is the total input for an industry and the numerator is one of the inter-industrial elements. In this case, the technical coefficients in input-output matrix are constructed as below:
A B C
108
18
120
30
180
45
A = 108
27
120
30
180
60
108
54
120
40
180
60
A B C
6
1
4
1
4
1
A = 4
1
4
1
3
1
2
1
3
1
3
1
(ii) The matrix (I – A) where I is the identity matrix of a given order.
A – the matrix of technical coefficients is called Leontif matrix in honour of the German Mathematician who invented it.
The projection requirement is: M = PM + N ……………… (1)
Where M - is the vector matrix for output of the various industries.
P - technical coefficient matrix. N – final demand matrix. From (I): M – PM = N M (I – P) = N M = (I – P)-1N i.e. the production for each industry = (inverse of Leontif Matrix) x (Final Demand)
Applying this expression, the production of the sectors A, B and C, given the matrix (I – A)-1, is computed as follows:
336 216 234 10
M = 109
1 288 372 294 2
396 348 486 20 8472 77.7248
i.e. A 4204 = 38.5688
B = 109
1 14376 131.8899
C
(iii) The Intermediated demand for the sectors in this case will be as follows: A - Sh. 77.7248 Billion B - Sh. 38.5688 Billion C - Sh. 131.8899 Billion.
QUESTION TWO
(a) The subject of calculus is an area of mathematics called the classical optimization model. It has many business and economic applications. It is concerned mainly with optimization i.e. the best combination of variables that will either maximize benefits or minimize losses. In the context of maximizing contribution, calculus is used in determining the level of production that will maximize profit or contribution. This is achieved through the concept of the rate of change in y, the dependent variable, given a change in x, the independent variable. Once a functional relationship between these two variables is established, then through the process of differentiation an optimal output level is determined.
In calculas if y = f(x), then at maximum or minimum the rate of change of „y‟
with respect to x will be Zero. For instance, suppose the Revenue “R” and Quantity “Q” are related by the
formula: R = 15q – ¾ q2 Then at optimal production level:
q
R
= 15 -
2
3q = 0.
15 = 2
3q or q =
3
15x2= 10
So that maximum revenue should be achieved at a production level of 10 units. The same concept is used for minimization of costs. Suppose costs are related
to the units produced as follows: C = x2 – 10x + 200.
Then x
C
= 2x – 10.
At minimum level, x
c
= 0, 2x – 10 = 0
2x = 10 X = 5 The cost will be minimized at a production level of 5 units. (b) (i) Profit will be maximized at a sales level in which marginal Revenue = Marginal Cost.
Company A sells goods at Sh. 200 per unit to company B. The unit selling price to B is fixed at Sh. 200/=. Let “q” be the optimal sales unit. Then the Revenue function for firm A is: R = q x 200 = 200q.
And Marginal Revenue, q
R
= 200.
The cost function for firm A is: C = 2q2 + 40q + 80
The marginal cost is then: 404qq
R
At maximum profit: q
c
q
R
i.e. 200 = 4q + 40 160 = 4q 40 = q. A will maximize its profit at a weekly sale of 40 units. Revenue function for firm B R = 1000q – 16q2. At weekly purchase of 40 units:
R = 1000 x 40 – 16 x (40)2 R = 40,000 – 25,600 R = 14,400 per week.
Weekly cost: C = 2q2 + 80q + 400 = 2 x (40)2 + 80 + 40 + 400 = 3,200 + 3,200 + 400 = 6,800. Add cost of sales: 40 units @200 = 8,000 Total cost 14,800 Less Revenue 14,400 Weekly loss (400)
(ii) Company B will maximize profit if its Marginal Revenue = Marginal Cost. Given the TR function R = 1000q – 16q2, the MR function is obtained as follows:
MR = q
R
= 1000 – 32q………….. (1)
Given the Total Cost, C = 2q2 + 80q + 400, the MC function is obtained as follows:
MC = q
C
= 4q + 80
Equating the two functions and solving for q:1000 – 32q = 4q + 80. 920 = 36q 25.6 = q q = 25 units A weekly sales of 25 units will maximize the profit of firm B.
