Quantitative Analysis | Business Statistics MGT 202

TU BBS MGT 202 Unit 11 board exam questions on decision making under uncertainty and risk (payoff tables, EMV, EVPI) and graphical linear programming (2078–2083).

Oct 28, 2026

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This unit covers quantitative analysis for decision making: decision environments, payoff and regret tables, decision making under uncertainty (maximax, maximin and minimax regret criteria) and decision making under risk (EMV, EPPI and EVPI), with worked stocking problems. Board exam questions, including graphical linear programming, are listed at the end by exam year, with marks in brackets.

Quantitative Analysis and Decision Making

The techniques used in analysing a problem which involve the use of numbers, symbols and mathematical expressions are called quantitative analysis or quantitative techniques.

Quantitative techniques are those statistical or programming techniques which provide the decision maker with a systematic and powerful means of analysis, based on quantitative data, in exploring policies for achieving pre-determined goals.

Decision making, also known as decision analysis, is the selection by the decision maker of the optimal strategy from the various options available to him.

Types of Decision-Making Environment

Decision making under certainty
Decision making under risk
Decision making under uncertainty

Payoff and Regret

Payoff: the benefit obtained from a given combination of decision alternative (strategy) and state of nature.

Regret value: a quantitative measure of loss obtained by subtracting a particular payoff from the highest payoff, both belonging to the same state of nature.

Decision Making under Uncertainty

When probabilities of the states of nature are not known, one of the following criteria is applied to the payoff (or regret) table.

Criterion	Attitude	Rule
Maximax	Optimistic	Find the maximum payoff of each strategy; select the strategy with the maximum of these maxima.
Maximin	Pessimistic	Find the minimum payoff of each strategy; select the strategy with the maximum of these minima.
Minimax regret	—	Build the regret table; find the maximum regret of each strategy; select the strategy with the minimum of these maxima.

Payoffs in Stocking Problems

When a trader stocks $S$ units and demand is $D$ units, the payoff is:

$$\begin{aligned} \text{Payoff} &= \text{Units sold} \times MP && \text{if } D \ge S \\ \text{Payoff} &= \text{Units sold} \times MP - \text{Units unsold} \times ML && \text{if } D < S \\ MP &= \text{Selling price (SP)} - \text{Cost price (CP)} \\ ML &= \text{Cost price (CP)} - \text{Salvage value (SV)} \end{aligned}$$

where $MP$ is the marginal profit and $ML$ is the marginal loss per unit. For example, with CP = Rs. 10, SP = Rs. 15 and SV = Rs. 1: $MP = 15 - 10 = \text{Rs. } 5$ and $ML = 10 - 1 = \text{Rs. } 9$.

Decision Making under Risk

When the probabilities of the states of nature are known, the strategy with the maximum expected monetary value is selected.

$$\begin{aligned} \text{EMV (strategy)} &= \sum (\text{Probability} \times \text{Payoff}) \\ \text{EPPI} &= \sum (\text{Probability} \times \text{Diagonal payoff}) \\ \text{EVPI} &= \text{EPPI} - \text{Maximum EMV} \end{aligned}$$

EMV — expected monetary value of a strategy.
EPPI — expected profit with perfect information: the best payoff for each state of nature (the diagonal of a stocking payoff table), weighted by its probability.
EVPI — expected value of perfect information: the maximum amount worth paying for perfect information.

Worked Examples

Example 1: Maximax, Maximin and Minimax Regret

Strategy	$N_1$	$N_2$	$N_3$	Maximum	Minimum
$S_1$	70	30	15	70 (max)	15
$S_2$	50	45	10	50	10
$S_3$	30	40	20	40	20 (max)

Regret table (column maxima 70, 45, 20):

Strategy	$N_1$	$N_2$	$N_3$	Maximum regret
$S_1$	0	15	5	15 (min)
$S_2$	20	0	10	20
$S_3$	40	5	0	40

Maximax: maximum of maxima is 70 → select $S_1$.
Maximin: maximum of minima is 20 → select $S_3$.
Minimax regret: minimum of maximum regrets is 15 → select $S_1$.