(iii) When the two firms merge, and in the absence of any tangible benefits to be brought by the merger, we simply add their total revenue and cost as if they were operating individually. We can then apply calculus to establish profit maximizing output and weekly profit as shown below:
Total Revenue: From A Ltd = 200q per week From B Ltd = 1000q – 16q2
Total R = 1200q – 16q2
MR = q
R
= 1200 – 32q
Total cost: A Ltd = 2q2 + 40q + 80 B Ltd = 2q2 + 80q + 400 Total cost C = 4q2 + 120q + 480
MC = q
c
= 8q + 120
At maximum q
R
q
c
8q + 120 = 1200 – 32q 40q = 1080 q = 27. Profit maximizing output = 27 units per week. The weekly profit is established by substituting q = 27 in the profit function P = TR – TC. First, Total Revenue = 1200q – 136q2 = 1200 x 27 – 16 x (27)2 = 20,736 And Total Cost = 4q2 + 120q x 480 = 4 x (27)2 x 120 x 27 + 480 = 6,636 Weekly profit 14,100
QUESTION THREE
(a) (i) This is a case of Normal distribution. A normal distribution is one suited to
describe continuous random variables. It is appropriate in many situations where measurement is involved, such as weight, volume, heights of people and the life times of electrical goods. The mean and the standard deviation are also given. Unlike Normal distribution, the PQ Binomial distribution has its mean given as μ = np.
Where P – probability of success. n – number of trials.
Variance for the Binomial distribution 2δ = npq.
Where q – is the probability of failure while p and n are as explained above. The poisson distribution has a constant mean “ λ ” per unit interval with the variance equal to the mean “x λ ”
(ii) For normally distributed function, N(μ , ), the standard values
Z =
μx
Where - is the standard deviation. μ - is the mean.
x - is the observation.
In this case, Z = 160
800600 = - 1.25
And from the normal tables: Required Area The required area = 0.5 – 0.3944 = 0.1034 -1.25 µ The area under the curve represents the probability or the proportion of events. In this case, the percentage components to be replaced is 10.3470.
(iii) The shortest life is 1% = 0.01 0.49 0.01 0.01
From the normal table, an area of 0.49 under the table corresponds to a Z value of 2.33. But since the Bulb with the shortest life is required, this is below the mean and must take a negative value. The value to be used as the guaranteed lifetime for bulbs, denoted as x, can be obtained from the expression for the Z value of -2.33 as follows:
Z = δ
μx
-2.33 = 160
800x
-2.33 x 160 = x – 800 800 – 2.33 x 160 = x 427.2 The guaranteed life time should be 427 Hours.
(iv) When all possible samples of size n are drawn from a population mean
with a mean μ and a standard deviation δ , then the means of the samples
have a probability distribution known as the sampling distribution of means with mean equal to the population mean μ , and a standard
deviation = n
δwhere δ is the population standard deviation from which
the samples are drawn.
The standard deviation of the sampling distribution of the mean is known as the standard error of the mean. In this case a sample of size
25 will have a mean 800x Hours and a standard error of the mean =
32.5
160
25
160
Then using the standard values, Z = 1.5632
800850
nδ
xx
or 2
We can establish the desired probability as follows: Area required 0 2 From the Normal table a Z value of 2 corresponds to an area = 0.4772, But we require the probability that it exceeds 850 Hours. This is the area to the Right of Z = 2 which is equal to 0.5 – 0.4772 = 0.023.
(b) (i) This is a case of Binomial distribution. This is because the interest is
success or failure and the situation deals with discrete variables and not continuous variables. The probability of success here, is P = 0.15, getting a perishable product. The probability of failure is then q = 1.00 – 0.15 = 0.85.
(ii) The commodity will be purchased if not more than 2 perishable items
are found in sample of 10. The probability of not more than 2 perishable can be written as:
Probability of 0 perishable, 1 perishable or 2 perishable. The probability of 0 perishable, 1 perishable and 2 perishable are obtained using the binomial expression as follows: The binomial probability expression states that the probability of event (x) written P(x) = nCxp
The probability of buying the consignment is 0.820 (c) (i) This is a case of Poisson distribution. It deals with discrete variables and
the probability of a given number of events happening within a specified time interval. The poisson distribution also deals with success or failure but within a specified time interval. This is poisson distribution with a mean “ λ ” of 2 per 10 seconds or 0.2 per second.
(ii) The probability of an event occurring is computed using the following
formula for a poisson distribution.
P(x) = x!
λe
xλ
The probability of more than 3 cars can be expressed as follows:
P(x>3) = 1 – probability of 0, 1, 2, and 3 cars passing. Using the Poisson formula, we get:
P(0) =
0!
e(1.2)0.20
0.818
P(1) =
1!
e(0.2)0.21
0.164
P(2) = !
.(
2
e0.2)0.22
= 0.016
P(3) = 3!
e(0.2)0.23
= 0.0011
0.9991 Probability of more than 3 cars = 1 – 0.9991 = 0.0009
QUESTION FOUR
(a) A hypothesis is some testable belief or opinion, and hypothesis testing is the process by which the belief is tested by statistical means. The process of hypothesis testing is also called significance testing. A null hypothesis is the statement of what we believe is the truth. On the other hand, an alternative hypothesis is the negation of what we believe.
For instance, assume a machine fills packets with spice which are supposed to
have a mean weight of 40 grams. A random sample of 36 packets is taken and the mean weight is found to be 42.4 grams with a standard deviation of 6 grams. We could be required to conduct a hypothesis test, also known as significance test, at say 5% level of significance. Once a level of significance is chosen then we are saying that our belief should be correct 95% of the times. We then set our hypothesis as:
Null hypothesis: H0; μ= 40 gm.
Alternative Hypothesis; HA; µ ≠ 40 gm. It is the null hypothesis which is tested. If we find it to be true, then we accept
it (H0) and reject alternative hypothesis (HA). If H0 is accepted then our conclusion is that the test is as expected and nothing unusual has happened. If on the other hand, we reject the Null hypothesis H0, then we must accept the alternative, which we did not believe in. We say the test is significant.
Assuming a normal distribution, 5% level of significance or 95% level of confidence means that out of our observations, 95% of them should be found with two standard deviation above or below the means.
In this case: Z = S.D.
μx
Z = 1
40x
± 1.96 ± x – 40 40± 1.96 = x 38.04 ≤ x ≤ 41.96
Recall: standard error of the estimate x
S = μ
δ=
36
6= 1.
This means that if our claim is true, i.e. H0 : µ = 40, then the mean of any
sample drawn should be found in the range of 38.04 ≤ x ≤ 41.96. Since the mean of the sample drawn in this case is 42, which is greater than
41.6, our claim is false, so we reject the Null hypothesis (H0) and accept the alternative hypothesis HA.
(b) The paired “t” statistics is used to test the difference between two means in
cases where the same persons are used to carry out the same job or same event but using different tools. In such a case, the difference between two means of the samples drawn from the same population cannot be used because the difference could be due to the faults in the tools. For example, assume that the table below gives the time taken by eight typists to type the same number of words using two different typewriters. We wish to find whether the data indicates any difference in speeds for the two typewriters.
Typist A B C D E F G H (X) Time in min:
using typewriter 1
6.3 4.5 7.1 8.4 3.7 3.9 4.7 5.2
(Y) Time in min: using typewriter 2
5.1 4.4 6.2 7.3 4.5 4.0 3.6 5.1
In this case, it would be wrong to calculate the mean for type variations between the two the typists could swamp any difference due to the machines. Instead we compute the difference “D” for each typist because if there is no difference between the machines then we would expect the mean difference “D” to be Zero. The computation is as shown below: Typist di = xi – yi di
2 A 1.2 1.44 B 0.1 0.01 C 0.9 0.81 D 1.1 1.21 E - 0.8 0.64 F - 0.1 0.01 G 1.1 1.21 H 0.1 0.01 3.6 5.34
d = 8
3.6= 0.45 min
sd =
2
8
3.6
8
5.34
= 0.68 min.
We then test: H0, μ= 0
H1 µ ≠ 0
tn-1 = 1.75.70.68
0.450
1nsd/
0.450
This is not significant at 5%. We retain H0 and conclude that there is no difference in speed between the two machines.
(c) A parametric equation is an equation indexed by a quantity called a parameter.
To illustrate let t = 4
5,y
5
2x
where t is a parameter. The term parameter is
therefore the unknown to be found. Parametric test therefore deals with situations where the unknown function has been estimated using some parameter and we wish to test whether the parameter really represents the unknown. For example, the sample mean being used to estimate the population mean or the sample standard deviation being used to estimate the population standard deviation. The parametric tests are those tests which are
carried out to draw inferences about the population using a parameter and the assumption is that the population from which the sample is drawn is normally distributed or is near normal.
On the other hand, where the experimenter does not know the exact form of
the distribution of the population about which the inference is about to be made, then one needs to design statistical techniques which are applicable regardless of the form of distribution. Such tests are called non-parametric tests, e.g. the likelihood ratio.
(d) The least squares method of regression is a technique of estimation in which
the aim is to minimize the error margin or the error term. If the function is
assumed linear between two variables then an equation in the form y = a + bx
can be used a predictor of the observed relationship. For instance assume that observed data for total cost and direct labour hours was as follows.