Example 2: Stocking Decision (Demand 10–14 Units)

Payoff table (Rs.), with stock $S$ and demand $N$:

Stock	$N$ = 10	11	12	13	14	Maximum	Minimum
$S_1$ (10)	400	400	400	400	400	400	400 (max)
$S_2$ (11)	380	440	440	440	440	440	380
$S_3$ (12)	360	420	480	480	480	480	360
$S_4$ (13)	340	400	460	520	520	520	340
$S_5$ (14)	320	380	440	500	560	560 (max)	320

Regret table (column maxima 400, 440, 480, 520, 560):

Stock	10	11	12	13	14	Maximum regret
$S_1$	0	40	80	120	160	160
$S_2$	20	0	40	80	120	120
$S_3$	40	20	0	40	80	80
$S_4$	60	40	20	0	40	60 (min)
$S_5$	80	60	40	20	0	80

Maximax: 560 → select $S_5$ (stock 14).
Maximin: 400 → select $S_1$ (stock 10).
Minimax regret: 60 → select $S_4$ (stock 13).

Example 3: CP Rs. 24, SP Rs. 31, Demand 33–36

Salvage value is 0, so $MP = 31 - 24 = \text{Rs. } 7$ and $ML = 24 - 0 = \text{Rs. } 24$. For example, $D = 33, S = 33$: $33 \times 7 = 231$; $D = 33, S = 34$: $33 \times 7 - 1 \times 24 = 207$.

Stock	$N$ = 33	34	35	36	Maximum regret
$S_1$ (33)	231	231	231	231	21
$S_2$ (34)	207	238	238	238	24
$S_3$ (35)	183	214	245	245	48
$S_4$ (36)	159	190	221	252	72

Regrets are found from column maxima 231, 238, 245, 252 (for example, $S_2$ at $N = 33$: $231 - 207 = 24$). Maximax selects $S_4$ (252), maximin selects $S_1$ (231), and minimax regret selects $S_1$ (21).

Example 4: SP Rs. 200, CP Rs. 140, Demand 12–17

$MP = 200 - 140 = \text{Rs. } 60$, $ML = 140 - 0 = \text{Rs. } 140$.

Stock	12	13	14	15	16	17	Maximum	Minimum	Maximum regret
$S_1$ (12)	720	720	720	720	720	720	720	720 (max)	300
$S_2$ (13)	580	780	780	780	780	780	780	580	240 (min)
$S_3$ (14)	440	640	840	840	840	840	840	440	280
$S_4$ (15)	300	500	700	900	900	900	900	300	420
$S_5$ (16)	160	360	560	760	960	960	960	160	560
$S_6$ (17)	20	220	420	620	820	1020	1020 (max)	20	700

Maximax selects $S_6$ (1020), maximin selects $S_1$ (720), and minimax regret selects $S_2$ (240).

Example 5: SP Rs. 25, CP Rs. 10, Demand 10–14

$MP = 25 - 10 = 15$, $ML = 10 - 0 = 10$.

Stock	10	11	12	13	14	Minimum	Maximum regret
$S_1$ (10)	150	150	150	150	150	150 (max)	60
$S_2$ (11)	140	165	165	165	165	140	45
$S_3$ (12)	130	155	180	180	180	130	30 (min)
$S_4$ (13)	120	145	170	195	195	120	30 (min)
$S_5$ (14)	110	135	160	185	210	110	40

Maximin selects $S_1$ (150). Minimax regret gives a tie at 30, so either $S_3$ or $S_4$ may be selected.

Example 6: SP Rs. 500, CP Rs. 300, Demand 45–50

$MP = 500 - 300 = \text{Rs. } 200$, $ML = 300 - 0 = \text{Rs. } 300$.

Stock	45	46	47	48	49	50	Maximum	Minimum	Maximum regret
$S_1$ (45)	9000	9000	9000	9000	9000	9000	9000	9000 (max)	1000
$S_2$ (46)	8700	9200	9200	9200	9200	9200	9200	8700	800
$S_3$ (47)	8400	8900	9400	9400	9400	9400	9400	8400	600 (min)
$S_4$ (48)	8100	8600	9100	9600	9600	9600	9600	8100	900
$S_5$ (49)	7800	8300	8800	9300	9800	9800	9800	7800	1200
$S_6$ (50)	7500	8000	8500	9000	9500	10000	10000 (max)	7500	1500

Optimistic approach (maximax) selects $S_6$ (10,000); pessimistic approach (maximin) selects $S_1$ (9,000); minimax regret selects $S_3$ (600).

Example 7: EMV, EPPI and EVPI

CP = Rs. 30, SP = Rs. 50, salvage value 0, so $MP = \text{Rs. } 20$ and $ML = \text{Rs. } 30$. Demand 10–14 units has probabilities 0.10, 0.15, 0.20, 0.25 and 0.30.