(y) (x) Months Total cost Direct Labour
Hrs 1 2 3 4 5
11,400 14,550 19,800 29,000 28,200
3,000 5,000 7,000 11,000 12,000
A relationship between x and y can be found in the form ŷ = a + bx in which
the error term (y – ŷ) is minimized. In essence, the equation is based on:
y - y = (y – ŷ) x (ŷ - y )
Squaring both sides and summing:
2
yy ∑(y – ŷ)2 + 2∑(y – ŷ) (ŷ - y ) + ∑(ŷ - y )2.
Then middle term is zero because ∑ (ŷ - y ) = 0
∑(y - y )2 = ∑(y – ŷ)2 + (ŷ - y )2 ………………………(2)
The error term, is (y – ŷ) which is to be minimized. The Regression Equation is obtained by solving simultaneously:
Y = a + bx …… (3) ∑y = na + b∑x …….. (4) ∑xy = a∑x + b∑x2 … (5) By solving (4) and (5) the regression equation is obtained.
QUESTION FIVE
(a) (i) A basic trend Quantitative forecasting models assume that the time series follow some patterns which can be extrapolated into the future. Such patterns that can be used for future extrapolations are known as basic trends. For instance, a linear trend model forecasts a straight line trend for any period in the future. Exponential trend forecasts that the amount of growth will increase continuously.
(ii) Seasonal fluctuations
If one were to closely study employment and output data in say the agricultural sector, one would easily note a definite relationship between output and season or between employment and the time of the year. Similarly, the sale of success cards will normally be associated with examination times and are therefore subject to large seasonal variations. Performances that vary with seasons are known as seasonal fluctuations.
(iii) Cyclical Fluctuations This consists of those fluctuations in a time series which do not repeat themselves periodically like seasonal variations. Cyclical fluctuations will show periods of rapid growth followed by those of slower growth and they will normally be of different duration and intensity.
(iv) Residual variations Residual variations are also known as random or irregular variations. This component of time series is due to purely random and irregular factors.
(b) (i) The trend and seasonal variations are calculated as shown below:
Using the two normal equations: ∑y = na + b∑x…………………………....…………… (1) ∑xy + a∑x + b∑x2…………………………………….. (2) 320.7 = 16a + 105b x 105 2,264.7 = 105a + 1015b x 16…………… (1) 33,675.5 = 1680 + 11,025b………….…… (2) 36235.2 = 1680 + 16,240b 2561.7 = 5,215b 0.4912 = b 16.82 = a ŷ = 16.82 + 0.4912x The estimated value of the demand is represented by equation; y = 16.82 + 0.4912x We now calculate estimated demand for hotel accommodation using the formula
ŷ = 16.82 + 0.49x, and express the actual accommodation realized as a percentage of the estimated accommodation as follows:
The estimated accommodations represents the trend in demand for hotel accommodation. The average seasonal variation of the series trend is obtained by re-arranging the quarterly estimated demand as shown below.
Quarters. Year Q1 Q2 Q3 Q4 1995 1996 1997 1998 Total Average
73.68 80.36 100.48 86.20 340.72 85.18
76.55 81.03 84.18 90.40 332.16 83.04
73.92 77.46 80.01 - 231.39 57.84
81.46 85.80 87.19 - 254.45 63.125
(ii) The average seasonal variation can be used in forecasting future demand in
hotel accommodation by using the estimated data for the past 4 years seasonally adjusted for trend variations. This is shown in the table below:
Year Quarters x Accommodation Seasonally Adjusted forecast
1 1 19.4 26.33 x 100
85.18
= 22.43 2 20.6 26.91 x 0.83
= 22.35 3 19.5 15.26 4 22.8 17.67 2 5 22.3 23.64 6 22.6 23.16 7 21.0 16.68 8 24.9 18.32 3 9 23.3 28.23 10 24.1 23.77 11 22.2 16.02 12 25.6 18.53 4 13 25.1 24.80 14 27.3 25.80 15 - - 16 - - To illustrate, the future estimate for the 1st quarter of year 1 is computed as
follows: = Trend estimate x percentage seasonal variation.
= 26.33 x 100
85.1
= 22.43 SECTION II QUESTION SIX
(a) The use of Network analysis is necessary because of the following reasons: (i)
Plan how to get started on the project and identifying the activities of the project.
Identify and determine the time duration for each activity‟ (ii)
Identify the critical activities of the project that require considerable attention.
Determine the total completion time to the project if every thing goes according to plan.
(ii) Normal cost = 46,100. (iii) Duration of the project is 33 days.
Activity Crash Time 1 - 3 3 - 4 4 - 7 6 - 7
3 8 1 ___6__
Total 18 days
(iv) Items of crash per unit time, we have: timeinChange
costinChange
Activity ∆C/∆ Time 1 – 2
26
4,00012,000
= 2,000
1 – 3 38
3,0006,000
= 600
2 – 4 47
2,8004,000
= 400
3 – 4 812
9,00011,000
= 500
4 – 6 13
10,00013,000
= 1,500
5 – 6 25
4,9007,000
= 700
3 – 5 37
1,8005,000
= 800
5 – 7 511
6,60012,000
= 900
6 -7 610
4,0008,400
= 1,100
The activities to be crashed are: 3-4, by 4 days @ 500 = 2,000 1-3, by 5 days @ 600 = 3,000 6-7 by 6 days @ 1,100 = 6,600 Total crash cost 11,600 Optimal duration of the project is 18 days. i.e. (33 – (4 + 5 + 6)) (v) The total cost is: Normal cost - 46,100 Add crash cost - 11,600 Direct cost. - 57,700 QUESTION SEVEN
(a) Sensitivity analysis is linear programming is concerned with the determination of the extent to which the contribution to the objective function and the constraints or the limiting resources of each constraint can change both adversely and favourably so as to sustain the current optimal solution.
e.g. Max Z = 8x1 + 6x2 ← Objective function Subject to (S.t) 2x1 + 3x2 ≤ 18 x1 + 5x2 ≤ 24 Let the optimal solution be x1 = 4, x2 = 3 The role of sensitivity analysis is to determine the extent to which: (i) Objective coefficient can change both favourably and adversely so as to
retain the existing solution.
(ii) The right hand side constraints, in this case, 18 and 24 can change both favourably and adversely to retain the existing solutions.
The dual price or the shadow price is the increase per unit limiting resource in the objective function if such limiting resources are increased by one unit. For instance, if a limiting resource in a given department is 120 hours with a shadow price of Sh. 200 then, this means that if the company increases the number of labour hours in the department by 1 unit (1 hour) then the objective function will also increase by Sh. 200 per hour. The dual price therefore is the maximum amount at which additional limiting resources can be acquired per unit.
(b)
UNIT USAGE
386 286 486 TOTAL
DESK TOP
DESK TOP
LAPTOP HOURS AVAILABLE
Production Hours Chip No. 80386 Chip No. 80286 Unit Profit
5 1 5,000
4 1 3,400
3 1 3,000
2,000 120 400
Let x1, x2 and x3 be the number of Desktop 386, Desktops 286, and Laptop 486 produced per month.
FORMULATION Max Z = 5,000x1 + 3,400x2 + 3,000x3 Subject to (S.t.) 5x1 + 4x2 + 3x3 ≥ 2,000 x1 ≥ 120 x2 + x3 ≥ 400 x1 + x2 ≥ 500 x3 ≥ 250 x1, x2 and x3 0 ≥ 0 ASSUMPTION The flowing assumptions relate to the linear programming problems:
(1) There is a linear relationship between the variables required to relate inequalities or the constraints.
(2) That the company is in a position to adjust the limiting resources within the required ranges established by the sensitive analysis.
(3) Any optimal solution obtained in the linear programming will satisfy all the constraints.
(c) (i) Optimal product mix x1 = 120 i.e. produce 120 Desktop 386 per month. x2 = 200 i.e. produce 200 Desktop 286 per month x3 = 200 i.e. produce 200 Desktop 486 per month Monthly profit to be obtained from the objective function Max Z = 5,000x1 + 3,400x2 + 3000x3 Monthly profit = 5,000 (120) + 3,400 (200) + 3,000 (200) = 600,000 + 680,000 + 600,000 = 1,880,000
Unused Resources – The resources are represented by the following inequalities:
5x1 + 4x2 + 3x3 + S1 = 2000 Where S1 = Slack variable Monthly usage. 5(120) + 4(200) + 3(200) + S1 = 2,000 S1 = 2,000 – 2,000 S1 = 0 S1 = 0, i.e. available monthly hours have been completely utilized.