Stock	10 (0.10)	11 (0.15)	12 (0.20)	13 (0.25)	14 (0.30)	EMV
$S_1$ (10)	200	200	200	200	200	200
$S_2$ (11)	170	220	220	220	220	215
$S_3$ (12)	140	190	240	240	240	222.5 (max)
$S_4$ (13)	110	160	210	260	260	220
$S_5$ (14)	80	130	180	230	280	205

$$\begin{aligned} \text{EMV}(S_3) &= 0.10 \times 140 + 0.15 \times 190 + 0.20 \times 240 + 0.25 \times 240 + 0.30 \times 240 = 222.5 \\ \text{EPPI} &= 0.10 \times 200 + 0.15 \times 220 + 0.20 \times 240 + 0.25 \times 260 + 0.30 \times 280 = 250 \\ \text{EVPI} &= \text{EPPI} - \text{Max. EMV} = 250 - 222.5 = 27.5 \end{aligned}$$

The maximum EMV (Rs. 222.5) corresponds to $S_3$, so profit is maximised by stocking 12 units. The expected value of perfect information is Rs. 27.5.

Important Questions

2078 Exam

Solve the following linear programming problem graphically: [10 marks]

$$\begin{aligned} \text{Maximize } Z &= 30x + 50y \\ \text{subject to } x + y &\le 30 \\ x + 2y &\le 40 \\ x,\ y &\ge 0 \end{aligned}$$

2079 Exam

Solve the following linear programming problem: [10 marks]

$$\begin{aligned} \text{Minimize } Z &= 2x + 3y \\ \text{subject to } 5x + y &\ge 10 \\ 2x + 2y &\ge 12 \\ x + 4y &\ge 12 \\ x,\ y &\ge 0 \end{aligned}$$
(a) From the given payoff table, give the decision according to (i) maximax approach, (ii) maximin approach, (iii) minimax regret approach. [10 marks]

State of nature $S_1$ $S_2$ $S_3$

$N_1$ 400 200 700

$N_2$ 100 600 300

$N_3$ 500 300 100

(b) Two fair dice are rolled at the same time. What is the probability that the two faces turn up to show (i) a sum of 8 or 9, (ii) a sum less than 5?

State of nature	\(S_1\)	\(S_2\)	\(S_3\)
\(N_1\)	400	200	700
\(N_2\)	100	600	300
\(N_3\)	500	300	100

2080 Exam

Solve the following problem graphically: [10 marks]

$$\begin{aligned} \text{Minimize cost } Z &= \text{Rs. } 20x + \text{Rs. } 30y \\ \text{subject to } 3x + 5y &\ge 45 \\ 2x + y &\ge 20 \\ x,\ y &\ge 0 \end{aligned}$$
The manager of a flower shop promises its customers delivery within four hours on all flower orders. All flowers are purchased on the previous day and delivered by 8.00 AM the next morning. The daily demand for roses is as follows: [15 marks]

Dozens of roses 70 80 90 100

Probability 0.1 0.2 0.4 0.3

The manager purchases roses for Rs. 10 per dozen and sells them for Rs. 30. All unsold roses are donated to a local hospital. Construct the payoff table. Also find EMV, EPPI and EVPI.

Dozens of roses	70	80	90	100
Probability	0.1	0.2	0.4	0.3

2082 Exam

From the given payoff table, give the decision according to (i) maximax approach, (ii) maximin approach. [2 marks]

State of nature $S_1$ $S_2$ $S_3$

$N_1$ 50 30 90

$N_2$ 70 80 40

$N_3$ 20 40 20
Solve the following linear programming problem graphically: [10 marks]

$$\begin{aligned} \text{Minimize } Z &= 3x + 5y \\ \text{subject to } x + 2y &\ge 40 \\ x + y &\ge 30 \\ x,\ y &\ge 0 \end{aligned}$$
A vendor buys magazines at the rate of Rs. 10 and sells them at the rate of Rs. 15. Unsold magazines are worthless. The number of copies demanded and their corresponding probabilities are given below: [10 marks]

Copies demanded 200 210 220 230 240

Probability 0.19 0.16 0.25 0.27 0.13

(a) Determine the optimum stock level. (b) Calculate the maximum expected monetary value (EMV).

State of nature	\(S_1\)	\(S_2\)	\(S_3\)
\(N_1\)	50	30	90
\(N_2\)	70	80	40
\(N_3\)	20	40	20

Copies demanded	200	210	220	230	240
Probability	0.19	0.16	0.25	0.27	0.13

2083 Exam

Solve the following linear programming problem graphically: [10 marks]

$$\begin{aligned} \text{Maximize } Z &= 40A + 70B \\ \text{subject to } 3A + 3B &\le 36 \\ 5A + 2B &\le 50 \\ 2A + 6B &\le 60 \\ A,\ B &\ge 0 \end{aligned}$$

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