The production hours is therefore a limiting resource hence the shadow price or a dual price of Sh. 20 (constraint No. 5). This means that if the company increases the production hours by 1 hour per month, then the contribution will increase by Sh. 20 per hour made available. x1 + S2 = 120 Monthly production = 120 + S2 = 120 S2 = 120 – 120 S2 = 0 The chips for making laptop 486 have been fully utilized and is a limiting resource with actual price of Sh. 150 with similar interpretation. x2 + x3 + S3 = 400
Monthly production = 200 + 200 + S3 = 400 S3 = 0 i.e. The chip for making Desktop 286 and Laptop 486 have been completely utilized and is a limiting resource with a shadow price of 90. All the available resources have been completely utilized. Unutilised resource is a resource that remains in excess of what is required to produce the optimal product mix. All unused resources have zero shadow price.
(ii) The lower and the upper limit for the basic variables as well as the
constraints are representatives of sensitivity analysis required to sustain an optimal product mix. For the basic variables x1, x2 and x3 the sensitivity analysis required is known as the objective coefficient range meaning that the unit contribution function can change both adversely and favourably up to a certain limit in order to sustain the current optimal product mix. Similarly, for the constraints, the sensitivity analysis required is known as Right Hand side Ranges which represent upper and lower limits for each constraint required to sustain the current level of optimal product mix.
If each of the constraints changes beyond the limits given in the sensitivity analysis, then the current optimal product mix will have to change.
(iii) The C.P.U 80386 is a limiting resource with a shadow price of Sh. 150. This means that if one more unit of the resource is made available, the objective function will increase by Sh. 150. The total increase in the objective function will therefore be equal to 10 more units @ 150 = 15,000.
QUESTION EIGHT
(a) (i) Consider a competitive situation for the share of a news paper market by publishers of this daily newspapers in a city. Here, every additional circulation gained by one of the firms is necessarily at the cost of the other. This situation is called Zero sum game. This is so because whatever is done by either competitor the sum of the net gain in the market share by the two competitors is Zero. Thus in a two – person zero sum game, the interest of the two persons are opposite and that the
sum of the gains of one is exactly equal to the sum of the losses of the other, or the sum of the gains for the two to add to zero. However, there are situations where the decisions which are taken by the firms may very well increase or reduce both the absolute size of their market and the total profits taken together. Such problems are called non-zero sum games.
(ii) The saddle point. This occurs in solving a game using the minimax and
maximin approach. It is the point at which the maximum among the minimum pay off is equal to the minimum of the maximum pay off. Example.
8 7 15 12
Let the matrix T = 9 14 8 10
10 12 14 13
be the pay off of player A and B. Player A takes strategies to minimize his losses for each row while player B aims at making the minimum of his maximum pay offs. i.e. Row minimum 8 7 15 12 7
9 14 8 10 8
10 12 14 13 - Max Row Column maximum 12 15 13
Min column. The maximum of the minimum Row is 10 and the minimum of the maximum columns is 10. Thus 10 is the saddle point. If the saddle point exists as in this case, the optimal strategies for player A and B is the Row and column containing the saddle point.
(iii) Pure and Mix strategy
10
10
The pure strategy occurs whenever there is a saddle point. If a saddle point exists, the each player must play a particular strategy so as to minimize the losses. Like in the above example, player A had to play strategy 3 and player B, strategy 1. Mixed strategy on the other hand occurs when there is no saddle point. Any player can adapt any strategy in order to win. Example: Row M in 20 8 -6 6
12 10 2 2
3 5 6 - Optimal
Column max: 20 10 optimal No saddle point. In this example, player A will decide not to use his 1st strategy but will choose his 2nd and 3rd strategy at random, each with probability of ½. Player B can also adopt such a move by never playing the first strategy.
(iv) The concept of Dominance
A game in which a player has more than two choices while the other player is limited to choices is denoted as a 2 x M or M x 2 game. The concept of dominance is said to occur if the elements in a column are greater than or equal to the corresponding elements in another column. The column with greater or equal to element is said to dominate. Check for Dominance in the matrix below:
(v) The Value of the Game The value of the game is the resultant solution by the players. For instance where there is saddle point, the value of the game is the saddle point. It is the result for each player. Alternatively, we can reduce a M x 2 or 2 x M to 2 x 2 and apply the probabilities to obtain the score for each player which now constitutes the value of the game. A graphical solution can also be used to obtain the game value.
(b) (Wife) W 1 2 10 5
3
6
Man 1 10 5 = M
2 10 2 10 2
Strategy 1 is for the man and strategy 2 is for the lady. (c) The assumptions and limitations of the game theory are: (i) Each participant knows about the pay off for each strategy.
(ii) That each participant will choose a strategy to minimize his losses since he is aware of the other person‟s strategy.
